Show that the surface area of a cone in terms of it’s height and radius is $\pi a \sqrt {a^2+h^2}$ by considering how a sector of a disc radius r angle $\phi $ can be folded to make a cone.
I cannot reach the surface area in the requested form, only in the form $2r (2 \pi - \phi)$
1 Answer
$\begingroup$Here’s a paper cone , of radius $a$ , and height $DG=h$.
By Pythagoras’ Theorem , the slant height $DF=\sqrt{a^2+h^2} $
Now , cut along $DH$ , and unravel the curved piece of paper. We obtain a circular sector .
Here , $H’$ and $H”$ both represent the same point $H$ . Therefore , we have $H’H”= 2 \pi a$ . The radius of the sector = $\sqrt{a^2+h^2} $. $\angle H’DH” = \frac{2 \pi a}{\sqrt{a^2+h^2}} $ , as $\theta = \frac{s}{r} $ .
$\therefore $ The area of the circular sector =$ \pi r^2 \cdot \frac{\theta}{2 \pi} = \boxed{\pi a \sqrt{a^2+h^2} }$
$\endgroup$
