1) Find the area of shaded area if the radius if circle is $\mathit{r}$.



Solution
Area of shaded region= Area of circle - area of square.

Area of circle = $\mathit{\pi }{\mathit{r}}^2$

Let $\mathit{a}$ be the sides of the square.

therefore ${\mathit{a}}^2+{\mathit{a}}^2={\mathit{r}}^2$
			$2{\mathit{a}}^2={\mathit{r}}^2$
			${\mathit{a}}^2=\frac{{\mathit{r}}^2}{2}$
Area of square= ${\mathit{a}}^2$ = $\frac{{\mathit{r}}^2}{2}$

Area of shaded region= ($\mathit{\pi }{\mathit{r}}^2$ - $\frac{{\mathit{r}}^2}{2}$).
					 = $\frac{\left(2\mathit{\pi }-1\right)}{2}{\mathit{r}}^2$.



2) Four equal circle of radius $\mathit{r}$ is arrange as shown in figure. Find the area of shaded area.

Solution

Join the center to the circle to form a square of sides $2\mathit{r}$.

Now the area of the shaded region = Area of square - 4 × area of minor sector
Area of square= side × side = $2\mathit{r}$×$2\mathit{r}$ = $4{\mathit{r}}^2$
Area of minor sector  = $\left(\frac{\mathit{\theta }}{360}\right)$ ×π ${\mathit{r}}^2$ 
					= $\left(\frac{90}{360}\right)$ ×π ${\mathit{r}}^2$
					= $\left(\frac{1}{4}\right)$ ×π ${\mathit{r}}^2$ 

Area of the shaded region = $4{\mathit{r}}^2$- 4 × $\left(\frac{1}{4}\right)$π ${\mathit{r}}^2$
Area of the shaded region = $4{\mathit{r}}^2$-π ${\mathit{r}}^2$
Ans:Area of the shaded region = $\left(4-\mathit{\pi }\right){\mathit{r}}^2$ 


3) Three equal circle of radius $\mathit{r}$ is arrange as shown in figure. Find the area of shaded area.

Solution
Join the center of the circle to form a equilateral triangle of sides $2\mathit{r}$
Area of shaded region= Area of triangle- 3 × area of minor sector 
Area of triangle= $\mathit{s}\left(\mathit{s}-\mathit{a}\right)\left(\mathit{s}-\mathit{b}\right)\left(\mathit{s}-\mathit{c}\right)$
where $\mathit{s}=\frac{\left(\mathit{a}+\mathit{b}+\mathit{c}\right)}{2}$
Since in equilateral trianlge $\mathit{a}=\mathit{b}=\mathit{c}$
Therefore  $\mathit{s}=\frac{\left(\mathit{a}+\mathit{a}+\mathit{a}\right)}{2}=\frac{3}{2}\mathit{a}$


Therefore  Area of equilateral triangle= $\sqrt{\mathit{s}{\left(\mathit{s}-\mathit{a}\right)}^3}$
								= $\sqrt{\frac{3}{2}\mathit{a}\times {\left(\frac{3}{2}\mathit{a}-\mathit{a}\right)}^3}$
								= $\sqrt{\frac{3}{2}\mathit{a}\times {\left(\frac{1}{2}\mathit{a}\right)}^3}$
								= $\sqrt{\frac{3}{16}\times {\mathit{a}}^4}$
								= $\frac{\sqrt{3}}{4}{\mathit{a}}^2$
Since the sides of the triangle is $2\mathit{r}$
Therefore Area of equilateral triangle = $\frac{\sqrt{3}}{4}{\left(2\mathit{r}\right)}^2$ = $\sqrt{3}{\mathit{r}}^2$                     

area of minor sector=$\left(\frac{\mathit{\theta }}{360}\right)$ ×π ${\mathit{r}}^2$ 
 				   =$\left(\frac{60}{360}\right)$ ×π ${\mathit{r}}^2$ 
 				   =$\left(\frac{1}{6}\right)$ ×π ${\mathit{r}}^2$ 
Area of shaded region= $\left(\sqrt{3}{\mathit{r}}^2-3\times \frac{1}{6}\times \mathit{\pi }{\mathit{r}}^2\right)$
Area of shaded region= $\left(\sqrt{3}{\mathit{r}}^2-\frac{1}{2}\times \mathit{\pi }{\mathit{r}}^2\right)$

4) Two tangent drawn from a point P and Q from a cirlce of radius $\mathit{r}$ make a angle of ${60}^0$  at T as shown in figure. 
The length of the tangent is $\mathit{a}$.Find the area enclosed as shown in the figure. 

Solution
Join Point of contact of tangent i.e P & Q with the center O, i.e OP and OQ 
Since ∠PTQ = ${60}^0$, so ∠POQ=(180-60)= ${120}^0$ 

Now, the area of shaded region = area of quadilateral(TPOQ) -area of sector (OPQR)  
Area of quadilateral(TPOQ) = area of triangle (TPQ) + area of triangle (OPQ) 

In  ∆TPS , $\mathit{s}\mathit{i}\mathit{n}\left({30}^0\right)$ =$\frac{\mathit{P}\mathit{S}}{\mathit{T}\mathit{P}}$
			$\frac{1}{2}$ = $\frac{\mathit{P}\mathit{S}}{\mathit{a}}$
			$\frac{\mathit{a}}{2}$ = $\mathit{P}\mathit{S}$
	Therefore PQ=2PS=$\mathit{a}$
			
    Similiarly, $\mathit{c}\mathit{o}\mathit{s}\left({30}^0\right)$ = $\frac{\mathit{T}\mathit{S}}{\mathit{T}\mathit{P}}$
				$\frac{\sqrt{3}}{2}$ = $\frac{\mathit{T}\mathit{S}}{\mathit{a}}$
				$\frac{\sqrt{3}\mathit{a}}{2}$ = $\mathit{T}\mathit{S}$



	area of triangle (TPQ) = $\frac{1}{2}\times \left(\mathit{P}\mathit{Q}\right)\times \left(\mathit{T}\mathit{S}\right)$
							=$\frac{1}{2}\times \mathit{a}\times \frac{\sqrt{3}\mathit{a}}{2}$
							=$\frac{\sqrt{3}}{4}{\mathit{a}}^2$
	
						
In  ∆OPS, $\mathit{c}\mathit{o}\mathit{s}\left({60}^0\right)$ =$\frac{\mathit{O}\mathit{S}}{\mathit{O}\mathit{P}}$
				$\frac{1}{2}$ =$\frac{\mathit{O}\mathit{S}}{\mathit{r}}$
				$\frac{\mathit{r}}{2}$ =$\mathit{O}\mathit{S}$
				
	Area of ∆OPQ = $\frac{1}{2}\times \left(\mathit{P}\mathit{Q}\right)\times \left(\mathit{O}\mathit{S}\right)$
				 = $\frac{1}{2}\times \left(\mathit{a}\right)\times \left(\frac{\mathit{r}}{2}\right)$
				 = $\frac{1}{4}\mathit{a}\mathit{r}$

				 
Area of quadilateral(TPOQ) = area of triangle (TPQ) + area of triangle (OPQ)
						   = $\frac{\sqrt{3}}{4}{\mathit{a}}^2$ + $\frac{1}{4}\mathit{a}\mathit{r}$


						   
area of sector (OPQR) = $\left(\frac{\mathit{\theta }}{360}\right)$ ×π ${\mathit{r}}^2$ 
					  = $\left(\frac{120}{360}\right)$ ×π ${\mathit{r}}^2$
					  = $\frac{\mathit{\pi }{\mathit{r}}^2}{3}$
					  


					  
Ans: Area of shade area = $\frac{\sqrt{3}}{4}{\mathit{a}}^2$ + $\frac{1}{4}\mathit{a}\mathit{r}$  -  $\frac{\mathit{\pi }{\mathit{r}}^2}{3}$ 

				
5) Two 1 rupees coins and two 50 paisa coins are arrange as shown in figure. 
Find the area of shaded area if the radius of 1 Rs coin is 5 cm and 50 paisa coin is 4 cm . 


Solution