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(B) Let the side of the equilateral triangle be $x$. Given that the rate of change of the side is $\frac{dx}{dt} = 2 \, cm/s$.The area $A$ of an equil

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

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(B) Let the side of the equilateral triangle be $x$. Given that the rate of change of the side is $\frac{dx}{dt} = 2 \, cm/s$.The area $A$ of an equilateral triangle with side $x$ is given by $A = \frac{\sqrt{3}}{4} x^2$.To find the rate of increase of the area,we differentiate $A$ with respect to time $t$:$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{\sqrt{3}}{4} x^2 ight) = \frac{\sqrt{3}}{4} \cdot 2x \cdot \frac{dx}{dt} = \frac{\sqrt{3}}{2} x \frac{dx}{dt}$.Given $x = 10 \, cm$ and $\frac{dx}{dt} = 2 \, cm/s$,we substitute these values into the expression:$\frac{dA}{dt} = \frac{\sqrt{3}}{2} 	imes 10 	imes 2 = 10 \sqrt{3} \, cm^2/s$.Thus,the rate of increase of the area is $10 \sqrt{3} \, cm^2/s$.

The side of an equilateral triangle expands at the rate of $2 \, cm/s$. The rate of increase of its area when each side is $10 \, cm$ is:

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