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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.

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