Question

s o l v e t h i s p l e a s e​

Answer 1

Answer:

Given:

h (vertical height)= 3m

r= 4m

width of tarpaulin= 250 cm

A) for slant height

${s}^{2} = {h}^{2} + {r}^{2} \\ = {3}^{2} + {4}^{2} \\ = 9 + 16 \\ = 25 \\ s = \sqrt{25} \\ s = 5m$

Slant height= 5m

B) Curved Surface are of cone=

$\pi \: r \: l$

C) Area of tarpaulin used= CSA is conical tent

= π r l

=

$\frac{22}{7} \times 4 \times 3 \\ = 37.71 \: {m}^{2}$

D) Area of base= πr²

E) Amount of Air present in tent= Volume of tent

Volume= (1/3) πr²h

$vol = \frac{1}{3} \times \frac{22}{7} \times 4 \times 4 \times 5 \\ = 4,693.33 \: {m}^{3}$

volume of air= 4,693.33 m³

Hope this helps

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