Problems on finding surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones. Frustum of a cone.

    In our day-to-day life we come across various solids which are combinations of two or more such solids. 
    We shall discuss problems on finding surface areas and volumes of such solids. 
    Before we proceed further let us recall the formulas for surface area and volume of some of the basic solids:

Right circular cylinder: For a right circular cylinder of base radius r and height h, we have -


Right circular cone: For a right circular cone of height h, slant height 
    l and radius of base r, we have,
    (i)     l2 = r2 + h2
    (ii)     Curved surface area = p r l sq units                    
    (iii)     Total surface area
            = Curved surface area + area of the base





Illustration 1
     The dimensions of a metallic cuboid are 100 cm x 80 cm ´ 64 cm. It is melted and recast into a cube. Find the surface area of the cube.
            We have 
            Volume of the metallic cuboid = 100  x 80 x 64 cm3 = 512000 cm3.
            Since the metallic cuboid is melted and is recast into a cube.
            Volume of the metallic cuboid = Volume of the cube.
            Let the length of each edge of the recasted cube be a cm, then,
            Volume of cube = Volume of the cuboid
            a3 = 512000
            a3 = 83 x 103 
            a = 8 x 10 cm = 80 cm

             Surface area of the cube = 6a2  cm2  = 6 x  (80)2 cm2 = 38400 cm2

 Illustration 2
            The radii of the bases of two right circular solid cones of same height are r1 and  r2 respectively. The cones are melted and recast into a solid sphere of radius R.

Problems involving converting one type of metallic solid into another and other mixed problems.

Surface area and volume of combination of solids      
    In this section we shall find the surface area and volume of solids which are combination of two or more solids.
    Consider a circus tent shown in the given figure.
    It consists of two parts I and II. Part-I is in the shape of a cone and Part-II is in the form of a cylinder.                                                               
    Total surface area of the circus tent = Curved surface area of part-I i.e., cone + Curved Surface area of part-II i.e., Cylinder.

Illustration 6
   A solid is composed of a cylinder with hemispherical ends. If the whole length of the  solid is 104 cm and the radius of each hemispherical end is 7         cm, find the

Solution : 
      The whole length of the solid = 104 cm
      The radius of each hemisphere = 7 cm
      Therefore, the length of the cylindrical part of the solid = (104 – 2 x 7) cm = 90 cm  Now, the total surface area of the solid 
     = 2 x (curved surface area of hemisphere) + (curved surface area of cylindrical part)

Volume of Combination of Solids :
 In the previous section, we have studied about the surface area of solids made of two or more solids. There we find that while calculating the surface area of a solid, some surface areas are not included. But, here we will find the total volume of a solid which is also the actual volume of two or more combined solid.

Illustration 7
 A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take p = 3.14)

Solution : 
Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere

Conversion of Solid from One Shape to Another
    Some solids like candle, clay etc. can be changed into any shape. But the volume of the both solid shapes are same. For example, if a candle which is generally in the shape of a cylinder can be changed into any shape, but the volume remains same.
    If a solid is transformed into a number of small identical solids of same or a different shape, then