In this guide
What the SAT gives you vs. what to memorize
The Digital SAT provides a reference sheet with basic area and volume formulas: the area of a circle, the Pythagorean theorem, the volume of a cylinder, and a handful of others. That's good news and a trap at once. Good, because you don't need to memorize those. A trap, because students waste time re-deriving what's already on the screen while forgetting the formulas that aren't provided.
The things the SAT does not give you (and that you must know cold) are the special right triangle ratios, the trig functions (SOH-CAH-TOA), the slope formula, and arc-length and sector-area relationships. This guide focuses on exactly those. Review the full list on the SAT reference sheet page so you know precisely what's on the screen during the test.
The LAMPS geometry method
Do not choose a formula just because it contains the numbers in the question. Match what you know to what the problem asks for. Use the same five-step check on every geometry problem:
- L: Label what is known. Mark the radius, diameter, central angle, triangle sides, and units. Convert hidden information immediately, such as
d = 18becomingr = 9. - A: Ask what the problem wants. Decide whether the answer is a length, area, angle, side, or ratio.
- M: Match the known information and requested quantity to a tool.
- Whole-circle length:
C = 2πr - Whole-circle area:
A = πr² - Arc length:
(θ/360)(2πr) - Sector area:
(θ/360)(πr²) - Two sides of a right triangle: the Pythagorean theorem
- A 30°, 45°, or 60° angle: special-triangle ratios
- An angle and side information: SOH-CAH-TOA
- Whole-circle length:
- P: Plug in only after writing the formula. This prevents mixing circumference with area or using diameter where the formula expects radius.
- S: Sanity-check size and units. Arc length must be smaller than circumference, sector area must be smaller than total area, the hypotenuse must be longest, and area requires square units.
LAMPS turns formula recall into a decision process: identify the information, identify the target, and only then choose the relationship connecting them.
Triangles and the Pythagorean theorem
For any right triangle, the Pythagorean theorem relates the two legs to the hypotenuse:
a² + b² = c² (c is the hypotenuse)
This is provided on the reference sheet, but you should still recognize the common Pythagorean triples on sight; they turn a calculation into instant recall: 3-4-5, 5-12-13, 8-15-17, and their multiples (like 6-8-10). If a triangle shows two of those numbers, the third is free.
Angle facts to keep handy
The angles in any triangle sum to 180°. Angles in a straight line sum to 180°, and angles around a point sum to 360°. Vertical angles (across an X intersection) are equal, and parallel lines cut by a transversal create equal corresponding and alternate angles. A large share of SAT geometry is just chasing these equalities around a figure.
Special right triangles
Two right triangles appear so often that their side ratios are worth burning into memory. Neither is on the reference sheet.
The 45-45-90 triangle
An isosceles right triangle. The two legs are equal, and the hypotenuse is a leg times √2:
sides in ratio 1 : 1 : √2
The 30-60-90 triangle
Half of an equilateral triangle. The sides follow a fixed ratio:
sides in ratio 1 : √3 : 2 (opposite 30° : 60° : 90°)
A 30-60-90 triangle has a hypotenuse of 10. Find the shorter leg.
The hypotenuse corresponds to the 2 in the ratio 1 : √3 : 2. So one "unit" is 10 ÷ 2 = 5.
The shorter leg corresponds to the 1, so it equals 5. The longer leg would be 5√3.
Circles: area, circumference, arcs
Area and circumference are provided, but you should still know them by heart:
Area = πr² Circumference = 2πr
A circular garden has a diameter of 18 meters. Find its circumference and area.
Label: the diameter is 18, so the radius is 18 ÷ 2 = 9.
Ask: circumference is a length; area measures the enclosed space.
Match: use C = 2πr and A = πr².
Plug in: C = 2π(9) = 18π meters
A = π(9)² = 81π square meters
Sanity-check: circumference uses linear units, while area uses square units. Using 18 as the radius would double the circumference and make the area four times too large.
What the SAT does not hand you is how a central angle carves the circle into pieces. A central angle of θ degrees cuts out a fraction θ/360 of the whole circle. That single idea gives you both arc length and sector area:
Arc length = (θ/360) · 2πr
Sector area = (θ/360) · πr²
Think of it as "the fraction of the circle times the whole." Once you see arcs and sectors as fractions of a full circle, these questions stop being intimidating.
A circle has radius 12, and a sector has a central angle of 150°. Find the sector's arc length and area.
Label: r = 12 and θ = 150°.
Ask: arc length is part of the circumference; sector area is part of the total area.
Match: first calculate the fraction of the circle: 150/360 = 5/12
Arc length: (5/12)(2π · 12) = (5/12)(24π) = 10π
Sector area: (5/12)(π · 12²) = (5/12)(144π) = 60π
Sanity-check: 150° is less than half of 360°, so both results should be less than half the full circumference 24π and area 144π. They are.
Radians
The Digital SAT can express angles in radians. Remember that 180° = π radians, so a full circle is 2π. To convert, multiply degrees by π/180. Keep this conversion in your back pocket: a question will occasionally slip a radian measure into an otherwise ordinary circle problem.
Right-triangle trigonometry
SAT trig rarely goes beyond the three basic ratios, remembered as SOH-CAH-TOA:
sin θ = Opposite / Hypotenuse
cos θ = Adjacent / Hypotenuse
tan θ = Opposite / Adjacent
A right triangle has an acute angle of 38° and a hypotenuse of 15. Find the leg adjacent to the 38° angle, to the nearest tenth.
Label: the known side is the hypotenuse, and the requested side is adjacent to the angle.
Ask: find the adjacent leg.
Match: cosine connects adjacent and hypotenuse: cos(38°) = adjacent / hypotenuse
Plug in and solve: cos(38°) = x/15
x = 15 cos(38°) ≈ 11.8
Sanity-check: a leg must be shorter than the hypotenuse, so 11.8 is reasonable. Make sure the calculator is in degree mode.
There's one relationship the test likes to exploit: the sine of an angle equals the cosine of its complement. Because the two non-right angles in a right triangle add to 90°, you get:
sin(x°) = cos(90° − x°)
So if a question tells you sin(x) = 0.6 and asks for cos(90 − x), the answer is simply 0.6: no triangle required. Spotting that identity turns a multi-step problem into a one-liner.
Practice this topic
Geometry rewards drawing. Sketch the figure, label everything you know, and mark equal angles and sides before you compute. Then practice until the special-triangle ratios come without hesitation.
- Drill the Geometry & Trig question bank.
- Read the full Geometry topic guide.
- Study exactly what's provided on the SAT reference sheet.