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.

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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step-by-step explanation:

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