The lengths of the sides of a triangle are 10 | vedclass

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(B) Let the sides of the triangle be $a = 20-2x^2$,$b = 10+x^2$,and $c = 10+x^2$.The semi-perimeter $s$ is given by:$s = \frac{a+b+c}{2} = \frac{(20-2

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$10$

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(B) Let the sides of the triangle be $a = 20-2x^2$,$b = 10+x^2$,and $c = 10+x^2$.The semi-perimeter $s$ is given by:$s = \frac{a+b+c}{2} = \frac{(20-2x^2) + (10+x^2) + (10+x^2)}{2} = \frac{40}{2} = 20$.The area of the triangle $\Delta$ is given by Heron's formula:$\Delta = \sqrt{s(s-a)(s-b)(s-c)}$$\Delta = \sqrt{20(20-(20-2x^2))(20-(10+x^2))(20-(10+x^2))}$$\Delta = \sqrt{20(2x^2)(10-x^2)(10-x^2)}$$\Delta = \sqrt{40x^2(10-x^2)^2} = 2\sqrt{10}|x(10-x^2)| = 2\sqrt{10}|10x-x^3|$.To maximize the area,we maximize $f(x) = 10x-x^3$ (assuming $x>0$ for side lengths to be positive).$f'(x) = 10-3x^2$.Setting $f'(x) = 0$,we get $10-3x^2 = 0$,which implies $x^2 = \frac{10}{3}$.Thus,at $x=k$,$k^2 = \frac{10}{3}$.Therefore,$3k^2 = 3 	imes \frac{10}{3} = 10$.

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

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