The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 - 2x2.

Question

The lengths of the sides of a triangle are 10 + x2, 10 + x2 and 20 - 2x2. If for x = k, the area of the triangle is maximum, then 3k2 is equal to :

Accepted Answer

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

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

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

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

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

$3{x^2} = 10$

$x = k$

$3{k^2} = 10$

Solutions · 1