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

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(D) Let $x_n$ be the side length of square $S_n$. Given that the side of $S_n$ equals the diagonal of $S_{n+1}$,we have $x_n = x_{n+1} \sqrt{2}$.This implies $x_{n+1} = \frac{x_n}{\sqrt{2}}$.Thus,$x_n = x_1 \left(\frac{1}{\sqrt{2}}\right)^{n-1}$.The area of $S_n$ is $A_n = x_n^2 = x_1^2 \left(\frac{1}{2}\right)^{n-1} = \frac{100}{2^{n-1}}$.We want $A_n  100$.Since $2^6 = 64$ and $2^7 = 128$,we require $n-1 \ge 7$,which implies $n \ge 8$.Therefore,for $n = 8, 9, 10, \dots$,the area is less than $1 \ cm^2$.

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 $10 \ cm$,then for which of the following values of $n$ is the area of $S_n$ less than $1 \ cm^2$?

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