Question

class 10 maths ncert chapter 13 example 4 doubt. why dont we add base area?

Answer 1

Answer:

It is given that the hemisphere is mounted on the cuboid, thus the hemisphere can take on complete as its diameter (which is maximum).

Thus the greatest diameter of the hemisphere is 7 cm.

Now, the total surface area of solid  =   Surface area of cube  +   Surface area of the hemisphere -  Area of the base of a hemisphere (as this is counted on one side of the cube)

The surface area of the cube is :      

=\ 6a^3

=\ 6\times 7^3\ =\ 294\ cm^2

Now the area of a hemisphere is      

=\ 2\pi r^2

=\ 2\times \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 77\ cm^2

And the area of the base of a hemisphere is

=\ \pi r^2\ =\ \frac{22}{7}\times \left ( \frac{7}{2} \right )^2\ =\ 38.5\ cm^2

Hence the surface area of solid is   = 294 + 77 - 38.5 = 332.5 \:cm^2