Question

Find the difference of the areas of two segments of a circle formed by a chord of length 7 cm subtending an angle of 60° at the centre

Answer 1

Answer:

Step-by-step explanation:

Let r be the radius of the circle and AB be the chord, which subtends angle of 90 degree at the center of the circle O. AB = 5 cm.

In the right angled triangle OAB , using Pythagoras Theorem.

OA² + OB² = AB²

⇒ r² + r² = 5²

⇒ 2r² = 25

⇒ r² = 25/2

⇒ r = 5/√2 cm

Therefore the area of the circle = πr²

= 22/7*5/√2*5/√2

= 39.2857 cm²

The area of minor segment = Area of the sector OAB - Area of triangle OAB

Area of sector OAB = πr²*90/360

⇒ 39.2857*1/4

Area of sector OAB = 9.8214 cm²

Area of triangle OAB = 1/2*OA*OB

= 1/2*r²  (Because OA and OB are the radii of the circle)

= 1/2*25/2

= 25/4

Area of triangle OAB = 6.25 cm²

Area of minor segment = 9.8214 - 6.25

= 3.5714 sq cm

Now,

Area of the major segment = Area of the circle - Area of minor segment  

= 39.2857 - 3.5714

Area of the major segment = 35.7143 sq cm

Difference between the areas of two segments = 35.7143 - 3.5714

Difference between areas of two segments = 32.1429 sq cm

Answer 2

Given : a chord of length 7 cm subtending an angle of 60 degree at the center​

To Find :  difference of the areas of two segments of a circle formed by   chord

Solution:

a chord of length 7 cm subtending an angle of 60 degree at the center

Hence chord will form am equilateral triangle with center

so radius of circle =  7 cm

Area of smaller segment = Area of sector with 60° angle - area of triangle

= (60/360)π(7)² -  (√3 / 4)7²

= 49 ( π/6 -  √3 / 4)

360° - 60° = 300°

Area of larger segment = Area of sector with 300° angle + area of triangle

= (300/360)π(7)² +  (√3 / 4)7²

= 49 ( 5π/6 + √3 / 4)

the difference of the areas of two segments of a circle

= 49 ( 5π/6 + √3 / 4) - 49 ( π/6 -  √3 / 4)

= 49 ( 4π/6 + 2√3 / 4)

=  49 ( 2π/3 +  √3 / 2)

= 49 * 2.96

≈  145 cm²

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