Question

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

Answer 1

Step-by-step explanation:

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

Answer 2

Answer: 49/2(π-2)sq.cm

Step-by-step explanation:

∆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