A cone used as an oil funnel has a radius of 5 in. and a height of 10 in. What volume of oil could it hold if the opening at the bottom is plugged?
Answers
Answered by
0
Answer:
part =
2
18
=9 cm
Radius (r) of lower circular end of frustum part = Radius of circular end of cylindrical part =
2
8
=4 cm
Height (h
1
) of frustum part =22−10=12 cm
Height (h
2
) of cylindrical part =10 cm
Slant height (l) of frustum part =
(R−r)
2
+h
1
2
=
(9−4)
2
+12
2
=
25+144
=
169
=13 cm
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
=π(R+r)l+2πrh
2
=
7
22
×(9+4)×13+2×
7
22
×4×10
=
7
22
[169+80]=
7
22×249
=782
7
4
cm
2
Similar questions