Question

A chord of length 21 CM substance and angle of 60 degree at the centre find the area of major and the minor sector

Answer 1

Answer:

minor sec area=210.59sqcm

and

major sec area=1052.95sqcm

Step-by-step explanation:

60° is 1/6 of 360°

chord is 21cm

so

circum=21*6=126cm

=2Πr

so

r=126/2Π

=20.06 cm

and area=Π*r*r

=1263.547sq cm

so major sector area

=1263.547*5/6

=1052.95 sq cm

and minor sector aea

=1263.547*1/6

=210.59 sq cm

CHECK:

1052.95+210.59

=1263.54 so it is correct.