What is Volume & solids on the Geometry Regents?

Volume & solids is a Solids & Modeling topic on the Geometry Regents. RegentsLab has 23 practice questions on it, each with a full worked solution.

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Volume & solids: Geometry Regents Practice

Practice Volume & solids for the Geometry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 49 · 2 pts

A cylinder has a radius of $3$ inches and a height of $8$ inches. What is the volume of the cylinder?

(A) $24\pi \text{ in}^3$
(B) $48\pi \text{ in}^3$
(C) $288\pi \text{ in}^3$
(D) $72\pi \text{ in}^3$ ✓

Worked solution

$V = \pi r^2 h$ Volume of a cylinder.

$V = \pi (3)^2 (8)$ Substitute $r=3$ and $h=8$.

$V = \pi (9)(8) = 72\pi$

Answer: $72\pi \text{ in}^3$

The volume of a cylinder is the base area $\pi r^2$ times the height $h$, so the radius must be squared.

Question 52 · 2 pts

A pyramid has a square base with a side length of $6$ and a height of $10$. What is its volume?

(A) $120$ ✓
(B) $20$
(C) $360$
(D) $80$

Worked solution

$B = 6^2 = 36$ Area of the square base.

$V = \tfrac{1}{3}Bh$

$V = \tfrac{1}{3}(36)(10) = 120$

Answer: $120$

A pyramid's volume is one-third the base area times the height, and a square base's area is the side length squared.

Question 13 · 2 pts

A cylinder has radius $3$ and height $10$. What is its volume? (Leave in terms of $\pi$.)

(A) $90\pi$ ✓
(B) $900\pi$
(C) $30\pi$
(D) $60\pi$

Worked solution

$V=\pi r^2 h$ Cylinder volume.

$=\pi(3)^2(10)=90\pi$

Answer: 90\pi

A cylinder's volume is the base area πr² times the height; square the radius first.

Question 50 · 2 pts

A cone has a radius of $6$ cm and a height of $10$ cm. To the nearest cubic centimeter, what is its volume?

(A) $1131 \text{ cm}^3$
(B) $1508 \text{ cm}^3$
(C) $377 \text{ cm}^3$ ✓
(D) $63 \text{ cm}^3$

Worked solution

$V = \tfrac{1}{3}\pi r^2 h$ Volume of a cone.

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

$V = \tfrac{1}{3}\pi (36)(10) = 120\pi$

$V \approx 376.99 \approx 377$ Round to the nearest cubic centimeter.

Answer: $377 \text{ cm}^3$

A cone's volume is one-third of the cylinder that shares its base and height.

Question 51 · 2 pts

A sphere has a radius of $6$ meters. What is its volume in terms of $\pi$?

(A) $288\pi \text{ m}^3$ ✓
(B) $144\pi \text{ m}^3$
(C) $48\pi \text{ m}^3$
(D) $216\pi \text{ m}^3$

Worked solution

$V = \tfrac{4}{3}\pi r^3$ Volume of a sphere.

$V = \tfrac{4}{3}\pi (6)^3$

$V = \tfrac{4}{3}\pi (216) = 288\pi$

Answer: $288\pi \text{ m}^3$

The volume of a sphere uses the cube of the radius with a coefficient of $\tfrac{4}{3}$.

Question 55 · 2 pts

A rectangle with side lengths $3$ and $5$ is rotated $360^\circ$ about the side of length $5$. What is the volume of the solid formed, in terms of $\pi$?

(A) $15\pi$
(B) $75\pi$
(C) $45\pi$ ✓
(D) $180\pi$

Worked solution

$\text{Rotating about the side of length } 5 \text{ forms a cylinder.}$ The radius is the other side, $3$; the height is $5$.

$V = \pi r^2 h = \pi (3)^2 (5)$

$V = \pi (9)(5) = 45\pi$

Answer: $45\pi$

Rotating a rectangle about one of its sides sweeps out a cylinder whose radius is the other side length and whose height is the axis side.

Question 56 · 2 pts

The volume of a cone is $75\pi$ cubic centimeters and the radius of its base is $5$ cm. What is the height of the cone?

(A) $9$ cm ✓
(B) $45$ cm
(C) $1$ cm
(D) $3$ cm

Worked solution

$V = \tfrac{1}{3}\pi r^2 h$ Volume of a cone.

$75\pi = \tfrac{1}{3}\pi (5)^2 h$

$75\pi = \tfrac{25}{3}\pi h$

$h = 75 \cdot \tfrac{3}{25} = 9$

Answer: $9$ cm

Substitute the known volume and radius into the cone formula, then isolate $h$ by multiplying both sides by $\tfrac{3}{r^2}$.

Question 19 · 2 pts

A cylindrical water tank has a radius of $3$ feet and a height of $8$ feet. What is the volume of the tank, to the nearest tenth of a cubic foot?

(A) $226.2$ ✓
(B) $904.8$
(C) $75.4$
(D) $150.8$

Worked solution

$V = \pi r^2 h$ Volume of a cylinder.

$V = \pi (3)^2 (8) = 72\pi$

$V \approx 226.19 \approx 226.2 \text{ ft}^3$

Answer: $226.2$

The volume of a cylinder is pi times the radius squared times the height: pi(9)(8) = 72 pi ≈ 226.2 cubic feet. The distractor 150.8 is the lateral surface area 2 pi r h, 904.8 uses the diameter instead of the radius, and 75.4 forgets to square the radius.

Question 29 · 4 pts · constructed response

A solid steel rod is in the shape of a cylinder with a diameter of $6$ centimeters and a height of $15$ centimeters. The density of the steel is $7.7$ grams per cubic centimeter. Determine, to the nearest gram, the mass of the rod.

Model answer: Approximately $3266$ grams. $V = \pi(3)^2(15) = 135\pi \approx 424.115$ cm$^3$; mass $= 424.115 \times 7.7 \approx 3266$ g.

Worked solution

$r = \frac{6}{2} = 3 \text{ cm}$ The radius is half the diameter.

$V = \pi r^2 h = \pi (3)^2 (15) = 135\pi \approx 424.115 \text{ cm}^3$

$\text{mass} = \text{density} \times \text{volume} = 7.7 \times 424.115$

$\text{mass} \approx 3265.7 \approx 3266 \text{ grams}$

Answer: Approximately $3266$ grams. $V = \pi(3)^2(15) = 135\pi \approx 424.115$ cm$^3$; mass $= 424.115 \times 7.7 \approx 3266$ g.

The volume of a cylinder is $\pi r^2 h$. With radius $3$ cm and height $15$ cm the volume is $135\pi \approx 424.115$ cm$^3$. Multiplying by the density $7.7$ g/cm$^3$ gives a mass of about $3266$ grams.

Question 7 · 2 pts

A cylindrical water tank has a radius of $4$ feet and a height of $10$ feet. What is the volume of the tank, in cubic feet, in terms of $\pi$?

(A) $40\pi$
(B) $80\pi$
(C) $160\pi$ ✓
(D) $320\pi$

Worked solution

$V = \pi r^2 h$ Volume of a cylinder.

$V = \pi (4)^2 (10)$

$V = \pi (16)(10) = 160\pi$

Answer: $160\pi$

The volume of a cylinder is $V = \pi r^2 h$. With $r = 4$ and $h = 10$, $V = \pi(16)(10) = 160\pi$ cubic feet. Choosing $80\pi$ comes from using $r^2 = 8$ (doubling instead of squaring), and $40\pi$ comes from using $r$ instead of $r^2$.

Question 30 · 4 pts · constructed response

A solid steel rod is in the shape of a cylinder with a diameter of $2$ inches and a length of $18$ inches. The density of the steel is $0.284$ pound per cubic inch. Determine and state, to the nearest tenth of a pound, the weight of the rod.

Model answer: The rod weighs about $16.1$ pounds.

Worked solution

$r = \frac{2}{2} = 1 \text{ in}$

$V = \pi r^2 h = \pi (1)^2 (18) = 18\pi \approx 56.5487 \text{ in}^3$

$W = V \cdot \text{density} \approx 56.5487 \times 0.284 \approx 16.06$

$W \approx 16.1 \text{ pounds}$

Answer: The rod weighs about $16.1$ pounds.

The volume of a cylinder is $\pi r^2 h$. With radius $1$ inch and length $18$ inches, the volume is $18\pi \approx 56.55$ cubic inches. Multiplying volume by the density, $0.284$ pound per cubic inch, gives a weight of about $16.1$ pounds.

Question 35 · 4 pts · constructed response

A machinist cuts a solid steel cylinder with a diameter of $10$ centimeters and a height of $15$ centimeters. The density of the steel is $7.7$ grams per cubic centimeter. Determine and state, to the nearest gram, the mass of the cylinder.

Model answer: The volume is $V = \pi(5)^2(15) = 375\pi \approx 1178.097$ cm$^3$, so the mass is $1178.097 \times 7.7 \approx 9071$ grams.

Worked solution

$r = \frac{10}{2} = 5 \text{ cm}$ The radius is half the diameter.

$V = \pi r^2 h = \pi (5)^2 (15) = 375\pi \approx 1178.097 \text{ cm}^3$

$\text{mass} = \text{density} \times \text{volume} = 7.7 \times 1178.097 \approx 9071.35$

$\text{mass} \approx 9071 \text{ grams}$ Round to the nearest gram; keep $\pi$ until the final step to avoid rounding error.

Answer: The volume is $V = \pi(5)^2(15) = 375\pi \approx 1178.097$ cm$^3$, so the mass is $1178.097 \times 7.7 \approx 9071$ grams.

Mass equals density times volume. The cylinder's volume is $\pi r^2 h = 375\pi \approx 1178.097$ cubic centimeters, and multiplying by the density of $7.7$ g/cm$^3$ gives approximately $9071$ grams.

Question 7 · 2 pts

A cylindrical water tank has a diameter of $6$ feet and a height of $10$ feet. What is the volume of the tank, to the nearest cubic foot?

(A) $283$ ✓
(B) $94$
(C) $1131$
(D) $188$

Worked solution

$r = \frac{6}{2} = 3 \text{ ft}$ The radius is half the diameter.

$V = \pi r^2 h = \pi (3)^2 (10)$

$V = 90\pi \approx 282.74$

$V \approx 283 \text{ cubic feet}$

Answer: $283$

The volume of a cylinder is pi times the radius squared times the height. The radius is 3 feet, half of the 6-foot diameter, so V = 90 pi, about 283 cubic feet. Using the diameter instead of the radius gives 1131, and using the cone formula gives 94.

Question 8 · 2 pts

A right circular cone has a radius of $6$ centimeters and a height of $10$ centimeters. What is the volume of the cone, in cubic centimeters, in terms of $\pi$?

(A) $40\pi$
(B) $120\pi$ ✓
(C) $360\pi$
(D) $60\pi$

Worked solution

$V = \frac{1}{3}\pi r^2 h$

$V = \frac{1}{3}\pi (6)^2 (10)$

$V = \frac{1}{3}\pi (360) = 120\pi$

Answer: $120\pi$

The volume of a cone is one third of pi times the radius squared times the height. Here that is one third of 360 pi, or 120 pi. Forgetting the one-third factor gives the cylinder volume 360 pi, and using the diameter or halving incorrectly gives the other distractors.

Question 19 · 2 pts

A right circular cylinder has a diameter of $10$ centimeters and a height of $12$ centimeters. What is the volume of the cylinder, to the nearest cubic centimeter?

(A) $3770$
(B) $942$ ✓
(C) $377$
(D) $1885$

Worked solution

$r = \frac{10}{2} = 5$ The radius is half the diameter.

$V = \pi r^2 h = \pi (5)^2 (12)$

$V = 300\pi \approx 942$

Answer: $942$

The volume of a cylinder is $V = \pi r^2 h$. The diameter is $10$, so the radius is $5$, giving $V = \pi(25)(12) = 300\pi \approx 942$ cubic centimeters. The answer $3770$ results from using the diameter as the radius, and $377$ comes from computing the lateral surface area $2\pi rh$ instead of the volume.

Question 29 · 4 pts · constructed response

A paper cup is in the shape of a right circular cone with a diameter of 6 centimeters and a height of 10 centimeters. Determine and state, to the nearest tenth of a cubic centimeter, the volume of water the cup can hold when completely full.

Model answer: The volume is $30\pi \approx 94.2$ cubic centimeters.

Worked solution

$r = \frac{6}{2} = 3 \text{ cm}$ The radius is half the diameter.

$V = \frac{1}{3}\pi r^2 h$

$V = \frac{1}{3}\pi (3)^2 (10) = 30\pi$

$V \approx 94.2 \text{ cm}^3$

Answer: The volume is $30\pi \approx 94.2$ cubic centimeters.

The volume of a cone is one third of the volume of the cylinder with the same base and height. With radius 3 cm and height 10 cm, $V = \frac{1}{3}\pi(9)(10) = 30\pi \approx 94.2$ cubic centimeters.

Question 33 · 4 pts · constructed response

A grain silo is shaped like a cylinder with a hemisphere on top. The cylinder has a radius of $4$ feet and a height of $30$ feet. The hemisphere has the same radius as the cylinder. Determine and state, to the nearest cubic foot, the total volume of the silo.

Model answer: The total volume is $480\pi + \frac{128}{3}\pi = \frac{1568}{3}\pi \approx 1642$ cubic feet.

Worked solution

$V_{cylinder} = \pi r^2 h = \pi (4)^2 (30) = 480\pi$ Volume of the cylindrical part

$V_{hemisphere} = \frac{1}{2} \cdot \frac{4}{3}\pi r^3 = \frac{2}{3}\pi (4)^3 = \frac{128}{3}\pi$ Half of a sphere of radius 4

$V_{total} = 480\pi + \frac{128}{3}\pi = \frac{1568}{3}\pi \approx 1641.99$ Add the two volumes

$V \approx 1642 \text{ cubic feet}$ Round to the nearest cubic foot

Answer: The total volume is $480\pi + \frac{128}{3}\pi = \frac{1568}{3}\pi \approx 1642$ cubic feet.

The silo is a composite solid, so its volume is the sum of the volume of the cylinder, $\pi r^2 h$, and the volume of the hemisphere, $\frac{2}{3}\pi r^3$. Using $r = 4$ and $h = 30$ gives $\frac{1568}{3}\pi$, which rounds to $1642$ cubic feet.

Question 24 · 2 pts

A cylinder has radius $3$ and height $10$. What is its volume? (Use $V = \pi r^2 h$.)

(A) $60\pi$
(B) $900\pi$
(C) $30\pi$
(D) $90\pi$ ✓

Worked solution

$V = \pi r^2 h = \pi (3)^2 (10)$ Substitute r = 3, h = 10.

$= \pi(9)(10) = 90\pi$

Answer: 90\pi

Cylinder volume is the base area (πr²) times the height. Square the radius before multiplying.

Question 28 · 2 pts

A right triangle with legs $3$ and $4$ is rotated about its leg of length $4$. What solid is generated?

(A) a cone with radius $3$ and height $4$ ✓
(B) a sphere of radius $4$
(C) a cone with radius $4$ and height $3$
(D) a cylinder with radius $3$ and height $4$

Worked solution

$\text{axis = leg of length } 4 \Rightarrow h=4$ The leg on the axis becomes the height.

$\text{other leg } 3 \Rightarrow r=3$

Answer: \text{cone, } r=3,\ h=4

Rotating a right triangle about a leg sweeps out a cone; the leg on the axis is the height, the other leg the radius.

Question 53 · 2 pts

A solid metal cylinder has a radius of $2$ cm and a height of $5$ cm. The metal has a density of $8$ grams per cubic centimeter. To the nearest gram, what is the mass of the cylinder?

(A) $63$ g
(B) $503$ g ✓
(C) $8$ g
(D) $251$ g

Worked solution

$V = \pi r^2 h = \pi (2)^2 (5) = 20\pi \approx 62.83$ Volume of the cylinder.

$\text{mass} = \text{density} \times \text{volume}$

$\text{mass} = 8 \times 62.83 \approx 502.65$

$\approx 503$ Round to the nearest gram.

Answer: $503$ g

Mass equals density times volume, so first compute the volume of the solid, then multiply by the density.

Question 54 · 2 pts

A solid is formed by placing a hemisphere on top of a cylinder. The cylinder has a radius of $3$ cm and a height of $8$ cm, and the hemisphere has a radius of $3$ cm. What is the total volume, in terms of $\pi$?

(A) $90\pi \text{ cm}^3$ ✓
(B) $72\pi \text{ cm}^3$
(C) $78\pi \text{ cm}^3$
(D) $108\pi \text{ cm}^3$

Worked solution

$V_{\text{cyl}} = \pi r^2 h = \pi (3)^2 (8) = 72\pi$

$V_{\text{hemi}} = \tfrac{1}{2}\cdot\tfrac{4}{3}\pi r^3 = \tfrac{2}{3}\pi (3)^3 = 18\pi$

$V = 72\pi + 18\pi = 90\pi$

Answer: $90\pi \text{ cm}^3$

The volume of a composite solid is the sum of its parts, and a hemisphere is half of a sphere's volume.

Question 23 · 2 pts

A cylindrical water tank has a diameter of $8$ feet and a height of $12$ feet. What is the volume of the tank, to the nearest cubic foot?

(A) $301$
(B) $603$ ✓
(C) $2413$
(D) $402$

Worked solution

$r = \frac{8}{2} = 4 \text{ feet}$ The radius is half the diameter.

$V = \pi r^2 h = \pi (4)^2 (12)$

$V = 192\pi \approx 603.19$

$V \approx 603 \text{ cubic feet}$

Answer: $603$

The volume of a cylinder is $V = \pi r^2 h$. The radius is half of the $8$-foot diameter, so $r = 4$, giving $V = \pi(4)^2(12) = 192\pi \approx 603$ cubic feet. Using the diameter instead of the radius, a very common error, gives $768\pi \approx 2413$.

Question 33 · 4 pts · constructed response

A cylindrical water tank has a radius of $1.5$ feet and a height of $8$ feet. The tank is completely filled with water. Water weighs $62.4$ pounds per cubic foot. Determine and state, to the nearest pound, the weight of the water in the fully filled tank.

Model answer: $3529$ pounds

Worked solution

$V = \pi r^2 h = \pi (1.5)^2 (8)$ Volume of a cylinder.

$V = 18\pi \approx 56.5487 \text{ ft}^3$

$\text{Weight} = 56.5487 \times 62.4 \approx 3528.6$

$\text{Weight} \approx 3529 \text{ pounds}$

Answer: $3529$ pounds

First find the volume of the cylinder, $V = \pi r^2 h = 18\pi \approx 56.5487$ cubic feet, keeping the full calculator value to avoid rounding error. Multiplying the volume by the density, $56.5487 \times 62.4 \approx 3528.6$, gives a weight of $3529$ pounds to the nearest pound.

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