Question

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

18cm 22cm 10cm 8cm 

Answer 1

Given figure is shown:

From the figure, we can write,

Height of the funnel $\left(\mathrm{H}\right)$ $=22\mathrm{cm}$

Height of cylindrical part $\left(\mathrm{h}\right)$ $=10\mathrm{cm}$

Height of the frustum of cone $\left({\mathrm{h}}_1\right)$ $=22-10=12\mathrm{cm}$

The radius of the upper part of the frustum of cone $\left({\mathrm{r}}_1\right)$ $=\frac{18}{2}=9\mathrm{cm}$

The radius of the lower part of the frustum of cone $\left({\mathrm{r}}_2\right)$ $=\frac{8}{2}=4\mathrm{cm}$

The radius of the cylindrical part $\left(\mathrm{r}\right)$$=\frac{8}{2}=4\mathrm{cm}$

We know that the slant height of the frustum $\left(\mathrm{l}\right)$$=\sqrt{{\left({\mathrm{h}}_1\right)}^2+{\left({\mathrm{r}}_1-{\mathrm{r}}_2\right)}^2}$

$$\begin{gathered} =\sqrt{{\left(12\right)}^2+{\left(9-4\right)}^2} \\ =\sqrt{144+25}=\sqrt{169}=13\mathrm{cm} \end{gathered}$$

Now the area of the tin sheet required to make the funnel $=\mathrm{CSA}\mathrm{of}\mathrm{frustum}\mathrm{of}\mathrm{cone}+\mathrm{CSA}\mathrm{of}\mathrm{the}\mathrm{cylinder}$

Step 1: Finding CSA of the frustum of a cone

We know that the CSA of the frustum $=\mathrm{\pi }\left({\mathrm{r}}_1+{\mathrm{r}}_2\right)\mathrm{l}$

$$\begin{gathered} =\mathrm{\pi }\left(9+4\right)\times 13 \\ =\mathrm{\pi }\times 13\times 13 \\ =169\mathrm{\pi }{\mathrm{cm}}^2 \end{gathered}$$

Step 2: Finding CSA of the cylinder

We know that the CSA of the cylinder $=2\mathrm{\pi rh}$

$$\begin{gathered} =2\mathrm{\pi }\times 4\times 10 \\ =80\mathrm{\pi }{\mathrm{cm}}^2 \end{gathered}$$

Therefore area of the tin sheet required to make the funnel $=(169\mathrm{\pi }+80\mathrm{\pi }){\mathrm{cm}}^2$

$$\begin{gathered} =249\mathrm{\pi } \\ =249\times \frac{22}{7}(\mathrm{where}\mathrm{\pi }=\frac{22}{7}) \\ =\frac{5478}{7} \\ =782.57{\mathrm{cm}}^2 \end{gathered}$$

Therefore the area of the tin sheet required to make the funnel is $=782.57{\mathrm{cm}}^2$.