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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

CK-12 Foundation: Special Right Triangle: 45-45-90

Watch the second half of this video.

James Sousa: Trigonometric Function Values of Special Angles

Now watch the second half of this video.

James Sousa: Solving Special Right Triangles

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}

CK-12 Foundation: Special Right Triangle: 45-45-90

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} .

Answers:

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}

Practice

  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.

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