question_answer

The sides of an equilateral triangle are increasing at the rate of 2 cm/s. How far is the area increasing when the side is 10 cm?       

Answer:

Let a be the side of triangle and A be its area, then \[A=\frac{\sqrt{3}}{4}{{a}^{2}}\] and \[\frac{da}{dt}=2\,\,cm/s\] and a = 10 cm Now,     \[A=\frac{\sqrt{3}}{4}{{a}^{2}}\] \[\Rightarrow \] \[\frac{dA}{dt}=\frac{\sqrt{3}}{4}\cdot 2a\frac{da}{dt}\]             \[=\frac{\sqrt{3}}{2}a\frac{da}{dt}=\frac{\sqrt{3}}{2}\times 10\times 2=10\sqrt{3}c{{m}^{2}}/s\] Thus, the area of triangle is increasing at the rate of \[10\sqrt{3}c{{m}^{2}}/s.\]