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Answer :
Given :
Radius (r) of conical cap = \(\frac{40}{2} cm = 20 cm = 0.2 m\)
Height (h) of conical cap = 1 m
Slant height (l) of the conical cap
\(\Rightarrow \) \({l}^2 = {h}^2 + {r}^2\)
\(\Rightarrow \) \(l^2= {1}^2 + {0.2}^2 {m}^2\)
\(\Rightarrow \) \(l^2= 1.04 {m}^2\)
\(\Rightarrow \) \({l}^2 = {1.02}^2 m\)
\(\Rightarrow \) l = 1.02 m
\(\therefore \) the slant height Of the tent is 1.02 m.
We know, CSA of tent
= \({\pi}rl\)
= [\(3.14 × 0.2 × 1.02\) \({m}^2\)]
= \(0.64056\) \({m}^2\)
So, CSA of 50 such cones
= \(50 × 0.64056\) \({m}^2\)
= 32.028 \({m}^2\).
Now, we have,
Cost of painting 1 \({m}^2\) area = Rs. 12
Cost of painting 32.02 \({m}^2\) area = Rs. (32.028 x 12)
= Rs. 384.336
= Rs 384.34 (approximately)
Therefore, it Will cost Rs. 384.34 in painting 50 such hollow cones.