Consider the above figure,

From here we can conclude that the portion or the segment below the chord AB is the minor segment and the segment above AB is major segment.

Also we know,

Area of minor segment = Area of sector – Area of ∆AOB → eqn1

Now, Area of sector

Where R = radius of the circle and θ = central angle of the sector

Given, R = 7 cm and θ = 90^°

Putting these values in the equation 2, we get

⇒ Area of sector = 38.5 cm²→ eqn3

Area of △AOB = 1/2 × base × height

As triangle is isosceles therefore height and base both are 7 cm.

⟹ Area of △AOB = 1/2×7×7

= 24.5 cm²→ eqn4

Putting values of equation 2 and 4 in equation 1 we get

Area of minor segment = 38.5 – 24.5

⇒ Area of minor segment = 14 cm²

Area of major segment = πR² – Area of minor segment → eqn5

Put the value of R, and Area of minor segment in equation 5

= π(7²) – 14

= 49π - 14

= (22×7) - 14

= 154 - 14

= 140 cm²

Area of minor segment is 14 cm² and of major segment is 140 cm².