The interior of a building is in the form of a cylinder of diameter 4.3m and height 3.8m, surmounted by
a cone whose vertical angle is the right angle.
Answers
Answer:
The interior of a building is in the form of cylinder of diameter 4.3 m and height 3.8 m, surmounted by a cone whose vertical angle is a right angle. Find the area of the surface and the volume of the building.
Step-by-step explanation:
ANSWER
We have
Radius of the base of the cylinder r
1
=
2
4.3
m=2.15m
Radius of base of the cone =r
1
=2.15m
Height of the cylinder h
1
=3.8m
In △VOA we have
sin45
o
=
VA
OA
⇒
2
1
=
VA
2.15
⇒VA=(
2
×2.15)m=3.04m
Clearly △VOA is an isosceles triangle
Therefore, VO=OA=2.15m
Thus, we have
height of the cone =h
2
=VO=2.15m
Slant height of the ocne l
2
=VA=3.04m
Surface area of the building = Surface area of the cylinder + Surface area of cone
=(2πr
1
h
1
+πr
2
l
2
)m
2
=(2πr
1
h
1
+πr
1
l
2
)m
2
=πr
1
(2h
1
+l
2
)m
2
=3.14×2.15×(2×3.8+3.04)m
2
=3.14×2.15×10.64m
2
=71.83m
2
Volume of the building = volume of the cylinder + volume of the cone
=(πr
1
2
h
1
+
3
1
πr
2
2
h
2
)m
3
=(πr
1
2
h
1
+
3
1
πr
1
2
h
2
)m
3
[∵r
2
=r
1
]
=πr
1
2
(h
1
+
3
1
h
2
)m
3
=3.14×2.15×2.15×(3.8+
3
2.15
)m
3
=65.55m
3