Due to heavy floods in a state, thousands were rendered homeless. 50 schools collectively offered to the state government to provide place and the canvas for 1500 tents to be fixed by the government and decided to share the whole expenditure equally. The lower part of each tent is cylindrical of base radius 2.8 m and height 3.5 m, with conical upper part of same base radius but of height 2.1 m. If the canvas used to make the tents costs Rs. 120 per sq. m, find the amount shared by each school to set up the tents. What value is generated by the above problem ? (Use `pi = (22)/(7)`)

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the Curved Surface Area (CSA) of the Cylinder
The formula for the curved surface area of a cylinder is given by:
\[ \text{CSA}_{\text{cylinder}} = 2 \pi r h \]
where \( r \) is the radius and \( h \) is the height.

Given:
- Radius \( r = 2.8 \, \text{m} \)
- Height \( h = 3.5 \, \text{m} \)

Substituting the values:
\[ \text{CSA}_{\text{cylinder}} = 2 \times \frac{22}{7} \times 2.8 \times 3.5 \]
Calculating this:
\[ = 2 \times \frac{22}{7} \times 2.8 \times 3.5 = 2 \times \frac{22 \times 2.8 \times 3.5}{7} \]
Calculating \( 22 \times 2.8 \times 3.5 \):
\[ = 22 \times 2.8 = 61.6 \]
\[ = 61.6 \times 3.5 = 216.6 \]
Now substituting back:
\[ \text{CSA}_{\text{cylinder}} = 2 \times \frac{216.6}{7} = \frac{433.2}{7} \approx 61.89 \, \text{m}^2 \]

### Step 2: Calculate the Curved Surface Area (CSA) of the Cone
The formula for the curved surface area of a cone is given by:
\[ \text{CSA}_{\text{cone}} = \pi r l \]
where \( l \) is the slant height of the cone.

To find the slant height \( l \), we use the Pythagorean theorem:
\[ l = \sqrt{r^2 + h^2} \]
Given:
- Radius \( r = 2.8 \, \text{m} \)
- Height \( h = 2.1 \, \text{m} \)

Calculating \( l \):
\[ l = \sqrt{(2.8)^2 + (2.1)^2} = \sqrt{7.84 + 4.41} = \sqrt{12.25} = 3.5 \, \text{m} \]

Now substituting \( l \) back into the CSA formula:
\[ \text{CSA}_{\text{cone}} = \frac{22}{7} \times 2.8 \times 3.5 \]
Calculating this:
\[ = \frac{22 \times 2.8 \times 3.5}{7} = \frac{216.6}{7} \approx 30.94 \, \text{m}^2 \]

### Step 3: Total Canvas Area for One Tent
The total canvas area for one tent is the sum of the CSA of the cylinder and the CSA of the cone:
\[ \text{Total Area}_{\text{tent}} = \text{CSA}_{\text{cylinder}} + \text{CSA}_{\text{cone}} = 61.89 + 30.94 = 92.83 \, \text{m}^2 \]

### Step 4: Total Canvas Area for 1500 Tents
Now, we calculate the total canvas area for 1500 tents:
\[ \text{Total Area}_{\text{1500 tents}} = 1500 \times 92.83 = 139245 \, \text{m}^2 \]

### Step 5: Calculate the Total Cost of the Canvas
The cost of the canvas is given as Rs. 120 per sq. m:
\[ \text{Total Cost} = \text{Total Area}_{\text{1500 tents}} \times \text{Cost per sq. m} = 139245 \times 120 \]
Calculating this:
\[ = 16709400 \, \text{Rs} \]

### Step 6: Amount Shared by Each School
There are 50 schools sharing the total cost equally:
\[ \text{Amount per school} = \frac{\text{Total Cost}}{50} = \frac{16709400}{50} = 334188 \, \text{Rs} \]

### Final Answer
The amount shared by each school to set up the tents is Rs. 334188.

### Values Generated by the Above Problem
The problem highlights the importance of community support during disasters, the value of collaboration among institutions, and the practical application of mathematical concepts in real-life scenarios.