Question

find the difference of the areas of 2 segments of a circle formed by a chord of length 5 cm subtending an angle of 90° at the centre

Answer 1

Let the chord be  AB. Center of circle be O.
ΔAOB is right angled at O.   AO = OB = radius
So:     AB = √2 AO = 5 cm
AO = radius = 5/√2  cm

area(ΔAOB) = 1/2 * 5/√2 * 5/√2 = 25/4 cm²
Since in the sector OAB, angle is π/4, its area is = 1/4 of total circle
    =  π*(5/√2)² / 4 = 25π/16 cm²

Minor segment:  25 π/16 - 25/4   cm²
Major segment =   π (5/√2)²  -  [25π/16 - 25/4] 
                          = 75π/16  + 25/4  cm²

Answer 2

Answer:

Let the chord be  AB. Center of circle be O.

ΔAOB is right angled at O.   AO = OB = radius

So:     AB = √2 AO = 5 cm

AO = radius = 5/√2  cm

area(ΔAOB) = 1/2 * 5/√2 * 5/√2 = 25/4 cm²

Since in the sector OAB, angle is π/4, its area is = 1/4 of total circle

   =  π*(5/√2)² / 4 = 25π/16 cm²

Minor segment:  25 π/16 - 25/4   cm²

Major segment =   π (5/√2)²  -  [25π/16 - 25/4]  

                         = 75π/16  + 25/4  cm²

Step-by-step explanation: