A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are and respectively, and the slant height of the top is , 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 ₹ per . (Note that the base of the tent will not be covered with canvas.)

Understand the Question The tent consists of two parts sharing the same circular base radius (): 1. A cylindrical base of height . 2. A conical top of slant height . Since the base of the tent rests on the ground, it does not require canvas. The total canvas needed is the sum of the curved surface areas: The total cost is calculated by multiplying the total canvas area by the rate of ₹. Step 1 · Identify Dimensions Given: Height of the cylindrical part, Diameter of the cylinder and cone Slant height of the conical top, Radius of cylinder and cone: Step 2 · Calculate the Area of Canvas Since the base is not covered with canvas: Substitute the values with : Step 3 · Calculate the Cost of Canvas Rate of canvas Answer > Area of canvas , Cost Common Mistakes Including Base Area: Adding for the circular base. The problem explicitly states that the base of the tent is not covered with canvas. Slant Height vs. Vertical Height: The value is the slant height (), not the vertical height () of the cone, so there is no need to calculate . Diameter vs. Radius: Using the diameter () directly in the formulas instead of the radius ().

Surface Areas and Volumes | Exercise 12.1

Question 7

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are $2.1\text{ m}$ and $4\text{ m}$ respectively, and the slant height of the top is $2.8\text{ m}$ , 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 $\text{m}^2$ . (Note that the base of the tent will not be covered with canvas.)

Solution

Understand the Question

The tent consists of two parts sharing the same circular base radius ( $r = 2\text{ m}$ $r = 2 m$ ):
1. A cylindrical base of height $h_c = 2.1\text{ m}$ .
2. A conical top of slant height $l = 2.8\text{ m}$ .
Since the base of the tent rests on the ground, it does not require canvas. The total canvas needed is the sum of the curved surface areas: $\text{Total Area} = \text{CSA of Cylinder} + \text{CSA of Cone} = 2\pi r h_c + \pi r l$
The total cost is calculated by multiplying the total canvas area by the rate of ₹ $500\text{ per m}^2$ .

Step 1 · Identify Dimensions

Given:

Height of the cylindrical part, $h_c = 2.1\text{ m}$
Diameter of the cylinder and cone $= 4\text{ m}$
Slant height of the conical top, $l = 2.8\text{ m}$

Radius of cylinder and cone:

\begin{aligned} r &= \dfrac{\text{diameter}}{2} \\[0.6em] &= \dfrac{4\text{ m}}{2} = 2\text{ m} \end{aligned}

Step 2 · Calculate the Area of Canvas

Since the base is not covered with canvas:

\begin{aligned} \text{Total Area} &= \text{CSA of cylinder} + \text{CSA of cone} \\[0.6em] &= 2\pi r h_c + \pi r l \\[0.6em] &= \pi r (2h_c + l) \end{aligned}

Substitute the values with $\pi = \dfrac{22}{7}$ :

\begin{aligned} \text{Total Area} &= \dfrac{22}{7} \times 2 \times (2 \times 2.1 + 2.8) \\[0.6em] &= \dfrac{22}{7} \times 2 \times (4.2 + 2.8) \\[0.6em] &= \dfrac{22}{7} \times 2 \times 7 \\[0.6em] &= 22 \times 2 \\[0.6em] &= 44\text{ m}^2 \end{aligned}

Step 3 · Calculate the Cost of Canvas

Rate of canvas $= \text{₹}500/\text{m}^2$

\begin{aligned} \text{Cost} &= \text{Total Area} \times \text{Rate} \\[0.6em] &= 44\text{ m}^2 \times \text{₹}500/\text{m}^2 \\[0.6em] &= 44 \times 500 \\[0.6em] &= 22{,}000 \end{aligned}

Answer

Area of canvas $= 44\text{ m}^2$ , Cost $= \text{₹}22{,}000$

Common Mistakes

Including Base Area: Adding $\pi r^2$ for the circular base. The problem explicitly states that the base of the tent is not covered with canvas.
Slant Height vs. Vertical Height: The value $2.8\text{ m}$ is the slant height ( $l$ ), not the vertical height ( $h$ ) of the cone, so there is no need to calculate $l = \sqrt{r^2 + h^2}$ .
Diameter vs. Radius: Using the diameter ( $4\text{ m}$ ) directly in the formulas instead of the radius ( $r = 2\text{ m}$ ).

Question 7

More questions in Exercise 12.1