Solution :   Area of the square lawn ABCD = 56 × 56 $m^2$  
			Let  OA =  OB = $x$ metres
				So, $x^2$  + $x^2$  =  562
				or, 2$x^2$  =  56 × 56
				or, $x^2$  =  28 × 56                                       (2)
				
				Now, area of sector OAB   = $90/360$ ×π$x^2$= $1/4$ ×π$x^2$
				                          =$1/4$ ×$22/7$ × 28 × 56
										  
				Also, area of ∆ OAB = $1/4$ × 56 × 56 $m^2$ ( Since ∠AOB = 90°)   

So,             area of flower bed AB =  $1/4$ ×$22/7$ × 28 × 56 -	$1/4$ × 56 × 56		

                                      = $1/4$ × 28 × 56 × $8/7$ $m^2$
									  
Similarly, area of the other flower bed =$1/4$ × 28 × 56 × $8/7$ $m^2$	

Therefore,                  total area = Area of the square lawn ABCD + 2 × area of flower bed AB	

							= 56 × 56 + 2 × ($1/4$ × 28 × 56 × $8/7$) $m^2$
							
							= 4032 $m^2$
							
Alternative  Solution  :
Total area =  Area of sector OAB + Area of sector ODC + Area of ∆ OAD + Area of ∆ OBC
			=$90/360$  × $22/7$ × 28 ×56 +  $90/360$  × $22/7$ × 28 ×56 + $1/4$ ×56 ×56 + $1/4$ ×56 ×56 $m^2$
			=  56 × 72 $m^2$  
			= 4032 $m^2$


Example: Find the area of the shaded region in Fig. , where ABCD is a square of side 14 cm.


	

Solution : Area of square ABCD = 14 × 14 $cm^2$  = 196 $cm^2$
Click
	

			Diameter of each circle = $14/2$ $cm$ = 7 $cm$
			
			So,    radius of each circle = $7/2$ $cm$
			
			So,    area of one circle = $πr^2$  =$22/7$×$7/2$×$7/2$ $cm^2$
			                           =$77/2$ $cm^2$

Therefore,   area of the four circles = 4× $77/2$ $cm^2$
                                       = 154 $cm^2$

Hence,    area of the shaded region = Area of square ABCD - area of the four circles
									=(196 – 154) $cm^2$
									= 42 $cm^2$.
			
										
Example: Find the area of the shaded design , where ABCD is a square  of  side  10  cm  
			and  semicircles  are  drawn  with  each  side  of  the  square  as diameter. (Use π = 3.14)
			
			
			
Solution : Let us mark the four unshaded regions as I, II, III and IV (see Fig. ).

		Area of I + Area of III =  Area of ABCD – Areas of two semicircles of each of radius 5 cm
 
								=$({10 ×10 }-2 × 1/2 × π×5^2  )$ $cm^2$
								=	21.5 $cm^2$
								
		Similarly, Area of II + Area of IV = 21.5 $cm^2$
		
		So, area of the shaded design = Area of ABCD – Area of (I + II + III + IV)

								=  (100 – 2 × 21.5) $cm^2$  = (100 – 43) $cm^2$  
								
								= 57 $cm^2$