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:
