Surface area and volume of solids on the ACT Mathematics test covers rectangular boxes, spheres, and similar solids.
Students must find volume after a linear scale factor is applied, recover a sphere's diameter from its volume, and compute the painted surface of a block when a face is left unpainted. Questions typically give dimensions in feet or inches, ask for a nearest-cubic-unit volume, or keep π in the given volume.
Traps include applying the scale factor only once rather than cubing it, mixing radius with diameter, and counting an unpainted face. Answer choices are numbers of square or cubic units, sometimes still involving π.
Sample questions
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Question 1Easier
A rectangular wooden block measures 12 inches by 7 inches by 5 inches. Every face except the 12-inch-by-7-inch bottom is painted. What is the area, in square inches, of the painted surface?
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Answer: C — 274
All 6 faces have area 2(12(7)+12(5)+7(5))=358 square inches. The unpainted bottom has area 12(7)=84, so the painted area is 358−84=274 square inches.
Question 2Mid
A rectangular container measures 5 by 3 by 2 feet. A similar container has every linear dimension multiplied by 1.6. To the nearest cubic foot, what is the larger container’s volume?
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Answer: C — 123
Volumes of similar solids scale by the cube of the linear scale factor. The new volume is 30(1.6)3=122.88, which is approximately 123 cubic units.
Question 3Mid
A sphere has a volume of 288π cubic centimeters. What is the diameter of the sphere, in centimeters?
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Answer: C — 12
Solving 34πr3=288π gives r3=216, so r=6 centimeters. The diameter is 2r=12 centimeters.
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