Question

The area of an arbitrary triangle is less than one-fourth the square of its semi-perimeter.

• A. True
• B. False

Answer 1

Answer: (A)

We have to prove $\sqrt{s(s-a)(s-b)(s-c)}<\frac{1}{4}{s}^2$
or $s(s-a)(s-b)(s-c)<\frac{1}{16}{s}^4$
or $(s-a)(s-b)(s-c)<\frac{1}{16}{s}^3\dots \dots \left(1\right)$
But $\frac{(s-a)+(s-b)+(s-c)}{3}$
$\ge \left[\right(s-a\left)\right(s-b\left)\right(s-c\left){]}^{\frac{1}{3}}\right[\because A.M.\ge G.M.]$
or $\frac{s}{3}\ge \left[\right(s-a\left)\right(s-b\left)\right(s-c){]}^{\frac{1}{3}}$
$[\because s-a+s-b+s-c=3s-(a+b+c)=3s-2s=s]$
or $(s-a)(s-b)(s-c)\le \frac{s^3}{27}<\frac{1}{16}{s}^3$