Question

The lengths of the sides of a triangle are $10+x^2$, $10+x^2$ and $20-2x^2$. If for $x=k$, the area of the triangle is maximum, then $3k^2$ is equal to:

• A. 5
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (C)

$a=20-2x^2,b=10+x^2,c=10+x^2$
$=\frac{a+b+c}{2}$
$=20$
$\Delta =\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{20\left(2x^2\right)\left(10-x^2\right)\left(10-x^2\right)}$
$=2\sqrt{10}\sqrt{x^2{\left(10-x^2\right)}^2}$
$=2\sqrt{10}\left|x\left(10-x^2\right)\right|$
$=2\sqrt{10}\left|10x-x^3\right|$
$S=10x-x^3$
$\frac{ds}{dx}=10-3x^2$
$\frac{ds}{dx}=0\Rightarrow x^2=\frac{10}{3}$
$3x^2=10$