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Question
A chord 10 cm long is drawn in a circle whose radius is `5sqrt(2)` cm. Find the areas of both the segments.
Solution
Let O be the centre of the circle and AB be the chord.
Consider Δ OAB.
OA = OB = `5sqrt(2) "cm"`
OA2 + OB = 50 + 50 = 100
Now,
`sqrt(100) = 10 "cm" = "AB"`
Thus, ΔOAB is a right isosceles triangle.
Thus we have :
Area of Δ OAB `=1/2xx5sqrt(2)xx5sqrt(2) = 25 "cm"^2`
Area of the minor segment = Area of the sector - Area of the triangle
`=(90/360xxpixx(5sqrt(2))^2) - 25`
= 14.25 cm2
Area of the major segment = Area of the circle -- Area of the minor segment
`= pixx(5sqrt(2))^2-14.25`
= 142.75 cm2
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