CDS Model Paper 2 Mathematics

In Δ‌AOB,AO=OB=r Using Pythagoras theorem,
AB2=OA2+OB2⇒(5)2=r2+r2
∴ r2=

25

2

‌cm
Now, area of sector
AOB=

θ

360°

×πr2=

90°

360°

×π×

25

2

=

25π

8

cm2
Now, area of minor segment
= Area of sector-Area of triangle
=

25π

8

−

r2

2

=

25π

8

−

25

4

=(

25π−50

8

)
Area of major segment = Area of circle − Area of minor segment
=πr2−(

25π−50

8

)=

25π

2

−

(25π−50)

8

=

100π−25π+50

8

=

75π+50

8

=

25

8

(3π+2)=

25

4

(

3π

2

+1)cm2