question_answer

The sides of an equilateral triangle are increasing at the rate of 2 cm/s. Find the rate at which the area increases, when the side is 10 cm.

Answer:

Let the side of triangle be a. Given, \[\frac{da}{dt}=2\,\,cm/s\] Now, area of equilateral triangle having side a, \[A=\frac{\sqrt{3}{{a}^{2}}}{4}\] On differentiating both sides w.r.t. t, we get \[\frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot (2a)\frac{da}{dt}\] On putting\[\frac{da}{dt}=2\,\,cm/s\]and \[a=10\,cm\], we get \[\frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2\times 10\times 2=10\sqrt{3}\,\,c{{m}^{2}}/s\]