Question

a chord of 10 cm long is drawn in a circle whose radius is 5√2 cm find the area of both the segments

Answer 1

Please see the attachment

Answer 2

for option a b c d check image


Given, a chord AB of length 10 cm and radius = OA = OB = 5√2 cm.

Construction: Draw OC perpendicular to AB.

Now, AC = BC = 10/2 = 5 cm [The perpendicular drawn from the centre of a circle to a chord always bisect the chord.]

In triangle OAC,

sin x = Perpendicular/hypotenuse = AC/ OA

= 5/5√2

sinx=1/√2
sinx=sin45
=>x=45
Similarly, ∠BOC = 45°

⇒ ∠AOB = ∠AOC + ∠BOC = 45°+ 45° = 90°

We know that, area of sector

πr2theta/360

Therefore, area of sector OAB
1 hey mete this part is in image no a
Again, in triangle OAC,

cos x = base/hypotenuse = OC/ OA

cos 45 =  OC/5√2

mate next is b
Therefore, area of triangle OAB

 

the next is c
Now, area of minor segment =  Area of sector OAB - Area of triangle OAB

= 39.28 cm2 - 25 cm2

= 14.28 cm2

Similarly, area of major segment = Area of circle - Area of minor segment

⇒ Area of major segment 

the next is d
I hope this will help you