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

**Solution:**

We use the concept of areas of sectors of circles to solve the question.

In a circle with radius r and the angle at the centre with degree measure θ:

(i) Area of the sector = θ/360 πr^{2}

(ii) Area of the segment = Area of the sector - Area of the corresponding triangle

Let's draw a figure to visualize the given question.

Here, radius, r = 12 cm, ∠AOB = θ = 120°

Visually it’s clear from the figure that AB is the chord that subtends 120° angle at the centre.

To find the area of the segment AYB, we have to find the area of the sector OAYB and the area of the ΔAOB

(i) Area of sector OAYB = θ/360° πr^{2}

(ii) Area of ΔAOB = 1/2 × base × height

For finding the area of ΔAOB, draw OM ⊥ AB then find base AB and height OM using the figure as shown above.

Area of sector OAYB = 120°/360° × πr^{2}

= 1/3 × 3.14 × (12 cm)^{2}

= 150.72 cm^{2}

Draw a perpendicular OM from O to chord AB

In ΔAOM and ΔBOM

AO = BO = r (radii of circle)

OM = OM (common side)

∠OMA = ∠OMB = 90° (perpendicular OM drawn)

∴ ΔAOM ≅ ΔBOM (By RHS Congruency)

⇒ ∠AOM = ∠BOM (By CPCT)

Therefore, ∠AOM = ∠BOM = 1/2 ∠AOB = 60°

In ΔAOM,

AM/OA = sin 60° and OM/OA = cos 60°

AM/12 cm = √3/2 and OM/12 cm = 1/2

AM = √3/2 × 12 cm and OM = 1/2 × 12 cm

AM = 6√3 cm and OM = 6 cm

⇒ AB = 2 AM

= 2 × 6√3 cm

= 12√3 cm

Area of ΔAOB = 1/2 × AB × OM

= 1/2 × 12√3 cm × 6 cm

= 36 × 1.73 cm^{2}

= 62.28 cm^{2}

Area of segment AYB = Area of sector OAYB - Area of ΔAOB

= 150.72 cm^{2} - 62.28 cm^{2}

= 88.44 cm^{2}

**Video Solution:**

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

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Question 7

**Summary:**

The area of the corresponding segment APB if a chord of a circle of radius12 cm subtends an angle of 120° at the centre is 88.44 cm^{2}.

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