CBSE Class 10

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/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 10
Chapter Name	Circles
Exercise	Ex 10.2
Number of Questions Solved	13
Category	NCERT Solutions

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/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 8
Chapter Name	Introduction to Trigonometry
Exercise	Ex 8.3
Number of Questions Solved	7
Category	NCERT Solutions

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.

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-13-ex-13-2/

Board	CBSE
Textbook	NCERT
Class	Class 10
Subject	Maths
Chapter	Chapter 13
Chapter Name	Surface Areas and Volumes
Exercise	Ex 13.2
Number of Questions Solved	8
Category	NCERT Solutions

NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)

Question 1.
A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of n.
Solution:
Radius of the hemisphere = 1 cm
Volume of the hemisphere = \(\frac { 2 }{ 3 }\)πr³ = \(\frac { 2 }{ 3 }\)π(1)³ = \(\frac { 2 }{ 3 }\)πcm³
Radius of base of the cone = 1 cm
Height of the cone = 1 cm

Volume of the cone = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)πr² x 1 = \(\frac { 1 }{ 3 }\)π cm³
Total volume of the solid = Volume of the hemisphere + Volume of the cone
= \(\frac { 2 }{ 3 }\)π cm³ + \(\frac { 1 }{ 3 }\)π cm³ = π cm³

Question 2.
Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
Solution:
Volume of air contained in the model = Total volume of the solid
Diameter of base of each cone = 3 cm
∴ Radius of base of each cone = \(\frac { 3 }{ 2 }\)
Height of each cone = 2 cm

Question 3.
A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure).

Solution:
Volume of one piece of gulab jamun
= Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8
Radius of each hemispherical portion = \(\frac { 2.8 }{ 2 }\) = 1.4 cm

Volume of both hemispherical ends = 2 x 5.74 cm³ = 11.48 cm³
Height of the cylindrical portion = (total height) – (radius of both hemispherical ends)
= 5 cm – 2(1.4) cm = 5 cm – 2.8 cm = 2.2 cm
Radius of the cylindrical portion = 1.4 cm
Volume of the cylindrical portion of gulab jamun = πr²h
= \(\frac { 22 }{ 7 }\) x (1.4)² x 2.2cm³
= \(\frac{22 \times 1.4 \times 1.4 \times 2.2}{7}\)cm³ = 13.55 cm³
Total volume of one gulab jamun = Volume of the two hemispherical ends + Volume of the cylindrical portion
= 11.48 cm³ + 13.55 cm³ = 25.03 cm³
Volume of sugar syrup = 30% of volume of gulab jamun
= \(\frac { 30 }{ 100 }\) x 25.03 cm³ = 7.50 cm³
∴ Volume of sugar syrup in 45 gulab jamuns
= 45 (volume of sugar syrup in one gulab jamun)
= 45 x 7.50 cm³ = 337.5 cm³ = 338 cm³ approx.

Question 4.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.
Find the volume of wood in the entire stand (see figure).

Solution:
Radius of one conical depression = 0.5 cm
Depth of one conical depression = 1.4 cm

Volume of cuboidal box = l x b x h
= 15 x 10 x 3.5 cm³
= 525 cm³
Remaining volume of box = Volume of cubical box – Volume of four conical depressions
= 525 cm³ – 1.464 cm³ = 523.5 cm³

Question 5.
A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.
Solution:

Question 6.
A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8 g mass. (Use π = 3.14)

Solution:
Given: radius of 1^st cylinder = 12 cm
and height of 1^st cylinder = 220 cm
∴ Volume of 1^st cylinder = πr²h
= π(12)² (220) cm³
= 144 x 220π cm³
= 144 x 220 x 3.14 cm³
= 99475.2 cm³ … (i)
Given: radius of 2^nd cylinder = 8 cm
and height of 2^nd cylinder = 60 cm
∴ Volume of 2^nd cylinder = πr²h
= π(8)² (60) cm³ = 64 x 60π cm³
= 64 x 60 x 3.14 cm³
= 12057.6 cm³ … (ii)
Total volume of solid = Volume of 1^st cylinder + Volume of 2^nd cylinder
= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³
Given: mass of 1 cm³ of iron = 8 g
∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g
= 892262.4 g = 892.262 kg

Question 7.
A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
Solution:
Radius of hemisphere = 60 cm

When solid (hemisphere + conical) is kept in cylindrical solid, then volume of water left in cylinder
= Volume of cylinder – (Volume of hemisphere + Volume of cone)
= [π(60)² x 180 – π(60)² x 80] cm³
= π(60)² [180 – 80]cm³ = π x 3600 x 100 cm³ = 1130400 cm³ = 1.130 m³

Question 8.
A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter, the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.
Solution:
Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)
Radius of the cylindrical part = \(\frac { 2 }{ 2 }\) = 1 cm

Height of the cylindrical part = 8 cm
∴ Volume of the cylindrical part = πr²h
= 3.14 x (1)² x 8 cm³
Diameter of the spherical part Radius = 8.5 cm
∴ Radius = \(\frac { 8.5 }{ 2 }\) cm = \(\frac { 85 }{ 20 }\)

Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part
= 25.12 cm³ + 321.39 cm³ = 346.51 cm³
Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, drop a comment below and we will get back to you at the earliest.