The two equal sides of an isosceles triangle with fixed base b are decreasing at the rate of 3 cm/sec. How fast is the area decreasing when two equal sides are equal to the base?

$-\sqrt{3}b\,cm^2/s$

The correct answer is Option (1) → $-\sqrt{3}b\,cm^2/s$

Let ABC be an isosceles triangle with sides AB = AC and base BC.

At any time, let AB = AC = a (say),

then $\frac{da}{dt}=-3\, cm/s$ (given).

Base $BC = b$ (constant).

Let A be the area of AABC at any time, then $A =\frac{1}{4}b\sqrt{4a^2 – b^2}$.

Diff. w.r.t. t, we get

$\frac{dA}{dt}=\frac{1}{4}b.\frac{1}{2}.(4a^2 – b^2)^{−1/2}.4.2a.\frac{da}{dt}=\frac{ab}{\sqrt{4a^2-b^2}}.\frac{da}{dt}$

When $a = b,\frac{dA}{dt}=\frac{b.b}{\sqrt{4b^2-b^2}}(-3) cm^2/sec = -\frac{3b^2}{\sqrt{3}b}\,cm^2/sec$

$=-\sqrt{3} b\, cm^2/sec$.

Hence, the area of an isosceles triangle is decreasing at the rate of $\sqrt{3} b\, cm^2/sec$ when two equal sides are equal to the base b.