Calculate the volume of a sheet required to develop the cone. If the values of radius, Height, slant height (length) are 6, 9, 12 respectively, the thickness of the metal sheet is 2mm.
Answers
Step-by-step explanation:
Answer
Let ABCD be the bucket which is frustum of a cone with vertex 0 .
Let ON=xcm
△OAB=△OMC
3+x
x
=
10
6
[since
OM
ON
=
MC
NB
]
=>
3+x
x
=
5
3
=>5x=3(3+x)
=>5x=9+3x
=>x=
2
9
Therefore,
ON=
2
9
cm and OM=
2
9
+3=
2
15
cm
=7.5
Therefore
The height of the cone = 7.5cm
Volume of the bucket=
3
1
π×10
2
×7.5−
3
1
π×6
2
×
2
9
[Volume of the large cone Volume of the small cone]
=
3
1
π(750−162)cm
3
=196πcm
3
Slant height of cone of radius 10cm
10
2
+7.5
2
cm
=
156.25
cm
=12.5cm
Slant height of cone of radius 6cm
2
9
2
+(6)
2
cm
=
4
81
+36
cm
=
2
15
cm
Therefore,
Slant height of bucket=(12.5−
2
15
)cm
=
2
10
cm
i.el=5cm
The area of the metal sheet=π×l(R+r)+πr
2
=π×5(10+6)+π6
2
cm
2
=80π+36πcm
2
=116πcm
2