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(A) The area $A$ of an equilateral triangle with side $a$ is given by:$A = \frac{\sqrt{3}}{4} a^2$Differentiating both sides with respect to time $t$,

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$10 \sqrt{3} 	ext{ cm}^2 	ext{ s}^{-1}$

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(A) The area $A$ of an equilateral triangle with side $a$ is given by:$A = \frac{\sqrt{3}}{4} a^2$Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2a \cdot \frac{da}{dt}$$\frac{dA}{dt} = \frac{\sqrt{3}}{2} a \cdot \frac{da}{dt}$Given that the rate of increase of the side is $\frac{da}{dt} = 2 	ext{ cm s}^{-1}$ and the side length is $a = 10 	ext{ cm}$,we substitute these values into the equation:$\frac{dA}{dt} = \frac{\sqrt{3}}{2} 	imes 10 	imes 2$$\frac{dA}{dt} = 10 \sqrt{3} 	ext{ cm}^2 	ext{ s}^{-1}$Thus,the area of the triangle increases at the rate of $10 \sqrt{3} 	ext{ cm}^2 	ext{ s}^{-1}$.

The sides of an equilateral triangle are increasing at the rate of $2 \text{ cm s}^{-1}$. How fast does its area increase when its side is $10 \text{ cm}$?

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