Question

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​

Answer 1

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