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Answer :
Here R = 20 cm, r = 8 cm and h = 16 cm
Capacity of the container = Volume of the frustum \(= \ \frac{1}{3} \pi h(R^2 \ + \ r^2 \ + \ Rr) \)
\( = \ \frac{1}{3} \ × \ 3.14 \ × \ 16(20^2 \ + \ 8^2 \ + \ 20 \ × \ 8) \ \)
\( = \ \frac{50.24}{3} \ × \ (400 \ + \ 64 \ + \ 160) \)
\(= \ \frac{50.24}{3} \ × \ 624 \ \)
\( = \ 50.24 \ × \ 208 \)
\(= \ \frac{10449.92}{1000} \) litres
Cost of milk @ Rs 20 per litre \( = \ 20 \ × \ \frac{10449.92}{1000} \)
\( = \ 208.99 \) \( \approx \) Rs \(209 \)
To find the slant height
\( l \ = \ \sqrt{16^2 \ + \ 12^2} \ \)
\( = \ \sqrt{256 \ + \ 144} \)
\( = \ \sqrt{400} \ \)
\( = \ 20 \) cm
Curved Surface Area of the container = \( \pi (R \ + \ r)l \)
\( = \ \frac{22}{7} \ × \ (20 \ + \ 8) \ × \ 20 \)
\( = \ \frac{22}{7} \ × \ 28 \ × \ 20 \ = \ 1758.4 \) cm2
Area of the bottom of the container = \( \pi r^2 \)
\( = \ \frac{22}{7} \ × \ (8)^2 \ = \ \frac{22}{7} \ × \ 64 \ = \ 200.96 \) cm2
\(\therefore \) Total area of metal required \(= 1758.4 \ + \ 200.96 \ = \ 1959.36 \) cm2
Cost of metal sheet used to manufacture the container @ Rs 8 per 100 cm2 \(= \ \frac{8}{100} \ × \ 1959.36 \ = \) Rs \(156.75 \)