Geometry · Chapter 31

Special Triangles

Equilateral, 45-45-90, 30-60-90 and 13-14-15 triangles plus a handful of Pythagorean triples: learn their ratios once and most right-triangle problems become arithmetic.

What's this chapter about?

31.1 Equilateral Triangle
31.2 Right Triangles
31.3 Practice Problems

You'll need

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31.10%

Equilateral Triangle

In this section you will be able to find the height and area of an equilateral triangle from its side, by cutting it into two 30-60-90 halves.

Cut an equilateral triangle down the middle and you get two 30-60-90 triangles. That one cut gives you its height and its area.

All three sides are equal and all three angles are $60^\circ$ . The height from any vertex lands at the midpoint of the opposite side.

So every length in the picture is a multiple of the side $a$ : half the base is $\frac{a}{2}$ and the height is $\frac{\sqrt{3}}{2}a$ . Double the side and every length doubles, but the area grows four times.

Look:When a problem has a

60^\circ

angle, look for an equilateral triangle to complete, or half of one.

Exploreplay first, formulas later

Try this: Step the side a to 2, 4, then 6. How much does the height grow each time?
Try this: Draw the height. Each half is a 30-60-90 triangle. Which of its sides is a/2?
Try this: Turn on Show the formula. Why is the area √3/4 · a² and not √3/2 · a²?

Name it

Theorem 31.1.1 · Equilateral Triangle

If the side length of an equilateral triangle is $a$ , then

\text{Height of the triangle} = \frac{\sqrt{3}}{2}a.

This follows directly from the 30-60-90 triangle.

\text{Area of the equilateral triangle} = \frac{\sqrt{3}}{4}a^2.

Contest cornerreal contest problems

Problem 31.3.1

AMC 10

Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB' = 3 \cdot AB$ . Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC' = 3 \cdot BC$ , and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA' = 3 \cdot CA$ . What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ ?

Problem 31.3.7(additional)

AMC 10

A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure, we have $3$ rows of small congruent equilateral triangles, with $5$ small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of $2003$ small equilateral triangles?

Problem 31.3.8(additional)

AMC 12

Let $A_0 = (0, 0)$ . Distinct points $A_1, A_2, \dots$ lie on the $x$ -axis, and distinct points $B_1, B_2, \dots$ lie on the graph of $y = \sqrt{x}$ . For every positive integer $n$ , $A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n \geq 100$ ?

Checkpoint for section 31.1

Get 2 of 3 right to master this section. One try each, no hints.

31.20%

Right Triangles

In this section you will be able to use the Pythagorean Theorem, the common triples, and the 45-45-90, 30-60-90 and 13-14-15 triangles to find lengths fast.

Three numbers, $3, 4, 5$ , helped build the pyramids. Learn six triples and three triangle shapes, and most right-triangle problems turn into arithmetic.

The Pythagorean Theorem links the three sides of any right triangle. The triples are the right triangles with whole-number sides, so spot one and you can skip the square roots.

Three shapes come up again and again:

45-45-90: half of a square. The sides are $a : a : a\sqrt{2}$ .
30-60-90: half of an equilateral triangle. The sides are $a : a\sqrt{3} : 2a$ .
13-14-15: two right triangles, $5, 12, 13$ and $9, 12, 15$ , glued along the height $12$ .

Look:No special triangle in sight? Drop a height. It often cuts an awkward triangle into two special ones.

Exploreplay first, formulas later

Try this: Press Rearrange the triangles. The four triangles didn't change. So which empty area equals c²?
Try this: Set the legs to 5 and 12, then to 10 and 24. Is c still a whole number? What happened to the triple?
Try this: Set the legs to 2 and 3. Why is c not a whole number now?

Name it

Theorem 31.2.1 · Pythagorean Theorem

A right triangle with legs $a$ and $b$ and hypotenuse $c$ satisfies

c^2 = a^2 + b^2.

Fact 31.2.2 · Important Pythagorean Triples

3, 4, 5 \qquad 5, 12, 13 \qquad 7, 24, 25

8, 15, 17 \qquad 9, 40, 41 \qquad 20, 21, 29

If all numbers in a Pythagorean triple are multiplied by a constant, the resulting numbers still form a Pythagorean triple. For example, these are all Pythagorean triples: $3, 4, 5$ ; $6, 8, 10$ ; $9, 12, 15$ ; $12, 16, 20$ ; $15, 20, 25$ .

Theorem 31.2.3 · 45-45-90 Triangle

If the side length (each leg) of a 45-45-90 triangle is $a$ , then

\text{hypotenuse} = \sqrt{2} \times \text{side length} = \sqrt{2}\,a

\text{Area} = \frac12 \times \text{side length}^2 = \frac12 a^2.

Theorem 31.2.4 · 30-60-90 Triangle

If the short leg of a 30-60-90 triangle is $a$ , then

\text{long leg} = \sqrt{3} \times \text{short leg} = \sqrt{3}\,a

\text{hypotenuse} = 2 \times \text{short leg} = 2a

\text{Area} = \frac{\sqrt{3}}{2} \times \text{short leg}^2 = \frac{\sqrt{3}}{2}a^2.

short leg a1

Drag the purple dot or use the stepper to resize. The ratios never change.

Area =

\frac{\sqrt{3}}{2}

≈ 0.87

a : a\sqrt3 : 2a,\quad \text{Area} = \tfrac{\sqrt3}{2}a^2

Equilateral: $\text{Area} = \tfrac{\sqrt3}{4}a^2$
45-45-90: $\text{Area} = \tfrac12 a^2$
30-60-90: $\text{Area} = \tfrac{\sqrt3}{2}a^2$

Theorem 31.2.5 · 13-14-15 Triangle

If the three sides of a triangle are $13$ , $14$ and $15$ , it can be divided into two right triangles with side lengths

5, 12, 13 \quad\text{and}\quad 9, 12, 15.

\text{Area of this triangle} = \frac12 \times 14 \times 12 = 84.

scale k1

Drag the purple dot or use the stepper to resize. The ratios never change.

Area =

84

5k,12k,13k \;+\; 9k,12k,15k:\quad \text{Area} = \tfrac12\cdot 14k\cdot 12k = 84k^2

Equilateral: $\text{Area} = \tfrac{\sqrt3}{4}a^2$
45-45-90: $\text{Area} = \tfrac12 a^2$
30-60-90: $\text{Area} = \tfrac{\sqrt3}{2}a^2$

Watch itpredict, then reveal

Example 31.1

AMC 10/12

Let $\triangle XOY$ be a right-angled triangle with $m\angle XOY = 90^\circ$ . Let $M$ and $N$ be the midpoints of legs $OX$ and $OY$ , respectively. Given that $XN = 19$ and $YM = 22$ , find $XY$ .

1
Call the legs $OX = 2x$ and $OY = 2y$ , so $OM = MX = x$ and $ON = NY = y$ . Triangle $XON$ has a right angle at $O$ . What is $4x^2 + y^2$ ?

Predict first:
Your prediction

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Example 31.2

AIME

In triangle $ABC$ , angles $A$ and $B$ measure $60$ degrees and $45$ degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$ , and $AT = 24$ . The area of triangle $ABC$ can be written in the form $a + b\sqrt{c}$ , where $a$ , $b$ , and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a + b + c$ .

1
The angles at $A$ and $B$ are $60^\circ$ and $45^\circ$ . How big is angle $C$ (in degrees)?

Predict first:
Your prediction

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Try it

Problem 31.3.3(additional)

AIME

In quadrilateral $ABCD$ , $\angle B$ is a right angle, diagonal $\overline{AC}$ is perpendicular to $\overline{CD}$ , $AB = 18$ , $BC = 21$ , and $CD = 14$ . Find the perimeter of $ABCD$ .

Contest cornerreal contest problems

Problem 31.3.2

AMC 10

Triangle $ABC$ has a right angle at $B$ , $AB = 1$ , and $BC = 2$ . The bisector of $\angle BAC$ meets $\overline{BC}$ at $D$ . What is $BD$ ?

Problem 31.3.4(additional)

AMC 10

In rectangle $ABCD$ , $\overline{AB} = 20$ and $\overline{BC} = 10$ . Let $E$ be a point on $\overline{CD}$ such that $\angle CBE = 15^\circ$ . What is $\overline{AE}$ ?

Problem 31.3.5(additional)

AMC 12

How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to $3$ times their perimeters?

Problem 31.3.6(additional)

AMC 10

Points $A, B, C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$ .

Checkpoint for section 31.2

Get 2 of 3 right to master this section. One try each, no hints.

31.3

Practice Problems

In this section you will be able to mix equilateral heights, triples and special triangles on a harder contest problem.

One last challenge: three distances from a point inside an equilateral triangle. The key is a $60^\circ$ turn and a triangle you already know.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 31.3.9(additional)

AMC 10Challenge

Suppose that $\triangle ABC$ is an equilateral triangle of side length $s$ , with the property that there is a unique point $P$ inside the triangle such that $AP = 1$ , $BP = \sqrt{3}$ , and $CP = 2$ . What is $s$ ?