Question

Let $ABCD$ be a square of a 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

Answer: (B)

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

$PQ\perp RS\Rightarrow$
$c-a=b-d.......\left(1\right)$
$PQ=\frac{3\sqrt{3}}{4}$
${PQ}^2=\frac{27}{16}$
$1+{(a-c)}^2=\frac{27}{16}.....\left(2\right)$
$RS=\sqrt{{(b-d)}^2+1}....\left(3\right)$
By equation (1),(2),(3)
$RS=\frac{3\sqrt{3}}{4}$