Question

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

Answer 1

Let ∆ABC be isosceles where BC is the base of fixed length b.
Let the length of the two equal sides of ∆ABC be a.
Draw AD⊥BC.

Now, in ∆ADC,by applying the Pythagoras theorem,we have:
AD=$\sqrt {a^2-\frac{b^2}{4}}$
∴Area of triangle $(A)=\frac{1}{2}b$ $\sqrt {a^2-\frac{b^2}{4}}$.
The rate of change of the area with respect to time (t) is given by
$\frac{dA}{dt}$=$\frac{1}{2}b$.$\frac{2a}{2\sqrt {a^2-\frac{b^2}{4}}}$ $\frac{da}{dt}$ = $\frac{ab}{\sqrt {4a^2-4b^2}}\frac{da}{dt}$
It is given that the two equal sides of the triangle are decreasing at the rate of 3cm per second.
$\frac{da}{dt}=3cm/s$
∴ $\frac{dA}{dt}=\frac{-3ab}{\sqrt{4a^2-b^2}}$
Then,when a=b, we have:
$\frac{dA}{dt}=\frac{-3ab}{\sqrt{4a^2-b^2}}$=-$\frac{3b^2}{\sqrt{3b^2}}$=$-{\sqrt3b}$
Hence, if the two equal sides are equal to the base, then the area of the triangle is decreasing at the rate of $-{\sqrt3b}\,cm^2/s.$