I don't know how a figure is drawn in this spaceLet M be the moiipdnt of DC & N be the moiipdnt of ABJoin MNLet the point of intrsection of the 2 arcs with centreat D & C be K (it is clear that K is in MN)Required area= area of the square-8*the area of the region bounded by the line segment AN & the arc AK (say Z)ie required area = area of the square -8ZBut Z = area of the rectangle ANMD -(area of the sector ADK + area of the triangle KMD) =50-(3.14*10*10*30/360+1/2*5*5root3 =50-(314/12 +25*1.732/2 =50-(26.167+21.65) =50-47.817 =2.183ie Z=2.183Required area =100-8*2.183 =100-17.464 =82.536 sq.cm nearlyThis problem may solve using another methodie Required area = area of the region bounded by the arcs ACs with centres B & D +area of the region bounded by the arcs DBs with centres A & C - area of the common region bouned by the 4 arcsArea of the common region is calculated by using the small square with the same corner points of the common region &4parts of smallregions(small region out side of the small square=sector-triangle) MURALEEDHARAN.C.R GVHSS VATTENAD

AFAICT you've coevred all the bases with this answer!