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Let $A B C D$ be a square of side length $1$ . Let $P, Q, R, S$ be points in the interiors of the sides $A D, B C, A B, C D$ respectively, such that $P Q$ and $R S$ intersect at right angles. If $P Q=\frac{3 \sqrt{3}}{4}$, then $R S$ equals
$\frac{2}{\sqrt{3}}$
$\frac{3 \sqrt{3}}{4}$
$\frac{\sqrt{2}+1}{2}$
$4-2 \sqrt{2}$
Solution
(b)
$A B C D$ is square
$A B=B C=C D=A D=1$
$P Q$ is perpendicular to $R S$
$\because$ Slope of $P Q \times$ Slope of $R S=-1$
$\therefore \quad q-p \times 1-0=-1$
$ \Rightarrow q-p =r-s $$ \Rightarrow (P Q)^2 =(1-0)^2+(q-p)^2 $
$\Rightarrow \quad(P Q)^2=(1-0)^2+(q-p)^2$
$\Rightarrow \quad\left(\frac{3 \sqrt{3}}{4}\right)^2=1+(q-p)^2$
$\Rightarrow \quad(q-p)^2=\frac{27}{16}-1=\frac{11}{16}$
$\Rightarrow \quad(r-s)^2=\frac{11}{16} \quad[\because q-p=r-s]$
$\Rightarrow \quad R S=\sqrt{(1-0)^2+(r-s)^2}$
$\Rightarrow \quad R S=\sqrt{1+\frac{11}{16}}$
$\therefore \quad R S=\sqrt{\frac{27}{16}}=\frac{3 \sqrt{3}}{4}$
