A chord of a circle of radius 12 cm subtends an angle of 120° at the center. Find the area of the corresponding segment of the circle. (Use π = 3.14 and \( \sqrt{3} = 1.73\))

Asked by Pragya Singh | 1 year ago |  67

1 Answer

Solution :-

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 = (\( \dfrac{θ}{360°}\))×πr2

= (\( \dfrac{120}{360}\))×(\( \dfrac{22}{7}\))×122

= 150.72 cm2

Consider the ΔAOB,

∠ OAB = 180°-(90°+60°) = 30°

Now, cos 30° = \( \dfrac{AD}{OA}\)

\( \dfrac{\sqrt{3}}{2}\) = \( \dfrac{AD}{12}\)

Or, AD =\(6 \sqrt{3}\) cm

We know OD bisects AB. So,

AB = 2×AD = \(12 \sqrt{3}\) cm

Now, sin 30° = \( \dfrac{OD}{OA}\)

Or, \( \dfrac{1}{2}=\) \( \dfrac{AD}{12}\)

OD = 6 cm

So, the area of ΔAOB = \( \dfrac{1}{2}\) × base × height

Here, base = AB = \( 12 \sqrt{3}\) and

Height = OD = 6

So, area of ΔAOB = \( \dfrac{1}{2}\)×\( 12 \sqrt{3}\)×6 = \( 36 \sqrt{3}\) cm = 62.28 cm2

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

= 150.72 cm2– 62.28 cm= 88.44 cm2

Answered by Abhisek | 1 year ago

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