A bucket is 20cm in diameter at the open end, 12cm in diameter at the bottom, and 16cm deep. To what depth would the bucket full
a cylindrical tin 28cm in diameter. 
1 answer:
The bucket has a base diameter of 12 and top opening diameter of 20 . This makes the base radius 6 and top radius 10 .
The bucket as shown has the bottom base of radius:
B C = 6
and the top opening of radius:
D E = 10
The depth is:
D B = 16
If we extend the lateral surface of the bucket down it will generate a cone that contains the bucket in it.
To calculate the volume of the bucket, we need to calculate the volume of the large cone with the depth of D A , then calculate the volume of the small cone with the depth of B A , and subtract the volume of the small cone from the volume of the large cone.
Let's let B A = x .The two triangles Δ A B C and Δ A D E are similar due to the Angle Angle theorem because they both share angle ∠ A D E = 90 ∘ and angle ∠ D A E .
Therefore, the ratio of their corresponding sides are equal.
B A /D A = B C /D E
x /16 + x = 6 /10
x /16 + x = 3/5
48 + 3 x = 5 x
2 x = 48
x = 24 cm
The volume of a cone is: V = 1 /3 h π r 2
The height ( h ) , which we are calling the depth, of the small cone is x = 24 c m and the depth of the large cone is x + 16 = 24 + 16 = 40 c m
V Large Cone = 1/ 3 ( 40 ) ( π ) ( 10 ) 2 = 4000 π/3 c m 3
V Small Cone = 1 /3 ( 24 ) ( π ) ( 6 ) 2 = 288 π c m 3
V Bucket = 4000 π /3 − 288 π = 3136 π /3 c m 3
Volume of a cylinder is:
V Cylinder = π r 2 h where h is the height(depth) and r is the radius of the base of the cylinder.
If the base of the cylinder has a diameter of 28 c m then the radius of the base is 14 c m
( π ) ( 14 ) 2 ( h ) = 3136 π /3
196 π h = 3136 π/ 3
196 h= 3136 /3
h = 3136 /3 /196 = 3136 /588 = 5.33 c m
Let's look at the figure below:
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