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Question Number 172195 by cherokeesay last updated on 24/Jun/22

Answered by som(math1967) last updated on 24/Jun/22

$$letsideofbiggestsquare=rad.of$$$$quartercircle=runit$$$$\therefore AB=AL=r$$$$\therefore sideof{S}_2=\frac{r}{\sqrt{2}}$$$$AC=r\sqrt{2}$$$$\therefore LC=r\sqrt{2}-r=r(\sqrt{2}-1)$$$$\therefore sideof{S}_1=\frac{r(\sqrt{2}-1)}{\sqrt{2}}$$$$$$$$\frac{S_1}{S_2}=\frac{ar.of{S}_1}{ar.of{S}_2}=\frac{\frac{r^2{(\sqrt{2}-1)}^2}{2}}{\frac{r^2}{2}}$$$$=\frac{3-2\sqrt{2}}{1}=17.15\mathrm{\%}\left(approx\right)$$

Commented by som(math1967) last updated on 24/Jun/22

Commented by cherokeesay last updated on 24/Jun/22

$$thanksir!$$

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