Spectrum Math Grade 8 Chapter 5 Posttest Answer Key

Students can use the Spectrum Math Grade 8 Answer Key Chapter 5 Posttest as a quick guide to resolve any of their doubts.

Spectrum Math Grade 8 Chapter 5 Posttest Answers Key

Check What You Learned

Geometry

Question 1.
What are the coordinates of the preimage?
A(_____), B(_____), C(_____), D(_____)
Answer: (2, 3), (5, 3), (4, 1), (1, 1)
The coordinates of the preimage are (2, 3), (5, 3), (4, 1), (1, 1)

Question 2.
What are the coordinates of the image?
A'(_____), B'(____), C'(____), D'(____)
Answer: (-1, 2), (-1, 5), (1, 4), (1, 1)
The coordinates of the image are (-1, 2), (-1, 5), (1, 4), (1, 1)

Question 3.
What transformation did you perform? _____
Answer: Rotation
The transformation performed was rotation.

Write the steps each figure must go through to be transformed from figure 1 to figure 2.

Question 4.
a.

Step 1: _____
Step 2: _____
Step 3: _____
Answer:
Step 1: The figure is rotated 90°.
Step 2: The figure is translated up and to the left.
Step 3: The figure is dilated by 2.

b.

Step 1: _____
Step 2: _____
Answer:
Step 1: The figure is reflected on the x-axis.
Step 2: The figure is translated to the right.

Draw similar right triangles to show that each line has a constant slope.

Question 5.
a.

Triangle 1 Legs:
_____ & _____
Triangle 2 Legs:
_____ & _____
Answer:

To test if the slope of the line is constant, draw a set of parallel lines that intersect the line.
Then, draw a line segment from each of the parallel lines to create a set of right triangles.
Find the length of the legs for each set of triangles.
Triangle 1 Legs:
3 & 2
Triangle 2 Legs:
6 & 4
Proportionality Test:
\(\frac{3}{2}\) = \(\frac{6}{4}\)
3 x 4 = 12 and 6 x 2 = 12
These leg lengths are proportional, so the line has a constant slope.

b.

Triangle 1 Legs:
_____ & _____
Triangle 2 Legs:
_____ & _____
Answer:

To test if the slope of the line is constant, draw a set of parallel lines that intersect the line.
Then, draw a line segment from each of the parallel lines to create a set of right triangles.
Find the length of the legs for each set of triangles.
Triangle 1 Legs:
3 & 1
Triangle 2 Legs:
6 & 2
Proportionality Test:
\(\frac{3}{1}\) = \(\frac{6}{2}\)
3 x 2 = 6 and 6 x 1 = 6
These leg lengths are proportional, so the line has a constant slope.

Answer each question using letters to name each line and numbers to name each angle.

Question 6.
Which 2 lines are parallel? _____
Answer: \(\overleftrightarrow{MN}\) and \(\overleftrightarrow{OP}\)
The two lines that are parallel are \(\overleftrightarrow{MN}\) and \(\overleftrightarrow{OP}\).

Question 7.
What is the name of the transversal? _____
Answer: \(\overleftrightarrow{QR}\)
The name of the transversal is \(\overleftrightarrow{QR}\)

Question 8.
Which angles are acute? _____
Answer: ∠1, ∠4, ∠5, ∠8
An acute angle measure less than 90 degrees.

Question 9.
Which angles are obtuse? ____
Answer: ∠2, ∠3, ∠6, ∠7
An acute angle measure greater than 90 degrees.

Question 10.
Which pairs of angles are vertical angles? _____
Answer: ∠1/∠4, ∠2/∠3, ∠5/∠8, ∠6/∠7

Question 11.
Which pairs of angles are alternate exterior angles? ____
Answer: ∠1/∠8, ∠2/∠7
A transversal is a line that intersects two or more lines at different points. The angles that are formed are called alternate interior angles and alternate exterior angles. When a transversal intersects parallel lines, corresponding angles are formed.

Question 12.
Which pairs of angles are alternate interior angles? _____
Answer: ∠3/∠6, ∠4/∠5
A transversal is a line that intersects two or more lines at different points. The angles that are formed are called alternate interior angles and alternate exterior angles. When a transversal intersects parallel lines, corresponding angles are formed.

Find the volume of each figure. Use 3.14 for π. Round answers to the nearest hundredth.

Question 13.
a.

V = ____ in.³
Answer: 1582.56 in.³
Volume is the amount of space a three-dimensional figure occupies. The volume of a cylinder by multiplying the area of the base by the height (Bh). The area of the base is the area of the circle, πr², so volume can be found using the formula: V = πr²h
The volume is expressed in cubic units, or units³.
The given values are r = 6 in. and h = 14 in.
Use 3.14 for π.
So, V = πr²h V = π(6² × 14) V = 1582.56 in.³

b.

V = ____ m³
Answer: 314 m³
Volume is the amount of space a three-dimensional figure occupies. The volume of a cone is calculated as \(\frac{1}{3}\)base × height.
This is because a cone occupies \(\frac{1}{3}\) of the volume of a cylinder of the same height. Base is the area of the circle, πr².
radius = 5 m
height = 13 m
V = \(\frac{1}{3}\)π5² 13  = 314 m³

c.

V = ____ ft.³ 
Answer: 381.51 ft³
Volume is the amount of space a three-dimensional figure occupies. The volume of a sphere is calculated as V = \(\frac{4}{3}\)πr³.
\(\frac{4}{3}\)πr³ Volume is given in cubic units or units³.
Given, d = 9ft
The radius of a sphere is half of its diameter. Find the radius, then calculate the volume.
r = \(\frac{1}{2}\)d = \(\frac{1}{2}\)(9) = 4.5
V = \(\frac{4}{3}\)πr(4.5)³ = 381.51 ft³

Use the Pythagorean Theorem to find the unknown lengths.

Question 14.
If a = 7 and b = 10, c = or about _____
Answer: \(\sqrt{149}\) or 12.2
The Pythagorean Theorem:
If a triangle is a right triangle, then a² + b² = c².
a = 7, b = 10 , c =?
a² + b² = c²
a² + b² = 7² + 10²  = 49 + 100 = 149
c² = a² + b²  = 149
\(\sqrt{c}\) = \(\sqrt{149}\) = 12.2

Question 15.
If a = 11 and c = 18, b = or about ____
Answer: \(\sqrt{203}\) or 14.2
The Pythagorean Theorem:
If a triangle is a right triangle, then a² + b² = c².
a = 11, b = ? , c =18
a² + b² = c²
c² –  a² = 18² – 11²  = 203
b² = c² – a² = 203
\(\sqrt{b}\) = \(\sqrt{203}\) = 14.2

Solve each problem.

Question 16.
A flagpole and a telephone pole cast shadows as shown in the figure. How tall are the poles?
The flagpole is ___ feet tall.

The telephone pole is ____ feet tall.
Answer: The flagpole is 30 feet tall.
The telephone pole is 20 feet tall.
Flagpole:
The Pythagorean Theorem:
If a triangle is a right triangle, then a² + b² = c².
a = 31.5, b = ? , c = 43.5
a² + b² = c²
c² –  a² = 43.5² – 31.5²  = 900
b² = c² – a² = 900
\(\sqrt{b}\) = \(\sqrt{900}\) = 30
Telephone:
The Pythagorean Theorem:
If a triangle is a right triangle, then a² + b² = c².
a = 21, b = ? , c = 29
a² + b² = c²
c² –  a² = 29² – 21²  = 400
b² = c² – a² = 400
\(\sqrt{b}\) = \(\sqrt{400}\) = 20