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

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

(A) Let the side of the equilateral triangle be $s$ and its area be $A$. The area of an equilateral triangle is given by $A = \frac{\sqrt{3}}{4} s^2$. Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} = \frac{\sqrt{3}}{2} s \cdot \frac{ds}{dt}$. Given that $\frac{ds}{dt} = 2 \text{ cm/sec}$ and $s = 10 \text{ cm}$. Substituting these values,we get $\frac{dA}{dt} = \frac{\sqrt{3}}{2} \times 10 \times 2 = 10 \sqrt{3} \text{ cm}^2/\text{sec}$.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

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The sides of an equilateral triangle are increasing at the rate of $2 \text{ cm/sec}$. The rate at which the area increases,when the side is $10 \text{ cm}$,is

MHT CET 2022 All MHT CET

A
$10 \sqrt{3} \text{ cm}^2/\text{sec}$
B
$20 \sqrt{3} \text{ cm}^2/\text{sec}$
C
$5 \sqrt{3} \text{ cm}^2/\text{sec}$
D
$2 \sqrt{3} \text{ cm}^2/\text{sec}$

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The sides of an equilateral triangle are increasing at the rate of $2 \text{ cm/sec}$. The rate at which the area increases,when the side is $10 \text{ cm}$,is

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