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NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-12-ex-12-2/

NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Radius of the sector (r) = 6 cm
Central angle of the sector = 60°

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Let radius of the circle = r
∴ Circumference of the circle = 2πr

Question 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 of the clock = 14 cm

Question 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 segment (Use ? = 3.14)
Solution:
Given: radius of the circle = 10 cm
Angle subtended by chord at centre = 90°
(i) Area of the minor segment

(ii) Area of the major segment = Area of the circle – Area of the minor segment
= πr² – 28.5 = 3.14 x 10 x 10-28.5
= 314-28.5 = 285.5 cm²

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) length of the arc.
(ii) area of the sector formed by the arc.
(iii) area of the segment formed by the corresponding chord.
Solution:
Radius of the circle = 21 cm
Angle at the centre = 60°

Question 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 \(\sqrt{3}\) = 1.73)
Solution:
Radius of the circle = 15 cm
Angle subtended by chord at centre = 60°
Area of the sector = \(\frac{\pi r^{2} \theta}{360^{\circ}}\) = 3.14 x \(\frac{15 \times 15 \times 60^{\circ}}{360^{\circ}}\) = 117.75 cm²
Area of the triangle formed by radii and chord = \(\frac { 1 }{ 2 }\)r²θ
= \(\frac { 1 }{ 2 }\)(15)² sin 60° = \(\frac { 1 }{ 2 }\) x 15 x 15 x \(\frac{\sqrt{3}}{2}\) = 97.31 cm²
Area of the minor segment = Area of the sector – Area of the triangle formed by radii and chord
= 117.75 – 97.31 = 20.44 cm²
Area of the circle = πr² = 3.14 x 15 x 15 = 706.5 cm²
Area of the circle – Area of the minor segment
= 706.5 – 20.44 = 686.06 cm²

Question 7.
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 \(\sqrt{3}\) = 1.73).
Solution:
Radius of the circle = 12 cm
Angle subtended by chord at centre = 120°

Area of the corresponding segment = Area of the sector – Area of A formed by radii and chord
= 150.72 – 62.28 cm² = 88.44 cm²

Question 8.
A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope (see figure). Find

(i) the area of that part of the field in which the horse can graze.
(ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (Use n = 3.14)
Solution:
(i) Length of the rope = Radius of the sector grazed by horse = 5 m
Here, angle of the sector = 90°

Length of the rope is increased from 5 m to 10 m
New radius of sector grazed by horse = 10 m

Question 9.
A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in figure.

Find:
(i) the total length of the silver wire required.
(ii) the area of each sector of the brooch.
Solution:
Length of one diameter = 35 mm
Total length of 5 diameters = 5 x 35 mm = 175 mm
Circumference of the circle = 2π = 2 x \(\frac { 22 }{ 7 }\) x \(\frac { 35 }{ 2 }\) = 110 mm
(i) Total length of the wire used = length of 5 diameters + circumference of brooch
= 175 + 110 = 285 mm

(ii) Total sectors are 10.

Question 10.
An umbrella has 8 ribs which are equally spaced (see figure). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
Radius of the circle = 45 cm
Number of ribs = 8

Question 11.
A car has two wipers which do not overlap.
Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Given: length of blade of wiper = radius of sector sweep by blade = 25 cm
Area cleaned by each sweep of the blade = area of sector sweep by blade
Angle of the sector formed by blade of wiper =115°

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (Use π = 3.14)
Solution:
Angle of the sector = 80°
Distance covered = 16.5 km
Radius of the sector formed = 16.5 km

Question 13.
A round table cover has six equal designs as shown in the figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm². (Use \(\sqrt{3}\) = 1.7)

Solution:
Radius of the cover = 28 cm
∵ There are six equal designs

Question 14.
Tick the correct answer in the following: Area of a sector of angle p (in degrees) of a circle with radius R is
(a) \(\frac{p}{180^{\circ}}\) × 2πR
(b) \(\frac{p}{180^{\circ}}\) × πR²
(c) \(\frac{p}{360^{\circ}}\) × 2πR
(d) \(\frac{p}{720^{\circ}}\) × 2πR²
Solution:
Sector angle isp in degrees
Radius of the circle = R
Area of the sector = \(\frac{\pi \mathrm{R}^{2} p}{361^{6}}\) = \(\frac{\left(\pi R^{2} p\right) 2}{720^{\circ}}\)
= \(\frac{p}{720^{\circ}}\) × 2πR²

We hope the NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-10-ex-10-2/

NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2

Question 1.
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
(a) 7 cm Sol.
(b) 12 cm
(c) 15 cm
(d) 24.5 cm
Solution:

Question 2.
In figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
(a) 60°
(b) 70°
(c) 80°
(d) 90°
Solution:

∠OPT = 90°
∠OQT = 90°
∠POQ = 110°
TPOQ is a quadrilateral,
∴ ∠PTQ + ∠POQ = 180° ⇒ ∠PTQ + 110° = 180°
⇒∠PTQ = 180°- 110° = 70°
Hence, correct option is (b).

Question 3.
If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then ∠POA is equal to
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Solution:
In AOAP and AOBP
OA = OB [Radii]
PA = PB

[Lengths of tangents from an external point are equal]
OP = OP [Common]
∴ ∆OAP ≅ ∆OBP [SSS congruence rule]
∠AOB + ∠APB = 180° ⇒ ∠AOB + 80° = 180°
⇒∠AOB = 180° – 80° = 100°
From eqn. (i), we get
⇒∠POA = \(\frac { 1 }{ 2 }\) x 100° = 50°
Hence, correct option is (a)

Question 4.
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Solution:
AB is a diameter of the circle, p and q are two tangents.
OA ⊥ p and OB ⊥ q
∠1 = ∠2 = 90°
⇒ p || q ∠1 and ∠2 are alternate angles]

Question 5.
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.
Solution:
XY tangent to the circle C(0, r) at B and AB ⊥ XY. Join OB.
∠ABY = 90° [Given]
∠OBY = 90°
[Radius through point of contact is perpendicular to the tangent]
∴ ∠ABY + ∠OBY = 180° ⇒ ABOiscollinear
∴ AB passes through centre of the circle.

Question 6.
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Solution:
OA = 5 cm, AP = 4 cm OP = Radius of the circle
∠OPA = 90° [Radius and tangent are perpendicular]

Question 7.
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.
Solution:
Radius of larger circle = 5 cm Radius of smaller circle = 3 cm
OP ⊥ AB
[Radius of circle is perpendicular to the tangent]
AB is a chord of the larger circle

Question 8.
A quadrilateral ABCD is drawn to circumscribe a circle (see figure). Prove that AB + CD = AD + BC.
Solution:
AP = AS … (i)
[Lengths of tangents from an external point are equal]
BP = BQ … (ii)
CR = CQ … (iii)
DR = DS … (iv)

Adding equations (i), (ii), (iii) and (iv), we get

AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
Hence proved.

Question 9.
In figure, XY and X’Y’ are two parallel tangents to a circle , x with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB = 90°.

Solution:
Given: Two parallel tangents to a circle with centre O. Tangent AB with point of contact C intersects XY at A and X’Y’ at B To Prove: ∠AOB = 90° with point of contact C intersects XY at A and X’Y’ at B

To Prove: ∠AOB = 90°
Construction: Join OA, OB and OC
Proof: In ∆AOP and ∆AOC

Question 10.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.
Solution:
PA and PB are two tangents, A and B are the points of contact of the tangents.

Question 11.
Prove that the parallelogram circumscribing a circle is a rhombus.
Solution:
Parallelogram ABCD circumscribing a circle with centre O.
OP ⊥ AB and OS ⊥ AD

Question 12.
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see figure). Find the sides AB and AC.

Solution:
BD = 8 cm and DC = 6 cm
BE = BD = 8 cm
CD = CF = 6 cm

Let AE = AF = x cm
In ∆ABC, a = 6 + 8 = 14 cm
b = (x + 6) cm
c = (x + 8) cm

Question 13.
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Solution:
AB touches at P.
BC, CD and DA touch the circle at Q, R and S.
Construction: Join OA, OB, OC, OD and OP, OQ, OR, OS.

Hence, opposite sides of quadrilateral circumscribing a circle subtend supplementary angles at the centre of a circle.

We hope the NCERT Solutions for Class 10 Maths Chapter 10 Circles Ex 10.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 10 Circles E 10.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-8-ex-8-3/

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

Solution:

Question 2.
Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) LHS = tan 48° tan 23° tan 42° tan 67°
= tan 48° tan 23° tan (90° – 48°) tan (90° – 23°)
= tan 48° tan 23° cot 48° cot 23° = tan 48° tan 23° .\(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)
= 1 = RHS

(ii) LHS = cos 38° cos 52° – sin 38° sin 52°
= cos 38° cos (90° – 38°) – sin 38° sin (90° – 38°)
= cos 38° sin 38°- sin 38° cos 38° = 0 = RHS

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = \(\frac { 108° }{ 3 }\)
∴ ∠ A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B ⇒ tan A = tan (90° – B) [ ∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B ⇒ A + B = 90° Proved

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = \(\frac { 110° }{ 5 }\)
A = 22°
∴ ∠ A = 22°

Question 6.
If A, Band Care interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)
Solution:

Question 7.
Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) = cos 23° + sin 15°

We hope the NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3, drop a comment below and we will get back to you at the earliest.