The length of canvas, 75 cm wide required to build a conical tent of height 14 m and the floor area `346.5 m^(2)` is

Options

A. 665 m
B. 860 m
C. 490 m
D. 770 m

Answer

To solve the problem of finding the length of canvas required to build a conical tent, we will follow these steps:

### Step 1: Find the radius of the base of the cone
We know that the floor area \( A \) of the conical tent is given by the formula:
\[ A = \pi r^2 \]
Given that the floor area \( A = 346.5 \, m^2 \), we can rearrange the formula to find the radius \( r \):
\[ r^2 = \frac{A}{\pi} \]
Using \( \pi \approx \frac{22}{7} \):
\[ r^2 = \frac{346.5 \times 7}{22} \]
Calculating this gives:
\[ r^2 = \frac{2425.5}{22} \approx 110.25 \]
Taking the square root:
\[ r = \sqrt{110.25} \approx 10.5 \, m \]

### Step 2: Find the slant height of the cone
The slant height \( l \) of the cone can be found using the Pythagorean theorem:
\[ l = \sqrt{r^2 + h^2} \]
Where \( h \) is the height of the cone, given as \( 14 \, m \):
\[ l = \sqrt{(10.5)^2 + (14)^2} \]
Calculating this gives:
\[ l = \sqrt{110.25 + 196} = \sqrt{306.25} \approx 17.5 \, m \]

### Step 3: Calculate the length of canvas required
The length of canvas required to cover the conical tent is given by the formula:
\[ \text{Length} = \pi r l \]
Substituting the values we have:
\[ \text{Length} = \pi \times 10.5 \times 17.5 \]
Using \( \pi \approx \frac{22}{7} \):
\[ \text{Length} = \frac{22}{7} \times 10.5 \times 17.5 \]
Calculating this gives:
\[ \text{Length} = \frac{22 \times 10.5 \times 17.5}{7} \]
Calculating the numerator:
\[ 22 \times 10.5 \times 17.5 = 22 \times 183.75 = 4057.5 \]
Now dividing by 7:
\[ \text{Length} = \frac{4057.5}{7} \approx 579.64 \, m \]

### Step 4: Convert the length to centimeters
Since the width of the canvas is given in centimeters, we need to convert the length from meters to centimeters:
\[ \text{Length in cm} = 579.64 \times 100 \approx 57964 \, cm \]

### Final Answer
The length of canvas required to build the conical tent is approximately **57964 cm**.

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