Objectives Justify and apply properties of 45°-45°-90° triangles.

A diagonal of a square divides it into two congruent isosceles right triangles. Since the base angles of an isosceles triangle are congruent, the measure of each acute angle is 45°. So another name for an isosceles right triangle is a 45°-45°-90° triangle. A 45°-45°-90° triangle is one type of special right triangle. You can use the Pythagorean Theorem to find a relationship among the side lengths of a 45°-45°-90° triangle.

Example 1A: Finding Side Lengths in a 45°- 45º- 90º Triangle Find the value of x. Give your answer in simplest radical form. By the Triangle Sum Theorem, the measure of the third angle in the triangle is 45°. So it is a 45°-45°-90° triangle with a leg length of 8.

Example 1B: Finding Side Lengths in a 45º- 45º- 90º Triangle Find the value of x. Give your answer in simplest radical form. The triangle is an isosceles right triangle, which is a 45°-45°-90° triangle. The length of the hypotenuse is 5. Rationalize the denominator.

Check It Out! Example 1b Find the value of x. Give your answer in simplest radical form. The triangle is an isosceles right triangle, which is a 45°-45°-90° triangle. The length of the hypotenuse is 16. Rationalize the denominator.

Check It Out! Example 1a Find the value of x. Give your answer in simplest radical form. By the Triangle Sum Theorem, the measure of the third angle in the triangle is 45°. So it is a 45°-45°-90° triangle with a leg length of x = 20 Simplify.

A 30°-60°-90° triangle is another special right triangle A 30°-60°-90° triangle is another special right triangle. You can use an equilateral triangle to find a relationship between its side lengths.

Example 3A: Finding Side Lengths in a 30º-60º-90º Triangle Find the values of x and y. Give your answers in simplest radical form. 22 = 2x Hypotenuse = 2(shorter leg) 11 = x Divide both sides by 2. Substitute 11 for x.

Check It Out! Example 3c Find the values of x and y. Give your answers in simplest radical form. 24 = 2x Hypotenuse = 2(shorter leg) 12 = x Divide both sides by 2. Substitute 12 for x.

Check It Out! Example 3b Find the values of x and y. Give your answers in simplest radical form. y = 2(5) y = 10 Simplify.

Check It Out! Example 3a Find the values of x and y. Give your answers in simplest radical form. Hypotenuse = 2(shorter leg) Divide both sides by 2. y = 27 Substitute for x.

Example 3B: Finding Side Lengths in a 30º-60º-90º Triangle Find the values of x and y. Give your answers in simplest radical form. Rationalize the denominator. y = 2x Hypotenuse = 2(shorter leg). Simplify.

Check It Out! Example 3d Find the values of x and y. Give your answers in simplest radical form. Rationalize the denominator. x = 2y Hypotenuse = 2(shorter leg) Simplify.

Example 4: Using the 30º-60º-90º Triangle Theorem An ornamental pin is in the shape of an equilateral triangle. The length of each side is 6 centimeters. Josh will attach the fastener to the back along AB. Will the fastener fit if it is 4 centimeters long? Step 1 The equilateral triangle is divided into two 30°-60°-90° triangles. The height of the triangle is the length of the longer leg.

Example 4 Continued Step 2 Find the length x of the shorter leg. 6 = 2x Hypotenuse = 2(shorter leg) 3 = x Divide both sides by 2. Step 3 Find the length h of the longer leg. The pin is approximately 5.2 centimeters high. So the fastener will fit.

Check It Out! Example 4 What if…? A manufacturer wants to make a larger clock with a height of 30 centimeters. What is the length of each side of the frame? Round to the nearest tenth. Step 1 The equilateral triangle is divided into two 30º-60º-90º triangles. The height of the triangle is the length of the longer leg.

Check It Out! Example 4 Continued Step 2 Find the length x of the shorter leg. Rationalize the denominator. Step 3 Find the length y of the longer leg. y = 2x Hypotenuse = 2(shorter leg) Simplify. Each side is approximately 34.6 cm.

Lesson Quiz: Part I Find the values of the variables. Give your answers in simplest radical form. 1. 2. 3. 4. x = 10; y = 20