Interior of a building is in the form of a cylinder of diameter 4.3 metre and height 3.8 mounted by a cone whose vertical angle is right angle find the area of the surface and volume of the building
Answers
Step-by-step explanation:
Given Interior of a building is in the form of a cylinder of diameter 4.3 metre and height 3.8 mounted by a cone whose vertical angle is right angle find the area of the surface and volume of the building
Radius of cone is equal to radius of cylinder.
= 4.3 / 2
= 2.15 m
Now in triangle ABC = 45 degree
So tan 45 = BC / AB
1 = 2.15 / AB
AB = 2.15
So height of the cone is 2.15 m.
Volume of building = volume of cylinder + volume of cone.
= π (2.15)^2 x 3.8 + 1/3 π (2.15)^2 x 2.15
= π (17.5655 + 3.3127)
= 3.14 x 20.8782
= 65.56 m^3
Now slant height of the cone is l = √2.15^2 + 2.15^2
= 3.04 m
Now surface area of the building = surface area of cylinder + surface area of cone
= 2 π x 2.15 x 3.8 + π x 2.15 x 3.04
= π (16.34 + 6.536)
= 3.14 x 22.876
= 71.83 sq m
Reference link will be
https://brainly.in/question/2365792
Answer
r 1=4.3/2
m=2.15m
Radius of base of the cone =r2
=2.15m
Height of the cylinder h1
=3.8m
In △VOA we have
sin45 o = VA/OA
⇒
21 = VA2.15
⇒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