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(C) Let $x$ be the length of each side of an equilateral triangle and $A$ be its area. The area of an equilateral triangle is given by $A = \frac{\sqr

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

Vedclass · Answer

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

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:

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