Question

Two circles, each of radius 7 cm,
intersect each other. The distance
between their centres is 7\sqrt{2} cm. Find the area of the portion common to both the circles.​

Answer 1

Hey There!!!

∆ ABC is an O scales and 45° - 45° - 90°

Triangle drop a perpendicular OA the the side BS is ∆ ABC

In ∆ AOB ,

< AOB = 90° , < ABO = 45° , AB = 7cm

sin 45° = AO/AB =

1/√2 = AO/7 (:- AO = 7/√2 cm)

OA is an bisector. in ∆ ABC

BC = 7√2 cm

BO = 7√2/2 = 7\√2 cm

Area of ∆ AOB= 1/2× 7/√2 × 7/×2

=> 49/4 cm^2

=> Area of sector ABO = 1/2 ×r^2× theeta = 1/2 × 7² × π/4 (:- 44° = π/4)

=> 49π/8 cm²

Area of ∆ AOB and ∆ AOC combined =49/4×2 = 49/2cm²

=> Area of sector of both sides combined 49π\8× 2 = 49π/4cm²

=> Area of overlap region = 2×(49π/4 -49/2 )cm²

=> the area common to both the circles is .49/2 (π-2)cm²