Question

Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm. Subtending an angle of 90
o
till the centre.​

Answer 1

Answer:

ANSWER

Let r be the radius of circle and AB be the chord, which makes 90

o

angle at centre.

AB=5cm

In right △OAB, using pythagoras theorem

OA

2

+OB

2

=AB

2

⇒r

2

+r

2

=5

2

⇒2r

2

=25

⇒r=

2

​

5

​

Area of circle =πr

2

=

7

22

​

×

2

​

5

​

×

2

​

5

​

=39.28cm

2

Area of minor segment =Areaofsector−Areaof△OAB

=

360

o

90

o

​

×πr

2

−

2

1

​

×

2

​

5

​

×

2

​

5

​

=

4

1

​

×39.28−

4

25

​

=

4

14.28

​

=3.57cm

2

Area of major segment =39.28−3.57=35.71cm

2

Required difference =35.71−3.57=32.14cm

2

Step-by-step explanation:

Answer 2

Answer:

AB = 5 cm. In the right angled triangle OAB , using Pythagoras Theorem. OA² + OB² = AB² ⇒ r² + r² = 5² ⇒ 2r² = 25

Step-by-step explanation:

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