Surface Area and Volume Test Review Answers: How to Solve Every Problem

To solve surface area and volume test review answers, apply the correct formula for each solid: prisms and cylinders use perimeter of base times height plus base areas, pyramids and cones use base area plus lateral triangle or sector area, and spheres use 4πr². For volume, multiply the base area by the height for prisms and cylinders, use ⅓ of that product for pyramids and cones, and use 4/3πr³ for spheres.

Surface Area Formulas

Surface area is the total area of every face and curved surface enclosing a solid. It is measured in square units — cm², m², in² — so always square the linear unit at the end.

Use each formula according to the base shape:

• Prisms (rectangular, triangular, any polygon): SA = 2B + Ph, where B is the area of one base, P is the perimeter of that base, and h is the height between bases.
• Cylinders: SA = 2πr² + 2πrh, which is the same 2B + Ph pattern with B = πr² and P = 2πr.
• Pyramids: SA = B + ½Pl, using the slant height l, not the vertical height.
• Cones: SA = πr² + πrl, again with slant height.
• Spheres: SA = 4πr², with no base term at all.

Slant Height Versus Height

This is the most common source of lost points. The vertical height h drops straight from apex to base center and belongs in volume formulas. The slant height l runs along a lateral face or the cone’s side and belongs in surface area formulas. Given r and h, find l with the Pythagorean theorem: l = √(r² + h²). Never substitute one for the other.

Sources like Khan Academy walk through the cone unrolling if the picture is unclear.

Volume Formulas

Volume is the space a solid occupies, measured in cubic units. Nearly every formula is “base area times height,” adjusted by a coefficient.

• Prisms and cylinders: V = Bh. For a cylinder, that becomes V = πr²h.
• Pyramids and cones: V = ⅓Bh. For a cone, V = ⅓πr²h.
• Spheres: V = 4/3πr³.

Choosing the right base is the trick. For a triangular prism, the base is the triangle, not the rectangle you happen to see first. Identify which face repeats parallel to itself — that pair is the base. In a cylinder the base is the circle, so the “height” is the distance between the circular ends, not the radius. If you catch yourself plugging a slant height into a volume formula, stop: volume only ever uses perpendicular height.

Keep π in your calculator rather than rounding to 3.14 unless the problem instructs otherwise, and only round the final answer.

Composite Solids

Composite solids are shapes built by joining two or more simple solids, or by cutting one solid out of another. The technique is always the same: split, solve, then combine.

For a shape like a cylinder topped by a hemisphere, compute each volume separately and add them. For a rectangular prism with a cylindrical hole drilled through it, compute both volumes and subtract the hole. The same logic applies to surface area, with one warning: when two solids are joined, the shared internal face disappears and must not be counted. Ask yourself which faces are actually exposed to the outside.

Working Through a Composite Example

Consider a silo: a cylinder of radius 3 m and height 10 m topped by a hemisphere of radius 3 m.

Cylinder volume = π(3²)(10) = 90π ≈ 282.7 m³. Hemisphere volume = ½ · 4/3π(3³) = 18π ≈ 56.5 m³. Total ≈ 339.2 m³.

For surface area, count only the exposed surfaces: the cylindrical side (2πrh = 60π), the bottom circle (πr² = 9π), and the hemispherical cap (2πr² = 18π). The circle where the cylinder meets the hemisphere is covered and excluded. Total = 87π ≈ 273.3 m².

That habit — testing each face for exposure — is worth more on a test than memorizing extra composite formulas, because there are no separate formulas for composite figures.

Worked Test Answers

These are the problem types that recur on review sheets, with the reasoning stated at each step.

Problem 1: Rectangular prism, 5 cm × 4 cm × 3 cm. Volume = 5 × 4 × 3 = 60 cm³. Surface area = 2(5·4) + 2(5·3) + 2(4·3) = 40 + 30 + 24 = 94 cm². Trap: triple-counting faces instead of using the paired-face structure.

Problem 2: Cone with r = 6 and h = 8. Slant height l = √(36 + 64) = √100 = 10. Volume = ⅓π(36)(8) = 96π ≈ 301.6 cubic units. Surface area = π(36) + π(6)(10) = 96π ≈ 301.6 square units. Trap: using h = 8 in the lateral area. Note the numbers match here because r² happens to equal rl for these values — do not assume that generally.

Problem 3: Sphere with diameter 10. Radius is 5. Volume = 4/3π(125) = 500π/3 ≈ 523.6 cubic units. Surface area = 4π(25) = 100π ≈ 314.2 square units. Trap: plugging the diameter into the formulas.

Problem 4: Cylinder with radius 2 and volume 100π. Solve 100π = π(4)h, so h = 25. Trap: forgetting to divide by the base area after substituting.

Always finish with a unit check: square units for area, cubic units for volume, and the quantity asked for in the prompt. The National Council of Teachers of Mathematics emphasizes showing the substitution step, since graders award method points even when arithmetic slips.

Key Takeaways

• Prisms and cylinders share the pattern SA = 2B + Ph and V = Bh; pyramids and cones are the same with ⅓ in the volume.
• Spheres are the exception: SA = 4πr² and V = 4/3πr³, using radius only.
• Use slant height for surface area, perpendicular height for volume — never swap them.
• For composite solids, split into simple shapes, then add or subtract; removed internal faces never count.
• Match the base to the shape’s repeated parallel face before computing anything.
• Finish every problem with a unit check and confirm you answered the quantity requested.