Grade 10 · Geometry & measurement

Surface area in context (SAT) practice

Surface area in context (SAT) is a grade 10 math skill aligned to Common Core standard HSG.GMD.A.3: use volume formulas for cylinders, pyramids, cones, and spheres to solve problems. Below are 8 practice questions with answers and step-by-step explanations, drawn from the 10 surface area in context (sat) problems our math games drill.

CCSS HSG.GMD.A.310 questions in the bank
Sample questions

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Question 1medium

A shipping company uses closed rectangular cartons that measure 66 centimeters by 44 centimeters by 55 centimeters. What is the total exterior surface area, in square centimeters, of one carton?

Use SA=2(lw+lh+wh)SA = 2(lw + lh + wh) with l=6l = 6, w=4w = 4, and h=5h = 5. Then SA=2(24+30+20)=2(74)=148SA = 2(24 + 30 + 20) = 2(74) = 148 square centimeters.

Question 2medium

A gardener builds an open-top planter box in the shape of a rectangular prism. The inside measures 88 inches long, 33 inches wide, and 44 inches deep, and the box has no lid. What is the exterior surface area, in square inches, of the planter?

First find the closed surface area: 2(8⋅3+8⋅4+3⋅4)=2(24+32+12)=1362(8 \cdot 3 + 8 \cdot 4 + 3 \cdot 4) = 2(24 + 32 + 12) = 136. The open top removes one 8×38 \times 3 face, so subtract 2424 to get 136−24=112136 - 24 = 112 square inches.

Question 3medium

A factory seals soup in closed cylindrical cans with radius 33 centimeters and height 55 centimeters. What is the total exterior surface area, in square centimeters, of one closed can?

A closed cylinder has two circular bases and a lateral surface: SA=2πr2+2πrhSA = 2\pi r^2 + 2\pi rh. With r=3r = 3 and h=5h = 5, SA=2π(9)+2π(3)(5)=18π+30π=48πSA = 2\pi(9) + 2\pi(3)(5) = 18\pi + 30\pi = 48\pi square centimeters.

Question 4medium

A paint store sells open-top metal cans with radius 22 inches and height 66 inches. The cans have a bottom but no lid. What is the exterior surface area, in square inches, of one open can?

An open can has one circular base and a lateral surface: SA=πr2+2πrhSA = \pi r^2 + 2\pi rh. With r=2r = 2 and h=6h = 6, SA=π(4)+2π(2)(6)=4π+24π=28πSA = \pi(4) + 2\pi(2)(6) = 4\pi + 24\pi = 28\pi square inches.

Question 5medium

A museum display is shaped like a closed square pyramid with a 66-centimeter-by-66-centimeter base. Each triangular face has slant height 55 centimeters. What is the total exterior surface area, in square centimeters, of the display?

The base area is 62=366^2 = 36. The four congruent triangular faces each have area 12(6)(5)=15\dfrac{1}{2}(6)(5) = 15, for a total of 4×15=604 \times 15 = 60. The surface area is 36+60=9636 + 60 = 96 square centimeters.

Question 6medium

A metal pipe is shaped like a cylinder with radius 22 centimeters and length 1010 centimeters. Both ends of the pipe are open. What is the exterior surface area, in square centimeters, of the pipe?

With both ends open, only the lateral surface is exterior: SA=2πrhSA = 2\pi rh. With r=2r = 2 and h=10h = 10, SA=2π(2)(10)=40πSA = 2\pi(2)(10) = 40\pi square centimeters.

Question 7hard

A rectangular pyramid tent has a 1010-foot-by-44-foot rectangular base. The two triangular faces along the 1010-foot sides each have slant height 66 feet, and the two triangular faces along the 44-foot sides each have slant height 55 feet. The tent has no floor. What is the exterior surface area, in square feet, of the tent fabric?

Without a floor, only the four triangular faces are counted. Along the 1010-foot sides: 2×12(10)(6)=602 \times \dfrac{1}{2}(10)(6) = 60. Along the 44-foot sides: 2×12(4)(5)=202 \times \dfrac{1}{2}(4)(5) = 20. The total is 60+20=8060 + 20 = 80 square feet.

Question 8hard

A ramp cover is shaped like a right triangular prism. The triangular faces are right triangles with leg lengths 33 meters and 44 meters, and the prism is 1010 meters long. What is the total exterior surface area, in square meters, of the cover?

The two triangular faces have total area 2×12(3)(4)=122 \times \dfrac{1}{2}(3)(4) = 12. The three rectangular faces have areas 3(10)=303(10) = 30, 4(10)=404(10) = 40, and 5(10)=505(10) = 50, where 55 is the hypotenuse. The total is 12+30+40+50=13212 + 30 + 40 + 50 = 132 square meters.

Tests that cover surface area in context (sat):
NAEPSATACTLEAP 2025NC EOG/EOC
More Grade 10 skills:
Special right trianglesSurface area and volumeSimilar right triangle trigonometryTrig ratios and unit circle valuesPerimeter and arc lengthMidpoint formulaEquation of a circleCircle theoremsAll Grade 10 practice →

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