Question

ROMPLE 7 Fina
6
Find the difference of the areas of two segments of a circle formed by a chord
of length 5cm sub
tending an angle of 90° at the centre.
INCERT EXEMPLAR]
ON Letrbe the radius of the circle​

Answer 1

Let r be the radius of the circle and AB be the chord, which subtends angle of 90 degree at the center of the circle O. AB = 5 cm.
In the right angled triangle OAB , using Pythagoras Theorem.
OA² + OB² = AB²
⇒ r² + r² = 5²
⇒ 2r² = 25
⇒ r² = 25/2
⇒ r = 5/√2 cm
Therefore the area of the circle = πr²
= 22/7*5/√2*5/√2
= 39.2857 cm²
The area of minor segment = Area of the sector OAB - Area of triangle OAB
Area of sector OAB = πr²*90/360
⇒ 39.2857*1/4
Area of sector OAB = 9.8214 cm²
Area of triangle OAB = 1/2*OA*OB
= 1/2*r²  (Because OA and OB are the radii of the circle)
= 1/2*25/2
= 25/4
Area of triangle OAB = 6.25 cm²
Area of minor segment = 9.8214 - 6.25
= 3.5714 sq cm
Now,
Area of the major segment = Area of the circle - Area of minor segment 
= 39.2857 - 3.5714
Area of the major segment = 35.7143 sq cm
Difference between the areas of two segments = 35.7143 - 3.5714
Difference between areas of two segments = 32.1429 sq cm
Answer.

Answer 2

Try to write area of required portion as sum , difference, product(sector,circle,here) first,then apply formula. Use other facts as they may be related to any property(Pythagoras theorem, here) that would be required in finding value of desired(here, length of chord)