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The Interior of a Building is in the Form of a Right Circular Cylinder of Diameter 4.2 M and Height 4 M Surmounted by a Cone of Same Diameter. the Height of the Cone is 2.8 M. Find the Outer - Mathematics

Sum

The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter.
The height of the cone is 2.8 m. Find the outer surface area of the building.

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Solution

We have,

Radius of the cylinder = Radius of the cone `= "r" = 4.2/2 = 2.1 "m",`

`"Height of the cylinder", "H" = 4  "m"` and

Height of the  cone, h = 2.8 m

Also,

The slant height of the cone, `"l" =sqrt("r"^2+"h"^2)`

`=sqrt(2.1^2 + 2.8^2)`

`=sqrt(4.41+7.84)`

`=sqrt(12.25)`

`=3.5 "m"`

Now,

The outer surface area of the building = CSA 0f the cylinder + CSA of the cone 

`=2pi"rH" + pi"rl"`

`=pir(2"H"+"l")`

`=22/7xx2.1xx(2xx4+3.5)`

`=6.6xx11.5`

= 75.9 m2

So, the outer surface area of the building is 75.9 m.

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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 19 Volume and Surface Area of Solids
Exercise | Q 33 | Page 916
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