Four congruent rectangles are placed as shown. The area of the outer square is $4$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side? $\mathrm{(A)}\ 3 \qquad$ $ \mathrm{(B)}\ \sqrt {10} \qquad$ $ \mathrm{(C)}\ 2 + \sqrt2 \qquad$ $ \mathrm{(D)}\ 2\sqrt3 \qquad$ $ \mathrm{(E)}\ 4$
Solution
Let the longer side be $a$ and the shorter side $b$. The outer square has side $a+b$ and the inner square has side $a-b$, so $(a+b)^2= 4 (a-b)^2$, $a+b=2(a-b)$, and $a=3b$.
Four congruent rectangles are placed as shown. The area of the outer square is $9$ times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side?
$\mathrm{(A)}\ 3 \qquad$ $ \mathrm{(B)}\ \sqrt {10} \qquad$ $ \mathrm{(C)}\ 2 + \sqrt2 \qquad$ $ \mathrm{(D)}\ 2\sqrt3 \qquad$ $ \mathrm{(E)}\ 4$ 
