Special Right Triangles Thank you, Mrs. Spitz, wherever you are. Adapted from internet version.

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Special Right Triangles Thank you, Mrs. Spitz, wherever you are. Adapted from internet version.

Objectives Find the side lengths of special right triangles. Use special right triangles to solve real-life problems that include finding the side lengths of the triangles.

Side lengths of Special Right Triangles Right triangles whose angle measures are 45°-45°-90° or 30°- 60°-90° are called special right triangles. The theorems that describe the relationships of side lengths of each of these special right triangles follow.

45°-45°-90° Triangle Theorem In a 45°-45°-90° triangle, the hypotenuse is √2 times as long as each leg. √2x 45 ° Hypotenuse = leg ∙√2

Ex. 1: Finding the hypotenuse in a 45°-45°-90° Triangle Find the value of x By the Triangle Sum Theorem, the measure of the third angle is 45°. The triangle is a 45°- 45°-90° right triangle, so the length x of the hypotenuse is √2 times the length of a leg. 33 x 45 °

Ex. 1: Finding the hypotenuse in a 45°-45°-90° Triangle Hypotenuse = leg ∙ √2 x = 3 ∙ √2 x = 3√2 33 x 45 ° 45°-45°-90° Triangle Theorem Substitute values Simplify

Ex. 2: Finding a leg in a 45°-45°-90° Triangle Find the value of x. Because the triangle is an isosceles right triangle, its base angles are congruent. The triangle is a 45°- 45°-90° right triangle, so the length of the hypotenuse is √2 times the length x of a leg. 5 x x

Ex. 2: Finding a leg in a 45°-45°-90° Triangle 5 x x Statement: Hypotenuse = leg ∙√2 5 = x ∙√2 Reasons: 45°-45°-90° Triangle Theorem 5 √2 x√2 √2 = 5 x= 5 x= 5√2 2 x= Substitute values Divide each side by √2 Simplify Multiply numerator and denominator by √2 Simplify

30°-60°-90° Triangle Theorem In a 30°-60°-90° triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is √3 times as long as the shorter leg. √3x 60 ° 30 ° Hypotenuse = 2 ∙ shorter leg Longer leg = shorter leg ∙ √3

Ex. 3: Finding side lengths in a 30°- 60°-90° Triangle Find the values of s and t. Because the triangle is a 30°- 60°-90° triangle, the longer leg is √3 times the length s of the shorter leg. 30 ° 60 °

Ex. 3: Side lengths in a 30°-60°-90° Triangle Statement: Longer leg = shorter leg ∙√3 5 = s∙√3 Reasons: 30°-60°-90° Triangle Theorem 5 √3 s√3 √3 = 5 s= 5 s= 5√3 3 s= Substitute values Divide each side by √3 Simplify Multiply numerator and denominator by √3 Simplify 30 ° 60 °

The length t of the hypotenuse is twice the length s of the shorter leg. Statement: Hypotenuse = 2 ∙ shorter leg Reasons: 30°-60°-90° Triangle Theorem t2 ∙ 5√3 3 = Substitute values Simplify 30 ° 60 ° t 10√3 3 =

Using Special Right Triangles in Real Life Example 4: Finding the height of a ramp. Tipping platform. A tipping platform is a ramp used to unload trucks. How high is the end of an 80 foot ramp when it is tipped by a 30° angle? By a 45° angle?

Solution: When the angle of elevation is 30°, the height of the ramp is the length of the shorter leg of a 30°-60°-90° triangle. The length of the hypotenuse is 80 feet. 80 = 2h 30°-60°-90° Triangle Theorem 40 = h Divide each side by 2.  When the angle of elevation is 30 °, the ramp height is about 40 feet.

Solution: When the angle of elevation is 45°, the height of the ramp is the length of a leg of a 45°-45°-90° triangle. The length of the hypotenuse is 80 feet. 80 = h∙√2 80 √2 =h Divide each side by √2 Use a calculator to approximate56.6 ≈ h  When the angle of elevation is 45 °, the ramp height is about 56 feet 7 inches. 45°-45°-90° Triangle Theorem

Ex. 5: Finding the area of a sign Road sign. The road sign is shaped like an equilateral triangle. Estimate the area of the sign by finding the area of the equilateral triangle. h 18 in. 36 in.

Ex. 5: Solution First, find the height h of the triangle by dividing it into two 30°-60°-90° triangles. The length of the longer leg of one of these triangles is h. The length of the shorter leg is 18 inches. h = 18∙√3 = 18√3 Long leg = short leg ∙√3 h 18 in. 36 in. Use h = 18√3 to find the area of the equilateral triangle.

Ex. 5: Solution Area = ½ bh = ½ (36)(18√3) ≈ h 18 in. 36 in.  The area of the sign is a bout 561 square inches.