Question

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

• A. 5
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (C)

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