A chord 10cm long is drawn in a circle whose radius is 5 root2 cm find the area of both segments.?
Given, a chord AB of length 10 cm and radius = OA = OB = 5√2 cm.
Construction: Draw OC perpendicular to AB.
Now, AC = BC = 10/2 = 5 cm [The perpendicular drawn from the centre of a circle to a chord always bisect the chord.]
In triangle OAC,
sin x = Perpendicular/hypotenuse = AC/ OA
Similarly, ∠BOC = 45°
⇒ ∠AOB = ∠AOC + ∠BOC = 45°+ 45° = 90°
We know that, area of sector
Therefore, area of sector OAB
Again, in triangle OAC,
cos x = base/hypotenuse = OC/ OA
cos 45 = OC/5√2
Therefore, area of triangle OAB
Now, area of minor segment = Area of sector OAB - Area of triangle OAB
= 39.28 cm2 - 25 cm2
= 14.28 cm2
Similarly, area of major segment = Area of circle - Area of minor segment
⇒ Area of major segment