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 \(-{\sqrt3b}\,cm^2/s.\)