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°

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

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

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

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

Therefore area of the sector = 132/7 cm²

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

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

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

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

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

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.