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

The correct option is 10√3 cm²/s.

Let us denote the side length of the equilateral triangle as s and its area as A.

We are given the following information:
1. The rate of increase of the sides is $\frac{ds}{dt}=2\text{cm/sec}$.
2. We need to find the rate of change of the area, $\frac{dA}{dt}$, at the instant when the side length $s=10\text{cm}$.

The formula for the area of an equilateral triangle in terms of its side length s is given by:

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

To find the rate of change of the area with respect to time, we differentiate both sides of the area equation with respect to t using the chain rule:

$\frac{dA}{dt}=\frac{d}{dt}\left(\frac{\sqrt{3}}{4}{s}^2\right)$

$\frac{dA}{dt}=\frac{\sqrt{3}}{4}·2s·\frac{ds}{dt}$

Simplifying the expression, we get:

$\frac{dA}{dt}=\frac{\sqrt{3}s}{2}·\frac{ds}{dt}$

Now, we substitute the given values $s=10\text{cm}$ and $\frac{ds}{dt}=2\text{cm/sec}$ into the differentiated equation:

$\frac{dA}{dt}=\frac{\sqrt{3}\left(10\right)}{2}·2$

$\frac{dA}{dt}=10\sqrt{3}{\text{cm}}^2\text{/s}$

Therefore, the rate at which the area of the equilateral triangle increases when its side is 10 cm is indeed 10√3 cm²/s.