**NCERT Solutions class 10 maths exercise 12.2 of chapter 12-Areas related to circles**

NCERT Solutions class 10 maths exercise 12.2 of chapter 12-Areas related to circles are the solutions of the unsolved questions from the NCERT text book of class 9 maths. All questions are solved by the step by step way by the expert so it is very easy to understand each NCERT solutions. You can study NCERT solutions of maths and science from class 9 to 12 , sample papers, solutions of previous year question papers, tips for competitive entrance exams and carrier in online jobs.

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**Pdf-NCERT solutions of class 10 maths chapter 12-Areas related to Circle**

**Q1.Find the area of a sector of a circle with a radius 6 cm if the angle of the sector is 60°[use π = 22/7]**

Ans.

We are given the radius of the circle,r = 6 cm, angle of the sector,θ = 60°

Therefore area of the sector = 132/7 cm²

**Q2.Find the area of a quadrant of a circle whose circumference is 22 cm.[use π = 22/7].**

Ans. We are given the circumference of the circle is = 22 cm

Circumference of the circle is = 2πr, where is the radius of the circle

According to question

2πr = 22

2×(22/7)r = 22

r = 7/2

r = 3.5

The quadrant of the circle is one-fourth of the circle, so it subtends the angle at the centre = 360/4 = 90°

Quadrant is the sector of the circle which subtends the angle 90° at the centre

= 9.625

Hence the area of the circle is = 9.625 sq.cm

**Q3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes. [use π = 22/7]**

In 60 minutes, minute hand creates = 360°

In 1 minute minute hand creates = 360°/60 = 6°

In 5 minute, minute hand creates = 6°×5= 30°

Area swept in 5 minutes by the minute hand

Therefore area swept in 5 minutes by the minute hand = 154/3 cm²

**Q4. 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]**

Ans.

(i) Ans. We are given a chord PQ subtending 90° at the centre

The area of the minor segment PQS= Area of the corresponding sector OPSQ-Area of the triangle POQ.

Area of sector is given by

Where θ= angle subtended by the chord, radius,r = 10 cm

Area of minor sector OPSQ

=3.14 × 5 × 5

= 78.5

Area of sector OPSQ = 78.5 cm²

Area of the triangle POQ

Area of triangle POQ is = 50 cm²

Hence area of the minor segment PQS

= 78.5 – 50 = 28.5 cm²

(ii) Angle subtended by the major arc PRQ at the centre = 360-90 = 270°

Area of major sector

= 235.5

Hence area of major sector is = 235.5 cm²

**Q5. 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**

Ans.

We are given radius, r= 21 cm, angle subtended by the arc ACB ,θ= 60°

(i)Let the length of the arc = l

= 22

Hence the length of the arc is = 22 cm

(ii) Area of the sector formed by the arc ,AOBC

= 11× 21 = 231

Area of the sector formed by the arc i.e arAOBC = 231 sq.cm

(iii) Area of the segment(ACB) formed by the corresponding chord

= Area of the sector AOBC – Area of the triangle AOB

In triangle AOB

OA = OB (radii of the circle)

∠OAB = ∠OBA (angles opposite to equal sides of the triangle)

∠OAB + ∠OBA + ∠AOB = 180°

2∠OAB + 60° = 180°

∠OAB = 60°

∠OAB =∠OBA =∠AOB = 60°

Therefore triangle AOB is an equilateral triangle

The area of equilateral triangle with sides ‘a’

a = 21 cm

Area of ΔAOB

Hence area of the segment(ACB) formed by the corresponding chord

**Q6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the area of the corresponding minor and major segments of the circle.[use π = 3.14 and √3 = 1.73]**

Ans.

We are given the radius of the circle ,r = 15 cm, angle subtended by the radii of the circle,θ = 60°

Area of the corresponding segment (minor segment) = Area of the corresponding sector – Area of the corresponding triangle

Area of the corresponding sector(AOBC)

=117.5

Area of sector AOBC is 117.5 cm²

Triangle AOB is an equilateral Δ(see Q.5)

= 0.4325 × 225 =97.3125 cm²

Area of the corresponding minor segment (AOBC)

= 117.5 – 97.3125 = 20.1875 cm²

Area of major segment = Area of circle – Area of minor segment

= πr² – 20.1875

= 3.14 × 15× 15 -20.1875

= 706.25 -20.1875

= 686.0625 cm²

Hence the area of minor and major segments of the circle are 20.1875 cm² and 686.0625 cm² respectively.

Hence the radius of the circle is 2 units, so the answer is (i) 2 units

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