If the semi vertical angle of a cone is $45^{\circ}$ and its height is $20.025 \mathrm{~cm}$, then the approximate value of its curved surface area (in sq. $\mathrm{cm}$ ) is

Let $r$ be the radius, $h$ be the slant height and $l$ be the slant height of a cone of semi vertical angle $45^{\circ}$, then, $r=h$ and $l=\sqrt{2} h$ Let $h=20$ and $h+\Delta h=20.025 \Rightarrow \Delta h=0.025$ Let $S$ be surface area, then $\left.S=\sqrt{2} \pi h^2 \quad \mid \because S=\pi r l \Rightarrow S=\pi h \times \sqrt{2} h\right\rceil$ $\Rightarrow \quad \frac{d S}{d h}=2 \sqrt{2} \pi h \Rightarrow\left(\frac{d s}{d h}\right)_{h=20}=40 \sqrt{2} \pi$ $\Rightarrow \quad \frac{d S}{d h}=2 \sqrt{2} \pi h $\therefore \quad \Delta S=\frac{d s}{d h} \Delta h \Rightarrow \Delta s=40 \sqrt{2} \pi \times 0.025=\sqrt{2} \pi$ Also, at $h=20$, We have, $S=400 \sqrt{2} \pi$ $\therefore S+\Delta S=400 \sqrt{2} \pi+\sqrt{2} \pi=401 \sqrt{2} \pi$