Question

A chord of a circle, of length 10 cm, subtends a night anglo at tho centre. Find the areas of the minor segment and the major segment formed by the chord. (????= 3.14)

Answer 1

1 ) In ∆AOB ,

<AOB = 90°,

OA = OB = r

AB = 10 cm

OA² + OB² = 10²

r² + r² = 100

2r² = 100

r² = 50

r = 5√2 cm

2 ) Area of the minor segment

= area of the sector OADB - area ∆OAB

= ( <AOB )/360° × πr² - ( OA × OB )/2

= ( 90/360 ) × 3.14 ×( 5√2)² - ( 5√2 )²/2

= ( 5√2 )² [ 3.14/4 - 1/2 ]

= 50 × 1.14/4

= 14.25 cm²

3 ) Area of the major segment OAEB

= Area of the circle - area of minor circle

= πr² - 142.5

= 3.14 × ( 5√2 )² - 14.25

= 3.14 × 50 - 14.25

= 157 - 14.25

= 142.75 cm²

I hope this helps you.

: )