A container is in the form of a frustum of a cone of height 30cm with radii of its lower and upper ends as 10cm and 20cm respectively.find the capacity and surface area of the container. also find the cost of milk which can completely fill the cointainer at the rate of rs.25 per litre
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Volume = 1/3πh[R² + r² + R*r]
= 1/3*22/7*30*[20² + 10² + 20*10]
= 660/21*[400 + 100 + 200]
= (660*700)/21
= 22000 cm^3 or capacity = 22 liters
Now, Slant height 'l' = √(R - r)² + h²
l = √(20 - 10)² + 30²
l = √10² + 30²
l = √1000
l = 31.62 cm
Slant height is 31.62 cm
Surface area = πR² + πr² + π(R + r)l
= π[R² + r² + (R + r)*l]
= 22/7*[20² + 10² + (20 + 10)*30]
= 22/7*[400 + 100 + 30*30]
= (22*1400)/7
Surface area = 4400 sq cm
Cost of 1 liter milk = Rs. 25
Total cost of 22 liter milk = 22*25
= Rs. 550...
Thank You!
= 1/3*22/7*30*[20² + 10² + 20*10]
= 660/21*[400 + 100 + 200]
= (660*700)/21
= 22000 cm^3 or capacity = 22 liters
Now, Slant height 'l' = √(R - r)² + h²
l = √(20 - 10)² + 30²
l = √10² + 30²
l = √1000
l = 31.62 cm
Slant height is 31.62 cm
Surface area = πR² + πr² + π(R + r)l
= π[R² + r² + (R + r)*l]
= 22/7*[20² + 10² + (20 + 10)*30]
= 22/7*[400 + 100 + 30*30]
= (22*1400)/7
Surface area = 4400 sq cm
Cost of 1 liter milk = Rs. 25
Total cost of 22 liter milk = 22*25
= Rs. 550...
Thank You!
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