This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Review Test detailed solutions for the textbook questions.

## Math in Focus Grade 7 Course 2 B Chapter 8 Review Test Answer Key

**Concepts and Skills**

For this review, r represents radius and h represents height. You may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth unless otherwise stated.

Question 1.

Find the volume of each cylinder to the nearest unit. Use the given dimensions,

a) r = 4.2 inches; h = 14 inches

Answer:

775 cubic units

Explanation:

The volume of the cylinder is =π r² h

π = 3.14, r = 4.2, h= 14 inches

volume of the cylinder = 3.14×4.2×4.2×14 = 775 cubic units

b) r = 7 centimeters; h = 12 centimeters

Answer:

1846 cubic units

Explanation:|

The volume of the cylinder is =π r² h

π = 3.14, r = 7, h= 12 centimeters

volume of the cylinder = 3.14×7×7×12 = 1846 cubic units

Question 2.

Find the volume of each cone to the nearest unit. Use the given dimensions.

a) r = 3 centimeters; h = 8 centimeters

Answer:

75 cm^{3}. ^{
}

Explanation:

The volume of each cone is 1/3hπr²

r = 3 centimeters; h = 8 centimeters

Therefore the volume of each cone is = 1/3×8×3.14×3×3 = 75 cm^{3}.

b) r = 8 inches; h = 15 inches

Answer:

1004 cm^{3}.

Explanation:

The volume of each cone is 1/3hπr².

Given r = 8 inches; h = 15 inches and π=3.14

Therefore the volume of each cone is = 1/3×8×3.14×15 ×15 = 1004 cm^{3}.

Question 3.

Find the volume of each pyramid. Use the given dimensions.

a) A square base with an edge length of 6 centimeters; h = 4 centimeters.

Answer:

We are given the cone:

l = 6

h = 4

Use the formula for the volume of a pyramid:

V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)l^{2}h

Substitute for l, h:

= \(\frac{1}{3}\) · 6^{2} · 4

Multiply:

= 48 cm^{3}

b) A rectangular base with length of 6 inches and width = 3.3 inches; h = 7 inches.

Answer:

We are given the cone:

l = 6

w = 3.3

h = 7

Use the formula for the volume of a pyramid:

V = \(\frac{1}{3}\)Bh = \(\frac{1}{3}\)lwh

Substitute for l, w, h:

= \(\frac{1}{3}\) · 6 · 3.3 · 7

Multiply:

= 46.2 in^{3}

Question 4.

Find the volume of each sphere to the nearest unit. Use the given dimensions,

a) r = 9.6 centimeters

Answer:

3705 cm^{3}

Explanation:

The volume of each sphere is = 4/3 πr³

Given r = 9.6 centimeters

Volume of the sphere is = 4/3×3.14×9.6×9.6×9.6 = 3705 cm^{3}.

b) d = 26 centimeters

Answer:

9202 cm^{3}.

Explanation:

The volume of each sphere is = 4/3 πr³

Given diameter is = 13 cm

Here we need a radius to calculate the volume of the sphere hence r = d/2 = 26/2 = 13 cm

Volume of the sphere is = 4/3×3.14×13×13×13=9202 cm^{3}.

Question 5.

Find the exact surface area of each solid.

Answer:

a) We are given the cylinder:

r = 3

h = 6

Use the formula for the surface area of a cylinder.

S = 2πr^{2} + 2πrh

Substitute for r, h:

= 2π · 3^{2} + 2π · 3 · 6

Multiply:

= 18π + 36π

Add:

= 54π ft^{2}

b) We are given the sphere:

d = 28

Determine the radius of the sphere:

r = \(\frac{d}{2}\) = \(\frac{28}{2}\) = 14

Use the formula for the surface area of a sphere:

S = 4πr^{2}

Substitute for r:

= 4π · 14^{2}

Multiply:

= 784π m^{2}

c) We are given the cone:

r = 6

l = 10

Use the formula for the surface area of a cone:

S = πr^{2} + πrl

Substitute for r, l:

= π · 6^{2} + π · 6 · 10

Multiply:

= 36π + 60π

Add:

= 96π in^{2}

Question 6.

Find the volume and surface area of each solid. Round to the nearest tenth.

a) A solid cone with a diameter of 5 feet, a slant height of 7 feet, and a height of 6.5 feet.

Answer:

We are given the cone.

d = 5

l = 7

h = 6.5

Find the radius:

r = \(\frac{d}{2}\) = \(\frac{5}{2}\) = 2.5

Use the formula for the volume of the cone:

V = \(\frac{1}{3}\) πr^{2}h

≈ \(\frac{1}{3}\) · 3.14 · 2.5^{2} · 6.5

≈ 42.5 ft^{3}

Use the formula for the surface area of the cone:

S = πr^{2} + πrl

= π · 2.5^{2} + π · 2.5 · 7

= 6.25π + 17.5π

= 23.75π

≈ 23.75 · 3.14

≈ 74.6 ft^{2}

b) A sphere with a radius of 28 millimeters.

Answer:

We are given the sphere:

r = 28

Use the formula for the volume of the sphere:

V = \(\frac{4}{3}\) πr^{3} = \(\frac{4}{3}\) · 3.14 · 28^{3}

≈ 91,905.7 mm^{3}

Use the formula for the surface area of the sphere:

S = 4πr^{2} ≈ 4 · 3.14 · 28^{2}

≈ 9,487 mm^{2}

c) A solid cylinder with a radius of 1.4 inches and a height of 4.2 inches.

Answer:

We are given the cylinder:

r = 1.4

h = 4.2

Use the formula for the volume of the cylinder:

V = πr^{2}h ≈ 3.14 · 1.4^{2} · 4.2

≈ 25.8 in^{3}

Use the formula for the surface area of the cylinder:

S = 2πr^{2} + 2πrh

= 2π · 1.4^{2} + 2π · 1.4 · 4.2

= 3.92π + 11.76π

= 15.68π

≈ 15.68 · 3.14

≈ 49.2 in^{2}

**Problem Solving**

**Solve. Show your work.**

Question 7.

The volume of a cone is 450 cubic centimeters and the radius of the base is 5 centimeters. What is the height of the cone to the nearest tenth?

Answer:

We are given:

V = 450

r = 5

Use the formula for the volume of the sphere:

V = \(\frac{1}{3}\) · πr^{2} · h

Substitute for V, πn r:

450 = \(\frac{1}{3}\) · 3.14 · 5^{2} · h

Multiply:

450 = \(\frac{78.5}{3}\) · h

Multiply:

450 = \(\frac{78.5}{3}\) · h

Multiply both sides by 3:

3 · 450 = 3 · \(\frac{78.5}{3}\) · h

Multiply:

1,350 = 78.5h

Divide both sides by 78.5

\(\frac{1,350}{78.5}\) = \(\frac{78.5h}{78.5}\)

Simplify

h ≈ 17.2 cm

Question 8.

The surface area of a sphere is 498.96 square centimeters. What is the radius of the sphere to the nearest tenth?

Answer:

We are given the sphere:

S = 498.96

Use the formula for the surface area of a sphere:

S = 4πr^{2}

Substitute S and π:

498.96 = 4 · 3.14 · r^{2}

Multiply:

498.96 = 12.56 · r^{2}

Divide each side by 12.56

\(\frac{498.96}{12.56}=\frac{12.56 r^{2}}{12.56}{78.5}\)

Evaluate:

39.7261 ≈ r^{2}

Find the square root of each side:

\(\sqrt[2]{39.7261}\) ≈ r^{2
}Round to the nearest tenth:

r ≈ 6.3 cm

Question 9.

A cone with a height of 6 inches and a slant height of 7.5 inches has a lateral surface with an area of approximately 106 square inches.

a) What is the radius? Round to the nearest tenth.

Answer:

We are given:

h = 6

l = 7.5

A = 106

Use the formula for the cone’s lateral surface area:

A = πrl

Substitute A, l, π to find r:

106 = 3.14 · r · 7.5

Multiply:

106 = 23.55r

Divide both sides by 23.55

\(\frac{106}{23.55}=\frac{23.55 r}{23.55}\)

r ≈ 4.5 in

b) What is the volume of the cone? Round to the nearest tenth.

Answer:

Use the formula for the cone’s volume:

V = \(\frac{1}{3}\) πr^{2}h

= \(\frac{1}{3}\) · 3.14 · 4.5^{2} · 6

≈ 127.2 in^{3}

Question 10.

A cylinder, a cone, and a sphere are shown below. Each solid has a radius of 1 inch and a height of 2 inches. Which of them has the greatest volume? Justify your answer.

Answer:

We are given:

r = 1

h = 2

Determine the cylinder’s volume:

V_{cylinder} = πr^{2}h = π · 1^{2} · 2 = 2π

Determine the cone’s volume:

V_{cone} = \(\frac{1}{3}\) πr^{2}h = \(\frac{1}{3}\) · π · 1^{2} · 2

= \(\frac{2 \pi}{3}\)

Determine the sphere’s volume:

V_{sphere} = \(\frac{4}{3}\) πr^{3} = \(\frac{4}{3}\) · π · 1^{3}

= \(\frac{4 \pi}{3}\)

We have:

\(\frac{2 \pi}{3}\) < \(\frac{4 \pi}{3}\) < 2π

V_{cone} < V_{sphere} < V_{cylinder
}The greatest volume is the cylinder’s

Question 11.

Two metal cubes each have an edge length of 4 centimeters. They are melted and recast into a square pyramid with a height of 5 centimeters. Find the area of the base of the pyramid.

Answer:

Determine the volume of the two cubes:

V = 2 · 4^{3} = 2 · 64 = 128 cm^{3}

Use the formula of a pyramids volume:

V = \(\frac{1}{3}\) Bh

Substitute for V, h:

128 = \(\frac{1}{3}\) · B · 5

Multiply both sides by 3:

3 · 128 = 3 · \(\frac{5B}{3}\)

384 = 5B

Divide by 5:

\(\frac{384}{5}\) = \(\frac{5B}{5}\)

B = 76.8 cm^{2}

Question 12.

A conical party hat has a diameter of 7 centimeters and a slant height of 14.5 centimeters. Paul wants to wrap the lateral surface of the party hat with plastic wrap. How much plastic wrap will he use? Round to the nearest square centimeter.

Answer:

We are given the cone:

d = 7

l = 14.5

Determine the radius of the cone:

r = \(\frac{d}{2}\) = \(\frac{7}{2}\) = 3.5

Determine the lateral surface of the cone:

A = πrl ≈ 3.14 · 3.5 · 14.5 ≈ 159 cm^{2}

= 159 cm^{2}

Question 13.

The composite solid shown is made up of a hemisphere attached to the top of a cube. The diameter of the hemisphere is the same as the edge length of the cube. Find the volume of the composite solid. Round to the nearest tenth.

Answer:

Determine the radius of the hemisphere:

r = \(\frac{9}{2}\) = 4.5

Determine the volume of the hemisphere:

V_{hemisphere} = \(\frac{1}{2}\) · \(\frac{4}{3}\) πr^{3}

≈ \(\frac{2}{3}\) · 3.14 · 4.5^{3}

≈ 190.8 cm^{3}

Determine the volume of the cube:

V_{cube} = 9^{3} = 729 cm^{3}

Determine the volume of the composite solid:

V_{solid} = V_{hemisphere} + V_{cube}

= 190.8 + 729

= 919.8 cm^{3}

Question 14.

The circumference of a fully inflated beach ball is 12π, or about 37.68 inches. What is the radius of the beach ball? What is the volume of

the beach ball? Round to the nearest tenth if necessary.

Answer:

The radius of the beach ball is 6 inches.

The volume of the beach ball is 151 inches.

Explanation:

The circumference of a fully inflated beach ball is 12π.

Circumference of a circle = 2πr

12π=2πr

r=6

Hence radius of the beach ball is 6 inches.

Now we need to calculate the volume of a beach ball

The volume of the sphere is 4/3 πr³

Volume = 4/3×3.14×6×6×6

The volume of the beach ball = 904 inches.

Question 15.

A pyramid made of clay has a square base of length 8 inches and a height of 12 inches. The pyramid is reshaped into a cylinder with a radius of 8 inches. What is the height of the cylinder? Round to the nearest inch

Answer:

The height of the cylinder is 1.27 inches

Explanation:

Given: The base(b) of the square pyramid as 8 inches length and height(h) of the square pyramid as 12 inches.

Now we need to calculate Volume of square pyramid to know the height of cylinder.

The volume of the square pyramid is given by = 1/3(b)^{2}h

Volume = 1/3 (8×8)^{2}×12

Volume = 1/3 (64)×12

Volume = 256 inches

Now pyramid is reshaped into a cylinder with a radius = 8 inches

Volume of cylinder is π r² h

Volume = 3.14×8×8×h

Volume is 256 inches

256 = 3.14×8×8×h

h = 1.27 inches