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

Toolbox:
• Let two equal sides be $x$
• $\perp$ from vertex of the isoceles $\Delta$ will divide the base into equal ratio.
• Area $A = \large\frac{1}{2}\: b \sqrt{x^2-\large\frac{b^2}{4}}$
Step 1
Given $\large\frac{dx}{dt}=3$
$\large\frac{dA}{dt}=?$ when $x = b$
Step 2
$A = \large\frac{1}{4} b \sqrt{4x^2-b^2}$
$\Large\frac{dA}{dt}=\Large\frac{bx}{\sqrt{4x^2-b^2}}.\large\frac{dx}{dt}$
Ans $\large\frac{dA}{dt} = \sqrt 3b$ when $x = b$