Question

MATHS CLASS - X

SURFACE AREAS AND VOLUMES

A hemispherical depression is cut out from one face of a cuboidal wooden block such that the diameter ${l}$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Answer 1

Given:

The diameter of the hemisphere is l .

The radius is Diameter / 2

Let r be the radius ......

==> r = l / 2

The edge of the cube is l .

We need to find the surface area after the hemisphere is cut.

Look at the figure I attached ......

We know that the surface area of a cube is $6 l^2$ where l is the edge of the cube ........

Also the surface area of the hemisphere will be counted.

Then we need to subtract the base area of the hemisphere because it has been cut out .

Surface area of the cube is :

$6 l^2$.....................................(1)

Curved surface area of the hemisphere is :

$2 \pi r^2$

But $r=\frac{l}{2}$

$\bf{Area} = 2 \pi (\frac{l}{2})^2$

$\implies 2 \pi \times \frac{l^2}{4}$

$\implies \pi \times \frac{l^2}{2}$..........................(2)

Area of the base of the hemisphere is:

$\pi \times r^2$

$\implies \pi \times \frac{l^2}{2^2}$

$\implies \frac{l^2}{4}\times \pi$ ..............................(3)

Adding (1) and (2) and then subtracting (3) will give us the answer;

Surface area of the remaining solid :-

$6 l^2 + \pi \times \frac{l^2}{2} - \pi \times \frac{l^2}{4}$

$\implies 6 l^2 + \pi l^2 ( \frac{1}{2} - \frac{1}{4} )$

$\implies 6 l^2 + \pi l^2 \times \frac{1}{4}$

$\implies l^2 ( 6 + \frac{\pi}{4})$

$\implies l^2 ( \frac{24 + \pi}{4})$

$\implies \frac{l^2}{4}( \pi + 24)$

The surface area of the remaining solid is:

$\boxed{\frac{l^2}{4}( \pi + 24) }$

Hope it helps you

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