Q. In the figure ab is the diameter of the circle with area pi sq. units another circle is drawn with c as centre which in on the given circle and passing through a and b find area of the shaded region.

Dear Student,

Please find below the solution to the asked query:

We form our diagram , As :

From theorem " The line joining the two points where two circle intersect is perpendicular to the line joining their centers . "  , So

CO is perpendicular on AB and we know if a perpendicular drawn from center to any chord that is also bisects that chord , So

AO  =  BO 

Given : Area of circle with diameter AB and ( We show center ' O ' ) is $\mathrm{\pi }$ square unit 

We know area of circle  = $\mathrm{\pi }$ r²  , Here r  = radius 

So,

$\mathrm{\pi }$ r²  = $\mathrm{\pi }$ ,

r²  = 1 ,

r = 1 unit , So

AO  =  CO  =  BO  =  1 unit                                                 --- ( 1 )

Now we apply Pythagoras theorem in triangle AOC and get :

AO² + CO² = AC²  , Substitute values from equation 1 and get :

1² + 1² = AC² ,

AC² = 2  ,

AC = $\sqrt{2}$ unit , That is radius of circle with center ' C '

We know angle in semicircle is always at 90$°$ , So

 $\angle$ ACB  = 90$°$
We know :  area of sector of circle  = $\frac{\mathrm{Center} \mathrm{angle}}{360°}\times \mathrm{\pi }{r}^2$ , So

Area of sector CAB  = $\frac{\angle \mathrm{ACB}}{360°}\times \mathrm{\pi }\times {\left(\sqrt{2}\right)}^2 = \frac{90°}{360°}\times \mathrm{\pi }\times 2= \frac{1}{4}\times \mathrm{\pi }\times 2= \frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{ }\mathbf{square}\mathbf{ }\mathbf{unit}$

As given : $\angle$ ACB  = 90$°$ , So we can find area of triangle ACB by using formula of triangle  = $\frac{1}{2}\times \mathrm{Base} \times \mathrm{height}$ , So

Area of triangle ACB  = $\frac{1}{2}\times \mathrm{AC} \times \mathrm{BC} = \frac{1}{2}\times \sqrt{ 2} \times \sqrt{ 2}=\frac{1}{2}\times 2= \mathbf{1}\mathbf{ }\mathbf{square}\mathbf{ }\mathbf{unit}$

Then ,

Area of small segment of circle with center ' c ' = Area of sector CAB  - Area of triangle ACB = $\frac{\mathrm{\pi }}{2}$ - 1 square unit

And area of semicircle  with diameter AB = $\frac{\mathrm{\pi }}{2}$ , 


Therefore,

Area of shaded region = Area of semicircle with diameter AB - Area of small segment of circle with center ' c '= $\frac{\mathrm{\pi }}{2}$ - ( $\frac{\mathrm{\pi }}{2}$ - 1 ) = $\frac{\mathrm{\pi }}{2}$ - $\frac{\mathrm{\pi }}{2}$ + 1 = 1 square unit                    ( Ans )

Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards