Question

The area of any cyclic quadrilateral ABCD is given by $A^2=(s-a)(s-b)(s-c)(s-d)$, where $2s=a+b+c+d,wherea,b,c,$ and d are the sides of the quadrilateral. Consider a cyclic quadrilateral ABCD of area $16$ sq. unit. The minimum perimeter of the quadrilateral is

• A. $4$
• B. $2$
• C. $1$
• D. none of these

Answer 1

The correct option is A $4$

applying $A.M\ge G.M$ to $(s-a),(s-b),(s-c)$ and $(s-d)$

$\frac{s-a+s-b+s-c+s-d}{4}\ge {\left[(s-a)(s-b)(s-c)(s-d)\right]}^{1/4}$

$\Rightarrow 4s-(a+b+c+d)\ge A^{\left(1/2\right)}$

$\Rightarrow 2s\ge 4$

$2s$ is the perimeter of the quadrilateral, so the minimum perimeter is $4$