Let S_1 be the sum of areas of the squares wh | vedclass

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Question

(a)

Here,

$S_1=a^2+\left(\frac{a}{2}\right)^2+\left(\frac{a}{4}\right)^2+\ldots$

$S_2=\left(\frac{a}{\sqrt{2}}\right)^2+\left(\frac{a}{2 \sqr

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(a)

Here,

$S_1=a^2+\left(\frac{a}{2}ight)^2+\left(\frac{a}{4}ight)^2+\ldots$

$S_2=\left(\frac{a}{\sqrt{2}}ight)^2+\left(\frac{a}{2 \sqrt{2}}ight)^2+\ldots$

$S_1=a^2+\frac{a^2}{4}+\frac{a^2}{16}+\ldots=\frac{a^2}{1-\frac{1}{4}}=\frac{4 a^2}{3}$

$S_2=\frac{a^2}{2}+\frac{a^2}{8}+\frac{a^2}{32}+\ldots=\frac{a^2 / 2}{1-\frac{1}{4}}=\frac{4 a^2}{6}$

$	herefore \quad \frac{S_1}{S_2}=\frac{\frac{4 a^2}{3}}{4 a^2 / 6}=2$

Let $S_1$ be the sum of areas of the squares whose sides are parallel to coordinate axes. Let $S_2$ be the sum of areas of the slanted squares as shown in the figure. Then, $\frac{S_1}{S_2}$ is equal to

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