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NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

NCERT Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.9

March 10, 2026

NCERT Class 9 Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.9

NCERT Solutions for Class 9 Maths Chapter 13 – Surface Areas and Volumes Topics and Sub Topics

Surface Areas And Volumes
Introduction
Surface Area of a Cuboid and a Cube
Surface Area of a Right Circular Cylinder
Surface Area of a Right Circular Cone
Surface Area of a Sphere
Volume of a Cuboid
Volume of a Cylinder
Volume of a Right Circular Cone
Volume of a Sphere
Summary

Surface Areas and Volumes Class 9 Maths Formulas

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.9

Ex 13.9 Class 9 Maths Question 1.
A wooden bookshelf has external dimensions as follows : Height = 110cm, Depth = 25cm, Breadth = 85cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing-is 20 paise per cm² and the rate of pointing is 10 paise per cm², find the total expenses required for palishing and painting the surface of the bookshelf.

Solution:
Here, l = 85 cm b = 25 cm and h = 110 cm
Area of the bookshelf of outer surface = 2 lb + 2bh + hl
= [2(85 x 25)+ 2(110 x 25)+ 85×110] cm2 = (4250 + 5500 + 9350) cm² = 19100 cm²
Cost of polishing of the outer surface of bookshelf
= 19100 x \(\frac { 20 }{ 100 }\) = ₹ 3820
Thickness of the plank = 5 cm
Internal height of bookshelf = (110 – 2 x 5) = 100 cm
Internal depth of bookshelf = (25 – 5) = 20 cm
Internal breadth of bookshelf = 85 – 2 x 5 = 75 cm
Hence, area of the internal surface of bookshelf
= 2(75 x 20)+ 2(100x 20)+ 75×100
= 3000 + 4000 + 7500 = 14500 cm2
So, cost of painting of internal surface of bookshelf
= 14500 x \(\frac { 10 }{ 100 }\) = ₹1450
Hence, total costing of polishing and painting = 3820 + 1450= ₹ 5270

Ex 13.9 Class 9 Maths Question 2.
The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are-used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm² and black paint costs 5 paise per cm².

Solution:
It is obvious, we have to subtract the cost of the sphere that is resting on the supports while calculating the cost of silver paint.
Surface area to be silver paint

Ex 13.9 Class 9 Maths Question 3.
The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?
Solution:
Let d be the diameter of the sphere

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes (Hindi Medium) Ex 13.9

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Science

March 10, 2026

NCERT Solutions for Class 9 Science are given for the students so that they can get to know the in text question and chapter end question answers in case they are not able to find it. It is important for all the students who are in Class 9th currently. Here we are providing the solutions to all the chapters of Class 9 Science Textbook for the students. They can refer to samajik vigyan class 9 Physics, Chemistry and Biology Solutions while they are solving the questions from the Science Book.

NCERT Solutions for Class 9 Science

Class 9 Science NCERT Solutions is given here. Students can click on the links of the particular chapter for which they are finding the solutions.

The students who are in class 9th and have the NCERT Science Books can check this page to know the answers for all the chapters of Science book. They need to ensure that they are checking the solutions for the chapter which they intend to check.

FAQs on NCERT Solutions for Class 9 Science

1. What’s the best way to study Class 9 Science CBSE?

The best way to study Class 9 Science is through NCERT. Don’t just mug up the concepts instead go through all the concepts thoroughly. Make notes and practice as much as you can.

2. Where can I get the NCERT solutions for Class 9 Science?

You need not panic as we will give you all the NCERT Solutions for Class 9 Science all at one place on our site. View or download them and practice as and when you want.

3. Which is the best book of science for Class 9 CBSE?

We will not specifically say one book is the best for preparing Class 9 Science. Instead, we will ay try covering all the concepts in NCERT Textbooks prescribed for Class 9 Science that can help in your Exams.

4. How do I score good marks in science class 9?

The only key for scoring more marks in the CBSE Class 9 Science Exam is through sheer practice. It’s the best way to score well in the exams as well as get command on the subject.

5. Where can I get CBSE Class 9 Sample Papers?

Aspirants can get the CBSE Class 9 Sample Papers that can ace up preparation for free of cost from our page. Check out the links available on our site and use them as a reference to score better grades in the exam.

6. What are the study tips for Class 9 Science Exam?

Some of the best tips that help you to score more in your Class 9 Science Exam are as follows

Make a proper plan and schedule to study
Focus on the basic concepts and try concentrating on weaker areas.
Dedicate time to each of the concepts.
Practice as much as you can to score well.

Important Questions For Class 9 Science

Importance of the Class 9 Physics, Chemistry, Biology NCERT Solutions

The NCERT Class 9 Science Textbook Solutions are important for the students to get good marks in their exams. Through these, they can get solutions to those questions on which they get stuck. These are the valid solutions and students can check these whenever they face any confusion in the questions.

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

NCERT Solutions for Class 9 Maths Number System Ex 1.6

March 9, 2026

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.6

Number systems
Introduction
Irrational Numbers
Real Numbers And Their Decimal Expansions
Representing Real Numbers On The Number Line
Operations On Real Numbers
Laws Of Exponents For Real Numbers
Summary
Chapter 1 Number System NCERT Solutions Ex 1.1
Chapter 1 Number System NCERT Solutions Ex 1.2
Chapter 1 Number System NCERT Solutions Ex 1.3
Chapter 1 Number System NCERT Solutions Ex 1.4
Chapter 1 Number System NCERT Solutions Ex 1.5
Extra Questions for Number Systems

Ex 1.6 Class 9 Maths Question 1.
Find:

Solution:

Ex 1.6 Class 9 Maths Question 2.
Find:

Solution:

Ex 1.6 Class 9 Maths Question 3.
Simplify:

Solution:

NCERT Solutions for Class 9 Maths Chapter 1 Number systems (Hindi Medium) Ex 1.6

NCERT Solutions for Class 9 Maths

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2

March 9, 2026

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2

Ex 5.2 Class 9 Maths Question 1.
How would you rewrite Euclid’s fifth postulate so that it would be easier to understand?
Solution:
Two distinct intersecting lines cannot be parallel to the same line.

Ex 5.2 Class 9 Maths Question 2.
Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Solution:
Yes.
According to Euclid’s fifth postulate when line x falls on line y and z such that ∠1+ ∠2< 180°. Then, line y and line z on producing further will meet in the side of ∠1 arid ∠2 which is less than 180°.

We find that the lines which are not according to Euclid’s fifth postulate. i.e., ∠1 + ∠2 = 180°, do not intersect.

NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry (युक्लिड के ज्यामिति का परिचय) (Hindi Medium) Ex 5.2

NCERT Solutions for Class 9 Maths

We hope the NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid Geometry Ex 5.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

March 9, 2026

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

The topics and sub-topics in NCERT Class 9 Maths Textbook Chapter 10 Circles:

Circles
Introduction
Circles And Its Related Terms: A Review
Angle Subtended By A Chord At A Point
Perpendicular From The Centre To A Chord
Circle Through Three Points
Equal Chords And Their Distances From The Centre
Angle Subtended By An Arc Of A Circle
Cyclic Quadrilaterals
Summary

Formulae Handbook for Class 9 Maths and Science Educational Loans in India

NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.6

Ex 10.6 Class 9 Maths Question 1.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Solution:
Given: Two circles with centres O and O’ which intersect each other at C and D.

To prove: ∠OCO’ = ∠ODO’
Construction: Join OC, OD, O’C and O’D
Proof: In ∆ OCO’and ∆ODO’, we have
OC = OD (Radii of the same circle)
O’C = O’D (Radii of the same circle)
OO’ = OO’ (Common)
∴ By SSS criterion, we get
∆ OCO’ ≅ ∆ ODO’
Hence, ∠OCO’ = ∠ODO’ (By CPCT)

Ex 10.6 Class 9 Maths Question 2.
Two chords AB and CD of lengths 5 cm and 11 cm, respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
Solution:
Let O be the centre of the given circle and let its radius be cm.
Draw ON ⊥ AB and OM⊥ CD since, ON ⊥ AB, OM ⊥ CD and AB || CD, therefore points N, O, M are collinear.

Let ON = a cm
∴ OM = (6 – a) cm
Join OA and OC.
Then, OA = OC = b c m
Since, the perpendicular from the centre to a chord of the circle bisects the chord.
Therefore, AN = NB= 2.5 cm and OM = MD = 5.5 cm
In ∆OAN and ∆OCM, we get
OA² = ON² + AN²
OC² = OM² + CM²
⇒ b² = a² + (2.5)²
and, b² = (6-a)² + (5.5)² …(i)
So, a² + (2.5)² = (6 – a)² + (5.5)²
⇒ a² + 6.25= 36-12a + a² + 30.25
⇒ 12a = 60
⇒ a = 5
On putting a = 5 in Eq. (i), we get
b² = (5)² + (2.5)²
= 25 + 6.25 = 31.25
So, r = \( \sqrt{31.25} \) = 5.6cm (Approx.)

Ex 10.6 Class 9 Maths Question 3.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre ?
Solution:
Let PQ and RS be two parallel chords of a circle with centre O such that PQ = 6 cm and RS = 8 cm.
Let a be the radius of circle.

Draw ON ⊥ RS, OM ⊥ PQ. Since, PQ || RS and ON ⊥ RS, OM⊥ PQ, therefore points 0,N,M are collinear.
∵ OM = 4 cm and M and N are the mid-points of PQ and RS respectively.
PM = MQ = \(\frac { 1 }{ 2 }\) PQ = \(\frac { 6 }{ 2 }\) = 3 cm
and RN = NS = \(\frac { 1 }{ 2 }\) RS = \(\frac { 8 }{ 2 }\) = 4 cm
In ∆OPM, we have
OP² = OM² + PM²
⇒ a² =4² + 3² = 16 + 9 = 25
⇒ a = 5
In ∆ORN, we have
⇒ OR² = ON² + RN²
⇒ a² = ON² + (4)²
⇒ 25 = ON² + 16
⇒ ON² = 9
⇒ ON = 3cm
Hence, the distance of the chord PS from the centre is 3 cm.

Ex 10.6 Class 9 Maths Question 4.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Solution:
Since, an exterior angle of a triangle is equal to the sum of the interior opposite angles.

∴ In ∆BDC, we get
∠ADC = ∠DBC + ∠DCB …(i)
Since, angle at the centre is twice at a point on the remaining part of circle.
∴ ∠DCE = \(\frac { 1 }{ 2 }\) ∠DOE
⇒ ∠DCB = \(\frac { 1 }{ 2 }\) ∠DOE (∵ ∠DCE = ∠DCB)
∠ADC = \(\frac { 1 }{ 2 }\) ∠AOC
∴ \(\frac { 1 }{ 2 }\) ∠AOC = ∠ABC + \(\frac { 1 }{ 2 }\) ∠DOE (∵ ∠DBC = ∠ABC)
∴ ∠ABC = \(\frac { 1 }{ 2 }\) (∠AOC – ∠DOE)
Hence, ∠ABC is equal to half the difference of angles subtended by the chords AC and DE at the centre.

Ex 10.6 Class 9 Maths Question 5.
Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.
Solution:
Given: PQRS is a rhombus. PR and SQ are its two diagonals which bisect each other at right angles.
To prove: A circle drawn on PQ as diameter will pass through O.

Construction: Through O, draw MN || PS and EF || PQ.
Proof : ∵ PQ = SR ⇒ \(\frac { 1 }{ 2 }\) PQ = \(\frac { 1 }{ 2 }\) SR
So, PN = SM
Similarly, PE = ON
So, PN = ON = NQ
Therefore, a circle drawn with N as centre and radius PN passes through P, O, Q.

Ex 10.6 Class 9 Maths Question 6.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Solution:
Since, ABCE is a cyclic quadrilateral, therefore

∠AED+ ∠ABC= 180°
(∵ Sum of opposite angle of a cyclic quadrilateral is 180°) .. .(i)
∵ ∠ADE + ∠ADC = 180° (EDC is a straight line)
So, ∠ADE + ∠ABC = 180°
(∵ ∠ADC = ∠ABC opposite angle of a || gm).. .(ii)
From Eqs. (i) and (ii), we get
∠AED + ∠ABC = ∠ADE + ∠ABC
⇒ ∠AED = ∠ADE
∴ In ∆AED We have
∠AED = ∠ADE
So, AD = AE
(∵ Sides opposite to equal angles of a triangle are equal)

Ex 10.6 Class 9 Maths Question 7.
AC and BD are chords of a circle which bisect each other. Prove that
(i) AC and BD are diameters,
(ii) ABCD is a rectangle.
Solution:
(i) Let BD and AC be two chords of a circle bisect at P.

In ∆APB and ∆CPD, we get
PA = PC ( ∵ P is the mid-point of AC)
∠APB = ∠CPD (Vertically opposite angles)
and PB = PD (∵ P is the mid-point of BD)
∴ By SAS criterion
∆CPD ≅ ∆APB
∴ CD= AB (By CPCT) …(i)

∴ BD divides the circle into two equal parts. So, BD is a diameter.
Similarly, AC is a diameter.
(ii) Now, BD and AC bisect each other.
So, ABCD is a parallelogram.
Also, AC = BD
∴ ABCD is a rectangle.

Ex 10.6 Class 9 Maths Question 8.
Bisectors of angles A, B and C of a ∆ABC intersect its circumcircle at D, E and F, respectively. Prove that the angles of the ∆DEF are 90° – \(\frac { 1 }{ 2 }\) A, 90° – \(\frac { 1 }{ 2 }\) B and 90° – \(\frac { 1 }{ 2 }\) C.
Solution:
∵ ∠EDF = ∠EDA + ∠ADF
∵ ∠EDA and ∠EBA are the angles in the same segment of the circle.
∴ ∠EDA = ∠EBA
and similarly ∠ADF and ∠FCA are the angles in the same segment and hence

Ex 10.6 Class 9 Maths Question 9.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Solution:
Let O’ and O be the centres of two congruent circles.

Since, AB is a common chord of these circles.
∴ ∠BPA = ∠BQA
(∵ Angle subtended by equal chords are equal)
⇒ BP = BQ

Ex 10.6 Class 9 Maths Question 10.
In any ∆ ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the ∆ABC.
Solution:
(i) Let bisector of ∠A meet the circumcircle of ∆ABC at M.
Join BM and CM.

∴ ∠MBC = ∠MAC (Angles in same segment)
and ∠BCM = ∠BAM (Angles in same segment)
But ∠BAM = ∠CAM (∵ AM is bisector of ∠A)…. .(i)
∴ ∠MBC = ∠BCM
So, MB = MC (Sides opposite to equal angles are equal)
So, M must lie on the perpendicular bisector of BC
(ii) Let M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ ABC.
Join AM.

Since, M lies on perpendicular bisector of BC.
∴ BM = CM
∠MBC = ∠MCB
But ∠MBC = ∠MAC (Angles in same segment)
and ∠MCB = ∠BAM (Angles in same segment)
So, from Eq. (i),
∠BAM = ∠CAM
AM is the bisector of A.
Hence, bisector of ∠A and perpendicular bisector of BC at M which lies on circumcircle of ∆ABC.