DWG NO. 08.2 — Lesson 2 of 5

Special Right Triangles

Unit 8: Right Triangles and Trigonometry · ~15–30 min

Objective

Find missing side lengths in 45-45-90 and 30-60-90 triangles directly from their fixed side ratios, without setting up the Pythagorean Theorem each time.

Why these two triangles are special

Two right triangles show up so often — from cutting a square along its diagonal, or an equilateral triangle down its height — that their side ratios are worth memorizing outright. Once you know the ratio, one given side unlocks the other two instantly.

Triangle	Angle measures	Side ratio (short : mid : hyp)
45-45-90	45°, 45°, 90°	x : x : x\(\sqrt{2}\)
30-60-90	30°, 60°, 90°	x : x\(\sqrt{3}\) : 2x

45-45-90 Triangle Theorem: the legs are congruent, and the hypotenuse equals a leg times \(\sqrt{2}\).
30-60-90 Triangle Theorem: the hypotenuse is twice the shorter leg (opposite the 30° angle), and the longer leg (opposite the 60° angle) equals the shorter leg times \(\sqrt{3}\).

Left: 45-45-90 with legs x and hypotenuse x\(\sqrt{2}\). Right: 30-60-90 with short leg x, long leg x\(\sqrt{3}\), hypotenuse 2x.

Worked Example 1 · 45-45-90, given a leg

ProblemA 45-45-90 triangle has a leg of 7. Find the hypotenuse.

1.Use the ratio x : x : x\(\sqrt{2}\), with x = 7.

2.Hypotenuse = x\(\sqrt{2}\) = 7\(\sqrt{2}\).

Hypotenuse = 7\(\sqrt{2}\) ≈ 9.9

Worked Example 2 · 30-60-90, given the hypotenuse

ProblemA 30-60-90 triangle has hypotenuse 10. Find both legs.

1.Hypotenuse = 2x, so 2x = 10 and x = 5. That's the short leg.

2.Long leg = x\(\sqrt{3}\) = 5\(\sqrt{3}\).

Short leg = 5, long leg = 5\(\sqrt{3}\) ≈ 8.66

Worked Example 3 · 45-45-90, given the hypotenuse

ProblemA 45-45-90 triangle has hypotenuse 8\(\sqrt{2}\). Find a leg.

1.Hypotenuse = x\(\sqrt{2}\), so x\(\sqrt{2}\) = 8\(\sqrt{2}\).

2.Divide both sides by \(\sqrt{2}\): x = 8.

Each leg = 8

Guided practice

1.A 45-45-90 triangle has a leg of 12. Find the hypotenuse.
Show answer
Hypotenuse = x\(\sqrt{2}\) = 12\(\sqrt{2}\) ≈ 17.0.
2.A 30-60-90 triangle has a short leg of 6. Find the long leg and hypotenuse.
Show answer
Long leg = 6\(\sqrt{3}\) ≈ 10.4; hypotenuse = 2(6) = 12.
3.A 30-60-90 triangle has a long leg of 9. Find the short leg.
Show answer
Long leg = x\(\sqrt{3}\) = 9, so x = 9/\(\sqrt{3}\) = 3\(\sqrt{3}\) ≈ 5.2.
4.A square has a diagonal of 10 cm. Find the side length of the square. (Hint: the diagonal splits the square into two 45-45-90 triangles.)
Show answer
The diagonal is the hypotenuse: x\(\sqrt{2}\) = 10, so x = 10/\(\sqrt{2}\) = 5\(\sqrt{2}\) ≈ 7.07 cm.
5.An equilateral triangle has sides of 14. Find the height of the triangle. (Hint: the height splits it into two 30-60-90 triangles.)
Show answer
The short leg is half the base, 7, and the height is the long leg: 7\(\sqrt{3}\) ≈ 12.1.

← Lesson 8.1 Unit 8 Overview Lesson 8.3 →