Question

Assertion :If $△$ be the area of a triangle with side lengths $a,b,c$ then $△\le \frac{1}{4}\sqrt{(a+b+c)abc}$ Reason: A.M$\ge$G.M

• A. Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion.
• B. Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion.

$\because △=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{\frac{s}{8}(b+c-a)(c+a-b)(a+b-c)}$
$\because$ sum of two sides is always greater then the third side.
$\therefore b+c-a,c+a-b,a+b-c>0$
$\Rightarrow (s-a)(s-b)(s-c)>0$
Let $s-a=x,s-b=y,s-c=z$
Now,
$x+y=s-a+s-b$
$=2s-(a+b)$
$=2s-(2s-c)$
$=2s-2s+c=c$
$\therefore x+y=c$
Again $y+z=s-b+s-c$
$=2s-(b+c)$
$=2s-2s-a=a$
$\therefore y+z=a$
and $z+x=s-c+s-a$
$=2s-(c+a)$
$=2s-2s-b=b$
$\therefore z+x=b$
$\because$A.M$\ge$G.M
$\Rightarrow \frac{x+y}{2}\ge \sqrt{xy},$
$\frac{y+z}{2}\ge \sqrt{yz}$ and $\frac{z+x}{2}\ge \sqrt{zx}$
$\Rightarrow \frac{(x+y)(y+z)(z+x)}{8}\ge xyz$
$\Rightarrow \frac{abc}{8}\ge (s-a)(s-b)(s-c)$
$\Rightarrow \frac{s\left(abc\right)}{8}\ge s(s-a)(s-b)(s-c)$
$\Rightarrow \frac{(a+b+c)abc}{16}\ge {△}^2$
$\therefore △\le \frac{1}{4}\sqrt{(a+b+c)abc}$