Two circles, each of radius 7 cm,
intersect each other. The distance
between their centres is 7 root 2 cm. Find
the area common to both the circles.(pls send image of the answer)
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Given:
Radius of the circle = 7
Distance between their centres = 7√2
To Find:
Common area of both circles
Solution:
In ∆ AOB ,
AB = 7cm
∠AOB = 90°
∠ABO = 45°
Now,
Sin 45° = AO/AB
= 1/√2 = AO/7
= AO = 7/√2
In ∆ ABC
OA is a bisector
BC = 7√2 cm
BO = 7√2/2
= 7\√2 cm
Area of ∆ AOB
= 1/2× 7/√2 × 7/√2
= 49/4 cm²
Area of sector ABO = 1/2 × r² × ∅
= 1/2 × 7² × π/4 (45° = π/4)
= 49π/8 cm²
Area of both ∆ AOB and ∆ AOC
=49/4 × 2
= 49/2cm²
Area of sector of both sides together =
= 49π\8× 2
= 49π/4cm²
Area of the overlapping region
= 2 × (49π/4 -49/2 )
= 49/2 (π-2)cm²
Answer: Area common to both the circles is 49/2 (π-2)cm²
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