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Two circles, each of radius 7 cm, intersect each other. the distance between their centers is 7√2 cm. find the area of the portion common to both the circles |
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Answer» ∆ABC is isasceles right ANGLE triangle Draw a perpendicular AO from A on side BC In∆ABO sin45° = AO/AB 1/√2 = AO/7 AO = 7/√2 cm AO is the perpendicular bisector of BC and ALSO,BC=7√2 ∆ABO= ∆AOC BO= 1/2 BC BO= 7√2/2 = 7/√2 Also,BO = CO AREA of ∆AOB = Area of ∆ACO = 1/2×CO×AO = 1/2×7/√2×7√2 =49/4 For area of sector ACD 2π angle has area= πr^2 π/4 angle has area = Area ACD = πr^2/2π×π/4 = πr^2/8 = π(49)/8 = 49π/8 sq.cm Area of portion AOD = [49π/8-49/4] sq.cm 49/4[π/2-1] sq.cm Total common area = 4× Area of AOD = 4×49/4[π/2-1] =49[π/2-1] =49/2[π-2] sq.cm ✌✌✌✌✌✌hope it helps u |
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