Asked by Allen
How do I solve: One leg of a right triangle is 4 m longer than the other. If the length of the hypotenuse is 18 m, how long are the other two legs?
Answers
Answered by
Damon
x^2 + (x+4)^2 = 18^2 = 324
2 x^2 + 8 x + 16 = 324
2 x^2 + 8 x - 308 = 0
x^2 + 4 x -154 = 0
x = [ -4 +/- sqrt (16 + 616)]/2
= [-4 +/- 2 sqrt 158 ] /2
= -2 +/- sqrt 158 use + root
10.57
14.57
2 x^2 + 8 x + 16 = 324
2 x^2 + 8 x - 308 = 0
x^2 + 4 x -154 = 0
x = [ -4 +/- sqrt (16 + 616)]/2
= [-4 +/- 2 sqrt 158 ] /2
= -2 +/- sqrt 158 use + root
10.57
14.57
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