Question

A chord 10 cm long is drawn in a circle whose radius is 5 root 2

Answer 1

Area of circle=πr²

=157.14(approx)

name the chord AB and centre o

as we know that perpendicular from centre bisect chord 

name the point of intersection of chord and radius p

in rt ∆apo

by Pythagoras theorm

H²=B²+P²

(5√2)²=5²+p2

50-25=P²

√25=p

5=p

in ∆apo

angle a=angle o....(1)(angle opposite to equal sides are equal)

A+P+O=180(angle sum property)

2O=180-90 (from 1)

O=90/2

O=45 degree

similarly in ∆PBO

angle o=45 degree

In ∆ABO

A+B+O=180

45+45+o=180

O=90 degree

hence the chord subtests 90 degree

area of minor segment=theta/360*πr² -1/2r²sin theta

=90/360*22/7*(5√2)² -1/2*(5√2)²(1)

=22*50/28 -50/2

=39.28-25

=14.28cm²(approx)

area of major segment=area of circle-area of minor segment

=157.14 -14.28

=142.86cm²(approx)

Answer 2

Please see the attachment