Let ABCD be a square of side length 1. Let P, | vedclass

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(B) Let the vertices of the square be $A(0,0)$,$B(1,0)$,$C(1,1)$,and $D(0,1)$.Since $P, Q, R, S$ lie on $AD, BC, AB, CD$ respectively,we can denote th

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$\frac{3\sqrt{3}}{4}$

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(B) Let the vertices of the square be $A(0,0)$,$B(1,0)$,$C(1,1)$,and $D(0,1)$.Since $P, Q, R, S$ lie on $AD, BC, AB, CD$ respectively,we can denote their coordinates as $P(0, p)$,$Q(1, q)$,$R(r, 0)$,and $S(s, 1)$,where $0 The slope of $PQ$ is $m_1 = \frac{q-p}{1-0} = q-p$.The slope of $RS$ is $m_2 = \frac{1-0}{s-r} = \frac{1}{s-r}$.Since $PQ \perp RS$,$m_1 \cdot m_2 = -1$,so $(q-p) \cdot \frac{1}{s-r} = -1$,which implies $q-p = r-s$.The length $PQ = \sqrt{(1-0)^2 + (q-p)^2} = \sqrt{1 + (q-p)^2}$.Given $PQ = \frac{3\sqrt{3}}{4}$,we have $\frac{27}{16} = 1 + (q-p)^2$,so $(q-p)^2 = \frac{11}{16}$.The length $RS = \sqrt{(s-r)^2 + (1-0)^2} = \sqrt{(r-s)^2 + 1}$.Since $(r-s)^2 = (q-p)^2 = \frac{11}{16}$,we have $RS = \sqrt{\frac{11}{16} + 1} = \sqrt{\frac{27}{16}} = \frac{3\sqrt{3}}{4}$.

Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

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