Math, asked by narenderkumar4675, 5 months ago

A container, open from the top and made of a metal sheet, is in the form of a frustum of a cone of height 16cm with radii of its lower and upper ends as 8cm and 20cm, respectively. Find the cost (in rupees) of the milk which can completely fill the container, at the rate of Rs 20 per litre. Also, Find the cost (in rupees) of metal sheet used to make the container, if it costs Rs 8 per 100cm^2. ( Take π = 3.14)​

Answers

Answered by nehaliganvit3
1

Step-by-step explanation:

Radius of lower end = r = 8 cm Radius of upper end = R = 20 cm Height of container frustum = h = 16 cm Cost of 100 cm2 metal sheet = Rs 10 So, Cost of 1 cm2 metal sheet = 10/100 = Rs 0.1 Let l be the slant height. l2 = (R-r)2 + h2 = (20-8)2 + 162 = 256 + 144 = 400 or l = 20 cm Surface area of frustum of the cone = πr2 + π(R + r)l cm2 = π[ (20 + 8 )20 + 82] = π[560 + 64] = 624 X ( 22/7 ) = 1961.14 cm2 (i) Find cost of metal sheet used to make the container if it costs Rs. 10 per 100 cm2 Cost of metal sheet per 100 cm2 = Rs. 10 Cost of metal for Rs. 1961.14 cm2 = (1961.14 x10)/100 = Rs. 196.114 (ii) Find the volume of frustum: Volume of frustum = 1/3 πh(r2 + R2 + rR) = 1/3 x 22/7 x 16(82 + 202 + 8×20) = 1/3 x 22/7 x 16(64 + 400 + 160) = 10345.4208 cm3 = 10.345 liters Using 1000 cm3 = 1 litre 1 cm3 = 1/1000 litre Cost of 1 liter milk = Rs. 35 Cost of 10.345 liter milk = Rs. 35 x 10.345 = Rs. 363 (approx)Read more on Sarthaks.com - https://www.sarthaks.com/715393/container-made-metal-sheet-open-the-top-is-of-the-form-of-frustum-of-cone-whose-height-is-16-cm

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