Question

The minimum value of the sum of the lengths of diagonals of a cyclic quadrilateral of area $a^2$ square units is

• A. $\sqrt{2}$ a
• B. 2$\sqrt{2}$ a
• C. 2a
• D. none of these

Answer 1

The correct option is B 2$\sqrt{2}$ a

Let $a,b,c,d$ be the length of the sides of a cyclic quadrilateral.

Given, Area of the quadrilateral is $a^2$ square units.

Let $e,f$ be the length of the diagonals.

For a cyclic quadrilateral, the relation between length of the sides and diagonals is given by

$e\ast f=ac+bd$

since, $A.M.\ge G.M.$

therefore,

$\frac{e+f}{2}\ge \sqrt{e\times f}$

$⟹e+f\ge 2\sqrt{ac+bd}$

But, $area=\frac{1}{2}(ac+bd)\sin \theta ⟹a^2=\frac{1}{2}(ac+bd)$

($\sin \theta =1$ i.e., max value for sine is chooesen so that $e+f$ should be greater than the maximum)

$⟹e+f\ge 2\sqrt{2a^2}$

$⟹e+f\ge 2\sqrt{2}a$