Question:
Side of an equilateral triangle expands at the rate of 2 cm/sec. The rate of increase of its area when each side is 10 cm is
(a) $10 \sqrt{2} \mathrm{~cm}^{2} / \mathrm{sec}$
(b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$
(c) $10 \mathrm{~cm}^{2} / \mathrm{sec}$
(d) $5 \mathrm{~cm}^{2} / \mathrm{sec}$
Solution:
(b) $10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$
Let $x$ be the side and $A$ be the area of the equilateral triangle at any time $t$. Then,
$A=\frac{\sqrt{3}}{4} x^{2}$
$\Rightarrow \frac{d A}{d t}=2 \times \frac{\sqrt{3}}{4} x \frac{d x}{d t}$
$\Rightarrow \frac{d A}{d t}=\frac{\sqrt{3}}{2} \times 2 \times 10$
$\Rightarrow \frac{d A}{d t}=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec}$