Question

If you are given a rectangular canvas of 1.5 m in width, what length of this canvas would you require to make a conical tent that is 48 m in diameter and 7 m in height? Note that 10% of the canvas is used (wasted) in folds and stitchings. 

Also, find the cost of the canvas at the rate of 24 per metre.

Answer 1

Given,

Diameter of conical tent = 48 m

Radius of conical tent \[(r) = \frac{48}{2} = 24\ \text{m}\]

Height of conical tent (h) = 7 m

Let l be the slant height of the conical tent.

By formula,

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

\[\Rightarrow l = \sqrt{(24)^2 + 7^2}\]

\[\Rightarrow l = \sqrt{576 + 49}\]

\[\Rightarrow l = \sqrt{625}\]

\[\Rightarrow l = 25\ \text{m}\]

By formula,

Surface area of conical tent \[(S) = \pi rl\]

\[S = \frac{22}{7} \times 24 \times 25\]

\[{} = \frac{13200}{7}\ \text{m}^2\]

Given,

\[10\%\] of the canvas is used in folds and stitching.

Thus, \[90\%\] of the total canvas area is used for making the tent.

⇒ \[\frac{90}{100} \times\] Total canvas area = Surface area of conical tent

⇒ \[\frac{90}{100} \times\] Total canvas area \[= \frac{13200}{7}\]

⇒ Total canvas area \[= \frac{13200}{7} \times \frac{100}{90}\]

⇒ Total canvas area \[= \frac{132000}{63}\ \text{m}^2\]

Thus,

⇒ Total canvas area = Length of canvas × Width of canvas

⇒ \[\frac{132000}{63}\] = Length of canvas \[\times\ 1.5\]

⇒ Length of canvas \[= \frac{132000}{63 \times 1.5} = \frac{132000}{94.5} = 1396.83\ \text{m}\]

Cost of canvas = Rate per m × Length

\[{} = 24 \times 1396.83\]

\[{} = \text{₹}\,33{,}523.92\]