This handy Math in Focus Grade 7 Workbook Answer Key Chapter 8 Lesson 8.5 Real-World Problems: Composite Solids detailed solutions for the textbook questions.

## Math in Focus Grade 7 Course 2 B Chapter 8 Lesson 8.5 Answer Key Real-World Problems: Composite Solids

### Math in Focus Grade 7 Chapter 8 Lesson 8.5 Guided Practice Answer Key

**Solve.**

Question 1.

A composite solid is made up of a cone and a cylinder. The slant height of the cone is 25 centimeters. The height of the cylinder is 15 centimeters and its radius is 12 centimeters. The height of the solid is 37 centimeters. Use 3.14 as an approximation for π.

a) Find the volume of the composite solid to the nearest cubic centimeter.

Volume of the cylinder:

πr^{2}h = ∙ ∙ Use the formula for volume of a cylinder.

= cm^{3} Evaluate .

Volume of the cone:

Height of cone = –

= cm

\(\frac{1}{3}\) πr^{2}h = \(\frac{1}{3}\) ∙ ∙ ∙ ∙ Use the formula for the volume of a cone.

= cm^{3} Multiply.

Volume of the composite solid:

+ = Add the volumes of the cylinder and the cone.

≈ cm^{3} Multiply and round.

The volume of the composite solid is about cubic centimeters.

Answer:

We are given:

l = 25

h_{cylinder} = 15

H = 37

r = 12

Use the formula for the volume of a cylinder:

V_{cylinder} = πr^{2}h_{cylinder}

= π · 12^{2} · 15

Evaluate:

= 2,160π cm^{3}

The height of the cone:

h_{cone} = H – h_{cylinder}

= 37 – 15

= 22 cm

Use the formula for the volume of a cone:

V_{cone} = \(\frac{1}{3}\)πr^{2}h_{cone}

= \(\frac{1}{3}\) · π · 12^{2} · 22

Multiply:

= 1,056π cm^{3
}Find the volume of the composite solid by adding the volumes of the cylinder and the cone:

V = V_{cylinder} + V_{cone}

= 2,160π + 1,056π

= 3,216π cm^{3}

Multiply and round:

≈ 10,098 cm^{3}

b) Find the surface area of the composite solid to the nearest square centimeter.

The surface area of the composite solid is about square centimeters.

Answer:

Use the formuLas for the surface areas:

S = πrl + 2πrh_{cylinder} + πr^{2}

Substitute for r, l, h_{cylinder}

= π · 12 · 25 + 2π · 12 · 15 + π · 12^{2}

Evaluate each term:

= 300π + 360π + 144π

Add:

= 804π

Multiply and round:

≈ 804 · 3.14 ≈ 2,525 cm^{2}

Question 2.

A birdhouse looks like a cube with a square pyramid on top. As shown, the birdhouse has a circular entrance with a diameter of 4 inches. Find the exterior surface area of the birdhouse.

Answer:

Determine the Lateral surface area of the pyramid:

S_{1} = 4 · \(\frac{1}{2}\) · 10 · 13 = 260 in^{2}

Determine the radius of the entrance:

r = \(\frac{4}{2}\) = 2

Determine the lateral surface area of the house:

S_{2} = 4 · 10 · 10 – π · 2^{2} ≈ 387.4 in^{2}

Compute the exterior surface area of the house:

S = S_{1} + S_{2} = 260 + 387.4 = 647.4 in^{2}

= 647.4 in^{2}

Question 3.

The solid shown is a cylinder with a cone-shaped hole. The diameter of the cylinder is 22 centimeters. Its height is 15 centimeters. The radius of the cone-shaped hole is 7 centimeters and the height is 10 centimeters. Find the volume of the solid. Use 3.14 as an approximation for π. Round your answer to the nearest cubic centimeter.

Answer:

Determine the radius R of the cylinder:

R = \(\frac{22}{2}\) = 11 cm

Determine the volume of the cylinder

V_{cylinder} = πR^{2}h = π · 11^{2 } · 15 = 1, 815π cm^{2}

Determine the volume of the cone:

V_{cone} = \(\frac{1}{3}\)πR^{2}h = \(\frac{1}{3}\) · π · 7^{2} · 10 ≈ 163π cm^{3}

Determine the volume of the solid:

V = V_{cylinder} – V_{cone}

= 1,815π – 163π

= 1,652π

≈ 3.14 · 1,652

≈ 5. 187 cm^{3}

### Math in Focus Course 2B Practice 8.5 Answer Key

**For this practice, you may use a calculator and use 3.14 as an approximation for π. Round your answers to the nearest tenth when you can.**

**Solve.**

Question 1.

Jack has a cylindrical block that has a radius of 0.6 cm and is 22 centimeters long. He puts together 8 such blocks to form the composite solid shown. What is the volume of the composite solid?

Answer:

We are given the cytinder:

r = 0.6

h = 22

Determine the voLume of one cyLinder:

V_{cylinder} = πr^{2}h ≈ 3.14 · 0.6^{2} · 22

≈ 24.87 cm^{3}

Determine the volume of the composite solid:

V_{solid} = 8V_{cylinder} = 8 · 24.87 ≈ 199 cm^{3}

Question 2.

The trophy for a basketball tournament is made up of a miniature basketball attached to a rectangular prism. The radius of the basketball is 5 centimeters. The prism measures 8 centimeters by 5 centimeters by 15 centimeters. What is the volume of the trophy to the nearest cubic centimeter?

Answer:

We are given:

r = 5

l = 8

w = 5

h = 15

Determine the volume of the sphere:

V_{sphere} = \(\frac{4}{3}\)πr^{3}

≈ \(\frac{4}{3}\) · 3.14 · 5^{3}

≈ 523 cm^{3}

Determine the volume of the prism:

V_{prism }= lwh = 8 · 5 · 15 = 600 cm^{3}

Determine the volume of the trophy:

V_{trophy} = V_{sphere} + V_{prism}

= 523 + 600

= 1,123 cm^{3}

Question 3.

A necklace is made up of 50 spherical beads. Each bead has a radius of 8 millimeters.

a) What is the volume of the necklace?

Answer:

We are given the sphere:

r = 8

Determine the volume of a spherical bead:

V_{bead} = \(\frac{4}{3}\)πr^{3}

≈ \(\frac{4}{3}\) · 3.14 · 8^{3}

≈ 2,143.573

Determine the volume of the necklace:

V_{necklace} = 50 · V_{bead}

≈ 50 – 2. 143.573 ≈ 107, 178.7 mm^{3}

b) What is the surface area of the necklace?

Answer:

Determine the surface area of a spherical bead:

S_{bead} = 4πr^{2} ≈ 4 · 3.14 · 8^{2}

≈ 803.84 mm^{2}

Determine the surface area of the necklace:

S_{necklace} = 50 · S_{bead}

= 50 · 803.84

≈ 40,192 mm^{2}

Question 4.

A crystal trophy is made up of a rectangular pyramid whose base is attached to the top of a rectangular prism. The base of the pyramid and the top of the prism are each 4 inches long and 3 inches wide. The height of the pyramid is 2.5 inches and the height of the prism is 9.5 inches. What is the volume of the crystal trophy?

Answer:

We are given:

l = 4

w = 3

h_{pyramid} = 2.5

h_{prism} = 9.5

Determine the volume of the pyramid:

V_{pyramid }=\(\frac{1}{3}\) · lwh_{pyramid }

= \(\frac{1}{3}\) · 4 · 3 · 2.5

= 10 in^{3}

Determine the volume of the prism:

V_{prism} = lwh_{prism} = 4 · 3 · 9.5

= 114 in^{3}

Determine the volume of the trophy:

V_{trophy} = V_{pyramid} + V_{prism}

= 10 + 114

= 124 in^{3}

Question 5.

At a food stand, you can buy a paper cone filled with slush made of frozen juice. The slush forms a hemisphere on top of the cone, as shown. What is the volume of the cone of slush?

Answer:

We are given:

d = 8

h = 10.8

Determine the radius:

r = \(\frac{d}{2}\) = \(\frac{8}{2}\) = 4

Determine the volume of the hemisphere:

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

≈ \(\frac{2}{3}\) · 3.14 · 4^{3}

≈ 134 cm^{3}

Determine the volume of the cone:

V_{cone} = \(\frac{1}{3}\) πr^{2}h

≈ \(\frac{1}{3}\) · 3.14 · 4^{2} · 10.8

≈ 180.9 cm^{3}

Determine the volume of the cone of slush:

V = V_{hemisphere} + V_{cone}

= 134 + 180.9

= 314.9 cm^{3}

Question 6.

A wooden paper towel holder is composed of two cylinders. The diameter of the base is 12 .centimeters and its height is 2 centimeters. The combined height of the two cylinders is 30 centimeters. What is the volume of the paper towel holder?

Answer:

We are given:

d_{1} = 3

h_{1 }= 30

d_{2 }= 12

h_{2} = 2

Determine the radius of the upper cylinder:

r_{1} = \(\frac{d_{1}}{2}\) = \(\frac{3}{2}\) = 1.5

Determine the volume of the upper cylinder:

V_{1} = πr_{1}^{2}h_{1} = π · 1 5^{2} · 30 = 67.5π cm^{3}

Determine the radius of the lower cylinder:

r_{2} = \(\frac{d_{2}}{2}\) = \(\frac{12}{2}\) = 6

Determine the volume of the lower cylinder:

V_{2} = πr_{2}^{2}h_{2} = π · 6^{2} · 2 = 72π cm^{3
}Determine the volume of the paper tower holder:

V = V_{1} + V_{2} = 67.5π + 72π

≈ 139.5π

≈ 139.5 · 3.14

≈ 438 cm^{3}

Question 7.

The edge of the base of a square pyramid is 11 inches. The pyramid has a height of 14 inches. What is the volume of the composite solid formed when two such pyramids are joined at the base, as shown in the diagram?

Answer:

We are given:

L = 11

h = 14

Determine the volume of one pyramid:

V_{1} = V_{2} = \(\frac{1}{3}\)l^{2}h = \(\frac{1}{3}\) · 11^{2} · 14 ≈ 564.67 in^{3}

Determine the volume of the composite solid:

V = V_{1 }+ V_{2} = 2V_{1} = 2 · 564.67

≈ 1,129.3 in^{3}

Question 8.

A clock in the shape of an hour glass is made up of two identical cones connected at their vertices. The radius of each cone is 7 centimeters. The combined height of the two cones is 11.8 centimeters. What is the volume of the clock?

Answer:

We are given:

r = 7

H = 11.8

Determine the height of each cone:

h = \(\frac{H}{2}\) = \(\frac{11.8}{2}\) = 5.9

Determine the volume of one cone:

V_{1} = V_{2} = \(\frac{1}{3}\)πr^{2}h = \(\frac{1}{3}\) · π · 7^{2 }· 59

≈ 96.3677π

Determine the volume of the clock:

V = V_{1} + V_{2} = 2V_{1} = 2 · 96.367π

≈ 192.734 · 3.14

≈ 605.2 cm^{3}

Question 9.

Jason made a long pole by joining three different lengths of cylindrical poles together. Each pole has the same diameter of 18 cm. The diagram below is not drawn to scale.

a) What is the volume of the long pole?

Answer:

We are given:

d = 18

h_{1} = 7

h_{2} = 11

h_{3} = 15

Determine the radius

r = \(\frac{d}{2}\) = \(\frac{18}{2}\) = 9

Determine the volume of the tong pole:

V = V_{1 }+ V_{2} + V_{3}

= πr^{2}h_{1} + πr^{2}h_{2 }+ πr^{2} h3

= πr^{2}(h_{1} + h_{2} + h_{3})

≈ 3.14 · 9^{2} ·_{ }(7 + 11 + 15)

≈ 8,393.2 cm^{3}

b) What is the surface area of the long pole?

Answer:

Determine the surface area of the long pole

= 2πr^{2} + 2πr(h_{1} + h_{2} + h_{3})

= 2πr(r + h_{1} + h_{2} + h_{3})

≈ 2 · 3.14 · 9(9 + 7 + 11 + 15)

≈ 2,373.8 cm^{2}

**Brain @ Work**

How can you make the following cross-sections by slicing a cube? Use a computer drawing program or pencil and paper to show your answers for a) and b).

a) An isosceles triangle

Answer:

We take a point on one edge, then take two congruent segments from one of the vertices on the same edge on the other two edges

We draw the cross-section in the cube in the shape of an isosceles triangle:

b) A regular hexagon

Answer:

To get a hexagon, we slice the cube with a plane going through alt six faces of the cube.

We draw the cross-section in the cube in the shape of a regular hexagon:

c) What other polygons can be cross-sections of a cube?

Answer:

The cross-section can be a quadrilateral when a plane crosses 4 faces of the cube:

The cross-section can also be a pentagon when a plane crosses 5 faces of the cube: