(A) Let the side of the equilateral triangle be $x$.The area $A$ of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} x^2$.Differentiating b

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(A) Let the side of the equilateral triangle be $x$.The area $A$ 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 $\frac{dx}{dt} = \sqrt{3} \text{ cm/sec}$ and $x = 12 \text{ cm}$.Substituting these values, $\frac{dA}{dt} = \frac{\sqrt{3}}{2} (12) (\sqrt{3}) = 6 \times 3 = 18 \text{ cm}^2/\text{sec}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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If the side of an equilateral triangle increases at the rate of $\sqrt{3} \text{ cm/sec}$, then the rate of change of its area when the side is $12 \text{ cm}$ is

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A
$18 \text{ cm}^2/\text{sec}$
B
$10 \text{ cm}^2/\text{sec}$
C
$12 \text{ cm}^2/\text{sec}$
D
$3\sqrt{3} \text{ cm}^2/\text{sec}$

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