A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ` 500 per m2. (Note that the base of the tent will not be covered with canvas.)
Step-by-Step Solution
Key Concept: The canvas covers only the curved (lateral) surface of the cylinder and the curved surface of the cone. Use the formulas:<br>- Lateral surface area of a cylinder: $2\pi r h$<br>- Curved surface area of a cone: $\pi r l$, where $l$ is the slant height.<br>Add the two areas to obtain the total canvas area, then multiply by the given cost per square metre.
1. Identify the dimensions\
- Diameter of cylindrical part $=4\,\text{m}$ \=> radius $r = \dfrac{4}{2}=2\,\text{m}$\
- Height of cylindrical part $h = 2.1\,\text{m}$\
- Slant height of conical top $l = 2.8\,\text{m}$\
2. Lateral surface area of the cylinder\
$$\text{Area}_{\text{cyl}} = 2\pi r h = 2\pi (2)(2.1) = 8.4\pi \;\text{m}^2$$\
3. Curved surface area of the cone\
$$\text{Area}_{\text{cone}} = \pi r l = \pi (2)(2.8) = 5.6\pi \;\text{m}^2$$\
4. Total canvas area (base is not covered)\
$$\text{Total area}=\text{Area}_{\text{cyl}}+\text{Area}_{\text{cone}} = 8.4\pi +5.6\pi = 14\pi \;\text{m}^2$$\
Using $\pi = \dfrac{22}{7}$ (as used in NCERT),\
$$\text{Total area}=14\times\frac{22}{7}=44\;\text{m}^2$$\
5. Cost of the canvas\
Rate = ` 500 per m$^2$\
$$\text{Cost}=44\times 500 = 22,000$$\
Hence, the canvas required is $44\,\text{m}^2$ and the cost is ` 22,000.
Correct Answer: Canvas area = $44\,\text{m}^2$; Cost = ` 22,000.