Question

A chord of a circle radius 12 cm subtends angle of 120° at the centre. Find the area of tge Corresponding segment of the circle.
(Use π=3.14 and √3= 1.73) ​

Answer 1

Answer

$150 {cm}^{2}$

Answer 2

$\huge\boxed{\fcolorbox{lime}{black}{⭐ANSWER~~}}$

Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB and it will bisect chord AB.

So, AD = DB

Ncert solution class 10 chapter 12-5

Now, the area of the minor sector = (θ/360°)×πr2

= (120/360)×(22/7)×122

= 150.72 cm2

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = AD/OA

√3/2 = AD/12

Or, AD = 6√3 cm

We know OD bisects AB. So,

AB = 2×AD = 12√3 cm

Now, sin 30° = OD/OA

Or, ½ = OD/12

∴ OD = 6 cm

So, the area of ΔAOB = ½ × base × height

Here, base = AB = 12√3 and

Height = OD = 6

So, area of ΔAOB = ½×12√3×6 = 36√3 cm = 62.28 cm2

∴ Area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 cm2– 62.28 cm2 = 88.44 cm2