Question

The sides of an equilateral triangle are increasing at the rate of 4 cm/sec. The rate at which its area is increasing, when the side is 14 cm.

• A. 42 $cm^2 / \sec$
• B. $10\sqrt{3} cm^2 / \sec$
• C. 14 $cm^2 / \sec$
• D. $14 \sqrt{3} cm^2 / \sec$

Answer 1

Answer: (B)

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

Answer 2

$\frac{dx}{dt}=4cm/ sec,$ $x = 14cm$ $A=\frac{\sqrt{3}}{4} x^{2}$ $\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2x \frac{d x}{d t}$ $=\frac{\sqrt{3}}{2}\times14\times4$ $=\sqrt{3}\times7\times4$ $=28\sqrt{3}$