Question

The area of an equilateral triangle of side ${}^{\prime }{a}^{\prime }$ feet is increasing at the rate of $4sq.ft./sec$. The rate at which the perimeter is increasing is

• A. $\frac{3\sqrt{8}}{2}$
• B. $\frac{8\sqrt{3}}{a}$
• C. $\frac{\sqrt{3}}{a}$
• D. $\frac{2\sqrt{3}}{a}$

Answer 1

The correct option is B $\frac{8\sqrt{3}}{a}$

The area of an equilateral triangle = $\frac{\sqrt{3}\times a^2}{4}$

If area is increasing at a rate of $4sq.ft./sec$,

$\Rightarrow 4=\frac{d}{dt}\frac{\sqrt{3}\times a^2}{4}$

$\Rightarrow \frac{4\times 4}{2\times \sqrt{3}}=a\frac{da}{dt}$

$\Rightarrow \frac{da}{dt}=\frac{8}{\sqrt{3}a}$

The perimeter of an equilateral triangle = $3\times a$

Therefore the perimeter is increasing at a rate = $3\frac{da}{dt}$

And $3\frac{da}{dt}=\frac{8\sqrt{3}}{a}$