Question

a chord of 10 cm long is drawn in a circle whose radius is√50 cm. find the atea of segments...

Answer 1

Answer: 142.85 sq cm

Step-by-step explanation:Given length of chord, AB = 10 cm

Radius of circle, OA = OB = √50 cm

Draw OC⊥AB

Hence AC = AB = 5 cm

In right ΔOCA,

OA2 = OC2 + AC2 [By Pythagoras theorem]

(√50)2 = OC2 + 52

OC2 = 25

OC = 5 cm

Thus area of triangle AOB = (1/2) x AB x OC

= (1/2) x 10 x 5 = 25 sq cm

From right triangle OCA, sin∠AOC = AC/OA

= 5/(√50) = 5/(5√2) = 1/√2

Thus ∠AOC = 45°

Similarly, we have ÐBOC = 45°

Therefore, ∠AOB = 90°

Area of minor sector = (θ/360) x πr2

= (90/360) x (22/7) x (√50)2

= 39.29 sq cm (nearly)

Area of minor sector = 39.29 – 25 = 14.29 sq cm (nearly)

Area of circle = (22/7) x (√50)2

= 157.14 sq cm (nearly)

Area of major segment = area of circle – area of minor segment

= 157.14 – 14.29 = 142.85 sq cm

Answer 2

So the answer is 142.85 sq cm