The semi-vertical angle of a cone is 45^°. If the height of the cone…

Let $r$ be the radius, $h$ be the 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 the lateral surface. Then,

$S=\sqrt{2} \pi h^2$              $[∵ S=\pi r l \Rightarrow S=\pi h \times \sqrt{2} h]$

$\Rightarrow \frac{d S}{d h}=2 \sqrt{2} \pi h$

$\Rightarrow \left(\frac{d S}{d h}\right)_{h=20}=40 \sqrt{2} \pi$

∴  $\Delta S=\frac{d S}{d h} \Delta h \Rightarrow \Delta S=40 \sqrt{2} \pi \times 0.025=\sqrt{2} \pi$

Also, at $h=20$, we have $S=400 \sqrt{2} \pi$

∴  $S+\Delta S=400 \sqrt{2} \pi+\sqrt{2} \pi=401 \sqrt{2} \pi$