$\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} $

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- 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:

It is given that $\large\frac{dx}{dt}$$=3cm/sec$

Now $\large\frac{dA}{dt}=\large\frac{b}{2}\times\frac{1}{2}\times \frac{1}{\sqrt{x^2-b^2/4}}\times 2x\times \large\frac{dx}{dt}$

$\Rightarrow \large\frac{dA}{dt}=\frac{bx}{2}\times \frac{2}{\sqrt{4x^2-b^2}}\times \large\frac{dx}{dt}$

$\large\frac{dA}{dt}=\frac{bx}{2}\times\frac{2}{\sqrt{4x^2-b^2}}\times \frac{dx}{dt}$

Step 3:

When $x=b$ we get,

$\large\frac{dA}{dt}=\frac{b\times b\times 2\times 3}{2\sqrt{4b^2-b^2}}$

$\qquad=\large\frac{3b^2}{\sqrt{3b}}$

$\qquad=\sqrt{3b}$

Hence the area is decreasing at $\sqrt {3b}cm^2/sec$

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