Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key

Practicing the Bridges in Mathematics Grade 4 Student Book Answer Key Unit 5 Module 3 will help students analyze their level of preparation.

Bridges in Mathematics Grade 4 Student Book Answer Key Unit 5 Module 3

 

Measuring Area

Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 1
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 2
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-1
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-2

Area Problems

Question 1.
Frank bought a rug for his room. It is 60 inches by 40 inches. What is the total area of the rug in square inches? Use labeled sketches, numbers, or words to solve this problem. Show all your work.
Answer:
2400 square inches.

Explanation:
Given,
Length = 60 inches
Width = 40 inches
Area of the rug = 60 × 40
= 2400 square inches.

Question 2.
The school gym is 80 feet by 50 feet. What is the total area of the gym floor in square feet? Use labeled sketches, numbers, or words to solve this problem. Show all your work.
Answer:
4000 square feet.

Explanation:
Given,
Length = 80 feet
Width = 50 feet
Area of the gym floor = 80 × 50
= 4000 square feet.

Question 3.
CHALLENGE Lisa’s room is 90 inches by 90 inches. She bought a rug for her floor that is 50 inches by 40 inches. How much of her floor is not covered by the rug? Use labeled sketches, numbers, or words to solve this problem. Show all your work.
Answer:
6100 square inches are not covered by the rug.

Explanation:
Given,
Lisa’s room is 90 inches by 90 inches.
Area of Lisa’s room = 90 × 90
= 8100 square inches.
She bought a rug for her floor that is 50 inches by 40 inches.
Area of the rug = 50 × 40
= 2000 square inches.
we need to find out how much of her floor is not covered by the rug.
8100 – 2000 = 6100 square inches.
Hence 6100 square inches are not covered by the rug.

Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Session 2 Answer Key

Measuring Perimeter

Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 3
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 4
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-3
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-4

Thinking About Area

Question 1.
Determine the area of each rectangle below. Write the area inside the rectangle.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 5
a. Look at the rectangles above. What happens to the area of the rectangle when one of the dimensions is doubled?
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-5

Explanation:
We have to determine the area of each rectangle given in the figure.
3 × 3 = 9
3 × 6 = 18
3 × 12 = 36
3 × 24  = 72

Question 2.
Determine the area of each rectangle below. Write the area inside the rectangle.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 6
a. Look at the rectangles above. What happens to the area of the rectangle when one of the dimensions is halved?
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-6

Explanation:
We have to determine the area of each rectangle given in the figure.
4 × 32 = 128
4 × 16 = 64
4 × 8 = 32
4 × 4 = 16

Question 3.
CHALLENGE What happens to the area of a rectangle when both dimensions are doubled? Start with this rectangle and then draw and label two more rectangles to show what happens.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 7
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-7

Explanation:
Given,
Length = 5 cm
Width = 3 cm
Now double the given dimensions
Length = 5 + 5 = 10 cm
Width = 3 + 3 = 6 cm
Area of the rectangle = 10 × 6 = 60 cm
Again double the given dimensions
Length = 10 + 10 = 20 cm
Width = 6 + 6 = 12 cm
Area of the rectangle = 12 × 20 = 240 cm.

Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Session 3 Answer Key

Area & Perimeter Formulas

Question 1.
Students are solving area and perimeter problems. Which kind of problems are they solving? Write “area” or “perimeter” next to each equation or statement below.
a. 25 + 76 + 25 + 76 ______________
Answer:
Perimeter problem.

Explanation:
Given,
the question is 25 + 76 + 25 + 76
length is 76
width is 25
From the given question the students are solving for the perimeter.
perimeter = 2 (l + b)
2 (25 + 76)
25 + 25 + 76 + 76

b. (2 × 25) + (2 × 76) ______________
Answer:
The given question is about the perimeter problem.

Explanation:
Given,
The question is (2 × 25) + (2 × 76)
here students are solving for perimeter problem.
perimeter = 2 (l + b)
= 2 (25 + 76)
= 2 × 25 + 2 × 76

c. 25 × 76 ______________
Answer:
The given question is about finding the Area.

Explanation:
Given,
the question is 25 × 76
which is a length of the figure 76 cm and a width is 25 cm
Now students are solving area problems for the given question.

d. 50 × 38 ______________
Answer:
The given question is about finding the Area.

Explanation:
Given,
the question is 50 × 38
which is a length of the figure 50 cm and width in 38 cm
Now students are solving area problems for the given question.

e. (20 × 76) + (5 × 76) _____________
Answer:
The given question is about finding the Area.

Explanation:
Given,
the question is (20 × 76) + (5 × 76)
Now students are solving area problems for the given question.

f. length × width ______________
Answer:
The given question is about finding the Area.

Explanation:
Given,
length × width
The formula is the area of the figure.

g. length + width + length + width ________________
Answer:
The given question is about finding the perimeter of the figure.

Explanation:
Given,
the question is length + width + length + width
which is the formula for the perimeter of the rectangle.

h. 2 × length + 2 × width _______________
Answer:
The given question is about finding the perimeter of the figure.

Explanation:
Given,
the question is
2 × length + 2 × width
which is the formula for the perimeter of the rectangle.
2 (l + b)

i. The answer is labeled in square units. ______________
Answer:
The area of the square is labeled in square units.

Explanation:
Multiplying the two sides of the square that has the same units is called the area of the square which is labeled in square units.

j. The answer is labeled in linear units. ______________
Answer:
The distance between any two points is labeled in linear units which is the perimeter of the square or rectangle.

Explanation:
The answer is labeled in linear units as the perimeter of the figure.

Question 2.
Teri was finding the area and perimeter of a rectangle. She spilled ketchup on her work.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 8
a. Use the information in the picture above to draw Teri’s rectangle and label its dimensions.
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-08

Explanation:
By using the information in the picture above I have drawn Teri’s rectangle and labelled its dimensions.

b. Write in the numbers that got covered with ketchup on Teri’s work. Also, circle the word to show whether each equation involves finding the area or the perimeter of the rectangle.
(2 × _________) + (2 × 99) = _____________ area or perimeter?
(100 × ____________) – (1 × 37) = ____________ area or perimeter?
Answer:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-8

Explanation:
The numbers that got covered with ketchup on Teri’s work are 37 cm. The area is 272 square cm and the perimeter is 3663 cm.

Question 3.
CHALLENGE Teri used an over strategy to multiply the dimensions of the rectangle. Show and describe a different multiplication strategy Teri could have used to find the area of the rectangle.
Answer:
The area of the rectangle is 3663 square cm.

Explanation:
The different multiplication strategy Teri could have used to find the area of the rectangle is multiplying the length and width of the rectangle.
The area of the rectangle is length × width
length = 99 cm and width = 37 cm
Area = 99 × 37
= 3663 square cm.

Question 4.
CHALLENGE The area of a rectangle is 4,500 square centimeters. One dimension is 30 cm and the perimeter is 360 cm. What is the other dimension? Show your work.
Answer:
The other dimension is 150 cm.

Explanation:
Given,
One dimension is 30 cm and the perimeter is 360 cm.
perimeter = 2 (l + b)
2 (30 + b) = 360 cm
30 + b = 360 ÷ 2
30 + b = 180
b = 180 – 30
= 150 cm.
The area of the rectangle is 30 × 150
= 4500 square centimeters.

Area & Perimeter Problems

Find the area and perimeter of each figure. Show your work.

Question 1.
a rectangle with dimensions 22 × 89 cm
a. area = _____________
b. perimeter = ______________
Answer:
Area = 1958 sq. cm and perimeter = 222 cm.

Explanation:
Given,
The dimensions are
Length = 22 cm
Width = 89 cm
Area = Length × width
= 22 × 89
= 1958 sq. cm
Perimeter = 2 (l + b)
= 2 (22 + 89)
= 2 (111)
= 222 cm.

Question 2.
a square with side length 44 cm
a. area = _____________
b. perimeter = ______________
Answer:
Area = 1936 sq cm and perimeter = 176 cm

Explanation:
Given,
Side of the square = 44 cm
Area of the square = 44 × 44
= 1936 sq cm
Perimeter = 4s
= 4 × 44
= 176 cm

Question 3.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 9
a. area = _____________
b. perimeter = ______________
Answer:
Area = 1170 sq. miles and perimeter = 278 miles.

Explanation:
Given,
area of the figure = 130 × 9
= 1170 sq miles
perimeter = 2 (l + b)
= 2 (130 + 9)
= 2 (139)
= 278 miles.

Question 4.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 10
a. area = _____________
b. perimeter = ______________
Answer:
Area = 625 sq km and perimeter = 260 km

Explanation:
Given,
length = 125 km
Width = 5 km
Area of the figure = 125 × 5
= 625 sq km
perimeter = 2 (l + b)
= 2 (125 + 5)
= 2 (130)
= 260 km

Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Session 4 Answer Key

Finding the Area & Perimeter of Complex Figures

Determine the area and perimeter of each figure below. Some of the figures are divided into rectangles for you with dotted lines, but some aren’t. You will need to figure out the missing lengths on some of the figures to find their perimeter. Show all your work.

Question 1.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 11
area = _____________
perimeter = ______________
Answer:
Area = 105 sq inches and perimeter = 44 inches.

Explanation:
Given,
Length of the figure = 15 inches
Width = 7 inches.
The area of the given figure is 15 × 7
= 105 sq inches.
The perimeter of the figure is 2 (l + b)
= 2 (15 + 7)
= 2 (22)
= 44 sq inches.

Question 2.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 12
area = _____________
perimeter = ______________
Answer:
Area = 210 square inches and perimeter = 74 inches.

Explanation:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-12

Question 3.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 13
area = _____________
perimeter = ______________
Answer:
Area = 700 square inches and perimeter =160 inches.

Explanation:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-13

Question 4.
Bridges in Mathematics Grade 4 Student Book Unit 5 Module 3 Answer Key 14
area = _____________
perimeter = ______________
Answer:
Area = 294 square inches and perimeter = 84 inches.

Explanation:
Bridges-in-Mathematics-Grade-4-Student-Book-Unit-5-Module-3-Answer-Key-14

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