Question

There are two squares $S_1$ and $S_2$ with areas 8 and 9 units, respectively. $S_1$ is inscribed within $S_2$, with one corner of $S_1$ on each side of $S_2$. The corners of the smaller square divides the sides of the bigger square into two segments, one of length ‘a’ and the other of length ‘b’, where $b > a$. A possible value of ‘b/a’, is:

Accepted Answer

D. Side of $S_2 = 3 \implies a + b = 3$. Side of $S_1 = \sqrt{8} = 2\sqrt{2} \implies a^2 + b^2 = 8$. $a^2 + (3-a)^2 = 8 \implies 2a^2 - 6a + 1 = 0 \implies a = \frac{3 - \sqrt{7}}{2}, b = \frac{3 + \sqrt{7}}{2}$. $\frac{b}{a} = \frac{3 + \sqrt{7}}{3 - \sqrt{7}} = \frac{(3+\sqrt{7})^2}{2} = 8 + 3\sqrt{7} \approx 8 + 7.937 = 15.93$.

Inscribed Squares Ratio

Explanation

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