
\(CD = \sqrt{(10+x^2)^2-(10-x^2)^2}\)
= \(2√10|x|\)
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): \(10\)
Let ABC be an equilateral triangle. A new triangle is formed by joining the middle points of all sides of the triangle ABC and the same process is repeated infinitely many times. If P is the sum of perimeters and Q is be the sum of areas of all the triangles formed in this process, then:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)