Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the center.

Length of the chord = 5 cm (Given)

Let r be the radius of the circle.

Then, OA = OB = r cm

Now, angle subtended at the center of the sector OABO = 90°

Angle subtended at the center of the sector OABO (in radians) = θ = π/2

∴ Triangle AOB is a right-angled triangle.

So, by Pythagoras theorem, (AB)² = (OA)² + (OB)²

⇒ 25 = 2r²

Also, AOB is an isosceles triangle.

Since, line segment OD is perpendicular on AB, therefore it divides Ab into two equal parts. Thus, AD = DB = 5/2 = 2.5 cm

Let AD = h cm

So, in right angled triangle AOD, by Pythagoras theorem,

(AO)² = (AD)² + (OD)²

= 25/4

⇒ h = 5/2 = 2.5 cm

∴ Area of the isosceles triangle AOB (1/2) × Base × Height

= (1/2) × 5 × (5/2) = 25/4 cm²

Now, Area of the minor sector = (1/2)r²θ

= = (25π/8) cm²

Area of the minor segment = Area of the minor sector – Area of the isosceles triangle

=

Area of the major segment = Area of the circle – Area of the minor segment

=

=

∴ Difference of the areas of two segments of a circle =

|Area of major segment – Area of minor segment|

Hence, the required difference of the areas of two segments is