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(B) Let $x$ be the side length and $A$ be the area of the equilateral triangle at time $t$.The area of an equilateral triangle is given by $A = \frac{

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$10\sqrt{3} \ cm^2/s$

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(B) Let $x$ be the side length and $A$ be the area of the equilateral triangle at time $t$.The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.Differentiating both sides with respect to $t$,we get:$\frac{dA}{dt} = \frac{\sqrt{3}}{4} 	imes 2x 	imes \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.Given that $\frac{dx}{dt} = 2 \ cm/s$ and $x = 10 \ cm$.Substituting these values into the derivative formula:$\frac{dA}{dt} = \frac{\sqrt{3}}{2} 	imes 10 	imes 2 = 10\sqrt{3} \ cm^2/s$.

The sides of an equilateral triangle are increasing at the rate of $2 \ cm/s$. Find the rate at which its area is increasing when each side is $10 \ cm$.

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