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# 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 ?

$\begin{array}{1 1} \sqrt {3b}cm^2/sec \\ \sqrt {-3b}cm^2/sec \\ \sqrt {3a}cm^2/sec \\ \sqrt {5b}cm^2/sec \end{array}$

Toolbox:
• Area of triangle $=\large\frac{1}{2}$$\times l\times h • \large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
Let $x$ be the equal side of isosceles triangle with fixed base $b$
$AB=AC=x$
$BC=b$
In right $\Delta ABL$,
$AL=\sqrt{AB^2-BL^2}$
$AL=\sqrt{x^2-\large\frac{b^2}{4}}$
Area of $\Delta ABC=A=\large\frac{1}{2}$$\times BC\times AL A=\large\frac{1}{2}$$\times b\times \sqrt{x^2-\large\frac{b^2}{4}}$
Step 2: