QJEEM22Mae065: JEE Main 2022 Mathematics - Class 11 JEEM22Mae065 Solution: The lengths of the sides of a triangle are 10 + x 2 , 10 + x 2…

$CD = \sqrt {{{(10 + {x^2})}^2} - {{(10 - {x^2})}^2}} = 2\sqrt {10} |x|$

Area $= {1 \over 2} \times CD \times AB = {1 \over 2} \times 2\sqrt {10} |x|(20 - 2{x^2})$

$A = \sqrt {10} |x|(10 - {x^2})$

${{dA} \over {dx}} = \sqrt {10} {{|x|} \over x}(10 - {x^2}) + \sqrt {10} |x|( - 2x) = 0$

$\Rightarrow 10 - {x^2} = 2{x^2}$

$3{x^2} = 10$

$x = k$

$3{k^2} = 10$