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)

$CD=\sqrt{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt{10}|x|$
$\text{ Area }=\frac{1}{2}\times CD\times AB=\frac{1}{2}\times 2\sqrt{10}|x|\left(20-2x^2\right)$
$A=\sqrt{10}|x|\left(10-x^2\right)$
$\frac{dA}{dx}=\sqrt{10}\frac{|x|}{x}\left(10-x^2\right)+\sqrt{10}|x|(-2x)=0$
$\Rightarrow 10-x^2=2x^2$
$3x^2=10$
$x=k$
$3k^2=10$