The height of a right circular cone is trisected by two planes drawn parallel to the base.
Show that the volumes of the three portions starting from the top are in the ratio 1:7:19.
Answers
Step-by-step explanation:
ANSWER
Let VAB be a right circular cone of height 3h and base radius r.
This cone is cut by planes parallel to its base at points O
′
and L such that VL=LO
′
=h
Since triangles VOA And VO
′
A
′
are similar
∴
VO
VO
′
=
OA
O
′
A
′
⇒
r
1
r
=
2h
3h
⇒r
1
=
3
2r
Also △VOA∼△VLC
∴
VL
VO
=
LC
OA
⇒
h
3h
=
r
2
r
⇒r
2
=
3
r
Let V
1
be the volume of the cone VCD. Then
V
1
=
3
1
πr
2
2
h=
3
1
π(
3
r
)
2
h=
27
1
πr
2
h
Let V
2
be the volume of the frustum A
′
B
′
D
′
C. Then
V
2
=
3
1
π(r
1
2
+r
2
2
+r
1
r
2
)h=
3
1
π(
9
4r
2
+
9
4r
2
+
9
2r
2
)h
⇒V
2
=
27
7
πr
2
h
Let V
3
be the volume of the frustum AB
′
A
′
. then,
V
3
=
3
1
π(r
1
2
+r
2
+r
1
r
)h=
3
1
(r
2
+
9
4r
2
+
3
2r
2
)h⇒V
3
=
27
19π
r
2
h
Required ratio V
1
:V
2
:V
3
=
27
1
πr
2
h:
27
7
πr
2
h:
27
19π
r
2
h
∴ V
1
:V
2
:V
3
=1:7:19 [Hence proved]
Answer:
Let VAB be a right circular cone of height 3h and base radius r.
This cone is cut by planes parallel to its base at points O
′
and L such that VL=LO
′
=h
Since triangles VOA And VO
′
A
′
are similar
∴
VO
VO
′
=
OA
O
′
A
′
⇒
r
1
r
=
2h
3h
⇒r
1
=
3
2r
Also △VOA∼△VLC
∴
VL
VO
=
LC
OA
⇒
h
3h
=
r
2
r
⇒r
2
=
3
r
Let V
1
be the volume of the cone VCD. Then
V
1
=
3
1
πr
2
2
h=
3
1
π(
3
r
)
2
h=
27
1
πr
2
h
Let V
2
be the volume of the frustum A
′
B
′
D
′
C. Then
V
2
=
3
1
π(r
1
2
+r
2
2
+r
1
r
2
)h=
3
1
π(
9
4r
2
+
9
4r
2
+
9
2r
2
)h
⇒V
2
=
27
7
πr
2
h
Let V
3
be the volume of the frustum AB
′
A
′
. then,
V
3
=
3
1
π(r
1
2
+r
2
+r
1
r
)h=
3
1
(r
2
+
9
4r
2
+
3
2r
2
)h⇒V
3
=
27
19π
r
2
h
Required ratio V
1
:V
2
:V
3
=
27
1
πr
2
h:
27
7
πr
2
h:
27
19π
r
2
h
∴ V
1
:V
2
:V
3
=1:7:19 [Hence proved