Geometry Quiz: Solving Problems With Volume Formulas

20 questions · exam conditions

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Solving Problems With Volume FormulasQuestion 1 of 20

A cone and a cylinder have the same base radius and height. If the volume of the cylinder is $432\pi$ cubic centimeters, what is the volume of the cone?

72\pi

cubic centimeters

144\pi

cubic centimeters

216\pi

cubic centimeters

324\pi

cubic centimeters

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Geometry Quiz

Geometry Quiz: Solving Problems With Volume Formulas

Practice Solving Problems With Volume Formulas in Geometry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Solving Problems With Volume Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A cone and a cylinder have the same base radius and height. If the volume of the cylinder is $432\pi$ cubic centimeters, what is the volume of the cone?

$72\pi$ cubic centimeters
$144\pi$ cubic centimeters (correct answer)
$216\pi$ cubic centimeters
$324\pi$ cubic centimeters

Explanation: The volume of a cylinder is

V_{cylinder} = \pi r^2 h

and the volume of a cone is

V_{cone} = \frac{1}{3}\pi r^2 h

. Since they have the same base radius and height,

V_{cone} = \frac{1}{3} V_{cylinder} = \frac{1}{3} \times 432\pi = 144\pi

cubic centimeters. Choice A (

72\pi

) represents

\frac{1}{6}

of the cylinder volume. Choice C (

216\pi

) represents

\frac{1}{2}

of the cylinder volume. Choice D (

324\pi

) represents

\frac{3}{4}

of the cylinder volume.

Question 2

A party cone is filled with candy. The cone has radius $6\text{ cm}$ and height $10\text{ cm}$ . Which calculation correctly applies the volume formula?

$V=\pi(6)^2(10)$
$V=\tfrac{1}{3}\pi(6)^2(10)$ (correct answer)
$V=\tfrac{4}{3}\pi(6)^3$
$V=2\pi(6)(10)+2\pi(6)^2$

Explanation: This problem involves finding the volume of a party cone filled with candy. The solid is a cone with radius 6 cm and height 10 cm. The volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. The correct calculation is V = (1/3)π(6)²(10), which matches option B. This formula gives one-third the volume of a cylinder with the same base and height. Option A incorrectly uses the cylinder formula without the 1/3 factor, while option D uses the surface area formula. To solve volume problems correctly, first identify whether the solid is a cone, cylinder, or sphere before selecting the appropriate formula.

Question 3

A spherical balloon has radius $7\text{ in}$ . What is the volume of the solid?

$\tfrac{4}{3}\pi(7)^3\text{ in}^3$ (correct answer)
$\pi(7)^2\text{ in}^3$
$\tfrac{1}{3}\pi(7)^2\text{ in}^3$
$\tfrac{4}{3}\pi(7)^3\text{ in}^2$

Explanation: This problem asks for the volume of a spherical balloon. The solid is a sphere with radius 7 inches. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. Applying the formula: V = (4/3)π(7)³ = (4/3)π(343) in³. The correct answer includes the proper cubic units (in³) for volume. Option B incorrectly uses πr², which is the area of a circle, not the volume of a sphere, while option D has the wrong units (in² instead of in³). When calculating sphere volume, remember to cube the radius and multiply by (4/3)π, not just π.

Question 4

A cylindrical container with radius 5 cm and height 20 cm is filled with water to a depth of 15 cm. A solid sphere is completely submerged in the water, causing the water level to rise to exactly 18 cm. What is the radius of the sphere?

$\sqrt[3]{\frac{225}{4}}$ cm
$\sqrt[3]{\frac{75}{\pi}}$ cm
$\sqrt[3]{\frac{225}{4\pi}}$ cm (correct answer)
$\sqrt[3]{\frac{300}{4\pi}}$ cm

Explanation: The volume of water displaced equals the rise in water level times the base area of the cylinder. The water level rises from 15 cm to 18 cm, a rise of 3 cm. Volume displaced =

\pi(5)^2(3) = 75\pi

cubic cm. This equals the volume of the sphere:

\frac{4}{3}\pi r^3 = 75\pi

. Solving:

\frac{4}{3}r^3 = 75

, so

r^3 = \frac{75 \times 3}{4} = \frac{225}{4}

, giving

r = \sqrt[3]{\frac{225}{4}}

cm. Choice A omits the

\pi

cancellation step. Choice B results from incorrectly setting

\frac{4}{3}\pi r^3 = 75

instead of

75\pi

. Choice D uses an incorrect volume calculation.

Question 5

A spherical balloon has diameter $14\text{ in}$ . What is the volume of the balloon (in cubic inches) when fully inflated?

$\frac{4}{3}\pi(14)^3\text{ in}^3$
$\frac{4}{3}\pi(7)^3\text{ in}^3$ (correct answer)
$4\pi(7)^2\text{ in}^2$
$\pi(7)^2(14)\text{ in}^3$

Explanation: Solving problems with volume formulas involves calculating the space occupied by three-dimensional solids using appropriate mathematical expressions. The solid in this problem is a sphere. The correct volume formula for a sphere is V = (4/3)πr³, where r is the radius. Applying the formula with diameter 14 in (r = 7 in) gives V = (4/3)π(7)³ in³, matching choice B. This result accurately computes the volume using the radius, not the diameter directly. A common distractor misconception is using the diameter in place of radius without halving, as in choice A, which overestimates the volume. To transfer this strategy, always identify the solid as a sphere before selecting and applying the volume formula.

Question 6

A scoop of ice cream is shaped like a hemisphere with radius $4\text{ cm}$ . What is the volume of the solid?

$\frac{128}{3}\pi\text{ cm}^3$ (correct answer)
$\frac{256}{3}\pi\text{ cm}^3$
$64\pi\text{ cm}^3$
$\frac{64}{3}\pi\text{ cm}^3$

Explanation: This problem requires finding the volume of a hemisphere (half-sphere) of ice cream. The solid is a hemisphere with radius 4 cm. The volume formula for a sphere is V = (4/3)πr³, so a hemisphere has half this volume: V = (1/2)(4/3)πr³ = (2/3)πr³. Applying the formula: V = (2/3)π(4)³ = (2/3)π(64) = (128/3)π cm³. The volume is (128/3)π cubic centimeters. Option B doubles the correct answer, possibly confusing hemisphere with full sphere. When dealing with partial solids like hemispheres, adjust the standard formula by the appropriate fraction.

Question 7

A cylindrical candle (radius $3\text{ cm}$ , height $14\text{ cm}$ ) has a conical hole drilled straight down from the top. The hole is a cone with the same radius $3\text{ cm}$ and depth (height) $6\text{ cm}$ . Which value represents the total volume of wax remaining?

$\pi(3)^2(14)-\tfrac{1}{3}\pi(3)^2(6)$ (correct answer)
$\tfrac{1}{3}\pi(3)^2(14)-\pi(3)^2(6)$
$\pi(3)^2(14)+\tfrac{1}{3}\pi(3)^2(6)$
$2\pi(3)(14)+2\pi(3)^2-\pi(3)(6)$

Explanation: This problem requires finding the volume of wax in a candle with a conical hole. The solid is a cylinder (radius 3 cm, height 14 cm) minus a cone (same radius 3 cm, depth 6 cm). The volume formula requires subtracting: V = cylinder volume - cone volume = πr²h - (1/3)πr²h. Applying: V = π(3)²(14) - (1/3)π(3)²(6) = 126π - 18π = 108π cm³. The subtraction accounts for the removed wax from the conical hole. Option C incorrectly adds the volumes instead of subtracting. When dealing with composite solids involving removal, subtract the volume of the removed portion from the original solid.

Question 8

A composite solid is formed by drilling a cylindrical hole straight through the center of a solid cube. The cube has side length $10\text{ cm}$ , and the drilled hole is a cylinder with radius $2\text{ cm}$ and height $10\text{ cm}$ . What is the volume of the remaining solid?

$1000-40\pi\text{ cm}^3$ (correct answer)
$1000-20\pi\text{ cm}^3$
$1000-8\pi\text{ cm}^3$
$600\pi\text{ cm}^3$

Explanation: Solving problems with volume formulas involves calculating the space occupied by three-dimensional solids using appropriate mathematical expressions. The solid in this problem is a composite formed by subtracting a cylinder from a cube. The correct approach is to find the cube's volume V_cube = s³ and subtract the cylinder's volume V_cyl = πr²h. Applying with cube side 10 cm (V_cube = 1000 cm³) and cylinder r = 2 cm, h = 10 cm (V_cyl = π(2)²(10) = 40π cm³) gives 1000 - 40π cm³, matching choice A. This result accurately represents the remaining volume after drilling. A common distractor misconception is using an incorrect cylinder volume, like halving the radius unnecessarily as in choice B. To transfer this strategy, always identify the solid as a composite before calculating and subtracting volumes.

Question 9

A cylindrical container and a spherical container are compared. The cylinder has radius $4\text{ cm}$ and height $12\text{ cm}$ . The sphere has radius $4\text{ cm}$ . Which value represents the total volume of both containers combined?

$\pi(4)^2(12)+\frac{4}{3}\pi(4)^3\text{ cm}^3$ (correct answer)
$\pi(4)(12)+\frac{4}{3}\pi(4)^3\text{ cm}^3$
$2\pi(4)(12)+4\pi(4)^2\text{ cm}^3$
$\frac{1}{3}\pi(4)^2(12)+\frac{4}{3}\pi(4)^3\text{ cm}^3$

Explanation: The skill involves solving volume problems for geometric solids. The solids are a cylinder and a sphere combined. The volume is the sum of the cylinder's volume πr²h and the sphere's (4/3)πr³. Substituting r = 4 cm and h = 12 cm for the cylinder, and r = 4 cm for the sphere, gives π(4)²(12) + (4/3)π(4)³ cm³. This result matches choice A, providing the total volume. A common distractor is choice C, which adds surface areas instead of volumes. To transfer this strategy, always identify the solid before calculating its volume.

Question 10

A candle is shaped like a cylinder with radius $2\text{ cm}$ and height $15\text{ cm}$ . A cylindrical hole of radius $0.5\text{ cm}$ is drilled straight through the center along the full height. Which value represents the total volume of wax remaining?

$\pi(2^2)(15) - \pi(0.5^2)(15)$ (correct answer)
$\pi(2^2)(15)$
$2\pi(2)(15) - 2\pi(0.5)(15)$
$\pi\big((2-0.5)^2\big)(15)$

Explanation: This problem involves finding the volume of a candle with a hole drilled through it. The solid is a cylinder with a cylindrical hole removed, where the outer cylinder has radius 2 cm and the inner hole has radius 0.5 cm, both with height 15 cm. To find the remaining volume, subtract the hole's volume from the original cylinder's volume: V = π(2²)(15) - π(0.5²)(15). This represents the volume of the outer cylinder minus the volume of the inner cylindrical hole. A common error is subtracting the radii first, calculating π((2-0.5)²)(15) = π(1.5²)(15), which incorrectly gives the volume of a solid cylinder with radius 1.5 cm. When dealing with hollow cylinders, always calculate volumes separately then subtract.

Question 11

A cylindrical log has radius $0.5\text{ m}$ and length $2\text{ m}$ . Which reasoning uses the correct formula?

A: Multiply the base area by the length: $V=\pi r^2h$ . B: Multiply the circumference by the length: $V=2\pi rh$ . C: Use the sphere formula because the ends are circles: $V=\frac{4}{3}\pi r^3$ . D: Use the cone formula because it narrows at the ends: $V=\frac{1}{3}\pi r^2h$ .

A (correct answer)
B
C
D

Explanation: This problem tests understanding of which volume formula applies to a cylindrical log. The solid is a cylinder with radius 0.5 m and length (height) 2 m. The correct volume formula for a cylinder is V = πr²h, which multiplies the circular base area (πr²) by the height. Option A correctly identifies this approach. Option B incorrectly uses the lateral surface area formula 2πrh, while option C wrongly applies the sphere formula despite the log being cylindrical. Option D incorrectly uses the cone formula, which would only apply if the log tapered to a point. To solve volume problems correctly, first identify the three-dimensional shape to select the appropriate formula.

Question 12

A solid is made by placing a hemisphere of radius $6\text{ cm}$ on top of a cylinder with the same radius $6\text{ cm}$ and height $10\text{ cm}$ . Which value represents the total volume of the solid (in cubic centimeters)?

$360\pi+288\pi$
$360\pi+144\pi$ (correct answer)
$60\pi+288\pi$
$360\pi+216\pi$

Explanation: This problem requires finding the volume of a composite solid made of a hemisphere and cylinder. The solid consists of a hemisphere with radius 6 cm placed on a cylinder with radius 6 cm and height 10 cm. The total volume is V_hemisphere + V_cylinder = (2/3)πr³ + πr²h. For the hemisphere: V = (2/3)π(6)³ = (2/3)π(216) = 144π cm³. For the cylinder: V = π(6)²(10) = π(36)(10) = 360π cm³. The total volume is 360π + 144π cubic centimeters. A common error is using the full sphere formula (4/3)πr³ instead of half that for a hemisphere. When working with composite solids, calculate each part's volume separately before adding.

Question 13

A cylindrical water tank has radius $3\text{ m}$ and height $10\text{ m}$ . What is the volume of the tank in cubic meters?

$90\pi\text{ m}^3$ (correct answer)
$60\pi\text{ m}^3$
$180\pi\text{ m}^3$
$78\pi\text{ m}^3$

Explanation: This problem requires finding the volume of a cylindrical water tank. The solid is a right circular cylinder with radius 3 m and height 10 m. The volume formula for a cylinder is V = πr²h, where r is the radius and h is the height. Applying the formula: V = π(3)²(10) = π(9)(10) = 90π cubic meters. This result represents the total capacity of the cylindrical tank. A common error would be using diameter instead of radius, which would give π(6)²(10) = 360π, or forgetting to square the radius, yielding π(3)(10) = 30π. When solving volume problems, always identify the shape first to select the correct formula.

Question 14

A cylindrical water tank has radius $4\text{ m}$ and height $9\text{ m}$ , as labeled in the diagram (not drawn to scale). What is the volume of the tank in cubic meters?

$2\pi(4)(9)$
$\pi(4^2)(9)$ (correct answer)
$\pi(4^2)+2\pi(4)(9)$
$\frac{1}{3}\pi(4^2)(9)$

Explanation: The skill involves solving problems with volume formulas. The solid is a cylinder. The volume of a cylinder is given by the formula V = π r² h. Substituting the radius r = 4 m and height h = 9 m yields V = π (4²) (9). This calculation provides the space inside the tank, correctly representing its capacity in cubic meters. Choice A confuses the volume with the lateral surface area by omitting the squared radius. To apply this to other problems, always identify the solid before selecting the volume formula.

Question 15

A cone-shaped paper cup has radius $3\text{ cm}$ and height $10\text{ cm}$ , as labeled in the diagram (not drawn to scale). Which expression represents the volume of the cup (in cubic centimeters)?

$\pi(3^2)(10)$
$\frac{1}{3}\pi(3^2)(10)$ (correct answer)
$\pi(3^2)+\pi(3)(10)$
$\frac{4}{3}\pi(3^3)$

Explanation: The skill involves solving problems with volume formulas. The solid is a cone. The volume of a cone is given by the formula V = (1/3) π r² h. Substituting the radius r = 3 cm and height h = 10 cm yields V = (1/3) π (3²) (10). This expression accounts for the tapering shape, equaling one-third the volume of a cylinder with the same base and height. Choice A represents the cylinder volume, overlooking the one-third factor for cones. To apply this to other problems, always identify the solid before selecting the volume formula.

Question 16

A storage bin is a rectangular prism with length $18\text{ in}$ , width $10\text{ in}$ , and height $12\text{ in}$ . What is the volume of the solid?

$2160\text{ in}^3$ (correct answer)
$480\text{ in}^3$
$112\text{ in}^3$
$2160\text{ in}^2$

Explanation: This problem requires finding the volume of a rectangular prism storage bin. The solid is a rectangular prism with length 18 inches, width 10 inches, and height 12 inches. The volume formula for a rectangular prism is V = lwh, where l is length, w is width, and h is height. Applying the formula: V = (18)(10)(12) = 180(12) = 2160 in³. The volume of the storage bin is 2160 cubic inches. Option D incorrectly uses square inches (area units) instead of cubic inches (volume units). Always verify that volume answers use cubic units.

Question 17

A cylindrical can has diameter $10\text{ cm}$ and height $12\text{ cm}$ . What is the volume of the can in cubic centimeters?

$600\pi\text{ cm}^3$
$120\pi\text{ cm}^3$
$300\pi\text{ cm}^3$ (correct answer)
$240\pi\text{ cm}^3$

Explanation: This problem involves finding the volume of a cylindrical can given its diameter. The solid is a right circular cylinder with diameter 10 cm and height 12 cm. Since the volume formula V = πr²h requires radius, we must first convert: radius = diameter/2 = 10/2 = 5 cm. Applying the formula: V = π(5)²(12) = π(25)(12) = 300π cubic centimeters. This represents the can's total capacity. A common mistake is using diameter directly in the formula, giving π(10)²(12) = 1200π, which is four times too large. Always convert diameter to radius before applying cylinder volume formulas.

Question 18

A solid metal sphere has radius $6\text{ cm}$ . Which calculation correctly applies the volume formula?

$(4\pi)(6^2)$
$\frac{4}{3}\pi(6^3)$ (correct answer)
$\pi(6^2)(6)$
$\frac{1}{3}\pi(6^2)(6)$

Explanation: This problem asks which calculation correctly applies the volume formula for a sphere. The solid is a sphere with radius 6 cm. The volume formula for a sphere is V = (4/3)πr³, where r is the radius. The correct calculation is V = (4/3)π(6³) = (4/3)π(216). This gives the volume in cubic centimeters. Option C represents the formula for a cylinder (πr²h), which is incorrect for a sphere. To solve volume problems accurately, first identify the three-dimensional shape before selecting the appropriate formula.

Question 19

A concrete pillar is a cylinder with diameter $12\text{ ft}$ and height $5\text{ ft}$ . What is the volume of the solid?

$180\pi\text{ ft}^3$ (correct answer)
$720\pi\text{ ft}^3$
$360\pi\text{ ft}^3$
$180\pi\text{ ft}^2$

Explanation: This problem requires finding the volume of a concrete pillar. The solid is a cylinder with diameter 12 ft and height 5 ft. The volume formula for a cylinder is V = πr²h, but we must first convert diameter to radius: r = 12/2 = 6 ft. Applying the formula: V = π(6)²(5) = π(36)(5) = 180π ft³. The volume represents the amount of concrete needed to form the pillar. A common mistake is using diameter directly in the formula instead of radius, which would give π(12)²(5) = 720π ft³. Always convert diameter to radius by dividing by 2 before applying cylinder volume formulas.

Question 20

An ice cream cone is a right circular cone with radius $4\text{ cm}$ and height $12\text{ cm}$ . Which calculation correctly applies the volume formula for the cone?

$\pi(4)^2(12)$
$\frac{1}{3}\pi(4)^2(12)$ (correct answer)
$4\pi(12)$
$2\pi(4)(12)$

Explanation: Solving problems with volume formulas involves calculating the space occupied by three-dimensional solids using appropriate mathematical expressions. The solid in this problem is a right circular cone. The correct volume formula for a cone is V = (1/3)πr²h, where r is the radius and h is the height. Applying the formula with r = 4 cm and h = 12 cm gives the expression (1/3)π(4)²(12), matching choice B. This expression correctly computes the volume by accounting for the conical shape's one-third factor of a cylinder's volume. A common distractor misconception is using the cylinder volume formula without the one-third, as in choice A, which overestimates the volume. To transfer this strategy, always identify the solid as a cone before selecting and applying the volume formula.