Class 10th, Area Related Circle Exe-12.2 (Solution)

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Maths NCERT Class 10 Chapter 12 – Areas Related to Circles

 Class 10 Maths Chapter 12 Exercise: 12.2

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1. Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.

Solution:

It is given that the angle of the sector is 60°

We know that the area of sector = (θ/360°)×π r2

∴ Area of the sector with angle 60° = (60°/360°) × π r2 cm2

= 36/6 π cm2

= 6 × 22/7 cm2 = 132/7 cm2

2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Circumference of the circle = 22 cm (given)

It should be noted that a quadrant of a circle is a sector which is making an angle of 90°.

Let the radius of the circle = r

As C = 2πr = 22,

R = 22/2π cm = 7/2 cm

∴ Area of the quadrant = (θ/360°) × π r2

Here, θ = 90°

So, A = (90°/360°)×π r2 cm2

= (49/16) π cm2

= 77/8 cm2 = 9.6 cm2

3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

Length of minute hand = radius of the clock (circle)

∴ Radius (r) of the circle = 14 cm (given)

Angle swept by minute hand in 60 minutes = 360°

So, the angle swept by the minute hand in 5 minutes = 360° × 5/60 = 30°

We know,

Area of a sector = (θ/360°)×π r2

Now, area of the sector making an angle of 30° = (30°/360°) × π r2 cm2

= (1/12) × π142

= (49/3) × (22/7) cm2

= 154/3 cm2

4. 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:

Here AB be the chord which is subtending an angle 90° at the center O.

It is given that the radius (r) of the circle = 10 cm

(i) Area of minor sector = (90/360°) × π r2

= (¼) × (22/7) × 102

Or, Area of minor sector = 78.5 cm2

Also, area of ΔAOB = ½ × OB × OA

Here, OB and OA are the radii of the circle i.e. = 10 cm

So, area of ΔAOB = ½ × 10 × 10

= 50 cm2

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

∴ 28.5 cm2

(ii) Area of major sector = Area of circle – Area of minor sector

= (3.14 × 102) – 78.5

= 235.5 cm2

5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

(i) the length of the arc

(ii) area of the sector formed by the arc

(iii) area of the segment formed by the corresponding chord

Solution:

Given,

Radius = 21 cm

θ = 60°

(i) Length of an arc = θ/360° × Circumference (2πr)

∴ Length of an arc AB = (60°/360°) × 2 × 22/7 × 21

= 1/6 × 2 × 22/7 × 21

Or Arc AB Length = 22cm

(ii) It is given that the angle subtend by the arc = 60°

So, area of the sector making an angle of 60° = (60°/360°) × π r2 cm2

= 441/6 × 22/7 cm2

Or, the area of the sector formed by the arc APB is 231 cm2

(iii) Area of segment APB = Area of sector OAPB – Area of ΔOAB

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔOAB is an equilateral triangle. So, its area will be √3/4 × a2

So, Area of segment APB = 231 – √3/4 × (OA)2

=> 231 – √3/4 × 212

Or, Area of segment APB = [ 231 – (441 × √3)/4 ]

6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle. (Use π = 3.14 and √3 = 1.73)

Solution:

Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = (60°/360°) × π r2 cm2

= 225/6 π cm2

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = √3/4 × a2

Or, √3/4 × 152

∴ Area of ΔAOB = 97.31 cm2

Now, area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, area of minor segment APB = (225/6 π – 97.31) cm2 = 20.43 cm2

And,

Area of major segment = Area of circle – Area of segment APB

Or, area of major segment = (π × 152) – 20.4 = 686.06 cm2

See next blog for more solution of this Exercise...............