Printable math worksheet for calculating the surface area of prisms and cylinders.

I'll solve each surface area problem step by step, working through all calculations carefully.

Problem 1: Rectangular Prism (6 ft × 8 ft × 12 ft)
- Surface Area = 2(lw + lh + wh)
- = 2(6×8 + 6×12 + 8×12)
- = 2(48 + 72 + 96)
- = 2(216)
- = 432 ft²

Problem 2: Triangular Prism
First, I need to find the missing side of the triangular base using Pythagorean theorem:
- The triangle has height 4 yd and base 6 yd
- Hypotenuse = √(4² + 6²) = √(16 + 36) = √52 ≈ 7.21 yd... but wait, let me check if this is a right triangle with different dimensions.

Looking at the diagram more carefully - it appears to be a right triangle with legs 4 yd and some other dimension, and hypotenuse shown as part of the prism. Let me reconsider: the triangle shows 4 yd height, and we need to determine the base. Actually, looking at typical problems, if one leg is 4 and we see "6 yd" on the bottom, this might be the base. But there's also what looks like another measurement.

Let me assume it's a right triangle with legs 4 yd and we need to find the third side. If the hypotenuse of the triangle face is labeled, I should use that. Without seeing exact labels clearly, for a standard problem with height 4 yd, base 6 yd, and length 11 yd:
- Triangle area = (1/2)(4)(6) = 12 yd²
- Two triangles = 24 yd²
- Three rectangular faces: 4×11, 6×11, and hypotenuse×11
- Hypotenuse = √(4² + 6²) = √52 ≈ 7.21 yd
- Rectangles: 44 + 66 + 79.31 = 189.31 yd²
- Total ≈ 213.31 yd²

But let me recalculate assuming cleaner numbers. If it's a 3-4-5 triangle scaled up or similar... Actually, I'll calculate with the visible measurements: base 6 yd, height 4 yd, prism length 11 yd.

Surface Area = 2(triangle area) + perimeter of triangle × length
- Triangle area = (1/2)(6)(4) = 12 yd²
- Perimeter = 6 + 4 + √(36+16) = 6 + 4 + √52 = 10 + 7.21 = 17.21 yd
- SA = 2(12) + 17.21(11) = 24 + 189.31 = 213.31 yd²

Rounding to nearest whole number: 213 yd²

Problem 3: Rectangular Prism (8 in × 10 in × 12 in)
- SA = 2(8×10 + 8×12 + 10×12)
- = 2(80 + 96 + 120)
- = 2(296)
- = 592 in²

Problem 4: Triangular Prism (right triangle)
Triangle has legs 9 in and 12 in, prism length 15 in
- First find hypotenuse: √(9² + 12²) = √(81 + 144) = √225 = 15 in
- Triangle area = (1/2)(9)(12) = 54 in²
- Two triangles = 108 in²
- Three rectangles: 9×15 + 12×15 + 15×15 = 135 + 180 + 225 = 540 in²
- Total SA = 108 + 540 = 648 in²

Problem 5: Cube (9 ft × 9 ft × 9 ft)
- SA = 6s² = 6(9²) = 6(81) = 486 ft²

Problem 6: Cylinder (radius 10 yd, height 16 yd)
- SA = 2πr² + 2πrh
- = 2π(10)² + 2π(10)(16)
- = 2π(100) + 2π(160)
- = 200π + 320π
- = 520π
- ≈ 520(3.14) = 1,632.8 yd²
- Rounded: 1,633 yd² (or keep as 520π yd² for exact answer)

Problem 7: Cylinder (radius 7 in, height 11 in)
- SA = 2πr² + 2πrh
- = 2π(7)² + 2π(7)(11)
- = 2π(49) + 2π(77)
- = 98π + 154π
- = 252π
- ≈ 252(3.14) = 791.28 in²
- Rounded: 791 in² (or 252π in² for exact)

Problem 8: Rectangular Prism (7 ft × 7 ft × 14 ft)
- SA = 2(7×7 + 7×14 + 7×14)
- = 2(49 + 98 + 98)
- = 2(245)
- = 490 ft²

Problem 9: Triangular Prism
Triangle with sides 10 yd, 13 yd, and we need to find the third side or it's given as 12 yd (based on typical problems). Looking at the diagram, it appears to show 10 yd base, 12 yd height (altitude), and 13 yd side, with prism length 14 yd.

If it's a right triangle with legs 5 and 12 (since 5-12-13 is a Pythagorean triple):
- Base would be 5 yd (half of 10 if 10 is the full base and 12 is altitude to it)
- Actually, if we have a triangle with base 10 yd and height 12 yd, and one side is 13 yd:
- Using area formula: Area = (1/2)(10)(12) = 60 yd²
- To find the other sides, we'd need more info, but if 13 yd is one side and it's not a right triangle with the 10 yd base...

Let me reconsider: if the triangle has sides 10 yd, 13 yd, and we can calculate the third using the height of 12 yd:
- If height is 12 yd to the 10 yd base, then we have two right triangles
- Half-base segments would give us the other sides
- This is getting complex. Let me assume it's showing a triangle with base 10 yd, height 12 yd, hypotenuse/side 13 yd, making it NOT a right triangle at the base.

Actually, checking: if we drop a perpendicular of 12 yd to a 10 yd base, and one side is 13 yd:
- One half of base: √(13² - 12²) = √(169 - 144) = √25 = 5 yd
- Other half: 10 - 5 = 5 yd
- Other side: √(5² + 12²) = √(25 + 144) = √169 = 13 yd

So it's an isosceles triangle with sides 13, 13, 10 yd!
- Triangle area = (1/2)(10)(12) = 60 yd²
- Two triangles = 120 yd²
- Three rectangles: 13×14 + 13×14 + 10×14 = 182 + 182 + 140 = 504 yd²
- Total SA = 120 + 504 = 624 yd²

Final Answer:
1) 432 ft²
2) 213 yd²
3) 592 in²
4) 648 in²
5) 486 ft²
6) 1,633 yd² (or 520π yd²)
7) 791 in² (or 252π in²)
8) 490 ft²
9) 624 yd²