Question

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

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{3\sqrt{3}}{4}$
• C. $\frac{\sqrt{2}+1}{2}$
• D. $4-2\sqrt{2}$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}$


Construct $MQ||AB\&RN||BC$
$\text{Consider}△OXQ,$
$\text{Let}\angle OQX=\theta$
$\text{then}\angle OXQ={90}^{\circ }-\theta$
$\therefore \text{In}△XYR,\angle XRY=\theta$

$\text{In}△PMQ,\cos \theta =\frac{1}{PQ}$
$\text{In}△SNR,\cos \theta =\frac{1}{RS}$
$\Rightarrow PQ=RS$
$\Rightarrow RS=\frac{3\sqrt{3}}{4}$