Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 1 - Problems - Page 75: 2

Answer

$h = \frac{P^2}{2P+24}$

Work Step by Step

Let the three sides of the triangle be $a,b,$ and $h$ Note that: $~~a^2+b^2 = h^2$ We can write two expressions for the area: $A = \frac{1}{2}(12)(h) = \frac{1}{2}ab$ Then: $ab = 12h$ We can express the length of the hypotenuse $h$ as a function of the perimeter $P$: $a+b+h = P$ $a+b = P-h$ $(a+b)^2 = (P-h)^2$ $a^2+2ab+b^2 = P^2-2Ph+h^2$ $2ab = P^2-2Ph$ $2(12h) = P^2-2Ph$ $24h = P^2-2Ph$ $24h+2Ph = P^2$ $h(24+2P) = P^2$ $h = \frac{P^2}{2P+24}$
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