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Volume and surface area

SAT Math Review · Geometry and Trigonometry

Volume and surface area

The formulas are printed on the test. Choosing one and rearranging it is the part that is scored.

What is the value of the volume?

A right circular cylinder has a height of 5 inches, and the circumference of its base is 20π inches. What is the volume, in cubic inches, of the cylinder?

The formulas are given to you; the choosing and the rearranging are not.

Every Digital SAT Math module opens with a reference sheet carrying the first five. The sixth is free: a cube is a prism with all three edges equal.

Solid	Volume	What you have to supply
rectangular prism	V = lwh	which number is which edge
cylinder	V = πr²h	the radius, often hidden in a circumference
cone	V = ⅓πr²h	the height, not the slant
sphere	V = ⁴⁄₃πr³	the radius, often hidden in a diameter
pyramid	V = ⅓lwh	the area of the base
cube	V = s³	one edge, from the surface area

So there is nothing to memorise here. What the SAT actually tests is whether you can pull the right measurement out of the wording and then solve for whichever letter is missing.

Surface area is the exception: the reference sheet does not list it. That is fine, because surface area is never a formula you recall — it is a count of the faces, which is step 5.

Base area × height, and exactly a third of that when the solid comes to a point.

Every one of the five volumes is one of those two sentences.

flat top and bottom

V = B × h

prism, cylinder, cube

comes to a point

V = ⅓ × B × h

cone, pyramid

The ⅓ is worth three times the answer

Drop it and you have computed the cylinder that the cone sits inside. Because that value is exactly triple the right one, it is almost always sitting there among the four choices waiting for you.

“Circumference 20π” means the radius is 10.

This one line is where more of these questions are lost than anywhere else.

2πr = 20π

the circumference the stem gave you

r = 10

divide both sides by 2π; the π cancels

V = π(10²)(5) = 500π

now the formula, with h = 5

500π cubic inches

The three ways a radius arrives in disguise

Circumference 20π → r = 10, because C = 2πr. Diameter 20 → r = 10, because the diameter is twice the radius. Base area 100π → r = 10, because A = πr². In all three the number in the stem is not the radius, and in all three the wrong value is one of the four choices.

Given the volume, put it in and solve for the letter that is missing.

A right circular cylinder has a volume of 252π cubic inches and a height of 7 inches. What is the radius, in inches, of the cylinder?

πr²h = 252π

the same formula, read right to left

πr²(7) = 252π

put in what you know

r² = 36

divide by 7π — every π on the page disappears

r = 6

a length, so take the positive root only

r = 6 inches

Let Desmos do the rearranging

The built-in calculator will do the rearranging for you: graph y = pi*x^2*7 and y = 252*pi, then click where the two cross and read x = 6 off the label. That is genuinely quicker when the numbers are not clean — a cone of volume 314 and radius 5, say, whose height is 11.99… rather than anything you would spot. Keep π as pi rather than 3.14 so the answer stays exact.

For surface area, count the faces. Do not try to recall a formula.

A box has six faces in three matching pairs, so add three areas and double.

front and back = 2(54) = 108

9 by 6

top and bottom = 2(36) = 72

9 by 4

the two ends = 2(24) = 48

6 by 4

total = 108 + 72 + 48 = 228

square units, not cubic

The same count run backwards handles a cube: six equal square faces, so a surface area of 96 gives 96 ÷ 6 = 16 per face, an edge of 4, and a volume of 4³ = 64. The cylinder counts the same way — two circles, πr² each, plus one rectangle that is the circumference 2πr wide and h tall.

The sphere and the pyramid are the two that carry a fraction.

A sphere cubes its radius; a pyramid needs the area of whatever its base happens to be.

sphere, volume

⁴⁄₃πr³

sphere, surface

4πr²

pyramid, volume

⅓Bh

sphere, r = 6: V = 43π(6³)

6³ = 216

V = 43(216)π = 288π

216 ÷ 3 = 72, then × 4

pyramid, 6 by 6 base, h = 7: B = 36

the base area first, always

V = 13(36)(7) = 84

cubic units

Cube the radius, do not cube the answer

In ⁴⁄₃πr³ only the r is cubed. With r = 6 that is 216, not (4 × 6)³ and not 6 × 3. And a sphere given by its diameter is the same trap as before: diameter 12 means r = 6, so the volume is 288π, while using 12 as the radius gives 2,304π — eight times too big.

Scale a solid and the volume moves by the cube of the factor.

“The radius is doubled — what happens to the volume?” is a standing SAT question.

any length

× k

surface area

× k²

volume

× k³

V = π(10²)(5) = 500π

the opening cylinder

double r only: π(20²)(5) = 2,000π

× 4, because only r is squared

double every length: π(20²)(10) = 4,000π

× 8 = 2³, the whole solid scaled

Read the stem carefully: k³ applies only when every dimension is scaled. If the radius alone doubles and the height is untouched, the factor is 2² = 4, because the height contributes nothing to the change.

Check: right radius, right formula, right units.

the radius 2π(10) = 20π it rebuilds the given circumference ✓

the formula flat ends, so no ⅓ a cylinder, not a cone ✓

the units in × in × in cubic inches, as asked ✓

Remember the flow

Find the radius · Pick the formula · Solve for what is missing

Circumference 20π is a radius of 10. Squares give area, cubes give volume.

Now you try

Practice problem

What is the value of the height?

A right circular cone has a volume of 48π cubic centimeters, and the radius of its base is 6 centimeters. What is the height, in centimeters, of the cone?

Hint: write the formula out with the cone's own numbers in it before you do any arithmetic, then divide the whole equation by π. Two of these four choices come from losing the ⅓ or losing the square, so check both before you answer.

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