Math

Right Triangles

5 min readMedium5-question drill

Right triangles are the gateway to half the SAT geometry section — Pythagorean theorem, special triangles, and trigonometry all build on the same setup: a triangle with a 90° angle.

Where this lesson fits

Right Triangles

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→Special Right Triangles →Trigonometry →Coordinate Geometry

A right triangle has one 90° angle. The side opposite the right angle (the longest side) is the hypotenuse; the other two sides are the legs.

The Pythagorean theorem is the foundation: for legs $a$ , $b$ and hypotenuse $c$ ,

a^2 + b^2 = c^2

Given any two sides, you can find the third.

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The four most common Pythagorean triples — and their multiples

Base triple	×2	×3	Watch for
3 – 4 – 5	6 – 8 – 10	9 – 12 – 15	Most common; small numbers
5 – 12 – 13	10 – 24 – 26	—	Mid-size; legs differ widely
8 – 15 – 17	—	—	Larger; appears in coord geom
7 – 24 – 25	—	—	Rare but tested

Memorize these Pythagorean triples — they show up constantly on the SAT:

$3$ – $4$ – $5$ (and multiples like $6$ – $8$ – $10$ , $9$ – $12$ – $15$ )
$5$ – $12$ – $13$ (and $10$ – $24$ – $26$ )
$8$ – $15$ – $17$
$7$ – $24$ – $25$

If you see two of those numbers in a problem, you don't even need to compute — recognize the triple and write down the third.

Two special right triangles with non-integer sides — also memorize:

45°–45°–90° (isosceles right triangle): legs are equal, hypotenuse = leg × $\sqrt{2}$ .

\text{legs} : \text{leg} : \text{hypotenuse} = 1 : 1 : \sqrt{2}

30°–60°–90°: short leg : long leg : hypotenuse = $1 : \sqrt{3} : 2$ . The short leg is opposite the 30°.

\text{short} : \text{long} : \text{hypotenuse} = 1 : \sqrt{3} : 2

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The two special right triangles — memorize these ratios

Triangle	Side ratio	If short side = $s$…
45-45-90 (isosceles right)	$1 : 1 : \sqrt{2}$	Legs both $s$; hypotenuse $s\sqrt{2}$
30-60-90	$1 : \sqrt{3} : 2$	Short leg $s$; long leg $s\sqrt{3}$; hypotenuse $2s$

SAT-typical setups:

A ladder leans against a wall — Pythagorean triple.
A square's diagonal — 45-45-90 triangle.
A cube or rectangular box — Pythagorean theorem twice (face diagonal then space diagonal).
A point on a coordinate grid — distance is hypotenuse of a right triangle whose legs are the coordinate differences.

Distance formula is just Pythagoras in coordinates:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Quick test: if a triangle problem gives you 5 and 12, expect 13. If it gives 8 and 15, expect 17. Recognizing triples saves seconds.

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Quick check

Quick check. Identify the hypotenuse (opposite the right angle), use Pythagoras or — if you spot a triple — write the third side directly.

A right triangle has legs of length 6 and 8. What is the hypotenuse?

Worked examples

Example 1

A 13-foot ladder leans against a wall, with its base 5 feet from the wall. How high up the wall does the ladder reach?

Example 2

An equilateral triangle has side length 6. What is its height?

Common pitfalls

Forgetting which side is the hypotenuse

The hypotenuse is the side OPPOSITE the right angle — the longest side. In Pythagoras, it's always $c$ in $a^2 + b^2 = c^2$ . Mixing up sides flips the equation.

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Mixing up special triangle ratios

45-45-90: $1 : 1 : \sqrt{2}$ . 30-60-90: $1 : \sqrt{3} : 2$ (short : long : hyp). Mixing these costs questions. The mnemonic: 30-60-90 is not symmetric, so the legs differ; 45-45-90 is symmetric, so the legs match.

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Not recognizing Pythagorean triples

If you see a problem with 5 and 13, the third side is almost certainly 12. With 8 and 17, expect 15. Memorize 3-4-5, 5-12-13, 8-15-17, 7-24-25 and their multiples.

Squaring negatives wrong

$(x_2 - x_1)^2$ — if the difference is $-3$ , $(-3)^2 = 9$ , NOT $-9$ . The distance formula always gives a positive distance because squares are positive. Don't drop signs and produce a wrong answer.

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Key takeaways

Pythagorean theorem: $a^2 + b^2 = c^2$ , where $c$ is the hypotenuse (opposite the right angle).
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Memorize the triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25 — and their multiples.
45-45-90 sides: $1 : 1 : \sqrt{2}$ (legs equal, hypotenuse leg × √2).
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30-60-90 sides: $1 : \sqrt{3} : 2$ (short opposite 30°, long opposite 60°, hyp opposite 90°).
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Distance formula is just Pythagoras in coordinates: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ .
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Try it yourself

5 practice questions on Right Triangles, drawn from the question bank. The tutor is one click away if you get stuck.

Where this lesson fits

Right Triangles

You are here

→Special Right Triangles →Trigonometry →Coordinate Geometry