If a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.
Answers
Answer:
The volume of two parts of the cone is 1 : 7 .
Step-by-step explanation:
SOLUTION :
Let r & R be the radius of the lower part of the frustum.
Height of a cone , AB’ = 10 cm
Height of a Smaller cone, AB = 5 cm
[Cut through the midpoint of its height]
From the figure,
AB = h = 5
AB’ = 2h = 10
BC = r
B'C = R
In ∆ABC & ∆AB’C’ ,
∠ABC = ∠AB’C’ (each 90°)
∠ACB = ∠AC’B’ (corresponding angles)
∆ABC ∼ ∆AB’C’ [By AA Similarity]
BC/B'C’ = AB/AB’
[Corresponding sides of a similar triangles are proportional]
r/R = 5 /10
r/R = ½
R = 2r
Volume of the upper part (Smaller cone) = ⅓ πr²h
Volume of solid cone = ⅓ π R²2h
= ⅓ π (2r)² 2h = ⅓ π × 4r² × 2h
Volume of solid cone = 8/3πr²h
Volume of lower part (frustum) = volume of solid cone - volume of Smaller cone
= 8/3πr²h - ⅓ πr²h = 7/3 πr²h
Volume of lower part (frustum) = 7/3 πr²h
Volume of the upper part (Smaller cone)/ Volume of lower part (frustum) = ⅓ πr²h / 7/3 πr²h
= 1/7
Hence, the ratio of volume of two parts of the cone is 1 : 7 .
HOPE THIS ANSWER WILL HELP YOU….
Answer:
Volume of cone=
3
1
πr
1
2
h
1
=
3
1
π(
2
r
)
2
×
2
h
=
3
1
π(
2
10
)
2
×
2
h
=
6
25πh
cm
3
Volume of frustum ABCD=
3
1
πh
2
(R
2
+r
2
+Rr)
=
3
1
π×
2
h
(10
2
+5
2
+10×5)
=
6
175πh
Required ratio=
6
175πh
6
25πh
=
175
25
=
7
1
- =1:7