If equal sides of an isosceles triangle with fixed base 10 cm are increasing at the rate of 4 cm / sec, how fast is the area of triangle increasing…

Let the equal side be ' $\displaystyle x$ ', then $\displaystyle \frac{d x}{d t}=4 \mathrm{~cm} / \mathrm{s}$ $$\begin{aligned} & A=5 \sqrt{x^{2}-25} \Rightarrow \frac{d A}{d t}=\frac{5 x}{\sqrt{x^{2}-25}} \frac{d x}{d t} \\ & \left.\frac{d A}{d t}\right]_{x=10}=\frac{40}{\sqrt{3}} \mathrm{~cm}^{2} / \mathrm{s} \end{aligned} $$