Question

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

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

Answer 2

Answer:

In the mentioned circle,

O is the centre and AO =BO = Radius = 10 cm

AB is a chord which subtents 90  

o

 at centre O, i.e., ∠AOB=90  

o

 

(i)

Area of minor segment APB (Shaded region) = Area of Sector AOB - Area of △AOB

=(  

4

π×10×10

​  

)−(0.5×10×10)

=78.5−50

=28.5cm  

2

 

(ii)

Area of Major sector = Area of circle - Area of Sector AOB

= (π×10×10)−(  

4

π×10×10

​  

)

=314−78.5

=235.5cm  

2