Thank you for reporting, we will resolve it shortly
Q.
If the semi vertical angle of a cone is $45^{\circ}$ and its height is $20.025 \,cm$, then the approximate value of its curved surface area (in sq.cm) is
TS EAMCET 2020
Solution:
Let $r$ be the radius, $h$ be the slant height and $\bar{l}$ 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
$S=\sqrt{2} \pi h^{2}[\because S=\pi r l \Rightarrow S=\pi h \times \sqrt{2} h]$
$\Rightarrow \frac{d S}{d h}=2 \sqrt{2} \pi h$
$ \Rightarrow \left(\frac{d s}{d h}\right)_{h=20}=40 \sqrt{2} \pi $
$\therefore \Delta S=\frac{d s}{d h} \Delta h$
$ \Rightarrow \Delta s=40 \sqrt{2} \pi \times 0.025=\sqrt{2} \pi$
Also, at $ h=20 $
We have, $ S=400 \sqrt{2} \pi $
$\therefore S+\Delta S=400 \sqrt{2} \pi+\sqrt{2} \pi=401 \sqrt{2} \pi$