Question

A tent of height $33m$ is in the form of a right circular cylinder of diameter $120m$ and height $22m$ surmounted by a right circular cone of the same diameter Find the cost of canvas of the tent at the rate of $1.50Rsper{m}^2$.

Answer 1

Step 1: Find the curve surface area of the cylindrical portion

Given that, height $h$ of the tent $=33m$

The height $\left(h\text{'}\right)$ of the cylindrical portion $=22m$

Therefore, height of the conical portion $=$height of the tent$-$height of the cylindrical portion

$$\begin{gathered} =33m-22m \\ =11m \end{gathered}$$

The diameter of cylinder is $120m$

Thus, radius $r$ of the cylinder $=\frac{120}{2}m$ $(Radius=\frac{Diameter}{2})$

$=60m$

The curved surface area of the cylindrical portion of radius $r$ and height $h\text{'}$ $=2\mathrm{\pi rh}\text{'}$

$$\begin{gathered} =2\times \frac{22}{7}\times 60\times 22(\mathrm{\pi }=\frac{22}{7}) \\ =8297.14m^2 \end{gathered}$$

Step 2: Find the curved surface area of the conical portion.

Let the slant height of conical portion be $l$

According to the Pythagoras theorem, $l^2=r^2+h^2$

$\Rightarrow l=\sqrt{r^2+h^2}$

$\Rightarrow l=\sqrt{{60}^2+{11}^2}$

$\Rightarrow l=\sqrt{3600+121}$

$$\begin{gathered} \Rightarrow l=\sqrt{3721} \\ \Rightarrow l=61m \end{gathered}$$

Curved surface area of the conical portion of radius $r$ and slant height $l$ $=\mathrm{\pi rl}$

$$\begin{gathered} =\frac{22}{7}\times 60\times 61(\mathrm{\pi }=\frac{22}{7}) \\ =11502.85{\mathrm{m}}^2 \end{gathered}$$

Step 3: Find the cost of canvas of the tent

The total surface area of the tent $=$ Curved surface area of the cylindrical portion $+$ Curved surface area of the conical portion

$$\begin{gathered} =8297.14m^2+11502.85m^2 \\ =19799.99m^2 \end{gathered}$$

Therefore, the canvas required to make the $=19800m^2$

Cost of $1m^2$ canvas $=Rs1.50$

Cost of $19800m^2$ canvas $=Rs1.50\times 19800$

$=Rs29700$

Hence, the cost of the canvas of the tent at the rate of $1.50Rspermsq=29700m^2$