A bucket open at the top has top and bottom radii of circular ends as 40 cm and 20 cm find the volume of the bucket if its depth is 21 cm and also area of tin used to make it.
Answers
The volume of the bucket if its depth is 21 cm is 61600 cm^3 and the area of tin used to make it is 10497 cm^2.
Given,
r1 = 40 cm
r2 = 20 cm
h = 21 cm.
Volume of the bucket
= 1/3 × π × h [ r1^2 + r2^2 + r1×r2 ]
= 1/3 × 22/7 × 21 [ 40^2 + 20^2 + 40 × 20 ]
= 22 [ 2800 ]
= 61600 cm^3.
Area of tin used
= π ( r1 + r2 ) l + π r1^2
where, l = √ [ h^2 + (r2 - r1)^2 ]
l = √ [ 21^2 + (20 - 40)^2 ]
l = 29
Area = π ( r1 + r2 ) l + π r1^2
= 22/7 (40 + 20) 29 + 22/7 × 40^2
= 38280/7 + 35200/7
= 73480/7
= 10497 cm^2
The volume of the bucket if its depth is 21 cm and also area of tin used to make it is 61600 cm³ and 10497 cm² respectively.
Step-by-step explanation:
Volume of the bucket is given by the formula:
V = 1/3 × π × h [ r₁² + r₂² + (r₁r₂)]
Where,
r₁ = Top radius = 40 cm
r₂ = Bottom radius = 20 cm
On substituting the values, we get,
V = 1/3 × 22/7 × 21 [40² + 20² + (40 × 20)]
V = 22 × 2800
∴ V = 61600 cm³
Area of tin used is given by the formula:
Area = (π(r₁ + r₂)l) + (πr₁²)
Where,
l = √(h² + (r₂ - r₁)²)
l = √(21² + (20 - 40)²)
∴ l = 29 cm
Now, the area becomes,
A = (22/7 × (40 + 20) × 29) + (22/7 × 40²)
A = 38280/7 + 35200/7
A = 73480/7
∴ A = 10497 cm²