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

Let the length of the two equal sides of $\mathrm{△}ABC$ be b

$\text{ Draw }AD\perp BC\cdot AD=\sqrt{a^2-\frac{b^2}{4}}$

$\therefore \text{ Area of triangle }A=\frac{1}{2}b\sqrt{a^2-\frac{b^2}{4}}$

$\frac{dA}{dt}=\frac{1}{2}b\cdot \frac{2a}{2\sqrt{a^2-\frac{b^2}{4}}}\frac{da}{dt}=\frac{ab}{\sqrt{4a^2-b^2}}\frac{da}{dt}$

$\frac{da}{dt}=-3\mathrm{cm}/\mathrm{s}\therefore \frac{dA}{dt}=\frac{-3ab}{\sqrt{4a^2-b^2}}\text{ When }a=b\text{, }$

$\text{ we have }\frac{dA}{dt}=-\sqrt{3}b$