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Geometry Help: Solving Problems With Volume Formulas

Review real example questions for Solving Problems With Volume Formulas in Geometry.

Question 1 / 10

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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?

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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.