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# 45-45-90 Right Triangles

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Practice 45-45-90 Right Triangles
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45-45-90 Right Triangles

What if you were given an isosceles right triangle and the length of one of its sides? How could you figure out the lengths of its other sides? After completing this Concept, you'll be able to use the 45-45-90 Theorem to solve problems like this one.

### Watch This

Watch the second half of this video.

Now watch the second half of this video.

### Guidance

A right triangle with congruent legs and acute angles is an Isosceles Right Triangle . This triangle is also called a 45-45-90 triangle (named after the angle measures).

$\triangle ABC$ is a right triangle with $m \angle A = 90^\circ$ , $\overline {AB} \cong \overline{AC}$ and $m \angle B = m \angle C = 45^\circ$ .

45-45-90 Theorem: If a right triangle is isosceles, then its sides are in the ratio $x:x:x \sqrt{2}$ . For any isosceles right triangle, the legs are $x$ and the hypotenuse is always $x \sqrt{2}$ .

#### Example A

Find the length of the missing side.

Use the $x:x:x \sqrt{2}$ ratio. $TV = 6$ because it is equal to $ST$ . So, $SV = 6 \cdot \sqrt{2} = 6 \sqrt{2}$ .

#### Example B

Find the length of the missing side.

Use the $x:x:x \sqrt{2}$ ratio. $AB = 9 \sqrt{2}$ because it is equal to $AC$ . So, $BC = 9 \sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18$ .

#### Example C

A square has a diagonal with length 10, what are the lengths of the sides?

Draw a picture.

We know half of a square is a 45-45-90 triangle, so $10=s \sqrt{2}$ .

$s \sqrt{2} &= 10\\s &= \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{10 \sqrt{2}}{2}=5 \sqrt{2}$

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### Guided Practice

Find the length of $x$ .

1.

2.

3. $x$ is the hypotenuse of a 45-45-90 triangle with leg lengths of $5\sqrt{3}$ .

Use the $x:x:x \sqrt{2}$ ratio.

1. $12 \sqrt{2}$ is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it splits the square into two 45-45-90 triangles. $12 \sqrt{2}$ would be the hypotenuse, or equal to $x \sqrt{2}$ .

$12 \sqrt{2} &= x \sqrt{2}\\12 &= x$

2. Here, we are given the hypotenuse. Solve for $x$ in the ratio.

$x \sqrt{2} &= 16\\x &= \frac{16}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}$

3. $x=5\sqrt{3}\cdot \sqrt{2}=5\sqrt{6}$

### Explore More

1. In an isosceles right triangle, if a leg is 4, then the hypotenuse is __________.
2. In an isosceles right triangle, if a leg is $x$ , then the hypotenuse is __________.
3. A square has sides of length 15. What is the length of the diagonal?
4. A square’s diagonal is 22. What is the length of each side?

For questions 5-11, find the lengths of the missing sides. Simplify all radicals.