Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Big Ideas Math Book Geometry Answer Key Chapter 11 Circumference, Area, and Volume

Circumference, Area, and Volume Maintaining Mathematical Proficiency

Find the surface area of the prism.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1

Answer:
The surface area of the prism = 158.

Explanation:
In the above-given question,
given that,
l = 5 ft, w = 8 ft, and h = 3 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(5×8 + 5×3 + 8×3).
surface area = 2(40 + 15 + 24).
surface area = 2(79).
surface area = 158.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2

Answer:
The surface area of the triangular prism = 68m.

Explanation:
In the above-given question,
given that,
l = 10 m, p = 4 m, and h = 10.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 8) + 4(10).
surface area = 2(14) + 40.
surface area = 28 + 40.
surface area = 68m.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3

Answer:
The surface area of the triangular prism = 42m.

Explanation:
In the above-given question,
given that,
w = 10 cm, p = 4 cm, and h = 5 cm, l = 6 cm.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(6 + 5) + 4(5).
surface area = 2(11) + 20.
surface area = 22 + 20.
surface area = 42cm.

Find the missing dimension.

Question 4.
A rectangle has a perimeter 0f 28 inches and a width of 5 inches. What is the length of the rectangle?

Answer:
The length of the rectangle = 9 in.

Explanation:
In the above-given question,
given that,
A rectangle has a perimeter of 28 inches and a width of 5 inches.
length of the rectangle = p/2 – w.
length = 28/2 – 5.
where perimeter = 28 in, and w = 5 in.
length = 14 – 5.
length = 9.
so the length of the rectangle = 9 in.

Question 5.
A triangle has an area of 12 square centimeters and a height of 12 centimeters. What is the base of the triangle?

Answer:
The base of the triangle = 2 cm.

Explanation:
In the above-given question,
given that,
A triangle has an area of 12 sq cm and a height of 12 cm.
The base of the triangle = 2(A)/h.
base = 2(12)/12.
base = 24/12.
base = 2cm.
so the base of the triangle = 2 cm.

Question 6.
A rectangle has an area of 84 square feet and a width of 7 feet. What is the length of the rectangle?

Answer:
The length of the rectangle = 12 ft.

Explanation:
In the above-given question,
given that,
A rectangle has an area of 84 sq ft and a width of 7 feet.
area of the rectangle = l x b.
84 = l x 7.
l = 84/7.
l = 12.
so the length of the rectangle = 12 ft.

Question 7.
ABSTRACT REASONING
Write an equation for the surface area of a Prism with a length, width, and height of x inches. What solid figure does the prism represent?

Answer:
The surface area of a prism = 2(lw + wh + lh).

Explanation:
In the above-given question,
given that,
length = l, width = w, and height = x inches.
the surface area of the prism = 2(lw + wh + lh).
the solid figure does the prism represent the rectangular prism.

Circumference, Area, and Volume Monitoring Progress

Draw a net of the three-dimensional figure. Label the dimensions.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4

Answer:
The surface area of the prism = 64 cm.

Explanation:
In the above-given question,
given that,
l = 2 ft, w = 4 ft, and h = 4 ft.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(2×4 + 4×4 + 4×2).
surface area = 2(8 + 16 + 8).
surface area = 2(32).
surface area = 64.

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 5

Answer:
The surface area of the prism = 392 m.

Explanation:
In the above-given question,
given that,
l =8 m, w = 12 m, and h = 5 m.
where l = length, w = width, and h = height.
The surface area of the rectanguler prism = 2(lw + lh + wh).
surface area = 2(8×12 + 12×5 + 5×8).
surface area = 2(96 + 60 + 40).
surface area = 2(196).
surface area = 392.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

Answer:
The surface area of the triangular prism = 170 in.

Explanation:
In the above-given question,
given that,
B = 10 in, p = 15 in, and h = 15 in, l = 10 in.
the surface area of the triangular prism = 2B + ph.
b = base, p = perimeter, and h = height.
surface area = 2(10) + 15(10).
surface area = 2(10) + 150.
surface area = 20 + 150.
surface area = 170 in.

11.1 Circumference and Arc Length

Exploration 1

Finding the Length of a Circular Arc

Work with a partner: Find the length of each red circular arc.

a. entire circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 7

Answer:

b. one-fourth of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8
Answer:

c. one-third of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9
Answer:

d. five-eights of a circle
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 10
Answer:

Exploration 2

Using Arc Length

Work with a partner: The rider is attempting to stop with the front tire of the motorcycle in the painted rectangular box for a skills test. The front tire makes exactly one-half additional revolution before stopping. The diameter of the tire is 25 inches. Is the front tire still in contact with the painted box? Explain.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Answer:

Communicate Your Answer

Question 3.
How can you find the length of a circular arc?
Answer:
The length of a circular arc = 2
LOOKING FOR REGULARITY IN REPEATED REASONING
To be proficient in math, you need to notice if calculations are repeated and look both for general methods and for shortcuts.
Answer:

Question 4.
A motorcycle tire has a diameter of 24 inches. Approximately how many inches does the motorcycle travel when its front tire makes three-fourths of a revolution?
Answer:

Lesson 11.1 Circumference and Arc Length

Monitoring Progress

Question 1.
Find the circumference of a circle with a diameter of 5 inches.

Answer:
Circumference C = πd
C = 3.14x 5 = 15.7 in

Question 2.
Find the diameter of a circle with a circumference of 17 feet.

Answer:
Diameter d = C/π
d = 17/π = 5.41 ft

Find the indicated measure.

Question 3.
arc length of \(\widehat{P Q}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Answer:
arc length of \(\widehat{P Q}\) is 5.887

Explanation:
\(\widehat{P Q}\) = \(\frac { 75 }{ 360 } \) . π(9)
= 5.887

Question 4.
circumference of ⊙N
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Answer:
arc length of LM/C = LM/360
61.26/C = 270/360
C = 81.68

Question 5.
radius of ⊙G
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Answer:
arc length of EF = \(\frac { 60 }{ 360 } \) • 2πr
10.5 = \(\frac { 1 }{ 6 } \) • 2πr
r = 10.02

Question 6.
A car tire has a diameter of 28 inches. How many revolutions does the tire make while traveling 500 feet?

Answer:
The car tire have to make 69 revolutions to travel 500 ft.

Explanation:
Circumference C = 2πr = πd
C = 28π
Distance travelled = number of revolutions x C
500 x 12 = number of revolutions x 28π
number of revolutions = 68.2

Question 7.
In Example 4. the radius of the arc for a runner on the blue path is 44.02 meters, as shown in the diagram. About how far does this runner travel to go once around the track? Round to the nearest tenth of a meter.

Answer:

Question 8.
Convert 15° to radians.

Answer:
15° = 15 . \(\frac { π radians }{ 180° } \) = \(\frac { π }{ 12 } \) radians

Question 9.
Convert \(\frac{4 \pi}{3}\) radians to degrees.

Answer:
\(\frac{4 \pi}{3}\) radians = \(\frac{4 \pi}{3}\) radians . \(\frac { 180° }{ π radians } \) = 240 degrees

Exercise 11.1 Circumference and Arc Length

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The circumference of a circle with diameter d is C = _______ .
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 1

Question 2.
WRITING
Describe the difference between an arc measure and an arc length.

Answer:
An arc measure is measured in degrees while an arc length is the distance along an arc measured in linear units.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 10, find the indicated measure.

Question 3.
circumference of a circle with a radius of 6 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 3

Question 4.
diameter of a circle with a circumference of 63 feet

Answer:
C = 63 ft
πd = 63
d = 20.05

Question 5.
radius of a circle with a circumference of 28π
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 5

Question 6.
exact circumference of a circle with a diameter of 5 inches

Answer:
C = πd
C = 5π = 15.707

Question 7.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 7

Question 8.
m\(\widehat{D E}\)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 16

Answer:
\(\frac { arc length of DE }{ 2πr } \) = \(\frac { DE }{ 360 } \)
\(\frac { 8.73 }{ 2π(10) } \) = \(\frac { DE }{ 360 } \)
DE = 50.01°

Question 9.
circumference of ⊙C
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 17
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 9

Question 10.
radius of ⊙R
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 18
Answer:
\(\frac { arc length of LM }{ 2πr } \) = \(\frac { LM }{ 360 } \)
\(\frac { 38.95 }{ 2πr } \) = \(\frac { 260 }{ 360 } \)
r = 8.583

Question 11.
ERROR ANALYSIS
Describe and correct the error in finding the circumference of ⊙C.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 19
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 11

Question 12.
ERROR ANALYSIS
Describe and correct the error in finding the length of \(\widehat{G H}\).
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 20

Answer:
\(\frac { arc length of GH }{ 2πr } \) = \(\frac { m GH }{ 360 } \)
\(\widehat{G H}\). = \(\frac { 5 }{ 24 } \) . 2π(10)
= 13.08

Question 13.
PROBLEM SOLVING
A measuring wheel is used to calculate the length of a path. The diameter of the wheel is 8 inches. The wheel makes 87 complete revolutions along the length of the path. To the nearest foot, how long is the path?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 13

Question 14.
PROBLEM SOLVING
You ride your bicycle 40 meters. How many complete revolutions does the front wheel make?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 22

Answer:
Circumference of the front wheel = 2π(32.5)
= 65π cm
Distance covered = 40 m = 40 x 100 = 4000 cm
Number of revolutions = \(\frac { 4000 }{ 65π } \) = 19.58

In Exercises 15-18 find the perimeter of the shaded region.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 23
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 24
Answer:
Two horizontal edges are 2 . 3 = 6
Circumference of circle = 2π(3) = 6π
The perimeter of the shaded region = 6 + 6π

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 25
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 26
Answer:

In Exercises 19 – 22, convert the angle measure.

Question 19.
Convert 70° to radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 19

Question 20.
Convert 300° to radians.

Answer:
300 • (\(\frac { π }{ 180 } \)) = \(\frac { 5π }{ 3 } \) radian

Question 21.
Convert \(\frac{11 \pi}{12}\) radians to degrees.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 21

Question 22.
Convert \(\frac{\pi}{8}\) radian to degrees.

Answer:
\(\frac { π }{ 8 } \) • \(\frac { 180 }{ π } \)) = 22.5°

Question 23.
PROBLEM SOLVING
The London Eye is a Ferris wheel in London, England, that travels at a speed of 0.26 meter per second. How many minutes does it take the London Eye to complete one full revolution?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 27
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 23

Question 24.
PROBLEM SOLVING
You are planning to plant a circular garden adjacent to one of the corners of a building, as shown. You can use up to 38 feet of fence to make a border around the garden. What radius (in feet) can the garden have? Choose all that apply. Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 28
(A) 7
(B) 8
(C) 9
(D) 10

Answer:
C = 38 ft
2πr = 38
r = 6.04

In Exercises 25 and 26, find the circumference of the circle with the given equation. Write the circumference in terms of π

Question 25.
x2 + y2 = 16
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 25

Question 26.
(x + 2)2 + (y – 3)2 = 9

Answer:
The radius of circle (x + 2)² + (y – 3)² = 9 is 3
C = 2πr = 2π(3) = 6π
The circumference of the circle is 6π units.

Question 27.
USING STRUCTURE
A semicircle has endpoints (- 2, 5) and (2, 8). Find the arc length of the semicircle.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 27

Question 28.
REASONING
\(\widehat{E F}\) is an arc on a circle with radius r. Let x° be the measure of \(\widehat{E F}\). Describe the effect on the length of \(\widehat{E F}\) if you (a) double the radius of the circle, and (b) double the measure of \(\widehat{E F}\).
Answer:

Question 29.
MAKING AN ARGUMENT
Your friend claims that it is possible for two arcs with the same measure to have different arc lengths. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 29

Question 30.
PROBLEM SOLVING
Over 2000 years ago, the Greek scholar Eratosthenes estimated Earth’s circumference by assuming that the Sun’s rays were Parallel. He chose a day when the Sun shone straight down into a well in the city of Syene. At noon, he measured the angle the Sun’s rays made with a vertical stick in the city of Alexandria. Eratosthenes assumed that the distance from Syene to Alexandria was equal to about 575 miles. Explain how Eratosthenes was able to use this information to estimate Earth’s circumference. Then estimate Earth’s circumference.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 29
Answer:

Question 31.
ANALYZING RELATIONSHIPS
In ⊙C the ratio of the length of \(\widehat{P Q}\) to the length of \(\widehat{R S}\) is 2 to 1. What is the ratio of m∠PCQ to m∠RCS?
(A) 4 to 1
(B) 2 to 1
(C) 1 to 4
(D) 1 to 2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 31

Question 32.
ANALYZING RELATIONSHIPS
A 45° arc in ⊙C and a 30° arc in ⊙P have the same length. What is the ratio of the radius r1 of ⊙C to the radius r2 of ⊙P? Explain your reasoning.
Answer:

Question 33.
PROBLEM SOLVING
How many revolutions does the smaller gear complete during a single revolution of the larger gear?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 30
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 33

Question 34.
USING STRUCTURE
Find the circumference of each circle.
a. a circle circumscribed about a right triangle whose legs are 12 inches and 16 inches long
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 1
c² = a² + b²
c² = 12²+ 16²= 400
c = 20 in
circumference c = dπ
= 20π = 62.83 in

b. a circle circumscribed about a square with a side length of 6 centimeters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 2
d² = 6²+ 6² = 72
d = 8.49 cm
C = dπ
C = 8.49π
C = 26.67 cm

c. a circle inscribed in an equilateral triangle with a side length of 9 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 3
r = \(\frac { a√3 }{ 3 } \)
r = \(\frac { 9√3 }{ 3 } \)
r = 3√3 = 5.2
C = 2πr
C = 2π (5.2) = 32.67 in

Question 35.
REWRITING A FORMULA
Write a formula in terms of the measure θ (theta) of the central angle in radians) that can he used to find the length of an arc of a circle. Then use this formula to find the length of an arc of a circle with a radius of 4 inches and a central angle of \(\frac{3 \pi}{4}\) radians.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 35

Question 36.
HOW DO YOU SEE IT?
Compare the circumference of ⊙P to the length of \(\widehat{D E}\). Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 31
Answer:

Question 37.
MAKING AN ARGUMENT
In the diagram. the measure of the red shaded angle is 30°. The arc length a is 2. Your classmate claims that it is possible to find the circumference of the blue circle without finding the radius of either circle. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 32
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 37

Question 38.
MODELING WITH MATHEMATICS
What is the measure (in radians) of the angle formed by the hands of a clock at each time? Explain your reasoning.
a. 1 : 30 P.M.

Answer:
3π/4

b. 3:15 P.M.

Answer:
π/24

Question 39.
MATHEMATICAL CONNECTIONS
The sum of the circumferences of circles A, B, and C is 63π. Find AC.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 33
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 39

Question 40.
THOUGHT PROVOKING
Is π a rational number? Compare the rational number \(\frac{355}{113}\) to π. Find a different rational number that is even closer π.

Answer:
π is not a rational number as it can not be represented as an equivalent fraction. π = 3.14 and 355/113 = 3.14. This fraction resembles that value of π. Therefore a more accurate fraction will be starting by the value of 7 decimla places of π, therefore 3.1415926 x x = a.

Question 41.
PROOF
The circles in the diagram are concentric and \(\overrightarrow{F G}\) ≅ \(\overrightarrow{G H}\) Prove that \(\widehat{J K}\) and \(\widehat{N G}\) have the same length.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 34
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 41.2

Question 42.
REPEATED REASONING
\(\overline{A B}\) is divided into four congruent segments, and semicircles with radius r are drawn.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 35
a. What is the sum of the four arc lengths?
Answer:

b. What would the sum of the arc lengths be if \(\overline{A B}\) was divided into 8 congruent segments? 16 congruent segments? n congruent segments? Explain your reasoning.
Answer:

Maintaining Mathematical Proficiency

Find the area of the polygon with the given vertices.

Question 43.
X(2, 4), Y(8, – 1), Z(2, – 1)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.1 Ques 43.1

Question 44.
L(- 3, 1), M(4, 1), N(4, – 5), P(- 3, – 5)

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 4
LP = √(-3 + 3)² + (-5 – 1)² = 6
PN = √(4 + 3)² + (-5 + 5)² = 7
MN = √(4 – 4)² + (-5 – 1)² = 6
LM = √(4 + 3)² + (1 – 1)²= 7
Area = 6 x 7 = 42 units

11.2 Areas of Circles and Sectors

Exploration 1

Finding the Area of a Sector of a Circle

Work with a partner: A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. Find the area of each shaded circle or sector of a circle.

a. entire circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 36
Answer:

b. one – fourth of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 37
Answer:

c. seven – eighths of a circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 38
Answer:

d. two – thirds of a circle
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 39
Answer:

Exploration 2

Finding the Area of a Circular Sector

Work with a partner: A center pivot irrigation system consists of 400 meters of sprinkler equipment that rotates around a central pivot point at a rate of once every 3 days to irrigate a circular region with a diameter of 800 meters. Find the area of the sector that is irrigated by this system in one day.
REASONING ABSTRACTLY
To be proficient in math, you need to explain to yourself the meaning of a problem and look for entry points to its solution.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 40
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a sector of a circle?
Answer:

Question 4.
In Exploration 2, find the area of the sector that is irrigated in 2 hours.
Answer:

Lesson 11.2 Areas of Circles and Sectors

Monitoring progress

Question 1.
Find the area of a circle with a radius of 4.5 meters.

Answer:
Circle area = πr²
A = π(4.5)² = 20.25π

Question 2.
Find the radius of a circle with an area of 176.7 square feet.

Answer:
Circle area = πr²
176.7 = πr²
r² = 56.24
r = 7.499

Question 3.
About 58,000 people live in a region with a 2-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 4615.49 people per square mile.

Explanation:
A = πr² = π • 2² = 4π
Population density = \(\frac { number of people }{ area of land } \)
= \(\frac { 58000 }{ 4π } \) = 4615.49

Question 4.
A region with a 3-mile radius has a population density of about 1000 people per square mile. Find the number of people who live in the region.

Answer:
The number of people who live in the region are 28274.

Explanation:
A = πr² = π • 3² = 9π
Population density = \(\frac { number of people }{ area of land } \)
Number of people = 1000 x 9π = 28274

Find the indicated measure

Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 41

Question 5.
area of red sector

Answer:
The area of red sector = 205.25

Explanation:
m∠FDE = 120°, FE = 120° and FGE = 360° – 120° = 240°
Area of red sector = \(\frac { FE }{ 360° } \) • πr²
= \(\frac { 120 }{ 360° } \) • π(14²)
= 205.25

Question 6.
area of blue sector

Answer:
Area of blue sector = 410.5

Explanation:
Area of blue sector = \(\frac { FGE }{ 360° } \) • πr²
= \(\frac { 240 }{ 360° } \) • π(14²) = 410.5

Question 7.
Find the area of ⊙H.

Answer:
Area of ⊙H = 907.92 sq cm

Explanation:
Area of sector FHG =\(\frac { FG }{ 360° } \) • Area of ⊙H
214.37 = \(\frac { 85 }{ 360° } \) • Area of ⊙H
Area of ⊙H = 907.92 sq cm

Question 8.
Find the area of the figure.

Answer:
Area of triangle = \(\frac { 1 }{ 2 } \) • 7 • 7
= 24.5 sq m
Area of semi circle = πr²/2
= π(3.5)²/2
= 19.242255
Area of the figure = 24.5 + 19.24 = 43.74 sq m

Question 9.
If you know the area and radius of a sector of a circle, can you find the measure of the intercepted arc? Explain.

Answer:

Exercise 11.2 Areas of Circles and Sectors

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ____________ of a circle is the region bounded by two radii of the circle and their intercepted arc.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 1

Question 2.
WRITING
The arc measure of a sector in a given circle is doubled. will the area of the sector also be doubled? Explain our reasoning.

Answer:
Yes

Explanation:
Area of sector with arc measure x and radius r is s = π/180(xr)
If x becomes doube, then s1 = π/180(2xr) = 2s
This means that if the arc measure doubles, area of the sector also doubles.

Monitoring Progress and Modeling with Mathematics

In Exercise 3 – 10, find the indicated measure,

Question 3.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 42
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 3

Question 4.
area of ⊙C
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 43

Answer:
Area A = πr²
A = π(10)² = 100π sq in

Question 5.
area of a circle with a radius of 5 inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 5

Question 6.
area of a circle with a diameter of 16 feet

Answer:
d = 2r
Circle area = πr² = (π/4)d²
= (π/4)16² = 64π

Question 7.
radius of a circle with an area of 89 square feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 7

Question 8.
radius of a circle with an area of 380 square inches

Answer:
A = πr²
380 = πr²
r = 10.99

Question 9.
diameter of a circle with an area of 12.6 square inches
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 9

Question 10.
diameter of a circle with an area of 676π square centimeters

Answer:
Area A = 676π square centimeters
(π/4)d² = 676π
d² = 2704
d = 52

In Exercises 11 – 14, find the indicated measure.

Question 11.
About 210,000 people live in a region with a 12-mile radius. Find the population density in people per square mile.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 11

Question 12.
About 650,000 people live in a region with a 6-mile radius. Find the population density in people per square mile.

Answer:
The population density is about 5747 people per square mile.

Explanation:
Area of region = π(6)² = 36π
Population density = \(\frac { Number of people }{ area of land } \)
= \(\frac { 650,000 }{ 36π } \) = 5747.2

Question 13.
A region with a 4-mile radius has a population density of about 6366 people per square mile. Find the number of people who live in the region.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 13

Question 14.
About 79,000 people live in a circular region with a population density of about 513 people per square mile. Find the radius of the region.

Answer:
The radius of the region is 7

Explanation:
Population density = \(\frac { Number of people }{ area of land } \)
513 = \(\frac { 79,000 }{ πr² } \)
πr² = 153.99
r = 7

In Exercises 15-18 find the areas of the sectors formed by∠DFE.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 44
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 45

Answer:
Area of sector = \(\frac { 104° }{ 360° } \) • π(14)²
= 177.88
Area of red region is 177.88 sq cm
Area of blue region = \(\frac { 256° }{ 360° } \) • π(14)²
= 437.86 sq cm

Question 17.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 46
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 17

Question 18.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 47

Answer:
Area of red region is 10.471 sq ft
Area of the blue region is 39.79 sq ft

Explanation:
Area of sector = \(\frac { 75° }{ 360° } \) • π(4)²
= 10.471
Area of red region is 10.471 sq ft
Area of blue region = \(\frac { 285° }{ 360° } \) • π(4)²
= 39.79 sq ft

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the circle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 48
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 19

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the area of sector XZY when the area of ⊙Z is 255 square feet.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 49

Answer:
Area of ⊙Z is 255 square feet
πr² = 255
r = 9
Area of sector XZY = \(\frac { 115 }{ 360 } \) • 255
n = 81.458 sq ft

In Exercises 21 and 22, the area of the shaded sector is show. Find the indicated measure.

Question 21.
area of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 50
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 21

Question 22.
radius of ⊙M
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 51

Answer:
radius of ⊙M = 3.98

Explanation:
Area of region = \(\frac { 89 }{ 360 } \) . Area of ⊙M
12.36 = \(\frac { 89 }{ 360 } \) . Area of ⊙M
Area of ⊙M = 49.99
πr² = 49.99
r = 3.98

In Exercises 23 – 28, find the area of the shaded region.

Question 23.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 52
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 23

Question 24.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 53

Answer:
The area of the shaded region is 85.840 sq in.

Explanation:
Area of square = 20² = 400
Diameter of one circle = 10
radius of one circle = 5 in
Area of one circle = π(5)² = 78.53
Areas of four circle = 314.159
Area of shaded region = 400 – 314.159 = 85.840

Question 25.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 54
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 25.2

Question 26.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 55

Answer:
The area of shaded region is 301.59

Explanation:
The radius of smaller circle is 8 cm
The radius of bigger circle is 16 cm
Area of smaller semicircle = \(\frac { 1 }{ 2 } \)(π(8)²) = 100.53
Area of lager semicircle = \(\frac { 1 }{ 2 } \)(π(16)²) = 402.123
Area of shaded region = 402.123 – 100.53 = 301.59

Question 27.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 56
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 27.1

Question 28.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 57
Answer:
Area of shaded region = 7.63

Explanation:
c² = 3² + 4² = 25
c = 5
Radius = 2.5
Circle area = π(2.5)² = 19.63
Area of triangle = (3 x 4)/2 = 6
Area of shaded region = 19.63 – 12 = 7.63

Question 29.
PROBLEM SOLVING
The diagram shows the shape of a putting green at a miniature golf course. One part of the green is a sector of a circle. Find the area of the putting green.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 58
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 29.1

Question 30.
MAKING AN ARGUMENT
Your friend claims that if the radius of a circle is doubled, then its area doubles. Is your friend correct? Explain your reasoning.

Answer:
The friend is not correct. doubling the radius quadruples the area.

Explanation:
Area of circle with radius r = πr²
Area of circle with radius 2r = π(2r)² = 4πr²

Question 31.
MODELING WITH MATHEMATICS
The diagram shows the area of a lawn covered by a water sprinkler.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 59
a. What is the area of the lawn that is covered by the sprinkler?
b. The water pressure is weakened so that the radius is 12 feet. What is the area of the lawn that will be covered?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 31

Question 32.
MODELING WITH MATHEMATICS
The diagram shows a projected beam of light from a lighthouse.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 60
a. What is the area of water that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 115 }{ 360 } \) x π(18)²
= 325.15 sq mi

b. What is the area of land that can be covered by the light from the lighthouse?
Answer:
Area = \(\frac { 245 }{ 360 } \) x π(18)²
= 692.72 sq mi

Question 33.
ANALYZING RELATIONSHIPS
Look back at the Perimeters of Similar Polygons Theorem (Theorem 8.1) and the Areas of Similar PoIyons Theorem (Theorem 8.2) in Section 8.1. How would you rewrite these theorems to apply to circles? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 33

Question 34.
ANALYZING RELATIONSHIPS
A square is inscribed in a circle. The same square is also circumscribed about a smaller circle. Draw a diagram that represents this situation. Then find the ratio of the area of the larger circle to the area of the smaller circle.

Answer:
We start by assigning a variable to the radius of the inner circle. It is r, therefore the area of the circle is πr²
It can be seen that the side length of square is twice this radius. Therefore it can be said that the side length of this square is 2r.
Next, it can be seen that the diagonal of the square is diameter of outer circle. Therefore, length of the diagonal of the circle d = 2r√2. outer circle radius = r√2
Area of outer circle 2πr²
The ratio of the area of larger circle to the smaller circle = 2.

Question 35.
CONSTRUCTION
The table shows how students get to school.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 61
a. Explain why a circle graph is appropriate for the data.
b. You will represent each method by a sector of a circle graph. Find the central angle to use for each sector. Then construct the graph using a radius of 2 inches.
c. Find the area of each sector in your graph.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 35.2

Question 36.
HOW DO YOU SEE IT?
The outermost edges of the pattern shown form a square. If you know the dimensions of the other square, is it possible to compute the total colored area? Explain.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 62
Answer:

Question 37.
ABSTRACT REASONING
A circular pizza with a 12-inch diameter is enough for you and 2 friends. You want to buy pizzas for yourself and 7 friends. A 10-inch diameter pizza with one topping Costs $6.99 and a 14-inch diameter pizza with one topping Costs $12.99. How many 10-inch and 14-inch pizzas should you buy in each situation? Explain.
a. You want to spend as little money as possible.
b. You want to have three pizzas. each with a different topping, and spend as little money as possible.
C. You want to have as much of the thick outer crust as possible.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 37.1

Question 38.
THOUGHT PROVOKING
You know that the area of a circle is πr2. Find the formula for the area of an ellipse, shown below.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 63

Answer:
Ellipse area = πab

Question 39.
MULTIPLE REPRESENTATIONS
Consider a circle with a radius of 3 inches.
a. Complete the table, where x is the measure of the arc and is the area of the corresponding sector. Round your answers to the nearest tenth.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 64
b. Graph the data in the table.
c. Is the relationship between x and y linear? Explain.
d. If parts (a) – (c) were repeated using a circle with a radius of 5 inches, would the areas in the table change? Would your answer to part (c) change? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 39.3

Question 40.
CRITICAL THINKING
Find the area between the three congruent tangent circles. The radius of each circle is 6 inches.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 65
Answer:

Question 41.
PROOF
Semicircles with diameters equal to three sides of a right triangle are drawn, as shown. Prove that the sum of the areas of the two shaded crescents equals the area of the triangle.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 66
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 41

Maintaining Mathematical proficiency

Find the area of the figure.

Question 42.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 67

Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
Area = \(\frac { 1 }{ 2 } \)(18 x 6) = 54 sq in

Question 43.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 68
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 43

Question 44.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 69
Answer:
Area = \(\frac { 1 }{ 2 } \)(base x height)
A = \(\frac { 1 }{ 2 } \)(13 x 9) = 58.5 sq in

Question 45.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 70
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.2 Ques 45

11.3 Areas of Polygons

Exploration 1

Finding the Area of a Regular Polygon

Work with a partner: Use dynamic geometry software to construct each regular polygon with side lengths of 4, as shown. Find the apothem and use it to find the area of the polygon. Describe the steps that you used.
a.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 71
Answer:

b.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 72
Answer:

c.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 73
Answer:

d.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 74
Answer:

Exploration 2

Writing a Formula for Area

Work with a partner: Generalize the steps you used in Exploration 1 to develop a formula for the area of a regular polygon.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
How can you find the area of a regular polygon?
Answer:

Question 4.
Regular pentagon ABCDE has side lengths of 6 meters and an apothem of approximately 4.13 meters. Find the area of ABCDE.
Answer:

Lesson 11.3 Areas of Polygons

Monitoring Progress

Question 1.
Find the area of a rhombus with diagonals d1 = 4 feet and d2 = 5 feet.

Answer:
Area of rhombus = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(4 x 5) = 5 sq ft

Question 2.
Find the area of a kite with diagonals d1 = 12 inches and d1 = 9 inches.

Answer:
Area of kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 9) = 27 sq in

In the diagram. WXYZ is a square inscribed in ⊙P.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 75

Question 3.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
P is the center, PY or PX is the radius, PQ is apothem, ∠XPY is the central angle.

Question 4.
Find m∠XPY, m∠XPQ, and m∠PXQ.

Answer:
m∠XPY = \(\frac { 360 }{ 4 } \) = 90
m∠XPQ = 90/2 = 45
m∠PXQ = 180 – (90 + 45) = 45

Find the area of regular polygon

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 76

Answer:
c = √(8² + 6.5²) = 10.3
a = 20.61
Area = 0.25(√5(5+2√5) a²
Area = 0.25(√5(5+2√5) 20.61² = 730.8

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 77
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
= \(\frac { 5(7²) }{ 2 } \)√(5+2√5)
Area = 55377

Exercise 11.3 Areas of Polygons

Vocabulary and Core Concept Check

Question 1.
WRITING
Explain how to find the measure of a central angle of a regular polygon.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 78
Find the radius of ⊙F. Answer:

Answer:
EF = radius = 6.8

Find the apothem of polygon ABCDE.

Answer:
GF = apothem = 5.5

Find AF.

Answer:
AF = √4² + 5.5²
AF = 6.8

Find the radius of polygon ABCDE.
Answer:
AF = radius = 6.8

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the area of the kite or rhombus.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 79
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 80

Answer:
d₁ = 6 + 6 = 12
d₂ = 2 + 10 = 12
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
= \(\frac { 1 }{ 4 } \)(12 x 12)
= 36

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 81
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 82
Answer:
area A = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(5 x 6) = 7.5

In Exercises 7 – 10, use the diagram

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 83

Question 7.
Identify the center of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 7

Question 8.
Identify a central angle of polygon JKLMN.

Answer:
∠NPM is the central angle of polygon JKLMN

Question 9.
What is the radius of polygon JKLMN?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 9

Question 10.
What is the apothem of polygon JKLMN?
Answer:
QP is the apothem of polygon JKLMN

In Exercises 11 – 14, find the measure of a central angle of a regular polygon with the given number of sides. Round answers to the nearest tenth of a degree, if necessary.

Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 84

Question 11.
10 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 11

Question 12.
18 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 18 } \) = 20

Question 13.
24 sides
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 13

Question 14.
7 sides

Answer:
The measure of central angle = \(\frac { 360 }{ 7 } \) = 51.42

In Exercises 15 – 18, find the given angle measure for regular octagon ABCDEFGH.

Question 15.
m∠GJH
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 15

Question 16.
m∠GJK

Answer:
m∠GJK = m∠GJH/2
m∠GJK = 22.5

Question 17.
m∠KGJ
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 17

Question 18.
m∠EJH
Answer:
m∠EJH = 3(45) = 135

In Exercises 19 – 24, find the area of the regular polygon.

Question 19.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 85
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 19

Question 20.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 86
Answer:
Area = \(\frac { 5a² }{ 2 } \)√(5+2√5)
Area = \(\frac { 5(6.84)² }{ 2 } \)√(5+2√5)
A = 359.9784

Question 21.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 87
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 21

Question 22.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 88
Answer:
Area = \(\frac { 3√3 a²}{ 2 } \)
A = \(\frac { 3√3 (7)²}{ 2 } \)
A = 127.30

Question 23.
an octagon with a radius of 11 units
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 23

Question 24.
a pentagon with an apothem of 5 units

Answer:
A = 90.75

Explanation:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 6

We know apothem a = and it divides pentagon into triangles, the central angle is divided into 360/5 = 72
After that, we halved this angle and got 2 right triangles with x = 44 and y = 36. Since we know one side and all three angles of the triangle, we can calculate p with the tangent function.
tan y = p/a
tan 36 = p/5
p = 3.63
Since p is just half of the length of the side, we have to multiply it by 2
2 . p = 2 . 3.63 = 7.26 = s
Area = \(\frac { a . s. n }{ 2 } \)
A = \(\frac { 15 x 7.26 x 5 }{ 2 } \)
A = 90.75

Question 25.
ERROR ANALYSIS
Describe and correct the error in finding the area of the kite.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 89
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 25

Question 26.
ERROR ANALYSIS
Describe and correct the error in finding. the area of the regular hexagon.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 90

Answer:
s = √15² – 13² = 7.48
Area = \(\frac { 1 }{ 2 } \)(a . ns)
A = \(\frac { 1 }{ 2 } \)(13 x 6 x 7.48)
A = 291.72

In Exercises 27 – 30, find the area of the shaded region.

Question 27.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 91
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 27

Question 28.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 92

Answer:
Area of the shaded region = 223.75

Explanation:
Square side = diagonal/√2
= 28/√2 = 19.79
Area of square = 19.79² = 392
Circle area = π(14)² = 615.75
Area of the shaded region = 615.75 – 392 = 223.75

Question 29.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 93
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 29

Question 30.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 94
Answer:

Question 31.
MODELING WITH MATHEMATICS
Basaltic columns arc geological formations that result from rapidly cooling lava. Giant’s Causeway in Ireland contains many hexagonal basaltic columns. Suppose the top of one of the columns is in the shape of a regular hexagon with a radius of 8 inches. Find the area of the top of the column to the nearest square inch.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 95
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 31

Question 32.
MODELING WITH MATHEMATICS
A watch has a circular surface on a background that is a regular octagon. Find the area of the octagon. Then find the area of the silver border around the circular face.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 96
Answer:

CRITICAL THINKING
In Exercises 33 – 35, tell whether the statement is true or false. Explain your reasoning

Question 33.
The area of a regular n-gon of a fixed radius r increases as n increases.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 33

Question 34.
The apothem of a regular polygon is always less than the radius.

Answer:
true, the radius always reaches the end of the circle but the apothem never does

Question 35.
The radius of a regular polygon is always less than the side length.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 35

Question 36.
REASONING
Predict which figure has the greatest area and which has the least area. Explain your reasoning. Check by finding the area of each figure.
(A) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 97
(B) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 98
(C) Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 99
Answer:
(B) has the highest area and C has the lowest area.

Explanation:
(A) area = π(6.5)² = 132.73
(B) area = 139.25
(C) area = \(\frac { 1 }{ 2 } \)(18 x 15) = 135

Question 37.
USING EQUATIONS
Find the area of a regular
pentagon inscribed in a circle whose equation is given by (x – 4)2 + y + 2)2 = 25.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 37.2

Question 38.
REASONING
What happens to the area of a kite if you double the length of one of the diagonals? if you double the length of both diagonals? Justify your answer.

Answer:
Area of a kite = \(\frac { 1 }{ 4 } \)(d₁d₂)
If you double the length of one diagonal, then d₁ = 2d₁
Area of kite = \(\frac { 1 }{ 2 } \)(d₁d₂)
If you double length of both diagonals
Area = \(\frac { 1 }{ 4 } \)(2d₁2d₂) = d₁d₂
If you double the length of one diagonal, then the area becomes halve. If you double length of both diagonals, then area becomes 4 times.

MATHEMATICAL CONNECTIONS
In Exercises 39 and 40, write and solve an equation to find the indicated lengths. Round decimal answers to the nearest tenth.

Question 39.
The area of a kite is 324 square inches. One diagonal is twice as long as the other diagonal. Find the length of each diagonal.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 39

Question 40.
One diagonal of a rhombus is four times the length of the other diagonal. The area of the rhombus is 98 square feet. Find the length of each diagonal.

Answer:
The length of each diagonal is 9.89, 2.47.

Explanation:
One diagonal of a rhombus is four times the length of the other diagonal.
d₁ = 4d₂
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
98 = \(\frac { 1 }{ 4 } \)(d₁(4d₁))
d₁ = 9.89
d₂ = 2.47

Question 41.
REASONING
The perimeter of a regular nonagon. or 9-gon, is 18 inches. Is this enough information to find the area? If so, find the area and explain your reasoning. If not, explain why not.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 41

Question 42.
MAKING AN ARGUMENT
Your friend claims that it is possible to find the area of any rhombus if you only know the perimeter of the rhombus. Is your friend correct? Explain your reasoning.

Answer:
No; A rhombus is not a regular polygon.

Question 43.
PROOF
Prove that the area of any quadrilateral with perpendicular diagonals is A = \(\frac{1}{2}\)d1d2, where d1 and d2 are the lengths of the diagonals.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 100
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 43.2

Question 44.
HOW DO YOU SEE IT?
Explain how to find the area of the regular hexagon by dividing the hexagon into equilateral triangles.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 101

Answer:

Question 45.
REWRITING A FORMULA
Rewrite the formula for the area of a rhombus for the special case of a square with side length s. Show that this is the same as the formula for the area of a square, A = s2.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 45

Question 46.
REWRITING A FORMULA
Use the formula for the area of a regular polygon to show that the area of an equilateral triangle can be found by using the formula A = \(\frac{1}{4}\)s2√3 where s is the side length.
Answer:

Question 47.
CRITICAL THINKING
The area of a regular pentagon is 72 square centimeters. Find the length of one side.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 47

Question 48.
CRITICAL THINKING
The area of a dodecagon, or 12-gon, is 140 square inches. Find the apothem of the polygon.

Answer:
Let the side length of dodecagon be 2x. The measure of each interior angle of a regular decagon is 150. This implies that the base angle C and A of the resulting isosceles triangle formed by the red sides is equal to 150/2 = 75. The adjacent to this angle is the length 2x/2 = x inches, while the opposite to it is the blue apothem in the right triangle BDC formed. Therefore a = x tan 75. Therefore, area of dodecagon is
140 = 1/2 (x tan75)(12 . 2x)
140 = 44.785 x²
x² = 3.126
x = 1.768

Question 49.
USING STRUCTURE
In the figure, an equilateral triangle lies inside a square inside a regular pentagon inside a regular hexagon. Find the approximate area of the entire shaded region to the nearest whole number.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 102
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 49.2

Question 50.
THOUGHT PROVOKING
The area of a regular n-gon is given by A = \(\frac{1}{2}\)aP. As n approaches infinity, what does the n-gon approach? What does P approach? What does a approach? What can you conclude from your three answers? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 103
Answer:

Question 51.
COMPARING METHODS
Find the area of regular pentagon ABCDE by using the formula A = \(\frac{1}{2}\)aP, or A = \(\frac{1}{2}\)a • ns. Then find the area by adding the areas of smaller polygons. Check that both methods yield the same area. Which method do you prefer? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 104
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 51.3

Question 52.
USING STRUCTURE
Two regular polygons both have n sides. One of the polygons is inscribed in, and the other is circumscribed about, a circle of radius r. Find the area between the two polygons in terms of n and r.

Answer:
The radius of the smaller polygon is equal to the apothem of the larger polygon. The central angle is 360/n, therefore the apothem makes an angle of 180/n. Use sine and cosine to find the apothem and side length of the smaller polygon.
asmall = r sin\(\frac { 180 }{ n } \)
ssmall = 2r cos\(\frac { 180 }{ n } \)
Use tangent to find the side length of the large polygon.
Slarge = 2r tan\(\frac { 180 }{ n } \)
Use the formula to find the area of the smaller polygon.
Asmall = 1/2 . asmall . n . ssmall
Asmall = 1/2 . r sin\(\frac { 180 }{ n } \) . n . 2r cos\(\frac { 180 }{ n } \)
Asmall = nr² sin \(\frac { 180 }{ n } \) cos\(\frac { 180 }{ n } \)
Use the formula to find the area of the larger polygon.
ALarge = 1/2 . alarge . n . slarge
= nr² tan\(\frac { 180 }{ n } \)
The area between the polygons is equal to the area of the larger polygon minus the area of the smaller polygon. Use some trig identities to simplify the expression.
A = Alarge – Asmall
A = nr² tan\(\frac { 180 }{ n } sin²[latex]\frac { 180 }{ n }

Maintaining Mathematical Proficiency

Determine whether the figure has line symmetry, rotational symmetry, both, or neither. If the
figure has line symmetry. determine the number of lines of symmetry. It the figure has rotational
symmetry, describe any rotations that map the figure onto itself.

Question 53.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 105
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 53

Question 54.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 106
Answer:
The figure has rotational symmetry.

Question 55.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 107
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.3 Ques 55

Question 56.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 108
Answer:
The figure has one line symmetry.

11.4 Three-Dimensional Figures

Exploration 1

Analyzing a Property of Polyhedra

Work with a partner: The five Platonic solids are shown below. Each of these solids has congruent regular polygons as faces. Complete the table by listing the numbers of vertices, edges, and faces of each Platonic solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 109
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 110
Answer:

Communicate Your Answer

Question 2.
What is the relationship between the numbers of vertices V, edges E, and faces F of a polyhedron? (Note: Swiss mathematician Leonhard Euler (1707 – 1783) discovered a formula that relates these quantities.)
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to reason inductively about data.
Answer:

Question 3.
Draw three polyhedra that are different from the Platonic solids given in Exploration 1. Count the numbers of vertices, edges, and faces of each polyhedron Then verify that the relationship you found in Question 2 is valid for each polyhedron.
Answer:

Lesson 11.4 Three-Dimensional Figures

Monitoring progress

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 1.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 111

Answer:
The solid is formed by polygons, so it is a polyhedron. The base is a square, it is a square pyramid.

Question 2.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 112
Answer:
The solid have curved faces. So it is not a polyhedron.

Question 3.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 113
Answer:
The solid is formed by polygons, so it is a polyhedron. iT has two triangles, one rectangle and two squares.

Describe the shape formed by the intersection of the plane and the solid.

Question 4.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 114
Answer:
The cross-section is a pentagon.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 115
Answer:
The cross-section is a hexagon.

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 116
Answer:
The cross-section is a circle.

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 117
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 8

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 118
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 119
Answer:

Exercise 11.4 Three-Dimensional Figures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
A(n) ___________ is a solid that is bounded by polygons.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 1

Question 2.
WHICH ONE DOESN’T BELONG?
Which solid does not belong with the other three? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 120

Answer:
Cone does not belong with the other three as it has a curved surface and others not.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, match the polyhedron with its name.

3. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 121 A. triangular Prism
4. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 122 B. rectangular pyramid
5. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 123 C. hexagonal pyramid
6. Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 124 D. Pentagonal prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 5

In Exercises 7 – 10, tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 125
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 7

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 126
Answer:
Yes, it is a polyhedron. It is a hexagonal prism.

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 127
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 128
Answer:
Yes, it is a polyhedron, truncated square pyramid.

In Exercises 11 – 14, describe the cross section formed by the intersection of the plane and the solid.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 129
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 130
Answer:
The cross-section is a square.

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 131
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 132
Answer:
The cross-section is a hexagon

In Exercises 15 – 18, sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 133
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 134
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 135
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 17

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 136
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 9

Question 19.
ERROR ANALYSIS
Describe and correct the error in identifying the solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 137
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 19

Question 20.
HOW DO YOU SEE IT?
Is the swimming pool shown a polyhedron? If it is, name the polyhedron. If not, explain why not.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 138
Answer:
It is an octagonal polyhedron.

In Exercises 21 – 26, sketch the polyhedron.

Question 21.
triangular prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 21

Question 22.
rectangular prism

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11

Question 23.
pentagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 23

Question 24.
hexagonal prism
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 12

Question 25.
square pyramid
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 25

Question 26.
pentagonal pyramid

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 13

Question 27.
MAKING AN ARGUMENT
Your Friend says that the polyhedron shown is a triangular prism. Your cousin says that it is a triangular pyramid. Who is correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 139
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 27

Question 28.
ATTENDING TO PRECISION
The figure shows a plane intersection a cube through four of its vertices. The edge length of the cube is 6 inches.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 140
a. Describe the shape of the cross section.
Answer:
The cross-section is a rectangle.

b. What is the perimeter of the cross section?
Answer:
The perimeter is 2(l + b)

c. What is the area of the cross section?
Answer:
Area is lb.

REASONING
In Exercises 29 – 34, tell whether it is possible for a cross section of a cube to have the given shape. If it is, describe or sketch how the plane could intersect the cube.

Question 29.
circle
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 29

Question 30.
pentagon
Answer:
yes, cross-section of the cube can be a pentagon.

Question 31.
rhombus
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 31

Question 32.
isosceles triangle
Answer:
Yes, the cross-section can be an isosceles triangle.

Question 33.
hexagon
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 33

Question 34.
scalene triangle
Answer:
Yes, the cross-section can be scalene triangle.

Question 35.
REASONING
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
a.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 141
b.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 142
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 35

Question 36.
THOUGHT PROVOKING
Describe how Plato might have argued that there are precisely five Platonic Solids (see page 617). (Hint: Consider the angles that meet at a vertex.)
Answer:

Maintaining Mathematical proficiency

Decide whether enough information is given to prove that the triangles are congruent. It so, state the theorem you would use.

Question 37.
∆ABD, ∆ CDB
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 143
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 37

Question 38.
∆JLK, ∆JLM
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 144

Answer:
∆JLK ≅ ∆JLM by SAS congruence theorem.

Question 39.
∆RQP, ∆RTS
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 145
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.4 Ques 39

11.1 – 11.4 Quiz

Find the indicated measure.

Question 1.
m[latex]\widehat{E F}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 146
Answer:
13.7 = \(\frac { m[latex]\widehat{E F}\) }{ 360 } [/latex] • 2π(7)
m\(\widehat{E F}\) = 112.13

Question 2.
arc length of \(\widehat{Q S}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 147
Answer:
arc length of \(\widehat{Q S}\) = \(\frac { 83 }{ 360 } \) • 2π(4)
= 5.79

Question 3.
circumference of ⊙N
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 148
Answer:
8 = \(\frac { 48 }{ 360 } \) • 2πr
C = 60

Question 4.
Convert 26° to radians and \(\frac{5 \pi}{9}\) radians to degrees.

Answer:
26° = 26 . \(\frac { π }{ 180 } \) = \(\frac { 13π }{ 90 } \) radians
\(\frac{5 \pi}{9}\) = \(\frac{5 \pi}{9}\) . \(\frac { 180 }{ π } \) = 100°

Use the figure to find the indicated measure.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 149

Question 5.
area of red sector

Answer:
area of red sector = \(\frac { 100 }{ 360 } \) . π(12)²
= 125.66

Question 6.
area of blue sector

Answer:
area of blue sector = \(\frac { 260 }{ 360 } \) . π(12)²
= 326.72

In the diagram, RSTUVWXY is a reuIar octagon inscribed in ⊙C.

Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 150

Question 7.
Identify the center, a radius, an apothem, and a central angle of the polygon.

Answer:
C is center, CY is radius, CZ is apothem, ∠YCR is central angle of the polygon

Question 8.
Find m∠RCY, m∠RCZ, and m∠ZRC.

Answer:
m∠RCY = 360/8 = 45
m∠RCZ = 45/2 = 22.5
m∠ZRC = 180 – (22.5 + 90) = 67.5

Question 9.
The radius of the circle is 8 units. Find the area of the octagon.

Answer:
Area of octagon = 0.5 x 8 x 8 sin 45 = 22.62

Tell whether the solid is a polyhedron. If it is, name the polyhedron.

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 151
Answer:
It is not a polyhedron.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 152
Answer:
The solid is a polyhedron. It is an octagonal pyramid.

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 153
Answer:
The solid is a polyhedron. It is a pentagonal prism.

Question 13.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 154

Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 14

Question 14.
The two white congruent circles just fit into the blue circle. What is the area of the blue region?
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 155
Answer:
White circle diameter = radius of the blue circle.
6 = radius of the blue circle.
Area of blue circle = π6² = 113.09
Area of white circle = π3² = 28.27
Area of blue region = 113.09 – 2(28.27) = 56.541

Question 15.
Find the area of each rhombus tile. Then find the area of the pattern.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 156

Answer:
Area of yellow tile = \(\frac { 1 }{ 4 } \)(15.7 x 11.4) = 44.745
area of red tile = \(\frac { 1 }{ 4 } \)(18.5 x 6) = 27.75
Area of pattern = 32(44.745) + 23(27.75) = 2070.09 sq mm

11.5 Volumes of Prisms and Cylinders

Exploration 1

Finding volume

Work with a partner: Consider a stack of square papers that is in the form of a right prism.

Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 157

a. What is the volume the prism?
Answer:

b. When you twist the slack of papers, as shown at the right, do you change the volume? Explain your reasoning.
Answer:

c. Write a carefully worded conjecture that describes the conclusion you reached in part (b).
ATTENDING TO PRECISION
To be proficient in math, you need to communicate precisely to others.
Answer:

d. Use your conjecture to find the volume of the twisted stack of papers.
Answer:

Exploration 2

Finding volume

Work with a partner: Use the conjecture you wrote in Exploration I to find the volume of the cylinder.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 158
Answer:

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 159
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a prism or cylinder that is not a right prism or right cylinder?
Answer:

Question 4.
In Exploration 1, would the conjecture you wrote change if the papers in each stack were not squares? Explain your reasoning.
Answer:

Lesson 11.5 Volumes of Prisms and Cylinders

Monitoring Progress

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 160
Answer:
Volume = Area * height
Area = \(\frac { 1 }{ 2 } \)(5 x 9) = 22.5
Volume = 22. 5 x 8 = 180 cubic m.

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 161

Answer:
Area of circle = πr² = π(8)² = 64π
Volume = 64π x 14 = 2814.86 cubic ft

Question 3.
The diagram shows the dimensions of a concrete cylinder. Concrete has a density of 2.3 grams per cubic centimeter. Find the mass of the concrete cylinder to the nearest gram.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 162
Answer:
Mass of the concrete cylinder = 32 x π(24)² = 18432π cubic in

Question 4.
WHAT IF?
In Example 4, you want the length to be 5 meters, the width to be 3 meters. and the volume to be 60 cubic meters. What should the height be?
Answer:
volume = lbh
60 = 5 x 3 x h
h = 4 m

Question 5.
WHAT IF?
In Example 5, you want the height to be 5 meters and the volume to be 75 cubic meters. What should the area of the base be? Give a possible length and width.
Answer:
volume V = base x height
75 = base x 5
Base = 15 sq m

Question 6.
Prism C and prism D are similar. Find the volume of prism D.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 163
Answer:
\(\frac { 12 }{ 3 } \) = \(\frac { 1536 }{ v } \)
v = 384
Volume of prism D = 384 cubic m

Question 7.
Find the volume of the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 164
Answer:
Volume = area x height
Volume = \(\frac { 1 }{ 2 } \)(10 x 3 x 6)
= 90 cubic ft

Exercise 11.5 Volumes of Prisms and Cylinders

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
In what type of units is the volume of a solid measured?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 1

Question 2.
COMPLETE THE SENTENCE
Density is the amount of _______ that an object has in a given unit of __________ .

Answer:
Density is the mass of the object divided by its volume.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the volume of the prism.

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 165
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 3

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 166
Answer:
Volume V = lbh
V = 1.5 x 2 x 4 = 12 m³

Question 5.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 167
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 5

Question 6.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 168
Answer:
Volume = 6 x 11 x 14
V = 924 m³

In Exercises 7 – 10. find the volume of the cylinder.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 169
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 170
Answer:
Volume = πr²h
V = π(13.4)² x 9.8
V = 1759.6π cm³

Question 9.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 171
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 9

Question 10.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 172
Answer:
Volume V = πr²h
Shoter side = 18/2 = 9
Height h = √18² – 9² = 15.588
V = π6² x 15.588
= 1763 m³

In Exercises 11 and 12. make a sketch of the solid and find its volume. Round your answer to the nearest hundredth.

Question 11.
A prism has a height of 11.2 centimeters and an equilateral triangle for a base, where each base edge is 8 centimeters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 11

Question 12.
A pentagonal prism has a height of 9 feet and each base edge is 3 feet.

Answer:
volume is 139.32 ft³

explanation:
Pentagon area = 15.48
Height h = 9 ft
Volume V = area x height
= 15.48 x 9 = 139.32

Question 13.
PROBLEM SOLVING
A piece of copper with a volume of 8.25 cubic centimeters has a mass of 73.92 grams. A piece of iron with a volume of 5 cubic centimeters has a mass of 39.35 grams. Which metal has the greater density?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 173
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 13

Question 14.
PROBLEM SOLVING
The United States has minted one-dollar silver coins called the American Eagle Silver Bullion Coin since 1986. Each coin has a diameter of 40.6 millimeters and is 2.98 millimeters thick. The density of silver is 10.5 grams per cubic centimeter. What is the mass of an American Eagle Silver Bullion Coin to the nearest grain?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 174
Answer:
V = πr²h
V =π(20.3)²  x (2.98)
V = 3856 mm³

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the cylinder.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 175
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 15

Question 16.
ERROR ANALYSIS
Describe and correct the error in finding the density of an object that has a mass of 24 grams and a volume of 28.3 cubic centimeters.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 176

Answer:
Density = mass / volume
Density = \(\frac { 24 }{ 28.3 } \)
Density = 0.8480

In Exercises 17 – 22, find the missing dimension of the prism or cylinder.

Question 17.
Volume = 500 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 177
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 17

Question 18.
Volume = 2700 yd3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 178

Answer:
Volume = 2700 yd³
12 x 5 x v = 2700
v = 15 yd

Question 19.
Volume = 80 cm3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 179
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 19

Question 20.
Volume = 72.66 in.3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 180
Answer:
Volume = 72.66 in.3
Area . x = 72.66
10.39 x = 72.66
x = 6.9 in

Question 21.
Volume = 3000 ft3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 181
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 21

Question 22.
Volume = 1696.5 m3
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 182
Answer:
Volume = 1696.5
πr²h = 1696.5
πz² x 15 = 1696.5
z² = 36.00
z = 6

In Exercises 23 and 24, find the area of the base of the rectangular prism with the given volume and height. Then give a possible length and width.

Question 23.
V= 154 in.3, h = 11 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 23

Question 24.
V = 27 m3,h = 3m

Answer:
V = Bh
27 = B x 3
B = 9

In Exercises 25 and 26, the solids are similar. Find the volume of solid B.

Question 25.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 183
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 25

Question 26.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 184
Answer:
\(\frac { 12 }{ 15 } \) = \(\frac { 4608π }{ V } \)
V = 5760π
Volume of cylinder B = 5760π

In Exercises 27 and 28, the solids are similar. Find the indicated measure.

Question 27.
height x of the base of prism A
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 185
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 27

Question 28.
height h of cylinder B
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 186
Answer:
height h of cylinder B is 40 ft

Explanation:
\(\frac { 7π }{ 5 } \) = \(\frac { 56π }{ h } \)
h = 40

In Exercises 29 – 32. find the volume of the composite solid.

Question 29.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 187
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 29

Question 30.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 188
Answer:
Volume V = 89.32

Explanation:
Volume of square = 4³ = 64
Volume of semicircle = π(2)² x 4 = 8π

Question 31.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 189
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 31

Question 32.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 190

Answer:
The volume of composite solid is 35 cubic ft

Explanation:
Volume of larger prism = 4 x 2 x 5 = 40
Volume of the smaller prism = 1 x 1 x 5 = 5
Volume of larger prism – volume of the smaller prism = 40 – 5 = 35 cubic ft

Question 33.
MODELING WITH MATHEMATICS
The Great Blue Hole is a cylindrical trench located off the coast of Belize. It is approximately 1000 feet wide and 400 feet deep. About how many gallons of water does the Great Blue Hole contain? (1 ft3 ≈ 7.48 gallons)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 191
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 33

Question 34.
COMPARING METHODS
The Volume Addition Postulate states that the volume of a solid is the sum of the volumes of all its non overlapping parts. Use this postulate to find the volume of the block of concrete in Example 7 by subtracting the volume of each hole from the volume of the large rectangular prism. Which method do you prefer? Explain your reasoning.
Answer:

REASONING
In Exercises 35 and 36, you are melting a rectangular block of wax to make candles. how many candles of the given shape can be made using a block that measures 10 centimeters by 9 centimeters by 20 centimeters?

Question 35.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 192
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 35

Question 36.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 193

Answer:
7 triangular prism candles with the given measures can be made.

Explanation:
Volume of block = 1800
The volume of triangular prism = 4 x 6 x 10 = 240
1800/240 = 7.5

Question 37.
PROBLEM SOLVING
An aquarium shaped like a rectangular prism has a length of 30 inches, a width of 10 inches, and a height of 20 inches. You fill the aquarium \(\frac{3}{4}\) fill with water. When you submerge a rock in the aquarium, the water level rises 0.25 inch.
a. Find the volume of the rock.
b. How many rocks of this size can you place in the aquarium before water spills out?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 37

Question 38.
PROBLEM SOLVING
You drop an irregular piece of metal into a container partially filled with water and measure that the waler level rises 4.8 centimeters. The square base of the container has a side length of 8 centimeters. You measure the mass of the metal to be 450 grams. What is the density of the metal?

Answer:
The density of metal is 1.4648

Explanation:
Density = \(\frac { Mass }{ Volume } \)
Volume V = 4.8 x 64 = 307.2
Density = \(\frac { 450 }{ 307.2 } \) = 1.4648

Question 39.
WRITING
Both of the figures shown arc made up of the same number of congruent rectangles. Explain how Cavalieri’s Principle can be adapted to compare the areas of these figures.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 194
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 39

Question 40.
HOW DO YOU SEE IT?
Each stack of memo papers contains 500 equally-sized sheets of paper. Compare their volumes. Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 195
Answer:

Question 41.
USING STRUCTURE
Sketch the solid formed by the net. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 196
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 41

Question 42.
USING STRUCTURE
Sketch the solid with the given views. Then find the volume of the solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 197

Answer:
Volume = 2.5 x 3.5 x 6
Volume = 52.5

Question 43.
OPEN-ENDED
Sketch two rectangular prisms that have volumes of 1000 cubic inches hut different surface areas. Include dimensions in your sketches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 43

Question 44.
MODELING WITH MATHEMATICS
Which box gives you more cereal for your money? Explain.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199

Answer:
First one gives more cerel for your money.

Explanation:
Bigger one volume = 16 x 4 x 10 = 640
Smaller one volume = 2 x 8 x 10 = 160
6 – 640 means 1 – 106.66
2 – 160 means 1 – 80

Question 45.
CRITICAL THINKING
A 3-inch by 5-inch index card is rotated around a horizontal line and a vertical line to produce two different solids. Which solid has a greater volume? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 199
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 45

Question 46.
CRITICAL THINKING
The height of cylinder X is twice the height of cylinder Y. The radius of cylinder X is half the radius of cylinder Y. Compare the volumes of cylinder X and cylinder Y. Justify your answer.

Answer:
Let the height of cylinder X be h, radius be r and its volume is πr²h
So, the height of cylinder Y is h/2 and radius is 2r, then the volume is 2πr²h
From both expressions, it can be seen that the volume of cylinder y is twice that of cylinder X.

Question 47.
USING STRUCTURE
Find the volume of the solid shown. The bases of the solid are sectors of circles.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 200
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 47

Question 48.
MATHEMATICAL CONNECTIONS
You drill a circular hole of radius r through the base of a cylinder of radius R. Assume the hole is drilled completely through to the other base. You want the volume of the hole to be half the volume of the cylinder. Express r as a function of R.

Answer:
r = √R²/2

Explanation:
The radius of a solid cylinder without a hole is R. So its volume is πR²h
As per the given condition, the volume of the hole must be half of that of the solid cylinder, hole volume is πR²h/2
Volume of cylinder V = πr²h
πR²h/2 = πr²h
R²/2 = r²
r = √R²/2
r = \(\frac { R√2 }{ 2 } \)

Question 49.
ANALYZING RELATIONSHIPS
How can you change the height of a cylinder so that the volume is increased by 25% but the radius remains the same?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 49

Question 50.
ANALYZING RELATIONSHIPS
How can you change the edge length of a cube so that the volume is reduced by 40%?

Answer:
Write the equation of volume of rectangular prism which can be used to evaluate the cube volume
Volume = s x s x s
The above equation shows that the volume of a cube is directly proportional to one of its side length, therefore, if the volume is to be reduced by 40%, then its the length of one of its side must be reduced by 40%, without changing the 2 other of its sides.

Question 51.
MAKING AN ARGUMENT
You have two objects of equal V0lume. Your friend says you can compare the densities of the objects by comparing their mass, because the heavier object will have a greater density. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 51

Question 52.
THOUGHT PROVOKING
Cavalieri’s Principle states that the two solids shown below have the same volume. Do they also have the same surface area? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 201
Answer:

Question 53.
PROBLEM SOLVING
A barn is in the shape of a pentagonal prism with the dimensions shown, The volume of the barn is 9072 cubic feel. Find the dimensions of each half of the root.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 202
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 53

Question 54.
PROBLEM SOLVING
A wooden box is in the shape of a regular pentagonal prism. The sides, top, and bottom of the box are 1 centimeter thick. Approximate the volume of wood used to construct the box. Round your answer to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 203
Answer:
Pentagon area = \(\frac { 5 }{ 2 } \)(4)² sin 72
A = 38
Volume = Area x height
Volume = 38 x 6 = 228

Maintaining Mathematical Proficiency

Find the surface area of the regular pyramid.

Question 55.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 204
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 55

Question 56.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 205
Answer:
Surface area = base area + 3bs
Surface area = 166.3 + 3(8)(10)
= 406.3

Question 57.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 206
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.5 Ques 57

11.6 Volumes of Pyramids

Exploration 1

Finding the Volume of a Pyramid

Work with a partner: The pyramid and the prism have the same height and the same square base.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 207
When the pyramid is filled with sand and poured into the prism, it takes three pyramids to fill the prism.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 208
Use this information to write a formula for the volume V of a pyramid.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Exploration 2

Finding the Volume of a Pyramid
Work with a partner: Use the formula you wrote in Exploration 1 to find the volume of the hexagonal pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 209
Answer:

Communicate Your Answer

Question 3.
How can you find the volume of a pyramid?
Answer:

Question 4.
In Section 11 .7, you will study volumes of cones. How do you think you could use a method similar to the one presented in Exploration 1 to write a formula for the volume of a cone? Explain your reasoning.
Answer:

Lesson 11.6 Volumes of Pyramids

Monitoring Progress

Find the volume of the pyramid.

Question 1.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 210
Answer:
The volume of the pyramid is 400 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(10 x 10 x 12)
V = 400

Question 2.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 211

Answer:
The volume of the pyramid is 2494.13 cm³

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh
V = \(\frac { 1 }{ 3 } \)(374.12 x 20)
V = 2494.13

Question 3.
The volume of a square pyramid is 75 cubic meters and the height is 9 meters. Find the side length of the square base.

Answer:
The side length of the square base is 5 m

Explanation:
Volume V = \(\frac { 1 }{ 3 } \)Bh = 75
\(\frac { 1 }{ 3 } \)B(9) = 75
B = 25
s = 5

Question 4.
Find the height of the triangular pyramid at the left.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 212

Answer:
The height of the triangular pyramid is 8 m

Explanation:
V = 24
\(\frac { 1 }{ 3 } \)Bh = 24
B = \(\frac { 1 }{ 2 } \)(3 x 6) = 9
\(\frac { 1 }{ 3 } \)(9)h = 24
h = 8

Question 5.
Pyramid C and pyramid D are similar. Find the volume of pyramid D.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 213

Answer:
The volume of pyramid D is 12 m³

Explanation:
\(\frac { volume of pyramid C }{ volume of pyramid D } \) = (\(\frac { pyramid C base }{ pyramid D base } \))³
\(\frac { 324 }{ V } \) = (\(\frac { 9 }{ 3 } \))³
V = 12

Question 6.
Find the volume of the composite solid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 214

Answer:
the volume of solid = 96

Explanation:
Volume of prism = Bh
B = 8 x 2 = 16
V = 16 x 5 = 80
\(\frac { 1 }{ 3 } \)Bh
= \(\frac { 1 }{ 3 } \)(16 x 3) = 16
the volume of solid = 16 + 80 = 96

Exercise 11.6 Volumes of Pyramids

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Explain the difference between a triangular prism and a triangular pyramid.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 1

Question 2.
REASONING
A square pyramid and a cube have the same base and height. Compare the volume of the square pyramid to the volume of the cube.

Answer:
Square pyramid = 1/3 Bh
Cube = BH
So, the volume of the square pyramid is 1/3 of the volume of the cube.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the volume of the pyramid.

Question 3.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 215
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 3

Question 4.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 216

Answer:
V = 6 in³

Explanation:
V =  \(\frac { 1 }{ 3 } \)Bh
B = 2 x 3 = 6
V = \(\frac { 1 }{ 3 } \)(6 x 3)

In Exercises 5 – 8, find the indicated measure.

Question 5.
A pyramid with a square base has a volume of 120 cubic meters and a height of 10 meters. Find the side length of the square base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 5

Question 6.
A pyramid with a square base has a volume of 912 cubic feet and a height of 19 feet. Find the side length of the square base.

Answer:
The side length of the square base is 12 ft

Explanation:
A pyramid with a square base has a volume of 912 cubic feet
h = 19
\(\frac { 1 }{ 3 } \)Bh = 912
\(\frac { 1 }{ 3 } \)B(19) = 912
B = 144
s = 12

Question 7.
A pyramid with a rectangular base has a volume of 480 cubic inches and a height of 10 inches. The width of the rectangular base is 9 inches. Find the length of the rectangular base.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 7

Question 8.
A pyramid with a rectangular base has a volume of 105 cubic centimeters and a height of 15 centimeters. The length of the rectangular base is 7 centimeters. Find the width of the rectangular base.

Answer:
The width of the rectangular base is 3 cm

Explanation:
A pyramid with a rectangular base has a volume of 105 cubic centimeters
h = 15
l = 7
\(\frac { 1 }{ 3 } \)Bh = 105
\(\frac { 1 }{ 3 } \)lbh = 105
\(\frac { 1 }{ 3 } \)(7 x 15 x b) = 105
b = 3

Question 9.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the pyramid.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 217
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 9

Question 10.
OPEN-ENDED
Give an example of a pyramid and a prism that have the same base and the same volume. Explain your reasoning.
Answer:
Let the rectangular prism have the base dimensions 4 x 2 nad a height of 5 so its volume is 4 x 2 x 5 = 40 cubic units
Therefore the base of the rectangular prism also have the dimensions of 4 x 2 and a height of 5 x 3 = 15 units so its volume V = 1/3 x 4 x 2 x 15 = 40 cubic units

In Exercises 11 – 14, find the height of the pyramid.

Question 11.
Volume = 15 ft3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 218
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 11

Question 12.
Volume = 224 in.3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 219

Answer:
The height of the pyramid is 10.5 in

Explanation:
Volume = 224
\(\frac { 1 }{ 3 } \)Bh = 224
B = 8² = 64
\(\frac { 1 }{ 3 } \)(64)h = 224
h = 10.5

Question 13.
Volume = 198 yd3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 220
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 13

Question 14.
Volume = 392 cm3
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 221

Answer:
The height of the pyramid is 12 cm

Explanation:
Volume = 392
\(\frac { 1 }{ 3 } \)Bh = 392
B = 14 x 7 = 98
\(\frac { 1 }{ 3 } \)(98)h = 392
h = 12

In Exercises 15 and 16, the pyramids are similar. Find the volume of pyramid B.

Question 15.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 222
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 15

Question 16.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 223

Answer:
Volume of A = 80

Explanation:
\(\frac { Volume of B }{ Volume of A } \) = (\(\frac { Side of B }{ side of A } \))³
\(\frac { V }{ 10 } \) = (\(\frac { 6 }{ 3 } \))³
V = 8 x 10
Volume of A = 80

In Exercises 17 – 20, find the volume of the composite solid.

Question 17.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 224
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 17

Question 18.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 225

Answer:
Composite solid volume = 306

Explanation:
Base area = \(\frac { 1 }{ 2 } \)bh = \(\frac { 1 }{ 2 } \)(12 x 9) = 54
Bottom solid volume V = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(54 x 10)
V = 180
Top solid volume v = \(\frac { 1 }{ 3 } \)(54 x 7) = 126
Composite solid volume = 180 + 126 = 306

Question 19.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 226
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 19

Question 20.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 227

Answer:
Composite solid volume = 1152

Explanation:
Volume of Box = 12 x 12 x 12 = 1728
Square pyramid volume = \(\frac { 1 }{ 3 } \)Bh = \(\frac { 1 }{ 3 } \)(144 x 12)
= 576
Composite solid volume = 1728 – 576 = 1152

Question 21.
ABSTRACT REASONING
A pyramid has a height of 8 feet and a square base with a side length of 6 feet.

a. How does the volume of the pyramid change when the base slays the same and the height is doubled?
b. How does the volume of the pyramid change when the height stays the same and the side length of the base is doubled?
C. Are your answers 10 parts (a) and (b) true for any square pyramid? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 21.2

Question 22.
HOW DO YOU SEE IT?
The cube shown is formed by three pyramids. each with the same square base and the same height. How could you use this to verify the formula for the volume of a pyramid?
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 228
Answer:

Question 23.
CRITICAL THINKING
Find the volume of the regular pentagonal pyramid. Round your answer to the nearest hundredth. In the diagram. m∠ABC = 35°
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 229
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 23.2

Question 24.
THOUGHT PROVOKING
A frustum of a pyramid is the part of the pyramid that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a square pyramid in terms of a, b, and h. (Hint: Consider the “missing” top of the pyramid and use similar triangles.)
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 230
Answer:

Question 25.
MODELING WITH MATHEMATICS
Nautical deck prisms were used as a safe way to illuminate decks on ships. The deck prism shown here is composed of the following three solids: a regular hexagonal prism with an edge length of 3.5 inches and a height of 1.5 inches, a regular hexagonal prism with an edge length of 3.25 inches arid a height of 0.25 inch, and a regular hexagonal pyramid with an edge length of 3 inches and a height of 3 inches. Find the volume of the deck prism.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 231
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 25.2

Maintaining Mathematical Proficiency

Find the value of X. Round your answer to the nearest tenth.

Question 26.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 232
Answer:
tan 35 = \(\frac { 9 }{ x } \)
0.7 = \(\frac { 9 }{ x } \)
x = 12.8

Question 27.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 233
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 27

Question 28.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 234
Answer:
tan 30 = \(\frac { x }{ 10 } \)
0.577 = \(\frac { x }{ 10 } \)
x = 5.77

Question 29.
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 235
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.6 Ques 29

11.7 Surface Areas and Volumes of Cones

Exploration 1

Finding the Surface Area of a Cone

Work with a partner: Construct a circle with a radius of 3 inches. Mark the circumference of the circle into six equal parts, and label the length of each part. Then cut out one sector of the circle and make a cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 236
a. Explain why the base of the cone is a circle. What are the circumference and radius
of the base?
Answer:

b. What is the area of the original circle? What is the area with one sector missing?
Answer:

c. Describe the surface area of the cone, including the base. Use your description to find the surface area.
Answer:

Exploration 2

Finding the Volume of a Cone

Work with a partner: The cone and the cylinder have the same height and the same circular base.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 237
When the cone is filled with sand and poured into the cylinder. it takes three cones to fill the cylinder.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 238
Use this information to write a formula for the volume V of a cone.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a cone?
Answer:

Question 4.
In Exploration 1, cut another sector from the circle and make a cone. Find the radius of the base and the surface area of the cone. Repeat this three times, recording your results in a table. Describe the pattern.
Answer:

Lesson 11.7 Surface Areas and Volumes of Cones

Monitoring progress

Question 1.
Find the surface area of the right cone.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 239

Answer:
The surface area of the right cone is 436.17 m²

Explanation:
r = 7.8
l = 10
S = πr² + πrl
S = π(7.8)² + π(7.8 x 10)
S = 436.17

Find the volume of the cone.

Question 2.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 240

Answer:
The volume of the cone is 2206.44 in³

Explanation:
r = 7, h = 13
l = √13² – 7²= 10.95
S = πr² + πrl
S = π7² + π(7 x 10.95)
S = 394.74
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 13)
V = 2206.44

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 241

Answer:
The volume of the cone is 163.4 m³

Explanation:
h = √8² – 5² = 6.24
Volume V = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 6.24)
V = 163.4

Question 4.
Cone C and cone D are similar. Find the volume of cone D.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 242

Answer:
Volume of cone D = 18.84 cm³

Explanation:
\(\frac { Volumeof cone C }{ Volume of cone D } \) = (\(\frac { height of C }{ height of D } \))³
\(\frac { 384π }{ Volume of cone D } \)= (\(\frac { 8 }{ 2 } \))³
Volume of cone D = 18.84

Question 5.
Find the volume of the composite solid.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 243

Answer:
Composite solid volume = 329.86 cm³

Explanation:
Volume of cylinder = πr²h = π(3)² x 10 = 90π
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
= \(\frac { 1 }{ 3 } \)(π x 3² x 5)
= 15π
Composite solid volume = 15π + 90π = 105π

Exercise 11.7 Surface Areas and Volumes of Cones

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe the differences between pyramids and cones. Describe their similarities.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 1

Question 2.
COMPLETE THE SENTENCE
The volume of a cone with radius r and height h is \(\frac{1}{3}\) the volume of a(n) __________ with radius r and height h.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the right cone.

Question 3.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 244
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 3

Question 4.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 245

Answer:
The surface area of cone is 219.44 sq cm.

Explanation:
S = πr² + πrl
S = π(5.5)² + π(5.5 x 7.2)
S = 219.44

Question 5.
A right cone has a radius of 9 inches and a height of 12 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 5

Question 6.
A right cone has a diameter of 11.2 feet and a height of 19.2 feet.

Answer:
The surface area is 421.52 sq ft.

Explanation:
r = 5.6
h = 19.2
l = √19.2² – 5.6² = 18.36
Surface area S = πr² + πrl
S = π(5.6)² + π(5.6 x 18.36)
S = 421.52

In Exercises 7 – 10, find the volume of the cone.

Question 7.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 246
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 7

Question 8.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 247

Answer:
The volume is 2.09 cubic meter

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π(1)² x 2)
V = 2.09

Question 9.
A cone has a diameter of 11.5 inches and a height of 15.2 inches.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 9

Question 10.
A right cone has a radius of 3 feet and a slant height of 6 feet.

Answer:
The volume is 56.54 cubic ft

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 6)
V = 56.54

In Exercises 11 and 12, find the missing dimension(s).

Question 11.
Surface area = 75.4 cm2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 11

Question 12.
Volume = 216π in.3

Answer:
The radius is 6.13 in

Explanation:
Volume = 216π in.3
\(\frac { 1 }{ 3 } \)(πr²h) = 216
\(\frac { 1 }{ 3 } \)(πr² x 18) = 216
r = 6.13

In Exercises 13 and 14, the cones are similar. Find the volume of cone B.

Question 13.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 248
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 13

Question 14.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 249

Answer:
Volume of cone B = 24.127

Explanation:
\(\frac { Volumeof cone A }{ Volume of cone B } \) = (\(\frac { height of A }{ height of B } \))³
\(\frac { 120π }{ Volume of cone B } \) = (\(\frac { 10 }{ 4 } \))³
Volume of cone B = 24.127

In Exercises 15 and 16, find the volume of the composite solid.

Question 15.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 250
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 15

Question 16.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 251

Answer:
Volume of the composite solid = 97.93 cubic m

Explanation:
Volume of box = lbh
V = 51 x 5.1 x 5.1 = 132.651
Cone volume = \(\frac { 1 }{ 3 } \)(πr²h)
v = \(\frac { 1 }{ 3 } \)(π x 2.55² x 5.1)
v = 34.72
Volume of the composite solid = 132.651 – 34.72 = 97.93

Question 17.
ANALYZING RELATIONSHIPS
A cone has height h and a base with radius r. You warn to change the cone so its volume is doubled. What is the new height if you change only the height? What is the new radius if you change only the radius? Explain.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 17

Question 18.
HOW DO YOU SEE IT
A snack stand serves a small order of popcorn in a cone-shaped container and a large order of popcorn in a cylindrical container. Do not perform any calculations.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 252
a. How many small containers of popcorn do you have to buy to equal the amount of popcorn in a large container? Explain.

Answer:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 3² x 8) = 75.39
Volume of cylinder = πr²h = π x 3² x 8 = 226.19
Volume of cylinder / Volume of cone = \(\frac { 226.19 }{ 75.39 } \) = 3
You have to buy 3 small containers of popcorn to equal the amount of popcorn in a large containe.

b. Which container gives you more popcorn for your money? Explain.
Answer:
$1.25 -> 75.39 i.e $1 = 60.312
$2.50 -> 226.19 i.e $1 = 90.47
So, large containers gives you more popcorn for your money

In Exercises 19 and 20. find the volume of the right cone.

Question 19.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 253
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 19

Question 20.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 254

Answer:
Volume of cone is 575.62 cubic yd

Explanation:
tan 32 = \(\frac { 7 }{ h } \)
h = 11.21
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 11.21)
V = 575.62

Question 21.
MODELING WITH MATHEMATICS
A cat eats hail a cup of food, twice per day. Will the automatic pet feeder hold enough food for 10 days? Explain your reasoning. (1 cup ≈ 14.4 in.3)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 255
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 21

Question 22.
MODELING WITH MATHEMATICS
During a chemistry lab, you use a funnel to pour a solvent into a flask. The radius of the funnel is 5 centimeters and its height is 10 centimeters. You pour the solvent into the funnel at a rate of 80 milliliters per second and the solvent flows out of the funnel at a rate of 65 milliliters per second. How long will it be before the funnel overflows? (1 mL = 1 cm3)
Answer:
17.45 seconds

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 5² x 10)
V = 261.8
\(\frac { 261.8 }{ 15 } \) = 17.45

Question 23.
REASONING
To make a paper drinking cup, start with a circular piece of paper that has a 3-inch radius, then follow the given steps. How does the surface area of the cup compare to the original paper circle? Find m∠ABC.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 256
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 23

Question 24.
THOUGHT PROVOKING
A frustum of a cone is the part of the cone that lies between the base and a plane parallel to the base, as shown. Write a formula for the volume of the frustum of a cone in terms of a, b, and h. (Hint: Consider the “missing” top of the cone and use similar triangles.)
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 257
Answer:
Volume V  = (1/3) * π * h * (r1² + r2² + (r1 * r2))

Question 25.
MAKING AN ARGUMENT
In the figure, the two cylinders are congruent The combined height of the two smaller cones equals the height of the larger cone. Your friend claims that this means the total volume of the two smaller cones is equal to the volume of the larger cone. Is your friend correct? Justify your answer.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 258
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 25

Question 26.
CRITICAL THINKING
When the given triangle is rotated around each of its sides. solids of revolution are formed. Describe the three solids and find their volumes. Give your answers in terms of π.
Big Ideas Math Answers Geometry Chapter 11 Circumference, Area, and Volume 259
Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 27.
area of a circle with a radius of 7 feet
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 27

Question 28.
area of a circle with a diameter of 22 centimeters

Answer:
d = 11
A = πr²
A = 121π

Question 29.
diameter of a circle with an area of 256 square meters
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.7 Ques 29

Question 30.
radius of a circle with an area of 529 π square inches
Answer:
A = πr²
529π = πr²
r = 23

11.8 Surface Areas and Volumes of Spheres

Exploration 1

Finding the Surface Area of a Sphere

Work with a partner: Remove the covering from a baseball or softball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 260
You will end up with two “figure 8” pieces of material, as shown above. From the amount of material it takes to cover the ball, what would you estimate the surface area S of the ball to be? Express your answer in terms of the radius r of the ball.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 261
Use the Internet or some other resource to confirm that the formula you wrote for the surface area of a sphere is correct.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to identify relevant external mathematical resources, such as content located on a website.
Answer:

Exploration 2

Finding the volume of a sphere

Work with a partner: A cylinder is circumscribed about a sphere, as shown. Write a formula for the volume V of the cylinder in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 262
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 263
When half of the sphere (a hemisphere) is filled with sand and poured into the cylinder, it takes three hemispheres to till the cylinder. Use this information to write a formula for the volume V of a sphere in terms of the radius r.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 264
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 265
Answer:

Communicate Your Answer

Question 3.
How can you find the surface area and the volume of a sphere?
Answer:

Question 4.
Use the results of Explorations 1 and 2 to find the surface area and the volume of a sphere with a radius of(a) 3 inches and (b) 2 centimeters.
Answer:

Lesson 11.8 Surface Areas and Volumes of Spheres

Monitoring Progress

Find the surface area of the sphere.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 266

Answer:
The surface area of the sphere is 5026.54 ft²

Explanation:
D = 40
r = 20
The surface area of the sphere = 4πr²
S = 4 x π x (20)²
S = 5026.54 ft²

Question 2.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 267

Answer:
The surface area of the sphere is 113.09 ft²

Explanation:
Circumference C = 6π
2πr = 6π
r = 3
The surface area of the sphere = 4πr²
S = 4π x 3²
S = 113.09

Question 3.
Find the radius of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 268

Answer:
The radius of the sphere is 2.73 m

Explanation:
The surface area of the sphere = 4πr²
30π = 4πr²
r = 2.73

Question 4.
The radius of a sphere is 5 yards. Find the volume of the sphere.

Answer:
The volume of the sphere is 523.59 yards³

Explanation:
r = 5
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 5³
V = 523.59 yards³

Question 5.
The diameter of a sphere is 36 inches. Find the volume of the sphere.

Answer:
The volume of the sphere is 24429.02 in³

Explanation:
D = 36
r = 18
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 18³
V = 24429.02

Question 6.
The surface area of a sphere is 576π square centimeters. Find the volume of the sphere.

Answer:
The volume of the sphere is 2304π cm³

Explanation:
The surface area of the sphere = 4πr²
576π = 4πr²
r = 12
The volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 12³
V = 2304π

Question 7.
Find the volume of the composite solid at the left.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 269

Answer:
The volume of the composite solid is 7.324 m³

Explanation:
The volume of cone = πr²\(\frac { h }{ 3 } \)
= π x 1² x \(\frac { 5 }{ 3 } \) = 5.23
The volume of sphere = \(\frac { 4 }{ 3 } \)πr³
= \(\frac { 4 }{ 3 } \)π x 1³ = 4.188
The volume of the composite solid = The volume of cone + The volume of sphere/2
= 5.23 + 4.188/2
= 7.324 m³

Exercise 11.8 Surface Areas and Volumes of Spheres

Question 1.
VOCABULARY
When a plane intersects a sphere. what must be true for the intersection to be a great circle?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 1

Question 2.
WRITING
Explain the difference between a sphere and a hemisphere.

Answer:
Hemisphere is a related term of the sphere. Sphere and hemisphere are three-dimensional solids. The volume of sphere is \(\frac { 4 }{ 3 } \)πr³ and hemisphere volume is \(\frac { 2 }{ 3 } \)πr³. The surface area of the sphere is 4πr² and hemisphere surface area is 3πr².

Monitoring progress and Modeling with Mathematics

In Exercises 3 – 6, find the surface area of the sphere.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 270
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 271

Answer:
The surface area of the sphere is 225π cm²

Explanation:
The surface area of the sphere = 4πr²
S = 4π x 7.5²
S = 225π

Question 5.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 272
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 273

Answer:
The surface area of the sphere is 8π ft²

Explanation:
C = 4π
2πr = 4π
r = 2
The surface area of the sphere = 4πr²
S = 4π x 2²
S = 8π

In Exercises 7 – 10. find the indicated measure.

Question 7.
Find the radius of a sphere with a surface area of 4π square feet.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 7

Question 8.
Find the radius of a sphere with a surface area of 1024π square inches.

Answer:
The radius of a sphere is 16 in

Explanation:
The surface area of the sphere = 1024π
4πr² = 1024π
r = 16

Question 9.
Find the diameter of a sphere with a surface area of 900π square meters.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 9

Question 10.
Find the diameter of a sphere with a surface area of 196π square centimeters.

Answer:
The diameter of a sphere is 14 cm

Explanation:
The surface area of the sphere = 196π
4πr² = 196π
r = 7
D = 2(7) = 14

In Exercises 11 and 12, find the surface area of the hemisphere.

Question 11.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 274
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 11

Question 12.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 275

Answer:
The surface area of the hemisphere is 108π in²

Explanation:
D = 12, r = 6
The surface area of the sphere = 3πr²
S = 3π x 6²
S = 108π

In Exercises 13 – 18. find the volume of the sphere.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 276
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 277

Answer:
The volume of the sphere is 268.08 ft³

Explanation:
r = 4 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4³
V = 268.08 ft

Question 15.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 278
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 279

Answer:
The volume of the sphere is 1436.75 ft³

Explanation:
D = 14 ft
r = 7 ft
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75 ft

Question 17.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 282
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 17

Question 18.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 280

Answer:
The volume of the sphere is 179.89 in³

Explanation:
C = 7π
2πr = 7π
r = 3.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 3.5³
V = 179.89 in

In Exercises 19 and 20, find the volume of the sphere with the given surface area.

Question 19.
Surface area = 16π ft2
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 19

Question 20.
Surface area = 484π cm2

Answer:
The volume of the sphere is 5575.27 cm³

Explanation:
Surface area = 484π
4πr² = 484π
r = 11
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 11³
V = 5575.27

Question 21.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 281
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 21

Question 22.
ERROR ANALYSIS
Describe and correct the error in finding the volume of the sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 283

Answer:
Diameter = 3
radius = 1.5
Volume of the sphere V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x (1.5)³
V = 14.137 cubic in

In Exercises 23 – 26, find the volume of the composite solid.

Question 23.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 284
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 23

Question 24.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 285

Answer:
Volume is 288π ft³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
volume of the cone = πr²\(\frac { h }{ 3 } \)
= π x 6² x \(\frac { 12 }{ 3 } \) = 144π
Area of circle = πr² = π x 6² = 36π
Volume of hemipshere + volume of the cone = 144π + 144π = 288π

Question 25.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 286
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 25

Question 26.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 287

Answer:
The volume of solid is 296π m³

Explanation:
Volume of hemipshere = \(\frac { 2 }{ 3 } \)πr³
= \(\frac { 2 }{ 3 } \)π x 6³ = 144π
Volume of cylinder = πr²h
= π x 6² x 14 = 504π
Volume of solid = 504π – 2(144π) = 296π

In Exercises 27 – 32, find the surface area and volume of the ball.

Question 27.
bowling ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
d = 8.5 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 27

Question 28.
basketball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 289
C = 29.5 in.

Answer:
The surface area is 277 in², volume is 43212.27 in³

Explanation:
C = 29.5
2πr = 29.5
r = 4.69
Surface area = 4πr²
S = 4π x 4.69² = 277
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 4.69³
V = 43212.27

Question 29.
softball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 290
C = 12 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 29

Question 30.
golf ball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 291
d = 1.7 in.

Answer:
The surface area is 9.07 in², volume is 2.57 in³

Explanation:
d = 1.7
r = 0.85
Surface area = 4πr²
S = 4π x 0.85² = 9.07
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 0.85³
V = 2.57

Question 31.
volleyball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 292
C = 26 in.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 31

Question 32.
baseball
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 293
C = 9 in.

Answer:
The surface area is 25.78 in², volume is 12.24 in³

Explanation:
C = 9
2πr = 9
r = 1.43
Surface area = 4πr²
S = 4π x 1.43²
S = 25.78
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.43³
V = 12.24

Question 33.
MAKING AN ARGUMENT
You friend claims that if the radius of a sphere is doubled, then the surface area of the sphere will also be doubled. Is our friend correct? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 33

Question 34.
REASONING
A semicircle with a diameter of 18 inches is rotated about its diameter. Find the surface area and the volume of the solid formed.

Answer:
The surface area is 1018 in², volume is 3054.02 in³

Explanation:
Diameter = 18
radius r = 9
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 9³
V = 3054.02
Surface area = 4πr²
S = 4π x 9²
S = 1018

Question 35.
MODELING WITH MATHEMATICS
A silo has the dimensions shown. The top of the silo is a hemispherical shape. Find the volume of the silo.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 294
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 35

Question 36.
MODELING WITH MATHEMATICS
Three tennis balls are stored in a cylindrical container with a height of 8 inches and a radius of 1.43 inches. The circumference of a tennis ball is 8 inches.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 295
a. Find the volume of a tennis ball.

Answer:
C = 8 in
2πr = 8
r = 1.27
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.27³ = 8.64
The volume of tennis ball = 8.64 in³

b. Find the amount of space within the cylinder not taken up by the tennis balls.
Answer:
The surface area of tennis ball S = 4πr²
S = 4π x 1.27² = 20.26
Area of cylinder s = 2πrh+2πr²
s = 2π x 1.43 x 8+2π x 1.43²
s = 84.72
Remaining space = 84.72 – 20.26 = 64.46 in²

Question 37.
ANALYZING RELATIONSHIPS
Use the table shown for a sphere.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 296
a. Copy and complete the table. Leave your answers in terms of π.
b. What happens to the surface area of the sphere when the radius is doubled? tripled? quadrupled?
c. What happens to the volume of the sphere when the radius is doubled? tripled? quadrupled?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 37

Question 38.
MATHEMATICAL CONNECTIONS
A sphere has a diameter of 4(x + 3) centimeters and a surface area of 784 π square centimeters. Find the value of x.

Answer:
x =11

Explanation:
Surface area = 4πr²
784π = πr²
r = 28
2r = diameter = 4(x + 3)
r = 2(x + 3)
28 = 2(x + 3)
x = 11

Question 39.
MODELING WITH MATHEMATICS
The radius of Earth is about 3960 miles. The radius of the moon is about 1080 miles.
a. Find the surface area of Earth and the moon.
b. Compare the surface areas of Earth and the moon.
c. About 70% of the surface of Earth is water. How many square miles of water are on Earth’s surface?
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 39

Question 40.
MODELING WITH MATHEMATICS
The Torrid Zone on Earth is the area between the Tropic of Cancer and the Tropic of Capricorn. The distance between these two tropics is about 3250 miles. You can estimate the distance as the height of a cylindrical belt around the Earth at the equator.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 297
a. Estimate the surface area of the Torrid Zone. (The radius of Earth is about 3960 miles.)
Answer:
Surface area of cylinder = 2πrh
S = 2π x 3960 x 3250 = 80875080
surface area of earth = 4πr²
= 4π x 3960² = 197086348.8

b. A meteorite is equally likely to hit anywhere on Earth. Estimate the probability that a meteorite will land in the Torrid Zone.
Answer:
Probability of meteorites hitting the torrid zone = 80875080/197086348.8 = 0.4104

Question 41.
ABSTRACT REASONING
A sphere is inscribed in a cube with a volume of 64 cubic inches. What is the surface area of the sphere? Explain your reasoning.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 41

Question 42.
HOW DO YOU SEE IT?
The formula for the volume of a hemisphere and a Cone are shown. If each solid has the same radius and r = h, which solid will have a greater volume? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 298
Answer:
The hemisphere has the highest volume.

Explanation:
Volume of hemisphere v = \(\frac { 2 }{ 3 } \)πr³
Volume of cone V = \(\frac { 1 }{ 3 } \)πr²h
If r = h
Volume of cone V = \(\frac { 1 }{ 3 } \)πr² x r = \(\frac { 1 }{ 3 } \)πr³
So, the hemisphere has the highest volume.

Question 43.
CRITICAL THINKING
Let V be the volume of a sphere. S be the surface area of the sphere, and r be the radius of the sphere. Write an equation for V in terms of r and S. (Hint: Start with the ratio \(\frac{V}{S}\).)
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 43

Question 44.
THOUGHT PROVOKING
A spherical lune is the region between two great circles of a sphere. Find the formula for the area of a lune.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 299
Answer:

Question 45.
CRITICAL THINKING
The volume of a right cylinder is the same as the volume of a sphere. The radius of the sphere is 1 inch. Give three possibilities for the dimensions of the cylinder.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 45.2

Question 46.
PROBLEM SOLVING
A spherical cap is a portion of a sphere cut off by a plane. The formula for the volume of a spherical cap is V = \(\frac{\pi h}{6}\) (3a2 + h2), where a is the radius of the base of the cap and h is the height of the cap. Use the diagram and given information to find the volume of each spherical cap.
Big Ideas Math Geometry Answer Key Chapter 11 Circumference, Area, and Volume 300
a. r = 5ft, a = 4ft
Answer:

b. r = 34 cm, a = 30 cm
Answer:

c. r = 13 m, h = 8 m
Answer:

d. r=75 in., h = 54in.
Answer:

Question 47.
CRITICAL THINKING
A sphere with a radius of 2 inches is inscribed in a right cone with a height of 6 inches. Find the surface area and the volume of the cone.
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.1
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.2
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 47.3

Maintaining Mathematical Proficiency

Solve the triangle. Round decimal answers to the nearest tenth.

Question 48.
A = 26°, C = 35°, b = 13

Answer:
B = 119°, a = 7.16, c = 9.5

Explanation:
B = 180 – (26 + 35) = 119
\(\frac { sin A }{ a } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 26 }{ a } \) = \(\frac { sin 119 }{ 13 } \)
a = 7.16
\(\frac { sin C }{ c } \) = \(\frac { sin B }{ b } \)
\(\frac { sin 35 }{ c } \) = \(\frac { sin 119 }{ 13 } \)
c = 9.5

Question 49.
B = 102°, C = 43°, b = 21
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 49

Question 50.
a = 23, b = 24, c = 20

Answer:
A = 62.2, B = 65.5, C = 49.4

Explanation:
a² = b² + c² – 2bc cos A
23² = 24²+ 20² – 2(24 x 20) cos A
A = 62.2
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin B }{ 24 } \)
B = 65.5
\(\frac { sin 62.2 }{ 23 } \) = \(\frac { sin C }{ 20 } \)
C = 49.4

Question 51.
A = 103°, b = 15, c = 24
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 11.8 Ques 51

Circumference, Area, and Volume Review

11.1 Circumference and Arc Length

Find the indicated measure.

Question 1.
diameter of ⊙P
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 301

Answer:
diameter of ⊙P is 29.99

Explanation:
Circumference = 94.24
πd = 94.24
d = 29.99

Question 2.
circumference of ⊙F
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 302

Answer:
circumference of ⊙F = 56.57

Explanation:
5.5 = \(\frac { 35 }{ 360 } \) . C
C = 56.57

Question 3.
arc length of \(\widehat{A B}\)
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 303

Answer:
arc length of \(\widehat{A B}\) = 26.09

Explanation:
arc length of \(\widehat{A B}\) = \(\frac { 115 }{ 360 } \) . 2π(13)
= 26.09

Question 4.
A mountain bike tire has a diameter of 26 inches. To the nearest foot, how far does the tire travel when it makes 32 revolutions?

Answer:
The tire travels 2613.80 inches.

Explanation:
D = 26 in
r = 13 in
Circumference C = 2π(13) = 81.68
32 revolutions = 32 x 81.68 = 2613.80

11.2 Areas of Circles and Sectors

Find the area of the blue shaded region.

Question 5.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 304

Answer:
Area = \(\frac { 240 }{ 360 } \) . π(9)²
= 169.64

Question 6.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 305

Answer:
Area of shaded region = 11.43

Explanation:
Area of rectangle = 6 x 4 = 24
Area of semicircle = π(2)² = 4π
Area of shaded region = 24 – 4π = 11.43

Question 7.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 306

Answer:
Area of shaded region = 173.13

Explanation:
Area of small region = 27.93 = \(\frac { 50 }{ 360 } \) . πr²
πr² = 201.096
r = 8
Area of shaded region = \(\frac { 310 }{ 360 } \) . π(8)²
= 173.13

11.3 Areas of Polygons

Find the area of the kite or rhombus.

Question 8.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 307
Answer:
Area = 65

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(13 x 20)
A = 65

Question 9.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 308
Answer:
Area = 48

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(16 x 12)
A = 48

Question 10.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 309
Answer:
Area = 52.5

Explanation:
Area = \(\frac { 1 }{ 4 } \)(d₁d₂)
A = \(\frac { 1 }{ 4 } \)(14 x 15)
A = 52.5

Find the area of the regular polygon.

Question 11.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 310
Answer:
Area = \(\frac { 3√3 }{ 2 } \)a²
A = \(\frac { 3√3 }{ 2 } \)(8.8)²
A = 201.195

Question 12.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 311
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(9 . 5.2 . 7.6)
A = 117.84

Question 13.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 312
Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(5 . 4 .  3.3)
A = 33

Question 14.
A platter is in the shape of a regular octagon with an apothem of 6 inches. Find the area of the platter.

Answer:
Area = \(\frac { 1 }{ 2 } \)(n . a. s)
A = \(\frac { 1 }{ 2 } \)(8 . 6 . sin 45)
A = 16.97

11.4 Three-Dimensional Figures

Sketch the solid produced by rotating the figure around the given axis. Then identify and describe the solid.

Question 15.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 313
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume

Question 16.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 314
Answer:
Big Ideas Math Geometry Answers Chapter 11 Circumference, Area, and Volume 15

Question 17.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 315
Answer:

Describe the cross section formed by the intersection of the plane and the solid.

Question 18.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 316
Answer:
The cross section is a rectangle.

Question 19.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 317
Answer:
The cross-section is a square.

Question 20.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 318
Answer:
the cross-section is a triangle.

11.5 Volumes of Prisms and Cylinders

Find the volume of the solid.

Question 21.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 319

Answer:
Volume = lbh
V = 3.6 x 2.1 x 1.5 = 113.4 m³

Question 22.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 320

Answer:
Volume = πr²h
V = π(2)² x 8 = 100.53 mm³

Question 23.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 321

Answer:
Pentagon area = 6.88
Volume = Area x height
V = 6.88 x 4 = 27.52 yd³

11.6 Volumes of Pyramids

Find the volume of the pyramid.

Question 24.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 322

Answer:
Volume V = Base area x height/3
Base Area = 9² = 81
V = 81 x 7/3 = 189 ft³

Question 25.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 323
Answer:
Volume V = Base area x height/3
Base Area = 4 x 15 = 60
V = 60x 20/3 = 400 yd³

Question 26.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 324

Answer:
Volume V = Base area x height/3
base area = 18 x 10 = 180
V = 180 x 5/3 = 300 m³

Question 27.
The volume of a square pyramid is 60 cubic inches and the height is 15 inches. Find the side length of the square base.

Answer:
The side length of the square base is 3.46 in

Explanation:
The volume of a square pyramid is 60 cubic inches
V = 60
s²h/3 = 60
s² x 15/3 = 60
s² = 12
s = 3.46

Question 28.
The volume of a square pyramid is 1024 cubic inches. The base has a side length of 16 inches. Find the height of the pyramid
Answer:
The volume of a square pyramid is 1024 cubic inches
s²h/3 = 1024
16²h = 3072
h = 12

11.7 Surface Areas and Volumes of Cones

Find the surface area and the volume of the cone.

Question 29.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 325
Answer:
Surface area is 678.58 cm²
volume is 1017.87 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 9² + π x 9 x 15
S = 678.58
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 9² x 12)
V = 1017.87

Question 30.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 326

Answer:
Surface area is 2513.27 cm²
volume is 8042.47 cm³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 16² + π x 16 x 34
S = 2513.27
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 16² x 30)
V = 8042.47

Question 31.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 327

Answer:
Surface area is 439.82 m²
volume is 562.102 m³

Explanation:
Surface area of cone S = πr² + πrl
S = π x 7² + π x 7 x 13
S = 439.82
h = √13² – 7² = 10.95
Volume of cone = \(\frac { 1 }{ 3 } \)(πr²h)
V = \(\frac { 1 }{ 3 } \)(π x 7² x 10.95)
V = 562.102

Question 32.
A cone with a diameter of 16 centimeters has a volume of 320π cubic centimeters. Find the height of the cone.

Answer:
The height of the cone = 15 cm.

Explanation:
r = 8
Volume V = 320π
\(\frac { 1 }{ 3 } \)(πr²h) = 320π
\(\frac { 1 }{ 3 } \)(π x 8² x h) = 320π
h = 15

11.8 Surface Areas and Volumes of Spheres

Find the surface area and the volume of the sphere.

Question 33.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 328

Answer:
The surface area is 615.75 in², volume is 1436.75 in³

Explanation:
Surface area S = 4πr²
S = 4π x 7²
S = 615.75
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 1436.75

Question 34.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 329

Answer:
The surface area is 907.92 ft², volume is 2572.44 ft³

Explanation:
d = 17
r = 8.5
Surface area S = 4πr²
S = 4π x 8.5²
S = 907.92
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 7³
V = 2572.44

Question 35.
Big Ideas Math Geometry Solutions Chapter 11 Circumference, Area, and Volume 330

Answer:
The surface area is 2827.43 ft², volume is 14137.16 ft³

Explanation:
C = 30π
2πr = 30π
r = 15
Surface area S = 4πr²
S = 4π x 15²
S = 2827.43
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 15³
V = 14137.16

Question 36.
The shape of Mercury can be approximated by a sphere with a diameter of 4880 kilometers. Find the surface area and the volume of Mercury.

Answer:
The surface area and the volume of Mercury is 23814400π, 19369045330π

Explanation:
d = 4880
r = 2440
Surface area S = 4πr²
S = 4π x 2440² = 23814400π
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 2440³
V = 19369045330π

Question 37.
A solid is composed of a cube with a side length of 6 meters and a hemisphere with a diameter of 6 meters. Find the volume of the composite solid.

Answer:
Volume of the composite solid = 272.52

Explanation:
Volume of cube = a³
= 6³ = 216
Volume of hemisphere = \(\frac { 4 }{ 6 } \)πr³
= \(\frac { 4 }{ 6 } \)π x 3³ = 18π
Volume of the composite solid = 216 + 18π = 272.52

Circumference, Area, and Volume Test

Find the volume of the solid.

Question 1.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 331

Answer:
Volume = 2577.29 m³

Explanation:
Volume = \(\frac { 3√3 }{ 2 } \)a²h
= \(\frac { 3√3 }{ 2 } \) x 8² x 15.5
= 2577.29

Question 2.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 332

Answer:
Volume is 17.157 ft³

Explanation:
d = 3.2
r = 1.6
Volume V = \(\frac { 4 }{ 3 } \)πr³
V = \(\frac { 4 }{ 3 } \)π x 1.6³
V = 17.157

Question 3.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 333

Answer:
Volume of sloid = 402.11 m³

Explanation:
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 4² x 3
= 50.26
Volume of cylinder = πr²h
= π x 4² x 6 = 301.59
Volume of sloid = 2(50.26) + 301.59 = 402.11

Question 4.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 334

Answer:
Volume of solid = 106.66

Explanation:
Volume of rectangular box = 5 x 2 x 8 = 80
Volume of pyramid = 80/3 = 26.66
Volume of solid = 80 + 26.66 = 106.66

Find the indicated measure.

Question 5.
circumference of ⊙F
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 335

Answer:
circumference of ⊙F is 109.71 in

Explanation:
64 = \(\frac { 210 }{ 360 } \) • C
C = 109.7

Question 6.
m\(\widehat{G H}\)
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 336

Answer:
m\(\widehat{G H}\) = 74.27

Explanation:
35 = \(\frac { x }{ 360 } \) • 2π x 27
x = 74.27

Question 7.
area of shaded sector
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 337

Answer:
Area is 142.41 in²

Explanation:
Area = \(\frac { 360 – 105 }{ 360 } \) • π x 8²
Area = 142.41

Question 8.
Sketch the composite solid produced by rotating the figure around the given axis. Then identify and describe the composite solid.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 338
Answer:

Question 9.
Find the surface area of a right cone with a diameter of 10 feet and a height of 12 feet.

Answer:
The surface area is 486.7 sq ft

Explanation:
l² = r² + h²
l² = 5² + 12²
l = 13
Surface area S = πr² + 2πrl
S = π x 5² + 2π x 5 x 13
S = 486.7

Question 10.
You have a funnel with the dimensions shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 339
a. Find the approximate volume of the funnel.

Answer:
Volume = \(\frac { 1 }{ 3 } \)πr²h
= \(\frac { 1 }{ 3 } \)π x 6² x 10
= 376.99

b. You use the funnel to put oil in a ear. Oil flows out of the funnel at a rate of 45 milliliters per second. How long will it take to empty the funnel when it is full of oil? (1 mL = 1 cm3)
Answer:

c. How long would it take to empty a funnel with a radius of 10 centimeters and a height of 6 centimeters if oil flows out of the funnel at a rate of 45 milliliters per second?
Answer:

d. Explain why you can claim that the time calculated in part (c) is greater than the time calculated in part (b) without doing any calculations.
Answer:

Question 11.
A water bottle in the shape of a cylinder has a volume of 500 cubic centimeters. The diameter of a base is 7.5 centimeters. What is the height of the bottle? Justify your answer.

Answer:
The height of the bottle is 11.3 cm

Explanation:
Volume of cylinder = 500
πr²h = 500
π(3.75)²h = 500
h = 11.3 cm

Question 12.
Find the area of a dodecagon (12 sides) with a side length of 9 inches.

Answer:
Area is 237.31

Explanation:
Area = \(\frac { 1 }{ 4 } \)πa²cot(π/n)
= \(\frac { 1 }{ 4 } \)π x 9² x cot(π/12)
= 237.31

Question 13.
In general, a cardboard fan with a greater area does a better job of moving air and cooling you. The fan shown is a sector of a cardboard circle. Another fan has a radius of 6 centimeters and an intercepted are of 150°. Which fan does a better job of cooling you?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 340

Answer:

Explanation:

Circumference, Area, and Volume Cumulative Assessment

Question 1.
Identify the shape of the cross section formed by the intersection of the plane and the
solid.

a.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 341
Answer:
The cross-section is a trapezoid.

b.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 342
Answer:
The cross-section is a pentagon.

c.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 343
Answer:
The cross-section is a rectangle.

Question 2.
In the diagram, Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 is tangent to ⊙P at Q and \(\overline{P Q}\) is a radius of ⊙P? What must be true about Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 344 and \(\overline{P Q}\)? Select all that apply.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 345
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 346
Answer:
PQ is perpendicular to RS.

Question 3.
A crayon can be approximated by a composite solid made from a cylinder and a cone.
A crayon box is a rectangular prism. The dimensions of a crayon and a crayon box containing 24 crayons are shown.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 348
a. Find the volume of a crayon.
Answer:
The volume of a crayon = πr²h + \(\frac { 1 }{ 3 } \)πr²h
= π x 4.25² x 80 + \(\frac { 1 }{ 3 } \)π x 3.25² x 10
= 4539.6 + 110.61
= 4650.21 mm³

b. Find the amount of space within the crayon box not taken up by the crayons.
Answer:
Volume of box = 94 x 28 x 71 = 186872
The volume of a crayon = 4650.21
Remaining space = 186872 – 24 x 4650.21
= 75266.96

Question 4.
What is the equation ol the line passing through the point (2, 5) that is parallel to the line x + \(\frac{1}{2}\)y = – 1?
(A) y = – 2x + 9
(B) y = 2x + 1
(C) y = \(\frac{1}{2}\)x + 4
(D) y = –\(\frac{1}{2}\)x + 6
Answer:
(A) y = – 2x + 9

Explanation:
x + \(\frac{1}{2}\)y = – 1
y = -2 – 2x
The slope of the line is -2
The euation of line is y – 5 = -2(x – 2)
y – 5 = -2x + 4
y = -2x + 9

Question 5.
The top of the Washington Monument in Washington, D.C., is a square pyramid, called a pyramidion. What is the volume of the pyramidion?
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 347
(A) 22,019.63 ft3
(B) 172,006.91 ft3
(C) 66,058.88 ft3
(D) 207,530.08 ft3

Answer:
(A) 22,019.63 ft3

Explanation:
Volume = a²\(\frac { h }{ 3 } \)
= 34.5² x \(\frac { 55.5 }{ 3 } \)
= 22019.62

Question 6.
Prove or disprove that the point (1, √3 ) lies on the circle centered at the origin and containing the point (0, 2).
Answer:
We consider the circle centered at the origin and containing the point (0, 2).
Therefore, we canconclude that rdaius is 2 and points be (0, 0), (1, √3)
distance = √(1 – 0)² + (√3 – 0)² = 2
As radius and distance are same. The point B(1, √3) lies on the circle.

Question 7.
Your friend claims that the house shown can be described as a composite solid made from a rectangular prism and a triangular prism. Do you support your friend’s claim? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 349
Answer:
Yes

Question 8.
The diagram shows a square pyramid and a cone. Both solids have the same height, h, and the base of the cone has radius r. According to Cavalieri’s Principle, the solids will have the same volume if the square base has sides of length ______ .
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 350
Answer:
2r/√2

Explanation:
Volume of square pyramid = a²\(\frac { h }{ 3 } \)
Square diagonal = √2a
radius = √2a/2
a = 2r/√2
Volume of cone = \(\frac { 1 }{ 3 } \)πr²h

Question 9.
About 19,400 people live in a region with a 5-mile radius. Find the population density in people per square mile.
Big Ideas Math Answer Key Geometry Chapter 11 Circumference, Area, and Volume 351

Answer:
The number of people per square mile is 247

Explanation:
S = πr²
= π x 5² = 78.5
Number of people per square mile = 19400/78.5 = 247

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