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 :
- 5
- 8
- 10
- 12
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): \(10\)
Top Questions on Triangles
In the adjoining figure, \(PQ \parallel XY \parallel BC\), \(AP=2\ \text{cm}, PX=1.5\ \text{cm}, BX=4\ \text{cm}\). If \(QY=0.75\ \text{cm}\), then \(AQ+CY =\)
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If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.
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