Question

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

Answer 1

Answer:

$10\sqrt{3} cm^2/s$, or, $17.32... cm^2/s$

Step-by-step explanation:

The area of an equilateral triangle of side a is given by:

$A = \frac{\sqrt{3}}{4} a^2$

Differentiating, with respect to time, we get:

$\frac{dA}{dt} = \frac{\sqrt{3}}{4} (2a\frac{da}{dt})$

Substituting the values of a and $\frac{da}{dt}$, we get:

$\frac{dA}{dt} = \frac{\sqrt{3}{2} (2 * 10) cm^2/s$

Or, $\frac{dA}{dt} = 10\sqrt{3} cm^2/s$

Or, the rate of change of area of the given equilateral triangle, when its side is 10cm, is $10\sqrt{3} cm^2/s$.