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 :

• A. 5
• B. 8
• C. 10
• D. 12

Answer 1

Answer: (C)

10

Answer 2

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

= $2√10|x|$

Area

= $\frac{1}{2} \times CD \times AB$

= $\frac{1}{2} \times 2\sqrt{10}|x| (20-2x^2)$

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

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

$⇒ 10 – x^2 = 2x^2$

$3x^2 = 10$

$x = k$

$3k^2 = 10$

Hence, the correct option is (C): $10$