Free video lesson

How to Solve Area and Volume Questions on the SAT

On the digital SAT, area and volume aren't a memorization test — the formulas are printed right on your reference sheet. The points come from setup, and one move trips up more students than anything else: using the diameter where the formula wants the radius.

Math · Area and volume3:20Published July 9, 2026

On YouTube: The Diameter Trap That Costs SAT Points — Area & Volume Made Simple

What this lesson covers

Area & Volume on the digital SAT — the formulas you're given, the diameter-vs-radius trap, composite shapes (whole minus hole), and a cone practice problem you can solve in your head.

Worked examples

The questions the video works, written out: the setup, each step, the answer and the trap.

0:50Example 1

A cylinder has a diameter of 6 and a height of 10. What is its volume?

  1. The reference sheet gives cylinder volume as V=πr2hV = \pi r^2 h. It wants the radius, and the problem gave the diameter.
  2. Halve the diameter: r=62=3\displaystyle r = \frac{6}{2} = 3.
  3. Substitute: V=π(32)(10)=π(9)(10)=90πV = \pi (3^2)(10) = \pi (9)(10) = 90\pi. Leave it in terms of pi unless the question asks for a decimal.
  4. Check: the base is a circle of area 9π9\pi, and stacking that base 10 high gives 90π90\pi. Using the diameter by mistake would give π(62)(10)=360π\pi (6^2)(10) = 360\pi, four times too big.

Answer: 90π90\pi

The formula is printed on the test, so the whole question is the setup: convert the diameter to a radius before anything else. The trap answer 360π360\pi comes from squaring 6 instead of 3, and because the radius is squared, that slip multiplies the volume by four rather than two.

1:20Example 2

A square plaque has side length 10. A circular hole of radius 3 is cut out of it. What is the area of the remaining plaque?

  1. Composite shapes follow one rule: area equals the whole minus the hole.
  2. Whole: the square, 102=10010^2 = 100.
  3. Hole: the circle, πr2=π(32)=9π\pi r^2 = \pi (3^2) = 9\pi.
  4. Subtract: 100−9π100 - 9\pi. That is the exact answer; do not force a decimal unless asked.
  5. Check: 9π≈28.39\pi \approx 28.3, so the remaining area is about 71.7, which is positive and less than 100, as a square with a hole in it must be.

Answer: 100−9π100 - 9\pi

Since the hole is removed, its area is subtracted from the square, giving 100−9π100 - 9\pi. The trap is adding the two areas, 100+9π100 + 9\pi, or halving the 3 as if it were a diameter; here 3 is already the radius, so read the given label before halving anything.

1:55Example 3

A paper cup is a cone with radius 2 and height 6. What is its volume?

  1. Reference sheet: cone volume is V=13πr2h\displaystyle V = \frac{1}{3}\pi r^2 h. Note the one third; a cone is one third of the cylinder that would enclose it.
  2. Substitute r = 2 and h = 6: V=13π(22)(6)=13π(4)(6)=13(24π)\displaystyle V = \frac{1}{3}\pi (2^2)(6) = \frac{1}{3}\pi (4)(6) = \frac{1}{3}(24\pi).
  3. 13(24π)=8π\displaystyle \frac{1}{3}(24\pi) = 8\pi.
  4. Check: the matching cylinder, π(4)(6)=24π\pi (4)(6) = 24\pi, is exactly three times the cone, as it should be.

Answer: 8π8\pi

With the radius given directly, the only thing to protect is the one third, which turns 24π24\pi into 8π8\pi. The trap answer 24π24\pi drops the fraction and computes a cylinder; the same slip on a sphere drops the four thirds.

Chapters

Lesson transcript

The narration of the video, word for word, under its chapter headings.

Welcome to SAT Climb. Area and volume feel like a formula memorization slog. Here's the secret: the digital SAT prints the formulas right on the test. The points come from setup, not memory.

Every one of these formulas is handed to you on the reference sheet. Circle, cylinder, sphere, cone, pyramid. Your only job is to pick the right one and plug in carefully. Start here: a cylinder has diameter six and height ten. Find its volume.

Cylinder volume is pi times radius squared times height. But watch the very first move. They gave you the diameter, six. The formula needs the radius. Half of six is three. Now plug in: pi times three squared times ten. Three squared is nine, times ten is ninety. Ninety pi. That radius step is where most points are won or lost.

The other half of this topic is composite shapes. Rule: area equals the whole minus the hole. Take a square plaque, side ten, with a circular hole of radius three. The square is ten squared, one hundred. The hole is pi times three squared, nine pi. Subtract. One hundred minus nine pi. Exact form is the answer, no calculator needed.

Your turn. A paper cup is a cone with radius two and height six. Find the volume. Remember the one third. One third times pi times two squared times six. If you got eight pi, that one third is now automatic.

Three traps to sidestep. One: plugging in the diameter where the formula wants the radius. Two: dropping the one third on a cone or pyramid, or the four thirds on a sphere. Three: mixing units. Fix your radius, keep your fractions, match your units, and this whole topic is setup.

Area and volume: solved. The formulas are given to you. The points are in the setup. Start practicing free at S.A.T. climb dot com. Your SAT is closer than you think.

Read the written version: the Area and volume strategy guide, then try 25 hard Area and volume questions with full explanations. Both are free, no account needed.

Watched it? Now drill it.

SAT Climb turns a 30-question diagnostic into a study plan across all 31 skills, with a 7-day free trial on every plan.

Start free trial →All video lessons