Let S_1, S_2, dots be squares such that for e | vedclass

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(D) Let $x_n$ be the side length of square $S_n$. The diagonal of $S_{n+1}$ is $x_{n+1}\sqrt{2}$.Given $x_n = x_{n+1}\sqrt{2}$,we have $x_{n+1} = \fra

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(D) Let $x_n$ be the side length of square $S_n$. The diagonal of $S_{n+1}$ is $x_{n+1}\sqrt{2}$.Given $x_n = x_{n+1}\sqrt{2}$,we have $x_{n+1} = \frac{x_n}{\sqrt{2}}$.This forms a geometric progression for the side lengths: $x_n = x_1 \cdot (\frac{1}{\sqrt{2}})^{n-1}$.The area of $S_n$ is $A_n = x_n^2 = x_1^2 \cdot (\frac{1}{2})^{n-1} = \frac{100}{2^{n-1}}$.We want $A_n  100$.For $n=8$,$2^{8-1} = 2^7 = 128 > 100$.For $n=9$,$2^{9-1} = 2^8 = 256 > 100$.For $n=10$,$2^{10-1} = 2^9 = 512 > 100$.Since all values $n=8, 9, 10$ satisfy the condition,the correct option is $(d)$.

Let $S_1, S_2, \dots$ be squares such that for each $n \ge 1$,the length of a side of $S_n$ equals the length of a diagonal of $S_{n+1}$. If the length of a side of $S_1$ is $10 \text{ cm}$,then for which of the following values of $n$ is the area of $S_n$ less than $1 \text{ cm}^2$?

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