Question

Two circles, each of radius 7 cm
intersect each other. The distance
between their centres is 72 cm. Find
the area of the portion common to both
the circles.​

Answer 1

Step-by-step explanation:

△ABC is isasceles right angle triangle.

Draw a perpendicular AO from A on side BC

In △ABO:

sin45

0

=

AB

AO

⇒

2

1

=

7

AD

⇒AO=

2

7

cm

∴AO is the perpendicular bisector of BC and also , BC=7

2

(given)

∴△ABO≅△AOC⇒BO=

2

1

BC⇒BO=

2

7

2

=

2

7

Also, BO=CO (cpct)

Area of △AOB= Area of △ACO=

2

1

×CO×AO

=

2

1

×

2

7

×7

2

=

4

49

For area of sector ACD:

2π angle has area →πr

2

⇒

4

π

angle has area =Area ACD=

2π

πr

2

×

4

π

=

8

πr

2

=

8

π(49)

=

8

49π

cm

2

Area of portion AOD=(

8

49π

−

4

49

)cm

2

=

4

49

(

2

π

−1)cm

2

Total common area =4× Area of AOD

(Due to symmetricity)

=4×

4

49

(

2

π

−1)

=49(

2

π

−1)

=

2

49

(π−2)cm

2