1953 AHSME Problems/Problem 33

Problem

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

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

Solution

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 $p^2(3-2\sqrt{2})$. $\boxed{C}$

See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 32
Followed by
Problem 34
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