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Let $\mathrm{ABCD}$ be a square of a side length 1 , Let $\mathrm{P}, \mathrm{Q}, \mathrm{R}, \mathrm{S}$ be points in the interiors of the sides $A D, B C, A B, C D$, respectively, such that $P Q$ and RS intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$ then RS equals
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Verified Answer
The correct answer is:
$\frac{3 \sqrt{3}}{4}$
$$
\mathrm{PQ} \perp \mathrm{RS} \Rightarrow
$$
$$
c-a=b-d
$$
$$
\begin{array}{l}
P Q=\frac{3 \sqrt{3}}{4} \\
P Q^{2}=\frac{27}{16} \\
1+(a-c)^{2}=\frac{27}{16} \\
R S=\sqrt{(b-d)^{2}+1}
\end{array}
$$
By equation $(1),(2)$ and (3)
$$
R S=\frac{3 \sqrt{3}}{4}
$$
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