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

The Correct Option is C

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

= \(2√10|x|\)

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): \(10\)