Solution

△ ABC is isosceles right-angle triangle. Draw a perpendicular AO from A on side BC In ΔΑΒΟ :   sin 45 = AO/AB 1/√2 =AO/7 AO =7/√2 cm AO is the perpendicular bisector of BC and, BC = 7√2 given triangle ABO = triangle AOC   BO= (1 /2) × BC     BO = 7√2/2 = 7/√2   Also, BO = CO (cpct) Area of Triangle AOB = Area of triangle ACO = 1/2 × CO×AO = 1/2 ×7/√2 ×7√2   = 49/4 For the area of sector ACD: 2 π angle has area → лг²   π /4 angle has area =Area ACD = (πr² / 2π) × π /4 = πr²/8  (put the value r). = 49 π /8   Area of portion AOD = (49 π /8-49/4) cm² = 49/4(π /2-1) cm²   Total common area = 4× Area of AOD (Due to symmetricity) = 4 x 49(π /2-1) = 49(π /2-1) = 49/2 (π – 2) cm² .