A bucket is in form of frustrum of a cone and holds 28.490 L of water. The radii of the top and bottom are 28cm and 21cm respectively. find the height of the bucket and the area of tin used in bucket
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R = 28 cm
r = 21 cm
Let height of frustum be h cm
∴ Volume held by bucket = 28.490 litres = 28490 cm³
Also,
Volume of frustum = π(R²+Rr+r²)h/3
=> 28490 = 22*(784+588+441)*h/(3*7)
=> 1813h = 28490*21/22
=> h = 27195/1813 = 15cm
r = 21 cm
Let height of frustum be h cm
∴ Volume held by bucket = 28.490 litres = 28490 cm³
Also,
Volume of frustum = π(R²+Rr+r²)h/3
=> 28490 = 22*(784+588+441)*h/(3*7)
=> 1813h = 28490*21/22
=> h = 27195/1813 = 15cm
Answered by
1
R = 28 cm
r = 21 cm
Let height of frustum be h cm
∴ Volume held by bucket = 28.490 litres = 28490 cm³
Also,
Volume of frustum = π(R²+Rr+r²)h/3
=> 28490 = 22*(784+588+441)*h/(3*7)
=> 1813h = 28490*21/22
=> h = 27195/1813 = 15cm
r = 21 cm
Let height of frustum be h cm
∴ Volume held by bucket = 28.490 litres = 28490 cm³
Also,
Volume of frustum = π(R²+Rr+r²)h/3
=> 28490 = 22*(784+588+441)*h/(3*7)
=> 1813h = 28490*21/22
=> h = 27195/1813 = 15cm
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