The radius of a circle is 5cm.The chord length is root 50cm is drawn in the circle.Find the area of the major segment? (use pi=22/7)
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The chord and the two radii form an isosceles triangle with equal sides equal to 5cm. If the chord bisector is drawn in we get two back to back congruent right-angled triangles, and we can find the angle at the centre, because it's half the bisected angle. The bisected chord makes a side of the right-angled triangle (sqrt(50))/2=5sqrt(2)/2. The sine of the angle is 5sqrt(2)/(2*5)=sqrt(2)/2, making the angle 45 degrees. Therefore the angle at the centre is 90 degrees making the area of the minor sector (1/4)(pi)(5^2). Therefore the area of the major sector is (3/4)(25(pi))=(75/4)(22/7)=58.93 approx.

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