Two circles, each of radius 7 cm,
intersect each other. The
The distance
between their centres is 7 V2 cm. Find
the area common to both the circles.
Answers
Answer:
∆ 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
MARK ME BRAINLIEST
The combined area of both intersection area is (1/8) of each circle, since the arc is 45 degrees that makes it 49*pi/8* 2or 49*pi/4 cm^2 for the combined area.