📖 generic · CBSE Class 10 ENGLISH MEDIUM · MATHEMATICS · Page 1example

V OLUMES · Part 4

Chapter 12: SURFACE AREAS AND VOLUMES · MATHEMATICS

cylinder, but the base of the cone is larger than the base of the cylinder. So, a part of the base of the cone (a ring) is to be painted. So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder =  rl +  r –  r  =  [( . × .

× . cm = . cm Now, the area to be painted yellow = CSA of the cylinder + area of one base of the cylinder =  r  h  +  ( r  ) =  r  ( h  + r  ) = ( . × .

cm Fig. . Example : Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end (see Fig. .

). The height of the cylinder is . m and its radius is cm. Find the total surface area of the bird-bath.

(Take  = Solution : Let h be height of the cylinder, and r the common radius of the cylinder and hemisphere. Then, the total surface area of the bird-bath = CSA of cylinder + CSA of hemisphere =  rh +  r =  r ( h + r ) ( ) cm = 33000 cm = . m EXERCISE . Unless stated otherwise, take  =  .

cubes each of volume cm are joined end to end. Find the surface area of the resulting cuboid. . A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder.

The diameter of the hemisphere is cm and the total height of the vessel is cm. Find the inner surface

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