Question

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

Answer 1

Let x be the side and A be the area of an equilateral triangle at any time t.

It is given that,

$\frac{dx}{dt}$ = 2 cm/sec

Area of the equilateral triangle, A = $\frac{\sqrt{3}}{4}{\left(\mathrm{Side}\right)}^2$ = $\frac{\sqrt{3}}{4}{x}^2$

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

Differentiating both sides with respect to t, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}\times \frac{d}{dt}{x}^2$

$\Rightarrow \frac{dA}{dt}=\frac{\sqrt{3}}{4}\times 2x\frac{dx}{dt}$

$\Rightarrow \frac{dA}{dt}=\frac{\sqrt{3}}{2}x\frac{dx}{dt}$

When x = 10 cm and $\frac{dx}{dt}$ = 2 cm/sec, we get

$\frac{dA}{dt}=\frac{\sqrt{3}}{2}\times 10\times 2$

$\Rightarrow \frac{dA}{dt}=10\sqrt{3}$ cm²/sec

Thus, the area of the equilateral triangle increases at the rate of $10\sqrt{3}$ cm²/sec.


The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when the side is 10 cm, is $\underline{10\sqrt{3}{\mathrm{cm}}^2/\sec }$.