In quadrilateral RSTU, and . Given: is not a right

a 45° 45° a a 2 Figure 5.32 It is better to memorize the sketch in Figure 5.32 than to repeat the steps of the “proof ” that precedes the 45-45-90 Theorem. Exs. 1–3 EXAMPLE 1 Find the lengths of the missing sides in each triangle in Figure 5.33. Reminder If two angles of a triangle are congruent, then the sides opposite these angles are congruent. 45° 45° a ? A C B 5 5 Figure 5.33 45° 45° b ? D E F 6 ? 45 5 5 45 a Figure 5.34 a a 45 2 a 45 b Solution a The length of hypotenuse is , the product of and the length of either of the equal legs. b Let a denote the length of and of . The length of hypotenuse is . Then , so . Simplifying yields Therefore, NOTE: If we use the Pythagorean Theorem to solve Example 1, the solution in part a can be found by solving the equation and the solution in part b can be found by solving . a 2 + a 2 = 6 2 5 2 + 5 2 = c 2 DE = EF = 3 12 L 4.24. = 3 12 = 6 12 2 a = 6 12 12 12 a = 6 12 a 1 2 = 6 a 1 2 DF EF DE 12 5 12 AB EXAMPLE 2 Each side of a square has a length of . Find the length of a diagonal. Solution The square shown in Figure 5.34a is separated into two 45°-45°-90° triangles. With each of the congruent legs represented by a in Figure 5.34b, we see that and the diagonal hypotenuse length is , so . a = 1 10 L 3.16 a 12 = 15 12 a = 1 5 15 쮿 쮿 Exs. 4–7 THE 30°-60°-90° RIGHT TRIANGLE The second special triangle is the 30°-60°-90° triangle. A 60° 30° a a B C Figure 5.35 A C D B 60° 30° b a 30° 6 0° 60° a 2 a 2 a In a triangle whose angles measure 30°, 60°, and 90°, the hypotenuse has a length equal to twice the length of the shorter leg, and the length of the longer leg is the product of and the length of the shorter leg. 13 THEOREM 5.5.2 왘 30-60-90 Theorem EXAMPLE 3 Study the picture proof of Theorem 5.5.2. See Figure 5.35a. PICTURE PROOF OF THEOREM 5.5.2 GIVEN: with , , , and PROVE: and PROOF: We reflect across to form an equiangular and therefore equilateral . As shown in Figures 5.35b and 5.35c, we have . To find b in Figure 5.35c, we apply the Pythagorean Theorem. So That is, . AC = a 1 3 b = a 13 b = 13 2a 2 b = 23a 2 b 2 = 3a 2 3a 2 = b 2 , 4a 2 = a 2 + b 2 2a 2 = a 2 + b 2 c 2 = a 2 + b 2 AB = 2a 䉭ABD AC 䉭ABC AC = a 1 3 AB = 2a BC = a m ∠C = 90° m ∠B = 60° m ∠A = 30° 䉭ABC 쮿

2 a

A B C 60 ° 30 ° c a b It would be best to memorize the sketch in Figure 5.36. So that you will more eas- ily recall which expression is used for each side, remember that the lengths of the sides follow the same order as the angles opposite them. Thus, Opposite the 30° smallest angle is a length of shortest side. Opposite the 60° middle angle is length of middle side. Opposite the 90° largest angle is 2a length of longest side. ∠ a 1 3 ∠ ∠ EXAMPLE 4 Find the lengths of the missing sides of each triangle in Figure 5.37. Exs. 8–10 60 ° 30 ° a 2 a 3 a Figure 5.36 60 ° 30 ° a ? 5 S T R ? Figure 5.37 60 ° 30 ° b V W U 20 ? ? Z c Y X 60 ° 30 ° ? ? 9