1960 AHSME Problems/Problem 21

Problem

The diagonal of square $I$ is $a+b$ . The perimeter of square $II$ with twice the area of $I$ is:

$\textbf{(A)}\ (a+b)^2\qquad \textbf{(B)}\ \sqrt{2}(a+b)^2\qquad \textbf{(C)}\ 2(a+b)\qquad \textbf{(D)}\ \sqrt{8}(a+b) \qquad \textbf{(E)}\ 4(a+b)$

Solution

Since the diagonal of square $I$ is $a+b$ units long, the side length of square $I$ is $\frac{a+b}{\sqrt{2}}$ , so the area of square $I$ is $\frac{(a+b)^2}{2}$ .

The area of square $II$ is twice as much as the area of square $I$ , so the area of square $II$ is $(a+b)^2$ . That means the side length of square $II$ is $a+b$ , so the perimeter of square $II$ is $4(a+b)$ , or answer choice $\boxed{\textbf{(E)}}$ .

1960 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1960 AHSME Problems/Problem 21

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