Question

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

Answer 1

There is a minor mistake in this question,

Correct Question :

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

Solution :

$\triangle ABC$ is isasceles right angle triangle.

Draw a perpendicular AO from A on side BC

In $\triangle ABO$ :

sin 45° = AO/AB

1/√2 = AD/7

AO = 7/√2 cm.

$\therefore AO$ is the perpendicular bisector of BC and also, BC = 7√2 (given)

$\therefore \triangle ABO$ ≅ $\triangle AOC$

BO = 1/2 BC

BO = 7√2/2 = 7/√2

Also, BO = CO (cpct)

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 = πr²

= π/4 angle has area = Area ACD = πr²/2π × π/4

= πr²/8

= π(49)/8 = 49π/8 cm²

= Area of portion AOD = (49π/8 - 49/4) cm²

= 49/4 (π/2 - 1) cm²

Total common area = 4 × Area of AOD

(Due to symmetricity)

= 4 × 49/4(π/2 - 1)

= 49(π/2 - 1)

= 49/2(π - 2) cm²

Therefore, Total common area = 49/2(π - 2) cm².