The remaining two sides of the right triangle are called the legs of the right triangle.

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1 10 MODULE 6. RADICAL EXPRESSIONS 6 Pythgoren Theorem The Pythgoren Theorem An ngle tht mesures 90 degrees is lled right ngle. If one of the ngles of tringle is right ngle, then the tringle is lled right tringle. It is trditionl to mrk the right ngle with little squre (see Figure 6.1). B A C Figure 6.1: Right tringle ABC hs right ngle t vertex C. Right tringle terminology. The longest side of the right tringle, the side diretly opposite the right ngle, is lled the hypotenuse of the right tringle. The remining two sides of the right tringle re lled the legs of the right tringle. Proof of the Pythgoren Theorem Eh side of the squre in Figure 6.2 hs een divided into two segments, one of length, the other of length. We n find the totl re of the squre y squring ny one of the sides of the squre. A =( + ) 2 A = Squre side to find re. Squring inomil pttern. Thus, the totl re of the squre is A = A seond pproh to finding the re of the squre is to sum the res of the geometri prts tht mke up the squre. We hve four ongruent right tringles, shded in light red, with se nd height. The re of eh of these tringles is found y tking one-hlf times the se times the height; i.e., the re of eh tringles is (1/2). In the interior, we hve smller squre

2 6C. PYTHAGOREAN THEOREM 11 Figure 6.2: Proving the Pythgoren Theorem. with side. Its re is found y squring its side; i.e., the re of the smller squre is 2. The totl re of the squre is the sum of its prts, one smller squre nd four ongruent tringles. Tht is: A = 2 +4 A = 2 +2 ( ) 1 2 Adding the re of the interior squre nd the re of four right tringles. Simplify: 4((1/2)) =2. The two expressions, nd 2 +2, othrepresentthetotlre of the lrge squre. Hene, they must e equl to one nother = = 2 Eh side of this eqution represents the re of the lrge squre. Sutrt 2 from oth sides. The lst eqution, = 2, is lled the Pythgoren Theorem. The Pythgoren Theorem. If nd re the legs of right tringle nd is its hypotenuse, then: = 2 We sy The sum of the squres of the legs of right tringle equls the squre of its hypotenuse. Good hint. Note tht the hypotenuse sits y itself on one side of the eqution = 2. The legs of the hypotenuse re on the other side. Let s put the Pythgoren Theorem to work.

3 12 MODULE 6. RADICAL EXPRESSIONS You Try It! EXAMPLE 1. Find the missing side of the right tringle shown elow. 3 4 Solution: First, write out the Pythgoren Theorem, then sustitute the given vlues in the pproprite ples = 2 Pythgoren Theorem. (4) 2 +(3 2 )= 2 Sustitute: 4 for, 3for = 2 Squre: 4 2 =16,3 2 =9. 25 = 2 Add: = 25. The eqution 2 = 25 hs two rel solutions, = 5 nd = 5. However, in this sitution, represents the length of the hypotenuse nd must e positive numer. Hene: = 5 Nonnegtive squre root. Thus, the length of the hypotenuse is 5. You Try It! EXAMPLE 2. An isoseles right tringle hs hypotenuse of length 8. Find the lengths of the legs. Solution: In generl, n isoseles tringle is tringle with two equl sides. In this se, n isoseles right tringle hs two equl legs. We ll let x represent the length of eh leg. x 8 x

4 6C. PYTHAGOREAN THEOREM 13 Use the Pythgoren Theorem, sustituting x for eh leg nd 8 for the hypotenuse = 2 x 2 + x 2 =8 2 2x 2 =64 Pythgoren Theorem. Sustitute: x for, x for, 8for. Comine like terms: x + x =2x. x 2 =32 Divide oth sides y 2. The eqution x 2 = 32 hs two rel solutions, x = 32 nd x = 32. However, in this sitution, x represents the length of eh leg nd must e positive numer. Hene: x = 32 Nonnegtive squre root. Rememer, your finl nswer must e in simple form. We must ftor out perfet squre when possile. x = 16 2 Ftor out perfet squre. x =4 2 Simplify: 16 = 4. Thus, the length of eh leg is 4 2. Applitions You Try It! EXAMPLE 3. A ldder 20 feet long lens ginst the grge wll. If the se of the ldder is 8 feet from the grge wll, how high up the grge wll does the ldder reh? Find n ext nswer, then use your lultor to round your nswer to the nerest tenth of foot. Solution: As lwys, we oey the Requirements for Word Prolem Solutions. 1. Set up vrile ditionry. We ll rete well-mrked digrm for this purpose, letting h represent the distne etween the se of the grge wll nd the upper tip of the ldder.

5 14 MODULE 6. RADICAL EXPRESSIONS 20 ft h 8ft 2. Set up n eqution. Using the Pythgoren Theorem, we n write: h 2 =20 2 Pythgoren Theorem h 2 =400 Squre: 8 2 =64nd20 2 = Solve the eqution. h 2 =336 Sutrt 64 from oth sides. h = 336 h will e the nonnegtive squre root. h = Ftor out perfet squre. h =4 21 Simplify: 16 = Answer the question. The ldder rehes 4 21 feet up the wll. 5. Look k. 20 ft 4 21 ft 8ft

6 6C. PYTHAGOREAN THEOREM 15 Using the Pythgoren Theorem: ( ) 2 21 = (4) 2 ( 21 ) 2 = (16) (21) = = = 400

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