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

Let us take either of the equal sides $AB\,\,\,\& AC$ of $\Delta ABC$ be $x$ at any instant of time $t$.
$\Rightarrow \cfrac{{dx}}{{dt}} = - 3cm/\sec$

If $A$ is the area of $\Delta ABC$ at time $t$, then

$A = \cfrac{1}{2}base \times height = \cfrac{1}{2}b\sqrt {A{B^2} - B{D^2}}$

$= \cfrac{b}{2}\sqrt {{x^2} - {{\left( {\cfrac{b}{2}} \right)}^2}} = \cfrac{b}{4}\sqrt {4{x^2} - {b^2}}$

Therefore,$\cfrac{{dA}}{{dt}} = \cfrac{b}{4} \times \cfrac{1}{2} \times \cfrac{{8x}}{{\sqrt {4{x^2} - {b^2}} }}\cfrac{{dx}}{{dt}} = \cfrac{{bx}}{{\sqrt {4{x^2} - {b^2}} }}\cfrac{{dx}}{{dt}}$

${\left( {\cfrac{{dA}}{{dt}}} \right)_{x = b}} = \left( {\cfrac{{b \times b}}{{\sqrt {4{b^2} - {b^2}} }}} \right)( - 3) = - \sqrt 3 b$ $\left[ {\therefore \cfrac{{dx}}{{dt}} = - 3} \right]$

Therefore area is decreasing at the rate of $b\sqrt 3 c{m^2}/sec$.