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# 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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## T Solutions for Class 10 Maths   Chapter 12-Areas related to circles

Exercise 12.1- Area Related to Circle

Exercise 12.2- Areas related to Circle

Exercise 12.3-Areas Related to the Circle

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

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

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°

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

$\boldsymbol{ =\frac{60^{\circ} }{360^{\circ}}\times \frac{22}{7}\times 6^{2}}$

$\boldsymbol{ =\frac{60^{\circ} }{360^{\circ}}\times \frac{22}{7}\times 6\times 6}$

$\boldsymbol{=\frac{1}{6}\times \frac{22}{7}\times 6\times 6=\frac{132}{7}}$

Therefore area of the sector = 132/7 cm²

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

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

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

$\boldsymbol{=\frac{90}{360}\times \frac{22}{7}\times 3.5\times 3.5}$

$\boldsymbol{=\frac{1}{4}\times 22\times 0.5\times 3.5}$

$\boldsymbol{=5.5\times 0.5\times 3.5}$

= 9.625

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

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

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°

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

Area swept in 5 minutes by the minute hand

$\boldsymbol{=\frac{30}{360}\times \frac{22}{7}\times 14\times 14}$

$\boldsymbol{=\frac{1}{12}\times 22\times 2\times 14}$

$\boldsymbol{=\frac{1}{3}\times 11\times 2\times 7}$

$\boldsymbol{=\frac{154}{3}}$

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

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

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

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

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

Area of minor sector OPSQ

$\boldsymbol{Area\: of \: OPSQ=\frac{90}{360}\times 3.14\times 10\times 10}$

$\boldsymbol{=\frac{1}{4}\times 3.14\times 10\times 10}$

$\boldsymbol{=\frac{1}{2}\times 3.14\times 5\times 10}$

=3.14 × 5 × 5

= 78.5

Area of sector OPSQ = 78.5 cm²

Area of the triangle POQ

$\boldsymbol{=\frac{1}{2}\times Base\times Altitude}$

$\boldsymbol{=\frac{1}{2}\times 10\times 10=50}$

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

$\boldsymbol{=\frac{270}{360}\times 3.14\times 10\times 10}$

$\boldsymbol{=\frac{3}{4}\times 3.14\times 10\times 10}$

$\boldsymbol{=\frac{3}{2}\times 3.14\times 5\times 10}$

$\boldsymbol{= 3\times 3.14\times 5\times 5}$

= 235.5

Hence area of major sector is = 235.5 cm²

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

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

$\boldsymbol{\left [use\: \pi =\frac{22}{7} \right ]}$

Ans.

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

(i)Let the length of the arc = l

$\boldsymbol{l=\frac{\Theta }{360}\times \boldsymbol{2}\pi r}$

$\boldsymbol{=\frac{60 }{360}\times \boldsymbol{2}\times \frac{22}{7}\times 21}$

$\boldsymbol{=\frac{1 }{6}\times \boldsymbol{2}\times \frac{22}{7}\times 21}$

$\boldsymbol{=\frac{1 }{3}\times \boldsymbol{2}\times \frac{11}{7}\times 21}$

= 22

Hence the length of the arc is = 22 cm

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

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

$\boldsymbol{=\frac{60}{360}\times \frac{22}{7}\times 21\times 21}$

$\boldsymbol{=\frac{1}{6}\times \frac{22}{7}\times 21\times 21}$

$\boldsymbol{=\frac{1}{3}\times \frac{11}{7}\times 21\times 21}$

= 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’

$\boldsymbol{=\frac{a^{2}}{4}\sqrt{3}}$

a = 21 cm

Area of ΔAOB

$\boldsymbol{=\frac{21^{2}}{4}\sqrt{3}}$

$\boldsymbol{=\frac{{441}}{4}\sqrt{3}\: cm^{2}}$

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

$\boldsymbol{=(231-\frac{{441}}{4}\sqrt{3})\: cm^{2}}$

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

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)

$\boldsymbol{Area =\frac{\Theta }{360^{\circ}}\times \pi r^{2}}$

$\boldsymbol{=\frac{60}{360}\times 3.14\times 15\times 15}$

$\boldsymbol{=\frac{1}{6}\times 3.14\times 15\times 15}$

$\boldsymbol{=\frac{1}{2}\times 3.14\times 5\times 15}$

$\boldsymbol{= 3.14\times 5\times 7.5}$

=117.5

Area of sector AOBC is 117.5 cm²

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

$\boldsymbol{=\frac{a^{2}}{4}\sqrt{3}}$

$\boldsymbol{=\frac{15\times 15}{4}\times 1.73}$

= 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

Q7.A chord of a circle of the 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).

Ans.We are given a chord AB which subtends ∠AOB = 120° at the centre

Drawing OC⊥ AB,since ΔAOB is an isoscell triangle therefore OC will bisect the ∠AOB

∴∠BOC = 60°

sin60° = BC/BO =BC/12

BC/12 = √3/2

BC = 12√3/2 = 6√3 cm

Since OC⊥ AB ,threfore OC bisects the chord AB

AB = 2×6√3 = 12√3 cm

cos60° = OC/BO =OC/12

OC/12 = 1/2

OC = 12/2 = 6 cm

arΔAOB = (1/2) AB×OC = (1/2)×12√3×6 = 36√3 =36×1.73 =62.28 cm²

Area of the sector AOBD =(θ/360°)πr² =(120°/360°)×3.14×12×12 =150.72 cm²

Area of the segment ADB = Area of the sector AOBD-arΔAOB =150.72-62.28 =88.44 cm²

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