Question

ABCD is a rectangle having length $(x+1)$ and breadth $(x-1)$ . If P, Q, R and S are the mid-points of side AB, BC, CD and DA respectively, then the perimeter of the quadrilateral PQRS is

• A. $\sqrt{\frac{4a(x^2-1)}{x}}$
• B. $\sqrt{\frac{4a(x^2-1)}{2}}$
• C. $2\sqrt{4x^2+1}$
• D. $2\sqrt{2(x^2+1)}$

Answer 1

The correct option is D $2\sqrt{2(x^2+1)}$

From given, we have,

In $△APS$

$S{P}^2=A{P}^2+A{S}^2$

$={\left(\frac{x+1}{2}\right)}^2+{\left(\frac{x-1}{2}\right)}^2$

$=\frac{x^2+1}{2}$

$\therefore SP=\sqrt{\frac{x^2+1}{2}}$

but, we have,

$SP=PQ=QR=RS$

Perimeter of $PQRS=4\left(SP\right)=4\times \sqrt{\frac{x^2+1}{2}}$

$=2\sqrt{2(x^2+1)}$