Two circles, of each of radius 7cm, intersect each other.The distance between the centre is 7√2 cm.Find the area common to both the circles.
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Thus by solving this, we can prove it.
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Step-by-step explanation:
∆ 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
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@GauravSaxena01
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