Refer to our Texas Go Math Grade 8 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 8 Lesson 9.3 Answer Key Volume of Spheres.

## Texas Go Math Grade 8 Lesson 9.3 Answer Key Volume of Spheres

**Texas Go Math Grade 8 Lesson 9.3 Explore Activity Answer Key**

**Modeling the Volume of a Sphere**

A sphere is a three-dimensional figure with all points the same distance from the center. The radius of a sphere is the distance from the center to any point on the sphere.

You have seen that a cone fills \(\frac{1}{3}\) of a cylinder of the same radius and height h. If you were to do a similar experiment with a sphere of the same radius, you would find that a sphere fills \(\frac{2}{3}\) of the cylinder. The cylinder’s height is equal to twice the radius of the sphere.

**Reflect**

Question 1.

Analyze Relationships A cone has a radius of r and a height of 2r. A sphere has a radius of r. Compare the volume of the sphere and cone.

Answer:

The volume of the cone with radius of r and a height of 2r is

V_{cone} = \(\frac{1}{3}\)πr^{2}h

V_{cone} = \(\frac{1}{3}\) ∙ π ∙ r^{2} ∙ (2r)

V_{cone} = \(\frac{2}{3}\) ∙ π ∙ r^{3}

The volume of the sphere with radius of r is

V_{sphere} = \(\frac{4}{3}\)πr^{3}

As we can see, the volume of the sphere is twice the volume of the cone.

**Example 1**

Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.

**Your Turn**

**Find the volume of each sphere. Round your answers to the nearest tenth. Use 3.14 for π.**

Question 2.

A sphere has a radius of 10 centimeters. __________________

Answer:

It is given that the radius of the sphere is 10 centimeters. Therefore, the volume of the sphere is:

V_{sphere} = \(\frac{4}{3}\)πr^{3}

V_{sphere} = \(\frac{4}{3}\) ∙ (3.14) ∙ (10)^{3}

V_{sphere} ≈ 4186.7 cm^{3}

Question 3.

A sphere has a diameter of 3.4 meters. __________________

Answer:

It is given that the diameter of the sphere is 3.4 meters, which means that the radius is 1.7 meters. Therefore, the volume of the sphere is:

V_{sphere} = \(\frac{4}{3}\)πr^{3}

V_{sphere} = \(\frac{4}{3}\) ∙ (3.14) ∙ (1.7)^{3}

V_{sphere} ≈ 20.6 m^{3}

**Example 2**

Soccer balls come in several different sizes. One soccer ball has a diameter of 22 centimeters. What is the volume of this soccer ball? Round your answer to the nearest tenth. Use 3.14 for π.

**Reflect**

Question 4.

What is the volume of the soccer ball in terms of π, to the nearest whole number multiple? Explain your answer.

Answer:

V_{sphere} = \(\frac{4}{3}\)πr^{3}

V_{sphere} = \(\frac{4}{3}\) ∙ π ∙ (11)^{3}

V_{sphere} = \(\frac{4}{3}\) ∙ π ∙ 1331

V_{sphere} = \(\frac{5324}{3}\) ∙ π

V_{sphere} ≈ 1775 π

The volume of the soccer ball in terms of π, to the nearest whole number multiple, is 1775 πcm^{3}.

Question 5.

Analyze Relationships The diameter of a basketball is about 1.1 times that of a soccer ball. The diameter of a tennis ball is about 0.3 times that of a soccer ball. How do the volumes of these balls compare to that of a soccer ball? Explain.

Answer:

The volume of the soccer ball is

V_{soccer} = \(\frac{4}{3}\)πr^{3}

Since the diameter of a basketball is about 1.1 times that of a soccer ball, its radius is also 1.1 times that of a soccer ball. Therefore, the volume of the basketball is

V_{basketball} = \(\frac{4}{3}\) ∙ π ∙ (1.1 r)^{3}

V_{basketball} = \(\frac{4}{3}\) ∙ π ∙ 1.331 ∙ r^{3}

V_{basketball} = 1.331 V_{soccer}

Since the diameter of a tennis ball is about 0.3 times that of a soccer ball, its radius is also 0.3 times that of a soccer ball. Therefore, the volume of the basketball is

V_{tennis} = \(\frac{4}{3}\) ∙ π ∙ (0.3r)^{3}

V_{tennis} = \(\frac{4}{3}\) ∙ π ∙ 0.027 ∙ r^{3}

V_{tennis} = 0.027 V_{soccer}

**Your Turn**

Question 6.

Val measures the diameter of a ball as 12 inches. How many cubic inches of air does this ball hold, to the nearest tenth? Use 3.14 for π.

Answer:

Here, the diameter of the ball = 12 inch

so,the radius of the ball = 6 inch

Volume of the sphere = \(\frac{4}{3}\) × π × r^{3}

= \(\frac{4}{3}\) × 3.14 × 6^{3}

= 904.3 in^{3}

**Texas Go Math Grade 8 Lesson 9.3 Guided Practice Answer Key**

Question 1.

Vocabulary A sphere is a three-dimensional figure with all points ___________ from the center. (Explore Activity)

Answer:

A sphere is a three-dimensional figure with all points the same distance from the center.

Question 2.

Vocabulary The _____________ is the distance from the center of a sphere to a point on the sphere. (Explore Activity)

Answer:

The radius is the distance from the center f the sphere to a point on the sphere.

**Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π. (Example 1)**

Question 3.

Answer:

Radius r = 1 in

Volume of the sphere = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ 1^{3}

V = 4.1867 in^{3}

V ≈ 4.12 in^{3}

Question 4.

Answer:

Diameter = 20 cm

Radius r = \(\frac{20}{2}\) cm

Radius r = 10 cm

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 10^{3}

Volume = 4186.66 cm^{3}

Volume ≈ 4186.7 cm^{3}

Question 5.

A sphere has a radius of 1.5 feet. ___________

Answer:

Radius r = 1.5 ft

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 1.5^{3}

Volume = 14.13 ft^{3}

Volume ≈ 14.1 ft^{3}

Question 6.

A sphere has a diameter of 2 yards. ___________

Answer:

Diameter = 2 yards

Radius r = \(\frac{2}{2}\) yards

Radius r = 1 yd

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 1^{3}

Volume = 4.1866 yd^{3}

Volume ≈ 4.2 yd^{3}

Question 7.

A baseball has a diameter of 2.9 inches. Find the volume of the baseball. Round your answer to the nearest tenth if necessary. Use 3.14 for π. (Example 2)

Answer:

Diameter of baseball = 2.9 in

Radius r = \(\frac{2.9}{2}\) in

Radius of baseball = 1.45 in

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 1.45^{3}

Volume = 12.763 in^{3}

Volume ≈ 12.8 in^{3}

Question 8.

A basketball has a radius of 4.7 inches. What is its volume to the nearest cubic inch? Use 3.14 for π. (Example 2)

Answer:

Radius of basketball r = 4.7 in

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 4.7^{3}

Volume = 434.6723 in^{3}

Volume ≈ 434.67 in^{3}

Question 9.

A company is deciding whether to package a ball in a cubic box or a cylindrical box. In either case, the ball will touch the bottom, top, and sides.

(Explore Activity)

a. What portion of the space inside the cylindrical box is empty? Explain.

Answer:

The volume of the cylinder is

V_{cylinder} = πr^{2}h

Since the ball touches the bottom, top, and sides, then the height of the cylinder will be equal to 2r.

V_{cylinder} = π ∙ r^{2} ∙ (2r)

V_{cylinder} = 2πr^{3}

On the other hand, the voLume of the sphere is

V_{sphere} = \(\frac{4}{3}\)πr^{3}

The volume of the empty space inside the cylindrical box is found by subtracting the volume of the sphere from the volume of the cylinder:

V_{cylinder} – V_{sphere} = 2πr^{3} – \(\frac{4}{3}\)πr^{3}

= (2 – \(\frac{4}{3}\))πr^{3}

= (\(\frac{6}{3}-\frac{4}{3}\))πr^{3}

= \(\frac{2}{3}\)πr^{3}

b. Find an expression for the volume of the cubic box.

Answer:

The volume of a cube with side a is

V_{cube} = a^{3}

Since the ball touches the bottom, top, and sides, then the side of the cube will be equal to 2r.

V_{cube} = (2r)^{3}

V_{cube} = 8r^{3}

c. About what portion of the space inside the cubic box is empty? Explain.

Answer:

The volume of the empty space inside the cubical box is found by subtracting the volume of the sphere from the volume of the cube:

V_{cube} – V_{sphere} = 8r^{3} – \(\frac{4}{3}\)πr^{3}

= (8 – \(\frac{4}{3}\)π)r^{3}

= (8 – \(\frac{4}{3}\) ∙ 3.14)r^{3}

≈ (8 – 4.2)r^{3}

≈ 3.8r^{3}

**Essential Question Check-In**

Question 10.

Explain the steps you use to find the volume of a sphere.

Answer:

Step 1: The radius of the sphere is found out

Step 2: The volume of the sphere is \(\frac{4}{3}\)πR^{3}; where R stands for the radius.

Step 3: Put the value of radius in the equation of volume.

Step 4 : Calculate the volume.

**Texas Go Math Grade 8 Lesson 9.3 Independent Practice Answer Key**

**Find the volume of each sphere. Round your answers to the nearest tenth if necessary. Use 3.14 for π.**

Question 11.

radius of 3.1 meters ____________

Answer:

The voLume of the sphere with a radius of 3.1 meters is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (3.1)^{3}

V ≈ 124.7 m^{3}

Question 12.

diameter of 18 inches __________

Answer:

The diameter of the sphere is 18 inches, which means that its radius is 9 inches. The volume of this sphere is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (9)^{3}

V = 3052.08

V ≈ 3052.1 inches^{3}

Question 13.

r = 6 in. _____________

Answer:

The volume of the sphere with a radius of 6 inches is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (6)^{3}

V = 904.32

V ≈ 904.3 inches^{3}

Question 14.

d = 36 m _____________

Answer:

Diameter d = 36 m

Radius r = \(\frac{36}{2}\) = 18 m

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (18)^{3}

V = 24416.64 m^{3}

Question 15.

Answer:

The volume of the sphere with a radius of 11 centimeters is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (11)^{3}

V ≈ 5572.5 cm^{3}

Question 16.

Answer:

The diameter of the sphere is 2.5 feet, which means that its radius is 1.25 feet. The volume of this sphere is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (1.25)^{3}

V ≈ 8.2 feet^{3}

The eggs of birds and other animals come in many different shapes and sizes. Eggs often have a shape that is nearly spherical. When this is true, you can use the formula for a sphere to find their volume.

Question 17.

The green turtle lays eggs that are approximately spherical with an average diameter of 4.5 centimeters. Each turtle lays an average of 113 eggs at one time. Find the total volume of these eggs, to the nearest cubic centimeter.

Answer:

The diameter of an egg (sphere) is 4.5 centimeters, which means that its radius is 2.25 centimeters. The volume of a single egg is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (2.25)^{3}

V ≈ 47.68875 cm^{3}

Therefore, the total volume of 113 eggs is

113 ∙ V = 113 ∙ 47.68875

= 5388.82875

≈ 5389 cm^{3}

The total volume of 113 eggs is approximately 5388.8 cm^{3}

Question 18.

Hummingbirds lay eggs that are nearly spherical and about 1 centimeter in diameter. Find the volume of an egg. Round your answer to the nearest tenth.

Answer:

The diameter of the egg (sphere) is 1 centimeter, which means that its radius is 0.5 centimeters. The volume of a single egg is

V = \(\frac{4}{3}\)πr^{3}

V = \(\frac{4}{3}\) ∙ 3.14 ∙ (0.5)^{3}

V ≈ 0.5233

V ≈ 0.5 cm^{3}

Question 19.

Fossilized spherical eggs of dinosaurs called titanosaurid sauropods were found in Patagonia. These eggs were 15 centimeters in diameter. Find the volume of an egg. Round your answer to the nearest tenth.

Answer:

Diameter of an egg (d) = 15 cm

Radius (r) = \(\frac{15}{2}\) = 7.5 cm

Volume = \(\frac{4}{3}\)πr^{3}

Volume = \(\frac{4}{3}\) × 3.14 × 7.5^{3}

Volume = 1766.25 cm^{3}

Volume ≈ 1766.3 cm^{3}

Question 20.

**Persevere in Problem Solving** An ostrich egg has about the same volume as a sphere with a diameter of 5 inches. If the eggshell is about \(\frac{1}{12}\) inch thick, find the volume of just the shell, not including the interior of the egg. Round your answer to the nearest tenth.

Answer:

The diameter of the egg, including the eggshell, is

d = 5 + (2 ∙ \(\frac{1}{12}\))

d = 5 + \(\frac{1}{6}\)

d = \(\frac{31}{6}\) inches

Therefore, the radius of the egg, including the eggshell, is

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

r = \(\frac{31}{12}\)

r ≈ 2.583 inches

The volume of the egg, including the eggshell, is

V_{1} = \(\frac{4}{3}\)πr^{3}

V_{1} = \(\frac{4}{3}\) ∙ 3.14 ∙ (2.583)^{3}

V_{1} ≈ 72.2 inches^{3}

The diameter of the egg, excluding the eggshell, is 5 inches, which means that its radius ¡s 25 inches The volume of the egg, excluding the eggshell, is

V_{2} = \(\frac{4}{3}\)πr^{3}

V_{2} = \(\frac{4}{3}\) ∙ 3.14 ∙ (2.5)^{3}

V_{2} ≈ 65.4 inches^{3}

To find the volume of just the shell, we subtract the volume of the egg excluding the shell from the volume of the egg including the shell.

V_{1} – V_{2} = 72.2 – 65.4

V_{1} – V_{2} = 6.8 inches^{3}

The volume of just the eggshell is approximately 6.8 inches^{3}.

Question 21.

**Multistep** Write the steps you would use to find a formula for the volume of the figure at right. Then write the formula.

Answer:

Radius of hemisphere = r

Radius of cyLinder = r

Height of cylinder = r

Step 1: Find the formula for volume of hemisphere.

Volume of hemisphere = \(\frac{\frac{4}{3} \pi r^{3}}{2}=\frac{2}{3} \pi r^{3}\)

Step 2: Find the formula for the volume of a cylinder.

Volume of cylinder = πr^{2}h

Height of cylinder is equal to radius, so

Volume of cylinder = πr^{3}

Step 3: Add both the volume expressions:

Total volume = \(\frac{2}{3}\)πr^{3} + πr^{3} = \(\frac{5}{3}\)πr^{3}

Question 22.

**Critical Thinking** Explain what happens to the volume of a sphere if you double the radius.

Answer:

Let radius = r

Volume V_{1} = \(\frac{4}{3}\)πr^{3}

Let radius = 2r

Volume V_{2} = \(\frac{4}{3}\)π(2r)^{3}

= 8 × \(\frac{4}{3}\)πr^{3}

= 8V_{1}

= 8 (initial volume)

By doubling the radius of sphere we make the volume 8 times the initial value.

Question 23.

**Multistep** A cylindrical can of tennis balls holds a stack of three balls so that they touch the can at the top, bottom, and sides. The radius of each ball is 1.25 inches. Find the volume inside the can that is not taken up by the three tennis balls.

Answer:

Radius of a tennis ball = 1.25 in

Diameter of one ball = 2 × 1.25 = 2.5 in

Height of the cylinder = 2.5 × 3 = 7.5 in

Radius of base of cylinder = 125 in

Volume of cylinder (V_{1})= πr^{2}h

V_{1} = 3.14 × 1.25^{2} × 7.5

V_{1} = 36.796875 in^{3}

V_{1} ≈ 36.8 in^{3}

Volume of a ball (all three) (V_{2}) 3 × \(\frac{4}{3}\)πr^{3}

V_{2} = 4 × 3.14 × 1.25^{3}

V_{2} = 24.53125 in^{3}

V_{2} ≈ 24.5 in^{3}

Volume of empty space = Volume of cylinder – Volume of a ball (all three)

= 36.8 – 24.5 = 12.3 in^{3}

**H.O.T. Focus on Higher Order Thinking**

Question 24.

**Critique Reasoning** A sphere has a radius of 4 inches, and a cube-shaped box has an edge length of 7.5 inches. J.D. says the box has a greater volume, so the sphere will fit in the box. Is he correct? Explain.

Answer:

Radius of sphere 4 in

Volume of sphere = \(\frac{4}{3}\)πr^{3} = \(\frac{4}{3}\)π4^{3} = 267.9467 in^{3}

Volume of sphere ≈ 267.9 in^{3}

Volume of cube-shaped box = 7.5 ^{3} = 421.875 in^{3}

Volume of cube-shaped box ≈ 421.9 in^{3}

Volume of cube > Volume of sphere

But the base of the cube has an area of (7.5 × 7.5) = 56.25 in^{2} while the cross-action area of the sphere is:

πr^{2} = 3.14 × 4^{2} = 50.24 in^{2}

The cross-section area of the sphere is less than that of a cube. J.D. is right and the ball (sphere) will fit in the cube.

Question 25.

**Critical Thinking** Which would hold the most water: a bowl in the shape of a hemisphere with radius r, a cylindrical glass with radius r and height r, or a cone-shaped drinking cup with radius r and height r? Explain.

Answer:

The volume of a sphere with radius r is

V_{sphere} = \(\frac{4}{3}\)πr^{3}

Therefore, the volume of a hemisphere is

V_{hemisphere} = \(\frac{V_{\text {sphere }}}{2}\)

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

The volume of a cylinder with radius r and height r is

V_{cylinder} = πr^{2}h

V_{cylinder} = πr^{3}

The volume of a cone with radius r and height r is

V_{cone} = \(\frac{1}{3}\)πr^{2}h

V_{cone} = \(\frac{1}{3}\)πr^{3}

As we can notice

V_{cone} < V_{hemisphere} < V_{cylinder}

Therefore, the cylindrical glass with radius r and height r will hold the most water.

Question 26.

**Analyze Relationships** Hari has models of a sphere, a cylinder, and a cone. The sphere’s diameter and the cylinder’s height are the same, 2r. The cylinder has radius r. The cone has diameter 2r and height 2r. Compare the volumes of the cone and the sphere to the volume of the cylinder.

Answer:

Diameter of sphere = 2r

Radius r = \(\frac{d}{2}\) = \(\frac{2r}{2}\) = r

Volume of sphere = \(\frac{4}{3}\)πr^{3}

Radius of cylinder = r

Height of cylinder 2r

Volume of cylinder = πr^{2}h = πr^{2}(2r) = 2πr^{3}

Diameter of cone = 2r

Radius of cone r = \(\frac{d}{2}\) = \(\frac{2r}{2}\) = r

Height of cone 2r

Volume of cone = \(\frac{1}{3}\)πr^{2}h = \(\frac{2}{3}\)πr^{2}r = \(\frac{2}{3}\)πr^{3}

Volume of cylinder > Volume of sphere > Volume of cone

2πr^{3} > \(\frac{4}{3}\)πr^{3} > \(\frac{2}{3}\)πr^{3}

Question 27.

A spherical helium balloon that is 8 feet in diameter can lift about 17 pounds. What does the diameter of a balloon need to be to lift a person who weighs 136 pounds? Explain.

Answer:

Diameter of heAium balloon = 8 ft

Radius = r = \(\frac{8}{2}\) = 4 ft

Weight it can lift = 17 pounds

Volume of a spherical helium balloon = \(\frac{4}{3}\)πr^{3} = 4^{3}(\(\frac{4}{3}\)π)

Volume of balloon which can lift 136 pounds is equal to \(\frac{4}{3}\)π8^{3}.

Radius of that balloon = 8 ft

Diameter = 8 × 2 = 16 ft

Diameter of helium balloon = 16 ft