Question

The side of an equilateral triangle is increasing at the rate of $5cm/\sec$. At what rate its area increasing when the side of the triangle is $10$ cm?

Answer 1

Say area of equilateral triangle is $A$ and side $a=10cm$ then

$A=\frac{\sqrt{3}}{4}{a}^2$

Differentiating the above equation with respect to time $t$, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2a\frac{da}{dt}$

Substituting the given values, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2\times 10\times 5$ [$\frac{da}{dt}=5cm/sec$ (Given)]

$\frac{dA}{d{t}_{(ata=10)}}=25\sqrt{3}c{m}^2/sec$

Its area will increase at the rate of $25\sqrt{3}c{m}^2/sec$ when the side of the triangle is $10cm$.