Question

A chord 10cm long is drawn in a circle whose radius is 5√2 cm. Find area of both the segments.(Take π=3.14)

Answer 1

Area of the minor segment = 14.29 cm²

Area of the major segment =  142.85 cm²

Step-by-step explanation:

Given radius = 10 cm. Length of chord = 5√2cm

In triangle OAB, OA = OB = 5√2cm

 Let's check OA² + OB² = (5√2)² + (5√2)² = 2 * 25 * 2 = 100

 OA = 10cm which is given.

So we find that angle OAB is 90 degrees.

Angle of the sector = 90 degrees.

 Area of the sector =  (θ/360)(πr²)  

  = 90/360 * 22/7 * 5√2 * 5√2

  = 39.29 cm²

Area of the triangle = 1/2 * height * base.

  So area of triangle  = 1/2 * 5√2 * 5√2

  = 25 cm²

Area of the circle = πr² = 22/7 * 5√2 * 5√2 = 157.14 cm²

Area of the minor segment = 39.29 - 25 = 14.29 cm²

Area of the major segment = 157.14 - 14.29 = 142.85 cm²

Answer 2

Answer:

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