Question

A chord of length 10 cm divides a circle of radius 5√2 cm in two segments. find the area of minor segment.

Answer:

Area of the shaded region is 100/7 sq cm.

In the given figure, OA = OB = 5√2 cm as they are the radii of the circle. Drop a perpendicular from O to the chord AB to instersect it at C. As we know, that a perpendicular from the center of a circle to a chord divides it into two parts of equal length, AC = CB = AB/2 = 5cm.
In right triangle OBC:
cos (angle OBA) = BC/OB = 5/5√2 = 1/√2 which gives us angle OBA = 45 degrees.
As OA = OB, angle opposite to them will be equal, i.e., angle OBA = angle OAB = 45 degrees.
Angle OAB + angle OBA + angle BOA = 180 (angle sum property of a triangle)
angle BOA = 180 - 2(45) = 180-90 = 90 degrees. Therefore, the angle of the sector is 90 degrees.
Therefore triangle OBA is a right triangle and so its area = 5√2*5√2/2 = 25*2/2 = 25 sq cm
Area of the sector = π*5√2*5√2*90/360 = 25π*2/4 = 25π/2 sq cm
Area of shaded region = Area of sector - Area of the triangle = 25π/2 - 25 = 25(π/2 - 1) = 25(11/7 - 1) = 25(4/7) = 100/7 sq cm


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