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# A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector. (Use π = 3.14)

**Solution:**

We use the formula for the area of a sector of the circle to solve the problem.

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

Area of the right triangle = 1/2 × base × height

Let's draw a figure to visualize the area to be calculated.

Here, Radius, r = 10 cm, θ = 90°

Visually it’s clear from the figure that,

AB is the chord that subtends a right angle at the centre.

(i) Area of minor segment APB = Area of sector OAPB - Area of right triangle AOB

(ii) Area of major segment AQB = πr² - Area of minor segment APB

Area of the right triangle ΔAOB = 1/2 × OA × OB

(i) Area of minor segment APB = Area of sector OAPB - Area of right ΔAOB

= θ/360° × πr^{2} - 1/2 × OA × OB

= 90°/360° × πr^{2} - 1/2 × r × r

= 1/4 πr^{2} -1/2r^{2}

= r^{2} (1/4 π - 1/2)

= r^{2 }(3.14 - 2)/4

= (r^{2} × 1.14)/4

= (10 × 10 × 1.14)/4 cm² (Since radius r is given as 10 cm)

= 28.5 cm²

(ii) Area of major sector AOBQ = πr^{2} - Area of minor sector OAPB

= πr^{2} - θ/360° × πr^{2}

= πr^{2} (1 - 90°/360°)

= 3.14 × (10 cm)^{2 }× 3/4

= 235.5 cm^{2}

**â˜› Check: **NCERT Solutions for Class 10 Maths Chapter 12

**Video Solution:**

## A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (i) minor segment (ii) major sector

NCERT Solutions Class 10 Maths Chapter 12 Exercise 12.2 Question 4

**Summary:**

The area of the minor segment APB and the area of major sector AOBQ if a chord of a circle of radius 10 cm subtends a right angle at the centre are 28.5 cm^{2} and 235.5 cm^{2} respectively.

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