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A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs 8 per 100 cm2.

Answer 1

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Answer 2

Given:

r1 = 20cm,

r2 = 8cm,

h = 16cm

Slant height =

$\longrightarrow$$\sf\red{l = \sqrt{ {h}^{2} + {(r1 - r2)}^{2} }}$

$\longrightarrow$$\sf\red{ \sqrt{ {(16)}^{2} + {(20 - 8)}^{2} } cm}$

$\longrightarrow$$\sf\red{\sqrt{256 + 144} cm}$

$\longrightarrow$$\sf\red{\sqrt{400} cm}$

$\longrightarrow$$\sf\red{20cm}$

Volume of the container =

$\longrightarrow$$\sf\red{v = \frac{1}{3} \pi \: h( {r1}^{2} + {r2}^{2} + r1r2}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16( {(20)}^{2} + {8}^{2} + 20 \times 8) {cm}^{3}}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16(400 + 64 + 16) {cm}^{3}}$

$\longrightarrow$$\sf\red{\frac{1}{3} \times 3.14 \times 16 \times 624 {cm}^{3}}$

$\longrightarrow$$\sf\red{10449.92 {cm}^{3}}$

$\longrightarrow$$\sf\red{ \frac{10449.92}{1000} litres = 10.45litres(approx.)}$

Cost of milk that fills the container at the rates of Rs. 20 per litre

= $\sf\red{Rs.20×10.45=Rs.209}$

TSA of container = CSA of the container + Base area

$\longrightarrow$$\sf\red{\pi(r1 + r2) l + \pi {r2}^{2}}$

$\longrightarrow$$\sf\red{(3.14 \times (20 + 8) \times 20 + 3.14 \times 8 \times 8) {cm}^{2} }$

$\longrightarrow$$\sf\red{(3.14 \times 28 \times 20 + 3.14 \times 64) {cm}^{2}}$

$\longrightarrow$$\sf\red{3.14 \times 624 {cm}^{2}}$

$\longrightarrow$$\sf\red{1959.36 {cm}^{2} }$

Cost of the metal sheet required at the rate of Rs. 8 per 100 cm square

= $\sf\red{Rs.\frac{8}{100} \times 1959.36 = Rs.156.75(approx.}$