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Surface Areas of a Combination of Solids

In our daily life we come across many combined solids like,

a cylinder over a hemisphere,

a capsule which is combination of a cylinder attached with two hemispheres on both the ends,

a toy shaped with a cone over a hemispherical bottom etc.

Some real life examples are given below for a immediate reference:
 


While calculating the surface area of the combined figures, we should only calculate the areas that are visible to our eyes. For example, if a cone is surmounted by a hemisphere and if we have to find out the total surface area of this solid, we need to just find out the curved surface area of the hemisphere and curved surface area of the cone separately and add them together. Note that we are leaving the base area of both the cone and the hemisphere in this case since both the bases are attached together and are not visible to our eyes.

 

Example

A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs.45 per m2.

Solution

Radius of the cylindrical base =  = 15 m
Height of the cylindrical portion = 5.5 m
Height of the conical portion = (8.25 – 5.5)m = 2.75m

 



Slant height (l) = 

                      =

                      =

                      = 15.25m
Surface area of the tent = Curved Surface area of the cylindrical portion + Curved      Surface area of the conical portion

                                = 2πrh + πrl

                                       = +

                                       = 

                                       =


                                 = 1237.5 m2
Therefore the cost of the canvas of the tent at the rate of Rs. 45

× 1237.5 = Rs. 55687.50.
 


 

Example

The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.

Solution

Radius of the cylindrical and conical base = 12 m
Height of the cylinder = 3.5 m
Slant height of the cone = 12.5 m
Height of the cone = 

                          =

                          =

                          = 3.5m

Internal curved surface area = Curved surface area of the cone + Curved surface area of the cylinder

                                        = πrl + 2πrh 

                                        =+

                                        =

                                        = 735.43 m2

Capacity of the building  = Volume of the cone + Volume of the cylinder

                                   = π r2h + πr2h

                                   =

                                   =

                                   =

                                   = 2112 m3.

Capacity of the building = 2112 m3

 


 

Example

A godown building is in the form as in figure. The vertical cross-section parallel to the width side of the building is a rectangle of size 7 m

× 3 m mounted by a semicircle of radius 3.5 m. The inner measurements of the cuboidal portion are 10 m × 7 m × 3 m. Find the volume of the godown and the total internal surface area excluding the floor.


          

 
Solution

Volume of the godown = Volume of the cuboid + Volume of the half cylinder

                                = lbh + × π r2h

                                = +

× 

                                = 210 + 192.5

                                = 402.5 m3

Total surface area of the godown
= Area of  the four walls + Surface area of the two semi circle + curved surface area of the half cylinder

= 2(l + b) h + (2 × π r2) +   × 2 × π rh

= [2 × 3 (10 + 7)] + [2

× × × 3.5 × 3.5] + [× 2 × × 3.5 × 10]

= 102 + 38.5 + 110
= 250.5 m2.
 


 

Example

A rocket is in the form of a cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of the radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area.

Solution

Radius of the cylindrical portion = r = 2.5 m

Height of the cylindrical portion = h = 21 m

 Surface area of the cylindrical portion = 2πrh

                                                        = 2 
× × 2.5 × 21 m2

                                                                   
= 330 m2

Radius of the conical portion = r = 2.5 m
Slant height of the conical portion = l = 8 m

 Curved surface area of the conical part = π r l
                                                            = 
× 2.5 × 8 m2
                                                           = 62.86 m2
Area of the base = 
π r2
                        = 
× 2.5 × 2.5 m2
                        = 19.64

 Total surface area of the rocket = Curved surface area of the cylindrical portion + Curved surface area of the conical portion + Area of the base
                                                = (330 + 62.86 + 19.64) m2
                                                412.5 m2





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