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

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(A) Let the side of the largest square be $a$. The squares with sides parallel to the coordinate axes have side lengths $a, \frac{a}{2}, \frac{a}{4},

Vedclass · Accepted Answer

$2$

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(A) Let the side of the largest square be $a$. The squares with sides parallel to the coordinate axes have side lengths $a, \frac{a}{2}, \frac{a}{4}, \dots$. The sum of their areas is $S_1 = a^2 + (\frac{a}{2})^2 + (\frac{a}{4})^2 + \dots = a^2 + \frac{a^2}{4} + \frac{a^2}{16} + \dots$. This is an infinite geometric series with first term $A = a^2$ and common ratio $r = \frac{1}{4}$. Thus,$S_1 = \frac{a^2}{1 - 1/4} = \frac{a^2}{3/4} = \frac{4a^2}{3}$. The slanted squares have side lengths $\frac{a}{\sqrt{2}}, \frac{a}{2\sqrt{2}}, \frac{a}{4\sqrt{2}}, \dots$. The sum of their areas is $S_2 = (\frac{a}{\sqrt{2}})^2 + (\frac{a}{2\sqrt{2}})^2 + (\frac{a}{4\sqrt{2}})^2 + \dots = \frac{a^2}{2} + \frac{a^2}{8} + \frac{a^2}{32} + \dots$. This is an infinite geometric series with first term $A = \frac{a^2}{2}$ and common ratio $r = \frac{1}{4}$. Thus,$S_2 = \frac{a^2/2}{1 - 1/4} = \frac{a^2/2}{3/4} = \frac{a^2}{2} 	imes \frac{4}{3} = \frac{2a^2}{3}$. Therefore,$\frac{S_1}{S_2} = \frac{4a^2/3}{2a^2/3} = 2$. The correct option is $A$.

Let $S_1$ be the sum of areas of the squares whose sides are parallel to the 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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