Question

Let ABCD be a square of a side length l, 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

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

$PQ\perp RS$

$\Rightarrow$$c-a=b-d\dots \dots \dots \left(1\right)$

$\Rightarrow$$PQ=\frac{3\sqrt{3}}{4}$

$\Rightarrow$$P{Q}^2=\frac{27}{16}$

$\Rightarrow$$1+(a-c{)}^2=\frac{27}{16}\dots \dots \dots (2)$

$\Rightarrow$$RS=\sqrt{(b-d{)}^2+1}\dots \dots \dots \left(3\right)$

By equation (1), (2) and (3)

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