1) Find the area of shaded area if the radius if circle is $r$.



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

Area of circle = $πr^2$

Let $a$ be the sides of the square.

therefore $a^2+a^2= r^2$
			$2a^2=r^2$
			$a^2=r^2/2$
Area of square= $a^2$ = $r^2/2$

Area of shaded region= ($πr^2$ - $r^2/2$).
					 = $(2π-1)/2r^2$.



2) Four equal circle of radius $r$ 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 $2r$.

Now the area of the shaded region = Area of square - 4 × area of minor sector
Area of square= side × side = $2r$×$2r$ = $4r^2$
Area of minor sector  = $(θ/360)$ ×π $r^2$ 
					= $(90/360)$ ×π $r^2$
					= $(1/4)$ ×π $r^2$ 

Area of the shaded region = $4r^2$- 4 × $(1/4)$π $r^2$
Area of the shaded region = $4r^2$-π $r^2$
Ans:Area of the shaded region = $(4-π)r^2$ 


3) Three equal circle of radius $r$ 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 $2r$
Area of shaded region= Area of triangle- 3 × area of minor sector 
Area of triangle= $ {s(s-a)(s-b)(s-c)}$
where $s=(a+b+c)/2$
Since in equilateral trianlge $a=b=c$
Therefore  $s=(a+a+a)/2 = 3/2a$


Therefore  Area of equilateral triangle= $√{s(s-a)^3}$
								= $√{3/2a×(3/2a-a)^3}$
								= $√{3/2a×(1/2a)^3}$
								= $√{3/16 ×a^4}$
								= $√3/4a^2$
Since the sides of the triangle is $2r$
Therefore Area of equilateral triangle = $√3/4(2r)^2$ = $√3r^2$                     

area of minor sector=$(θ/360)$ ×π $r^2$ 
 				   =$(60/360)$ ×π $r^2$ 
 				   =$(1/6)$ ×π $r^2$ 
Area of shaded region= $(√3r^2 - 3×1/6 ×π r^2)$
Area of shaded region= $(√3r^2 - 1/2 ×π r^2)$

4) Two tangent drawn from a point P and Q from a cirlce of radius $r$ make a angle of $60^0$  at T as shown in figure. 
The length of the tangent is $a$.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 , $sin(30^0)$ =${PS}/{TP}$
			$1/2$ = ${PS}/{a}$
			$a/2$ = ${PS}$
	Therefore PQ=2PS=$a$
			
    Similiarly, $cos(30^0)$ = ${TS}/{TP}$
				$√3/2$ = ${TS}/{a}$
				${√3a}/2$ = ${TS}$



	area of triangle (TPQ) = $1/2 ×(PQ)×(TS) $
							=$1/2 × a × {√3a}/2 $
							=$√3/4a^2$
	
						
In  ∆OPS, $cos(60^0)$ =${OS}/{OP}$
				$1/2$ =${OS}/{r}$
				$r/2$ =${OS}$
				
	Area of ∆OPQ = $1/2 ×(PQ)×(OS) $
				 = $1/2 ×(a)×(r/2) $
				 = $1/4ar $

				 
Area of quadilateral(TPOQ) = area of triangle (TPQ) + area of triangle (OPQ)
						   = $√3/4a^2$ + $1/4ar $


						   
area of sector (OPQR) = $(θ/360)$ ×π $r^2$ 
					  = $(120/360)$ ×π $r^2$
					  = ${πr^2}/3$
					  


					  
Ans: Area of shade area = $√3/4a^2$ + $1/4ar $  -  ${πr^2}/3$ 

				
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