Question

a chord of a circle of radius10 cm subtends a right angle at the centre .Find the area of the corresponding 1)minor segment 2)major segment

1

Answer 1

Solution :

radius of a circle, AO = 10 cm ∠AOC = 90°

Area of AAOC = ½ x OA x OC

= ½ x 10 x 10

= 50 cm²

Area of sector OAECO = Ø/360° x πr²

= 90°/360 x 3.14 x (10)²

= 314/4

= 78.5 cm²

i) Area of minor segment AECDA

area of sector OAECO - area of ΔAOC

= 78.5 - 50

= 28.5 cm²

ii) Area of major sector OAFGCO

Area of circle - Area of sector OAECO

= 3.14 x (10)² - 78.5

= 314 - 78.5

= 235.5 cm²

Answer 2

Answer:

GIVEN

a chord of a circle of radius10 cm subtends a right angle at the centre .Find the area of the corresponding 1)minor segment 2)major segment

To find

1. Minor segment

2. Major segment

Solution

Radius of circle AO = 10 and Angle AOC = 90

Area of AOC = ½ × OA × OC

= ½ × 10 × 10

= 50 sq. cm

Area of sectors OAECO = ∅/360⁰ × πr²

= 90/360 × 3.14 × 10²

= 314/4

= 78.5 sq.cm

$\huge \fbox {AREA OF MINOR SEGMENT}$

Area of sectors OAECO - Area of ∆ AOC

78.5 - 50

= 23.5

$\huge \fbox {AREA OF MAJOR SEGMENT}$

Area of circle - Area of sectors OAECO

= 3.14 × 10² - 78.5

= 314-78.5

= 235.5 sq.cm

______________________________

So, the minor segment = 23.5 sq.cm

The major segment = 235.5 sq.cm

______________________________

$\huge \fbox {HOPE IT HELPS}$