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

Question

The sides of an equilateral triangle are increasing at the rate of $2 \mathrm{~cm} \mathrm{~s}^{-1}$. How fast does its area increase when its side is $10 \mathrm{~cm}$ ?, $10 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}$, $5 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}$, $\sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}$, $2 \sqrt{3} \mathrm{~cm}^2, \mathrm{~s}^{-1}$

MARKS App · Accepted Answer

Area of an equilateral triangle is, $A=\frac{\sqrt{3} a^2}{4}$ Differentiating with respect to time we get, $\Rightarrow \quad \frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a \cdot \frac{d a}{d t}$ Here, $\frac{d a}{d t}=2 \mathrm{~cm}^{-1}$ when $a=10 \mathrm{~cm}$ So, $\left.\frac{d A}{d t}\right|_{a=10 \mathrm{~cm}}=\frac{\sqrt{3}}{4} \times 2 \times 10 \times 2$ $=10 \sqrt{3} \mathrm{~cm}^2 \cdot \mathrm{s}^{-1}$

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