Question

A bucket open at the top has top and bottom radii of circular ends as
40 cm and 20 cm respectively. Find the volume of the bucket if its depth
is 21 cm. Also find the area of the tin sheet required for making the
bucket. Usepi=22/7​

Answer 1

Step-by-step explanation:

r1=40

r2=20

h=21

volume of bucket=1/3*πh(r1^2+r2^2+r1r2)

1/3*22/7*21*1600+400+800

volume of bucket=36400cm^2

Answer 2

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Let the radius of the top part be 'R' and radius of the bottom part be 'r' respectively.

Diameter (Top) = 40 cm

Radius (R) = 40/2 = 20 cm

10 cm = 1 dm

20 cm = 2dm

Diameter (Bottom) = 20 cm

Radius (r) = 20/2 = 10 cm

10 cm = 1 dm

Height = 12 cm = 1.2 dm

Slant height ('l') = √H² + (R - r)²

= √(1.2)² + (2 - 1)²

= √1.44 + 1

= √2.44

= 1.56

So, slant height (l) is 1.56 dm

Area of the tin sheet used for bucket = Curved surface area of the bucket + Area of the base

= π(R + r)*l + πr²

⇒ 3.14*(2 + 1)*1.56 + 3.14*1²

⇒(3.14*3*1.56) + 3.14

⇒ 14.6952 + 3.14

⇒ 17.8352 sq dm

CSA of the tin sheet used = 17.8352 sq dm

Rate of tin sheet = Rs. 1.2 per sq dm

Total cost of the tin sheet used for making the bucket

= 17.8352*1.2

= Rs. 21.40

Volume = 1/3πh(R² + r² + R*r)

= 1/3*3.14*1.2*(2² + 1² + 2*1)

= 1/3*3.14*1.2*7

= 26.376/3

Volume = 8.792 cu dm

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