Question

An oil funnel made of tin sheet consists of a $10cm$ long cylindrical portion attached to a frustum of a cone. If the total height is $22cm$, diameter of the cylindrical portion is $8cm$ and the diameter of the top of the funnel is $18cm$, find the area of the tin sheet required to make the funnel .

Answer 1

Given:
Radius $\left(R\right)$ of upper circular end of frustum part = $\frac{18}{2}=9$ cm
Radius $\left(r\right)$ of lower circular end of frustum part = Radius of circular end of cylindrical part =$\frac{8}{2}=4$ cm

Height $\left(h_1\right)$ of frustum part $=22-10=12$ cm
Height $\left(h_2\right)$ of cylindrical part $=10$ cm

Slant height (l) of frustum part = $\sqrt{(R-r{)}^2+h_1^2}=\sqrt{(9-4{)}^2+{12}^2}=\sqrt{25+144}$
$=\sqrt{169}=13$ cm

Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
$=\pi (R+r)l+2\pi r{h}_2$

$=\frac{22}{7}\times (9+4)\times 13+2\times \frac{22}{7}\times 4\times 10$

$=\frac{22}{7}[169+80]=\frac{22\times 249}{7}$

$=782\frac{4}{7}$ cm${}^2$