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 :
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 :
Updated On: Mar 21, 2026
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The Correct Option is C
Solution and Explanation
\(CD = \sqrt{(10+x^2)^2-(10-x^2)^2}\)
= \(2√10|x|\)
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): \(10\)
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