3D Solid Volumes - Math 3

A spherical balloon has a volume of $36\pi$ cubic inches. If the balloon is inflated until its radius doubles, by what factor does the surface area increase?
Answer: 4 Original volume: $\frac{4}{3}\pi r^3 = 36\pi$ , so $r^3 = 27$ and $r = 3$ inches. Original surface area: $4\pi r^2 = 4\pi(3^2) = 36\pi$ square inches. After doubling the radius to 6 inches: New surface area: $4\pi(6^2) = 144\pi$ square inches. Factor of increase: $\frac{144\pi}{36\pi} = 4$ . Choice A is the radius factor, not surface area. Choice C confuses volume and surface area relationships. Choice D is the volume factor increase.
A rectangular prism has dimensions 6 inches by 8 inches by 10 inches. A smaller rectangular prism with dimensions 2 inches by 3 inches by 4 inches is removed from one corner. What is the volume of the remaining solid?
Answer: 456 cubic inches The volume of the large prism is $6 \times 8 \times 10 = 480$ cubic inches. The volume of the removed small prism is $2 \times 3 \times 4 = 24$ cubic inches. The remaining volume is $480 - 24 = 456$ cubic inches. Choice B results from miscalculating the small prism volume as 20 instead of 24. Choice C comes from calculating the small prism volume as 16. Choice D is the original volume without subtracting the removed piece.
A cylindrical water tank has a radius of 3 meters and a height of 8 meters. Due to maintenance, the water level is kept at 75% of the tank's capacity. If water is being pumped out at a rate that reduces the volume by $\frac{1}{6}$ every hour, what volume of water remains after 2 hours?
Answer: $45\pi$ cubic meters First, find the tank's total volume: $V = \pi r^2 h = \pi(3^2)(8) = 72\pi$ cubic meters. At 75% capacity: $0.75 \times 72\pi = 54\pi$ cubic meters initially. Each hour, $\frac{1}{6}$ of the current volume is removed, so $\frac{5}{6}$ remains. After 2 hours: $54\pi \times (\frac{5}{6})^2 = 54\pi \times \frac{25}{36} = 45\pi$ cubic meters. Choice A assumes linear reduction rather than exponential. Choice B uses only one hour of reduction. Choice D is the initial volume.
A regular tetrahedron has edge length $a$ . If the volume of this tetrahedron is $18\sqrt{2}$ cubic units, what is the volume of a cube with the same edge length $a$ ?
Answer: $216$ cubic units This problem tests your ability to work with 3D geometric formulas and use given information to find unknown quantities. When you see a question connecting different geometric shapes with the same dimensions, focus on finding the shared parameter first. Start with the volume formula for a regular tetrahedron: $V = \frac{a^3\sqrt{2}}{12}$ , where $a$ is the edge length. Since the tetrahedron's volume is $18\sqrt{2}$ , you can set up the equation: $\frac{a^3\sqrt{2}}{12} = 18\sqrt{2}$ Divide both sides by $\sqrt{2}$ : $\frac{a^3}{12} = 18$ Multiply by 12: $a^3 = 216$ Therefore $a = 6$ . Now find the cube's volume using $V = a^3 = 6^3 = 216$ cubic units. Looking at the wrong answers: Choice A (108) represents half the correct volume, likely from forgetting to cube the edge length or making an arithmetic error. Choice C (72) might result from incorrectly using $a^2 \cdot 6$ instead of $a^3$ , mixing up area and volume formulas. Choice D (144) could come from calculation errors when solving for $a^3$ or incorrectly manipulating the tetrahedron formula. The correct answer is B: 216 cubic units. Strategy tip: When problems give you one shape's properties to find another's, always solve for the shared dimension first. Master the standard volume formulas for common 3D shapes, especially cubes ( $a^3$ ) and regular tetrahedra ( $\frac{a^3\sqrt{2}}{12}$ ), as these appear frequently together in geometry problems.
A cube has the same volume as a sphere with radius 6 cm. What is the edge length of the cube to the nearest tenth of a centimeter?
Answer: 9.1 cm When you encounter problems involving equal volumes between different shapes, you need to set up an equation using their respective volume formulas and solve for the unknown dimension. Start by finding the sphere's volume using $V = \frac{4}{3}\pi r^3$ . With radius 6 cm: $V = \frac{4}{3}\pi (6)^3 = \frac{4}{3}\pi (216) = 288\pi$ cubic cm. Since the cube has the same volume, and a cube's volume is $V = s^3$ where $s$ is the edge length, you can set up: $s^3 = 288\pi$ . Taking the cube root: $s = \sqrt[3]{288\pi} = \sqrt[3]{288 \times 3.14159} = \sqrt[3]{904.78} \approx 9.1$ cm. Looking at the wrong answers: Choice A (7.8 cm) results from incorrectly using the sphere's surface area formula instead of volume, or making calculation errors early in the process. Choice B (8.4 cm) comes from forgetting the $\frac{4}{3}$ coefficient in the sphere volume formula and just using $\pi r^3$ . Choice C (9.7 cm) occurs when students make rounding errors during intermediate steps or use an imprecise value of π. The correct answer is D (9.1 cm). Study tip: For volume equality problems, always write out both volume formulas completely before substituting values. Double-check that you're using volume formulas, not surface area ones, and be careful with π calculations—use at least 3.14159 for accuracy in intermediate steps before final rounding.
A sphere has a surface area of $144\pi$ square centimeters. What is the volume of this sphere in cubic centimeters?
Answer: $288\pi$ cubic centimeters First, find the radius using surface area formula $SA = 4\pi r^2$ . So $144\pi = 4\pi r^2$ , which gives $r^2 = 36$ , so $r = 6$ cm. Then use the volume formula $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(6)^3 = \frac{4}{3}\pi(216) = 288\pi$ cubic centimeters. Choice A results from incorrectly using $V = \frac{2}{3}\pi r^3$ . Choice B assumes the volume equals the surface area numerically. Choice C comes from using $V = \pi r^3$ instead of $\frac{4}{3}\pi r^3$ .
A hollow cylinder has an outer radius of 8 cm, inner radius of 5 cm, and height of 12 cm. If this cylinder is filled with water and then a solid sphere with radius 4 cm is completely submerged in it, what volume of water overflows?
Answer: $0$ cubic centimeters Hollow cylinder volume: $\pi(8^2 - 5^2)(12) = \pi(64-25)(12) = 468\pi$ cubic cm. Sphere volume: $\frac{4}{3}\pi(4^3) = \frac{256\pi}{3} \approx 268.1\pi$ cubic cm. Since the sphere volume ( $268.1\pi$ ) is less than the cylinder capacity ( $468\pi$ ), no water overflows. Choice A is the sphere volume. Choice B results from incorrect calculation. Choice D uses wrong sphere volume formula.
A cone and a cylinder have the same base radius of 5 meters and the same height of 9 meters. How many times greater is the volume of the cylinder than the volume of the cone?
Answer: 3 times greater The volume of a cylinder is $V_{cylinder} = \pi r^2 h = \pi(5)^2(9) = 225\pi$ . The volume of a cone is $V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi(5)^2(9) = 75\pi$ . The ratio is $\frac{225\pi}{75\pi} = 3$ . Choice A results from using $\frac{1}{2}$ instead of $\frac{1}{3}$ in the cone formula. Choice C comes from incorrectly comparing surface areas or using wrong formulas. Choice D uses the height value as the multiplier.
A cylindrical water tank has a diameter of 8 feet and a height of 12 feet. If the tank is currently filled to 75% of its capacity, how many cubic feet of water are in the tank?
Answer: $144\pi$ cubic feet The volume of a cylinder is $V = \pi r^2 h$ . With diameter 8 feet, the radius is 4 feet. The full volume is $V = \pi(4)^2(12) = 192\pi$ cubic feet. At 75% capacity, the volume is $0.75 \times 192\pi = 144\pi$ cubic feet. Choice B gives the full tank volume without applying the 75% factor. Choice C incorrectly uses diameter instead of radius: $\pi(8)^2(12) \times 0.75 = 576\pi \times 0.75 = 432\pi$ , but this doesn't match any calculation. Choice D uses diameter instead of radius for the full volume: $\pi(8)^2(12) = 768\pi$ , then incorrectly calculates $768\pi \times 0.75 = 576\pi$ .
Two spheres have radii in the ratio 2:3. If the smaller sphere has a volume of $32\pi$ cubic units, what is the volume of the larger sphere?
Answer: $108\pi$ cubic units When radii are in ratio 2:3, volumes are in ratio $2^3:3^3 = 8:27$ . If the smaller volume is $32\pi$ , then $\frac{32\pi}{8} = 4\pi$ per unit ratio. The larger volume is $4\pi \times 27 = 108\pi$ cubic units. Alternatively, volume scales as the cube of linear dimensions: $V_{large} = V_{small} \times (\frac{3}{2})^3 = 32\pi \times \frac{27}{8} = 108\pi$ . Choice A uses ratio 3:2 instead of $(\frac{3}{2})^3$ . Choice B uses ratio squared instead of cubed. Choice D uses $32\pi \times \frac{27}{4}$ .
A sphere is inscribed in a cube with edge length 6 cm. What is the ratio of the volume of the sphere to the volume of the cube?
Answer: $\frac{\pi}{6}$ An inscribed sphere touches all faces of the cube, so its diameter equals the cube's edge length. The sphere's radius is $r = 3$ cm. Sphere volume: $V_s = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3^3) = 36\pi$ cubic cm. Cube volume: $V_c = 6^3 = 216$ cubic cm. Ratio: $\frac{36\pi}{216} = \frac{\pi}{6}$ . Choice B results from using $\frac{4}{3}\pi r^3$ incorrectly as $\pi r^3$ . Choice C comes from confusing diameter with radius. Choice D uses the wrong sphere volume formula.
A cylindrical container has a radius of 6 inches and contains water to a height of 8 inches. If a solid sphere with radius 3 inches is completely submerged in the water, by how many inches will the water level rise?
Answer: 1 inch The volume of the sphere is $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(3)^3 = 36\pi$ cubic inches. This volume will displace an equal volume of water in the cylinder. The cross-sectional area of the cylinder is $\pi(6)^2 = 36\pi$ square inches. The rise in water level is $\frac{\text{displaced volume}}{\text{cross-sectional area}} = \frac{36\pi}{36\pi} = 1$ inch. Choice A uses radius 3 instead of volume calculation. Choice C incorrectly uses $\frac{4}{3} \times 1$ . Choice D doubles the correct answer.
A hemispherical bowl has an inner radius of 15 cm. It is filled with water to a depth of 9 cm from the bottom. What is the volume of water in the bowl?
Answer: $972\pi$ cubic centimeters This requires finding the volume of a spherical cap with height $h = 9$ cm and sphere radius $R = 15$ cm. The formula is $V = \frac{\pi h^2}{3}(3R - h) = \frac{\pi (9^2)}{3}(3 \times 15 - 9) = \frac{81\pi}{3}(45 - 9) = 27\pi \times 36 = 972\pi$ cubic cm. Choice A uses $\pi h^2 R/3$ . Choice B uses $\frac{\pi h^2(2R)}{3}$ . Choice C uses $\pi h^3$ .
A rectangular prism has dimensions 4 cm × 6 cm × 9 cm. If each dimension is increased by the same factor $k$ , and the new volume is 1728 cubic cm, what is the value of $k$ ?
Answer: 2 Original volume: $V_1 = 4 \times 6 \times 9 = 216$ cubic cm. After scaling by factor $k$ , the new dimensions are $4k$ , $6k$ , and $9k$ . New volume: $V_2 = 4k \times 6k \times 9k = 216k^3 = 1728$ . Solving: $k^3 = \frac{1728}{216} = 8$ , so $k = 2$ . Choice B results from incorrectly thinking $k^3 = 3$ . Choice C comes from $\sqrt{8}$ instead of $\sqrt[3]{8}$ . Choice D is the value of $k^3$ , not $k$ .
A square pyramid has a base with side length 10 feet and a height of 12 feet. What is the volume of the pyramid in cubic feet?
Answer: 400 cubic feet The volume of a pyramid is $V = \frac{1}{3}Bh$ where $B$ is the base area and $h$ is the height. The base area is $10^2 = 100$ square feet. So $V = \frac{1}{3}(100)(12) = \frac{1200}{3} = 400$ cubic feet. Choice A uses only base area times height divided by 10. Choice B uses $\frac{1}{3} \times 10 \times 12 \times 9$ , confusing side length with area. Choice D forgets the $\frac{1}{3}$ factor entirely.

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