Answer:
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse, AC =
√
3
2
+
4
=
9
16
25
5
cm
Area of
Δ
A
B
C
1
×
⇒
D
6
So, DB =
12
2.4
The volume of double cone = Volume of cone 1 + Volume of cone 2
π
r
h
[
]
3.14
30.14
c
m
The surface area of double cone = Surface area of cone 1 + Surface area of cone 2
l
7
52.75
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Answers & Comments
Answer:
The double cone so formed by revolving this right-angled triangle ABC about its hypotenuse is shown in the figure.
Hypotenuse, AC =
√
3
2
+
4
2
=
√
9
+
16
=
√
25
=
5
cm
Area of
Δ
A
B
C
=
1
2
×
A
B
×
A
C
⇒
1
2
×
A
C
×
D
B
=
1
2
×
4
×
3
⇒
1
2
×
5
×
D
B
=
6
So, DB =
12
5
=
2.4
cm
The volume of double cone = Volume of cone 1 + Volume of cone 2
=
1
3
π
r
2
h
1
+
1
3
π
r
2
h
2
=
1
3
π
r
2
[
h
1
+
h
2
]
=
1
3
π
r
2
[
D
A
+
D
C
]
=
1
3
×
3.14
×
2.4
2
×
5
=
30.14
c
m
3
The surface area of double cone = Surface area of cone 1 + Surface area of cone 2
=
π
r
l
1
+
π
r
l
2
=
π
r
[
4
+
3
]
=
3.14
×
2.4
×
7
=
52.75
cm
2