Question

If $\Delta$ is the area of a triangle with sides of lengths a, b, c, then

• A. $\Delta \le \frac{1}{4}\sqrt{(a+b+c)abc}$
• B. $\Delta =\frac{1}{4}\sqrt{(a+b+c)abc}$ if $a=b=c$
• C. $\Delta \ge \frac{1}{4}\sqrt{(a+b+c)abc}$
• D. $\Delta =\frac{1}{4}abc\sqrt{(a+b+c)}$

Answer 1

Answer: (A), (B)

The correct options are
A $\Delta \le \frac{1}{4}\sqrt{(a+b+c)abc}$
B $\Delta =\frac{1}{4}\sqrt{(a+b+c)abc}$ if $a=b=c$

We have $a^2\ge a^2-(b-c{)}^2=(a+b-c\left)\right(a-b+c)$

$\Rightarrow a^2\ge (2s-2c)(2s-2b)=4(s-b)(s-c)$

Similarly $b^2\ge 4(s-c)(s-a)$

$c^2\ge 4(s-a)(s-b)$

So that $a^2{b}^2{c}^2\ge 64\left[\right(s-a\left)\right(s-b\left)\right(s-c){]}^2$

$\Rightarrow abc\ge 8(s-a)(s-b)(s-c)$

$\Rightarrow (a+b+c)\left(abc\right)\ge 8\times 2s(s-a)(s-b)(s-c)=16{\Delta }^2$

$\Rightarrow \Delta \le \frac{1}{4}\sqrt{(a+b+c)abc}$

and if $a=b=c$, the triangle is equilateral and its area is

$=\frac{1}{4}\sqrt{(a+b+c)abc}=\frac{1}{4}\sqrt{3a\times a^3}=\frac{\sqrt{3}{a}^2}{4}$