1953 AHSME Problems/Problem 33

The perimeter of an isosceles right triangle is $2p$ . Its area is:

$\textbf{(A)}\ (2+\sqrt{2})p \qquad \textbf{(B)}\ (2-\sqrt{2})p \qquad \textbf{(C)}\ (3-2\sqrt{2})p^2\\ \textbf{(D)}\ (1-2\sqrt{2})p^2\qquad \textbf{(E)}\ (3+2\sqrt{2})p^2$

Given leg length $x$ , we can write the perimeter of this triangle to be $2x+x\sqrt{2}=2p$ . Thus, $x(2+\sqrt{2})=2p$ . Divide to get $x=\frac{2p}{2+\sqrt{2}}$ . Multiply by the conjugate and simplify to get $x=p(2-\sqrt{2})$ . Square and divide by two to get the area of the triangle, or $\frac{p^2*(2-\sqrt{2}^2)}{2}$ , or $p^2(3-2\sqrt{2})$ . $\boxed{C}$

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1953 AHSME Problems/Problem 33