NCERT Solutions for Class 9 Maths

Download NCERT Solutions for Class 9 Maths for UP Board (High School) and CBSE Board in PDF format free in Hindi Medium and English Medium.  NCERT solutions for session 2019-20 is now available to download in PDF form. Download NCERT Solutions Apps for class 9 Hindi & English Medium Free. These Apps works well without internet also.




Table of Contents

NCERT Solutions for Class 9 Maths

Chapter 1: Number Systems Solutions

Chapter 2: Polynomials Solutions

Chapter 3: Coordinate Geometry Solutions

Chapter 4: Linear Equations in Two Variables Solutions

Chapter 5: Introduction to Euclid’s Geometry Solutions

Chapter 6: Lines and Angles Solutions




Chapter 7: Triangles Solutions

Chapter 8: Quadrilaterals Solutions

Chapter 9: Areas of Parallelograms and Triangles Solutions

Chapter 10: Circles Solutions

Chapter 11: Constructions Solutions

Chapter 12: Heron’s Formula Solutions

Chapter 13: Surface Areas and Volumes Solutions

Chapter 14: Statistics Solutions

Chapter 15: Probability Solutions

NCERT Solutions for class 9 Maths all chapters in PDF form are given below. A separate PDF of all exercises are given to download. Download Apps for class 9 Maths in Hindi and 9 Maths in English Free. These Apps works well without internet also. NCERT Books as well as Exemplar problems books both are equally important for the exams. Chapter wise assignments Test Papers, Previous year Question Papers issued by CBSE and other schools, Chapter wise tests, Syllabus for the academic year 2019 – 2020 and other online study material.



Exemplar books are designed to enhance the practice and improve the knowledge and concepts about each chapter. So, students are advised to go through NCERT Exemplar book after doing NCERT books. Vedic Maths is good for improving calculations faster and easier. We are also providing help in solving holiday homework, if you are facing problem in doing holiday homework, upload in our website and get the solution with a week. The STUDY ONLINE option is given for NCERT solutions, if you don’t want to download. Download Apps for class 9 Maths in Hindi and 9 Maths in English Free. These Apps works well without internet also.

Chapter 1: संख्या पद्धति के हल

Chapter 2: बहुपद के हल

Chapter 3: निर्देशांक ज्यामिति के हल

Chapter 4: दो चरों वाले रैखिक समीकरण के हल

Chapter 5: यूक्लिड की ज्यामिति का परिचय के हल

Chapter 6: रेखाएँ और कोण के हल

Chapter 7: त्रिभुज के हल

Chapter 8: चतुर्भुज के हल



Chapter 9: समांतर चतुर्भुजों और त्रिभुजों के क्षेत्रफल के हल

Chapter 10: वृत के हल

Chapter 11: रचनाएँ के हल

Chapter 12: हीरोन का सूत्र के हल

Chapter 13: पृष्ठीय क्षेत्रफल और आयतन के हल

Chapter 14: सांख्यिकी के हल

Chapter 15: प्रायिकता के हल

Assignments for Practice





Download Apps for class 9 all subjects Free. These Apps works well without internet also. The first draft of solutions of class 9 maths is prepared in English and then it will be translated into Hindi Medium. Considering global thinking, it was decided to prepare the solutions first in English. But as the demand of Hindi Medium is also increasing (according to feedback submitted by students), so we are going to prepare Hindi medium solutions. Download NCERT Books for class 9 all subjects.
A panel of experts from Tiwari Academy held a workshop of two days and thoroughly discussed the content quality and made some suggestions. Amendments were carried out in the English as well as Hindi draft accordingly.



The final draft was thus prepared. The solutions are prepared according to new syllabus for 2019-20 having classical approach of constructions and geometrical chapters. उत्तर प्रदेश में भी NCERT की किताबें  लगने के बाद हम लोग UP बोर्ड के लिए भी अध्धयन सामग्री बना रहे हैं। अगर UP Board NCERT Solutions प्राप्त करने में विद्यार्थियों को किसी प्रकार की परेशानी हो रही है तो हमें Contact करें या फोन पर भी बता सकते हैं। UP Board Secondary Education (High School) के लिए अध्ययन सामग्री, Sample Papers, Notes तथा पश्नों के हल यहाँ से प्राप्त करें।

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Table of Contents

एक क्रिकेट मैच में, एक महिला बल्लेबाज खेली गई 30 गेंदों में 6 बार चौका मारती है। चौका न मारे जाने की प्रायिकता ज्ञात कीजिए।

कुल गेंदों की संख्या = 30
गेंदें, जिनपर चौका मारा गया = 6
इसलिए, गेंदें, जिनपर चौका नहीं मारा गया = 30 – 6 = 24
P(चौका नहीं मारा गया )
= (24)/30 = 4/5 = 0.8
अतः, चौका न मारे जाने की प्रायिकता 0.8 है।

Give five examples of data that you can collect from your day-to-day life.

Five examples of data that we can collect from our day-to-day life:
1. Height of our classmates or weight of our classmate.
2. Height of first 100 plants near by our locality.
3. Maximum or minimum temperature of a particular month.
4. Time spend for watching TV in a particular week.
5. Rainfall in our city in last 10 years.

उस लंब वृत्तीय शंकु का आयतन ज्ञात कीजिए जिसकी त्रिज्या 6 cm और ऊँचाई 7 cm है।

शंकु के आधार की त्रिज्या r = 6 cm और
ऊँचाई h = 7 cm है।
शंकु का आयतन = 1/3 πr^2 h
= 1/3 ×22/7 × 6 × 6 ×7
= 264 cm^3
अतः, लंब वृत्तीय शंकु का आयतन 264 cm^3 है।

It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2, find inner curved surface area of the vessel.

Cost of painting the inner curved surface of cylindrical vessel = ₹ 2200
and height h = 10 m.
Let, the inner radius of cylindrical vessel = r m
The inner curved surface area of cylindrical vessel = 2πrh
The cost of painting is at the rate of ₹ 20 per m2 = ₹ 20 × 2πrh
According to question,
₹ 20 × 2πrh = ₹ 2200
⇒ 2πrh = 2200/20 = 110

Hence, the inner curved surface area is 110 m^2.

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Here, the sides of triangle are
a = 15 m, b = 11 m and c = 6 m.
So, the semi-perimeter of triangle is given by
s = (a + b + c)/2
= (15 + 11 + 6)/2
= 32/2
=16 m
Therefore,
using Heron’s formula, the area of triangle
= √(s(s – a)(s – b)(s – c) )
= √(16(16 – 15)(16 – 11)(16 – 6) )
= √(16(1)(5)(10) )
= √(4 × 4 ×(1)(5)(5 × 2) )
= 4 × 5 √2
= 20 √2 m^2

Hence, the area painted in colour is 20√2 m^2.

What should we have for constructing a geometrical figure?

(i) A graduated scale, on one side of which centimetres and millimetres are marked off and on the other side inches and their parts are marked off.
(ii) A pair of set – squares, one with angles 90°, 60° and 30° and other with angles 90°, 45° and 45°.
(iii) A pair of dividers (or a divider) with adjustments.
(iv) A pair of compasses (or a compass) with provision of fitting a pencil at one end.
(v) A protractor.

याद कीजिए कि दो वृत्त सर्वांगसम होते हैं, यदि उनकी त्रिज्याएँ बराबर हों। सिद्ध कीजिए कि सर्वांगसम वृत्तों की बराबर त्रिज्याएँ उनके केन्द्रों पर बराबर कोण अंतरित करती हैं।

दिया है: C(A, r) और C(P, r) दो वृत्त सर्वांगसम हैं तथा BC = QR है।
सिद्ध करना है: ∠BAC = ∠QPR
उपपत्ति: ABC और PQR में,
BC = QR [∵ दिया है] AB = PQ [∵ सर्वांगसम वृत्तों की त्रिज्याएँ] AC = PR [∵ सर्वांगसम वृत्तों की त्रिज्याएँ] अतः, ABC ≅ PQR [∵SSS सर्वांगसमता नियम] ∠BAC = ∠QPR [∵ सर्वांगसम त्रिभुज के संगत भाग बराबर होते हैं]

Show that the diagonals of a parallelogram divide it into four triangles of equal area.

Diagonals of parallelogram bisect each other.
Therefore, PO = OR and SO = OQ
In ΔPQS, PO is median. [∵ SO = OQ] Hence, ar(PSO) = ar(PQO) … (1)
[∵ A median of a triangle divides it into two triangles of equal areas.]

Similarly, in ΔPQR, QO is median. [∵ PO = OR] Hence, ar(PQO) = ar(QRO) … (2)

And in ΔQRS, RO is median. [∵ SO = OQ] Hence, ar(QRO) = ar(RSO) … (3)

From the equations (1), (2) and (3), we get
ar(PSO) = ar(PQO) = ar(QRO) = ar(RSO)
Hence, in parallelogram PQRS, diagonals PR and QS divide it into four triangles in equal area.

ABCD एक चतुर्भुज है जिसमें P, Q, R और S क्रमशः भुजाओं AB, BC, CD और DA के मध्य-बिंदु हैं । AC उसका एक विकर्ण है। दर्शाइए कि SR || AC और SR=1/2 AC है।

ACD में,
S भुजा DA का मध्य-बिंदु हैं [∵ दिया है] R भुजा DC का मध्य-बिंदु हैं [∵ दिया है] अतः, SR || AC और SR = 1/2 AC
[∵ मध्य-बिंदु प्रमेय]

ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.

Draw perpendicular bisectors of AB, BC and AC, which intersects each other at G. Point G is equidistant from the three vertices of ΔABC.

What do you mean by collinear points or non-collinear points?

If three or more points lie on the same line, they are called collinear points; otherwise they are called non-collinear points.

What are the Euclid’s five postulates?

Euclid’s postulates were :
Postulate 1 : A straight line may be drawn from any one point to any other point.
Postulate 2 : A terminated line can be produced indefinitely.
Postulate 3 : A circle can be drawn with any centre and any radius.
Postulate 4 : All right angles are equal to one another.
Postulate 5 : If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right
angles.

k का मान ज्ञात कीजिए जबकि x = 2, y = 1 समीकरण 2x + 3y = k का एक हल हो।

दिया है: x = 2, y = 1

2x + 3y = k में x = 2 और y = 1 रखने पर,
2 × 2 + 3 × 1 = k
⇒ k = 7

उस बिंदु का नाम बताइए जहाँ x-अक्ष और y-अक्ष प्रतिच्छेदित होती हैं।

मूलबिंदु

सीधे गुणा किए बिना निम्न्लिखित मान ज्ञात कीजिए: 103×107

103 × 107
= (100 + 3)(100 + 7)
= (100)^2 + (3 + 7)100 + 3 × 7
[∵〖(x + a)(x + b) = x〗^2 + (a + b)x + ab] = 10000 + 1000 + 21
= 11021

क्या सभी धनात्मक पूर्णांकों के वर्गमूल अपरिमेय होते हैं? यदि नहीं, तो एक ऐसी संख्या के वर्गमूल का उदाहरण दीजिए जो एक परिमेय संख्या है।

सभी धनात्मक पूर्णांकों के वर्गमूल अपरिमेय नहीं होते हैं। जैसे कि √4 = 2, जो एक परिमेय संख्या है।

3 thoughts on “NCERT Solutions for Class 9 Maths”

  1. Manohar Re jo based on NCERT hai uska solution de dijiye sir ji
    Aur 2^nd math ka pramey available kara dijiye please

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