Chapter 1 - Problems - Page 76: 2

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

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 $P$: $a+b+h = P$ $a+b = P-h$ $(a+b)^2 = (P-h)^2$ $a^2+2ab+b^2 = P^2-2Ph+h^2$ $2ab = P^2-2Ph$ $2(12h) = P^2-2Ph$ $24h = P^2-2Ph$ $24h+2Ph = P^2$ $h(24+2P) = P^2$ $h = \frac{P^2}{2P+24}$