The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 60o. If the area of the quadrilateral is 4√3, then the perimeter of the quadrilateral is :

Area(ΔABCD)=4√3-given=A(ΔABC)+A(ΔADC)InΔABC,Area=12xABΔBCsinB=12×2×5×sin60o=5√32∴Area(ΔADC=4√2󔼼√3=32√3 (I)InΔADC,Area=12×AD×CD×sin120o32√x=1x×c×d×√32from (i) cd=6. (ii)InΔABC,cosB=22+52−AC22×2×5..cosive rule12=4+25−AC220(AC)2=19. (iii)InΔADC,cos120o=c2+d2−(AC)22cd...cosive rule𕒵2=c2d2𕒵92×6(iii)−(AC)2(ii)−cdc2+d2=13. (iv)(iv+2(ii)=c2+d2+2cd=13+2×6=13+12=25(c+d)2=25⇒Perimeter□ABCD=2+5+c+d=7+5=12