# 3. 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 ?

It is given that the base of the triangle is b
and let the side of the triangle be x cm  , $\frac{dx}{dt} = -3 cm/s$
We know that the area of the triangle(A) = $\frac{1}{2}bh$
now, $h = \sqrt{x^2-(\frac{b}{2})^2}$
$A= \frac{1}{2}b \sqrt{x^2-(\frac{b}{2})^2}$
$\frac{dA}{dt}=\frac{dA}{dx}.\frac{dx}{dt}= \frac{1}{2}b\frac{2x}{2\sqrt{x^2-(\frac{b}{2})^2}}.(-3)$
Now at x = b
$\frac{dA}{dx} = \frac{1}{2}b\frac{2b}{\frac{\sqrt3b}{2}}.(-3)=-\sqrt3b$
Hence,  the area decreasing when the two equal sides are equal to the base  is $\sqrt3b$ $cm^2/s$

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