Question

The least perimeter of a cyclic quadrilateral of a given area A square units is

• A. $\sqrt{A}$
• B. 2$\sqrt{A}$
• C. 3$\sqrt{A}$
• D. 4$\sqrt{A}$

Answer 1

Answer: (D)

4$\sqrt{A}$

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

Given area is $A$ sq. units

Let the perimeter be $P=a+b+c+d$

Area of a cyclic quadrilateral is given by $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ (Brahmagupta's formula)

where $s=\frac{a+b+c+d}{2}$

therefore, $A=\sqrt{(s-a)(s-b)(s-c)(s-d)}$

we know that $A.M.\ge G.M.$

$⟹\frac{(s-a)+(s-b)+(s-c)+(s-d)}{4}\ge \sqrt[4]{4(s-a)(s-b)(s-c)(s-d)}$

$⟹\frac{4s-(a+b+c+d)}{4}\ge \sqrt[4]{4(s-a)(s-b)(s-c)(s-d)}$

$⟹\frac{2s}{4}\ge \sqrt[4]{4(s-a)(s-b)(s-c)(s-d)}$

$⟹\frac{a+b+c+d}{4}\ge (\sqrt{(s-a)(s-b)(s-c)(s-d)}{)}^{\frac{1}{2}}$

$⟹\frac{P}{4}\ge (A{)}^{\frac{1}{2}}$

$⟹P\ge 4\sqrt{A}$

therefore the least perimeter of the cyclic quadrilateral with area $A$ sq units is $4\sqrt{A}$