Question

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A chord of a circle of radius 10 cm subtends a right angle at the centre.
Find :
a. area of the minor segment
b. area of minor sector
c. area of major sector
d. area of major segment

Answer 1

Hey,


Radius of the circle = 10 cm

Major segment is making 360° - 90° = 270°

Area of the sector making angle 270° 

                                 = (270°/360°) × π r2 cm2

                                           = (3/4) × 102π  = 75 π cm2 
                                = 75 × 3.14 cm2 = 235.5 cm2 
∴ Area of the major segment = 235.5 cm2

Height of ΔAOB = OA = 10 cm

Base of ΔAOB = OB = 10 cm

Area of ΔAOB = 1/2 × OA × OB

                         = 1/2 ×10 × 10 = 50 cm2

Major segment is making  90°

 Area of the sector making angle 90°        

= (90°/360°) × π r2 cm2

                                          = (1/4) × 102π  = 25 π cm2 
                               = 25 × 3.14 cm2 = 78.5 cm2 
Area of the minor segment = Area of the sector making angle 90° - Area of ΔAOB

                                            = 78.5 cm2 -  50 cm2 = 28.5 cm2

HOPE IT HELPS YOU:-))

Answer 2

Radius of the circle = 10 cm

•Major segment is making 360° - 90° = 270°

Area of the sector making angle 270° 

= (270°/360°) × π r² cm²

= (¾) × (10)²π  = 75 π cm²

= 75 × 3.14 cm² = 235.5 cm²

∴ Area of the major segment = 235.5 cm²

Height of ΔAOB = OA = 10 cm

Base of ΔAOB = OB = 10 cm

Area of ΔAOB = ½ × OA × OB

                         = ½ ×10 × 10 = 50 cm²

Major segment is making  90°

Area of the sector making angle 90°        

= (90°/360°) × π r² cm²

= (¼) × 102π  = 25 π cm²

= 25 × 3.14 cm² = 78.5 cm²

Area of the minor segment =Area of the (sector making angle 90° - ΔAOB)

= 78.5 cm² -  50 cm² = 28.5 cm²

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