NCERT Solutions for Class 9 Mathematics Chapter 7- Triangles

Mathematics is the study of numbers and their applications. It is mainly associated with accounting and various operations of Mathematics. It is broadly categorised into two sections; Algebra and Geometry. Hence, to master Mathematics, one should focus on both the sections in detail.

The main topics covered in the Chapter 7 of Class 9 Mathematics Triangles are congruent triangles, SSS, SAS, ASA and AAS criteria for congruence, why are SSA and AAA rules not valid, properties of isosceles triangles and inequalities in triangles.

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You will find all the properties of the congruence of triangles listed in a structured way in our NCERT solutions for Class 9 Mathematics Chapter 7. It has all the sections and subsections properly divided and highlighted. All the key points in each section are being written in the format required in CBSE examinations. Thus, it proves to be a great guide for students to step up their preparation to get excellent results.

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Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 7

You have read about the basics of triangles in earlier classes. You are already familiar with different types of triangles, their properties and how to refer to a triangle in a figure.

NCERT Class 9 Mathematics Chapter 7 briefly talks about their congruency and all the properties associated with it. You will know about the rules associated with congruence and its application in geometry. . You can look up NCERT Solutions for Class 9 Mathematics Chapter 7 on our Extramarks’ website where you can find the chapter being covered in detail which clears their concepts and helps them to think analytically to solve a problem..

We at Extramarks believe in going beyond the call of our duty to facilitate students with great learning experience ; be it note making, guiding the students, designing the mock tests or analysing the results. Hence, parents, teachers and students have faith and trust in Extramarks NCERT Solutions for reliable and authentic notes which has further strengthened over the years.

This chapter is mostly based on theorems, formulas and properties of triangles . Detailed study of this chapter will help students to do the exercises and find solutions effectively and easily.

Introduction

This section helps you recall all the basics of triangles covered in the lower classes. Further, it gives you detailed insights of the topics you will study in this chapter. The Extramarks subject matter experts have meticulously covered each and every exercise and solution of the chapter in our NCERT Solutions for Class 9 Mathematics Chapter 7. Students can practice with NCERT books and cross check their answers and if required they may even go ahead and solve other sample papers to strengthen their base.

Congruence of Triangles

The figures which are the same in shape and size or are equal in all aspects are known as congruent. When two triangles are of the same shape and of the same size they are called congruent triangles.

In this section, you will learn how to interpret congruence of triangles by analysing their sides and angles. All your doubts regarding congruence will be cleared with the help of the illustrations given in the chapter.

Criteria for Congruence of Triangles

There are certain criterias one needs to note while looking for congruence in two triangles. They are as follows:

Side-Angle-Side or SAS congruence rule –

Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

Angle-Side-Angle or ASA congruence rule-

Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other triangle.

Some properties of a Triangle

In this section, you will learn how to apply the criterion you have read about in the last section when two sides are given equal. You will also get to know about some of the theorems and their uses in various Geometrical concepts with the help of some examples given in the NCERT Solutions for Class 9 Mathematics Chapter 7.

Some More Criteria for Congruence of Triangles

You will read about some more criteria for congruence of triangles in this section. They are listed below:

Side-Side-Side or SSS congruence rule-

If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

Right angle-Hypotenuse-Side or RHS congruence rule-

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Inequalities in a triangle

Whenever there is a relation between the unequal sides or angles of a triangle it is called as Inequalities of a triangle. There are various activities and examples included from this section in our NCERT Solutions for Class 9 Mathematics Chapter 7 r in such a way that they can get to the point answers without wasting much time on a single topic. It gives a better understanding of all the concepts associated and helps them to solve any difficult or advanced level questions with ease.

NCERT Solutions for Class 9 Mathematics Chapter 7: Exercise & Solutions

Testing is the cornerstone of any preparation.; since it shows what you have learnt and how much you can retain. You may gauge your performance by solving the questions. However, the essential prerequisite is that you must have the right solutions and answers for you to rectify your mistakes and shortcomings.

Therefore, NCERT Solutions for Class 9 Mathematics Chapter 7 have been compiled in a systematic and organised manner to revise and cross-check your answers. You can access it from Extramarks’ website as per your convenience to ace the examinations with flying colours.

The following links has exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 7:

Chapter 7: Exercise 7.1 Question and answers
Chapter 7: Exercise 7.2 Question and answers
Chapter 7: Exercise 7.3 Question and answers
Chapter 7: Exercise 7.4 Question and answers
Chapter 7: Exercise 7.5Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 7, students can explore NCERT Solutions on our Extramarks’ website for all primary as well as secondary classes.

NCERT Solutions Class 1
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NCERT Solutions Class 5
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NCERT Solutions Class 9
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NCERT solutions Class 11
NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics

NCERT Exemplar is the collection of NCERT related- questions. It helps in laying the foundation to all the basic as well as advanced level concepts in the manner required for different competitive examinations. It is especially designed by subject matter experts. As a result, teachers and mentors suggest students to include NCERT Exemplar books in their study material. Studying NCERT textbooks alone won’t help you to crack the tests. We therefore recommend students to start using it early to beat the competition and stay ahead of the pack.

The book gives insights of all the topics from the NCERT Class 9 Mathematics textbook. You can find questions ranging from basic to advanced level; thereby making students capable of solving questions of varying degrees to strengthen their base.

Students can be rest assured that not a single topic has been left untouched in the chapter and hence are definitely going to nail the exams by referring to NCERT Solutions and NCERT Exemplar. You can get NCERT Exemplar for Class 9 Mathematics easily from the Extramarks website.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 7

In order to score well, you have to learn well and to learn well, you have to read well. Hence, NCERT Solutions for Class 9 Mathematics Chapter 7 has complete academic notes to study from. . The following are the key features: :

It has well-versed short notes to read and learn and revise quickly and easily.
You can also find a detailed overview of which topics to revise the most in our NCERT Solutions.
Gradually, it will boost your confidence and hence be able to leverage your performance.
After completing this chapter, you will have a complete idea of the congruency of the triangles and the rules needed to be applied where necessary.

The notes are prepared by highly qualified and experienced faculty who meticulously follow the NCERT textbooks and CBSE guidelines to provide authentic and reliable study material.

Q.1

$\begin{array}{l} In quadrilateral ACBD, AC = AD and AB bisects ∠ A \\ (see figure below) . Show that ΔABC ≅ ΔABD . What can \\ you say about BC and BD . \end{array}$

Ans.

$\begin{array}{l} Given : ABCD is a quadrilateral, AC = BD, AB bisects ∠ A. \\ To prove: Δ ABC ≅ Δ ABD \\ Proof:In Δ ABC and Δ ABD \\ AC = AD [Given] \\ ∠ BAC = ∠ ABD [Given] \\ AB = AB [Common] \\ ∴ Δ ABC ≅ Δ ABD [By SAS] \\ BC = BD [By CPCT] \end{array}$

Q.2

$\begin{array}{l} ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA \\ (see figure below) . Prove that \\ (i) ΔABD ≅ ΔBAC \\ (ii) BD = AC \\ (iii) ∠ ABD = ∠ BAC . \end{array}$

Ans.

$\begin{array}{l} Given : ABCD is a quadrilateral, AD = BC, ∠ D AB = ∠ C BA . \\ To prove: \\ (i) Δ ABD ≅ Δ B AC \\ (i i) B D = A C \\ (i i i) ∠ ABD = ∠ BAC \\ Proof: (i) In Δ ABD and Δ B AC \\ AB = B A [Common] \\ ∠ B A D = ∠ A B C [Given] \\ A D = B C [Given] \\ ∴ Δ ABD ≅ Δ BAC [B y SAS] \\ (i i) B D = A C [B y CPCT] \\ (i i i) ∠ ABD = ∠ BAC [B y CPCT] \end{array}$

Q.3 AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

Ans.

$\begin{array}{l} Given : AD and BC are equal perpendiculars to a line segment AB. \\ To Prove: CD bisects AB i.e., CO = OD \\ Proof : In ΔBOC ≅ ΔAOD \\ ∠ CBO = ∠ DAO [Given] \\ ∠ BOC = ∠ AOD [Vertical opposite angles] \\ BC = AD [Given] \\ ∴ ΔBOC ≅ ΔAOD [By AAS] \\ ∴ BO = OA [By C.P.C.T.] \\ \Rightarrow CD bisects line segment AB. \end{array}$

Q.4

$\begin{array}{l} l and m are two parallell ines intersected by another pair of \\ parallel lines p and q (see figure below) . Show that Δ ABC ≅ ΔCDA . \end{array}$

Ans.

$\begin{array}{l} Given : line l and m are parallel and \overset{⇀}{p} ∥ \overset{⇀}{q} . \\ To prove: Δ ABC ≅ ΔCDA \\ Proof: In Δ ABCand ΔCDA \\ ∠ BAC = ∠ DCA [Alternate interior angles] \\ AC = CA [Common] \\ ∠ BCA = ∠ DAC [Alternate interior angles] \\ Δ ABC ≅ ΔCDA [By ASA] \end{array}$

Q.5

$\begin{array}{l} Line l is the bisector of an angle ∠A and ∠B is any point on l. \\ BP and BQ are perpendiculars from B to the arms of ∠A \\ (see figure below) . Show that : \\ (i) ΔAPB ≅ ΔAQB \\ (ii) BP = BQ or B is equidistant from the arms of ∠ A . \end{array}$

Ans.

$\begin{array}{l} Given : Line l is bisector of ∠ A, BP and BQ are perpendiculars to \\ the arms of angle. \\ To prove: (i) Δ APB ≅ ΔAQB \\ (ii) BP = BQ or B is equidistant from the arms of∠A. \\ Proof: (i) In Δ ABQ and ΔAPB \\ ∠ BAQ = ∠ BAP [Given] \\ AB = AB [Common] \\ ∠ AQB = ∠ APB [each 90°] \\ Δ ABQ ≅ ΔAPB [By AAS] \\ (ii) BQ = BP [By C.P.C.T.] \\ \Rightarrow BP = BQ \end{array}$

Q.6 In figure below, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Ans.

$\begin{array}{l} Given : In Δ ABC and Δ ADE, AC = AE, AB = AD and ∠ BAD = ∠ EAC \\ To prove:BC=DE \\ Proof: In Δ ABC and ΔADE \\ AB = AD [Given] \\ ∠ BAC = ∠ DAE [∠ BAD + ∠ DAC = ∠ DAC + ∠ EAC] \\ AC = AE [Given] \\ Δ ABC ≅ ΔADE [By SAS] \\ BC = DE [By C.P.C.T.] \end{array}$

Q.7

$\begin{array}{l} AB is a line segment and P is its mid - point . D and E are \\ points on the same side of AB such that ∠ BAD = ∠ ABE \\ and ∠ EPA = ∠ DPB (see figure below) . Show that \\ (i) ΔDAP ≅ ΔEBP \\ (ii) AD = BE \end{array}$

Ans.

$\begin{array}{l} Given : P is mid-point of AB, ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB \\ To prove: (i) ΔDAP ≅ ΔEBP \\ (ii) AD = BE \\ Proof: (i) In Δ DAPand ΔEBP \\ ∠ DPA = ∠ EPB [∠ APE + ∠ EPD = ∠ BPD + ∠ DPE] \\ A P = PB [Given] \\ ∠ DAP = ∠ EBP [Given] \\ Δ DAP ≅ ΔEPB [By SAS] \\ (ii) AD = BE [By C.P.C.T.] \end{array}$

Q.8

$\begin{array}{l} In right triangle ABC, right angled at C, M is the mid - point \\ of hypotenuse AB . C is joined to M and produced to a point \\ D such that DM = CM . Point D is joined to point B \\ (see figure below) . Show that : \\ (i) ΔAMC ≅ ΔBMD \\ (ii) ΔDBC is a right angle . \\ (iii) ΔDBC ≅ ΔACB \\ (iv) CM = \frac{1}{2} AB \end{array}$

Ans.

$\begin{array}{l} Given : In right Δ ABC, ∠ C = 90 ° and M is mid-point on \\ hypotenuse AB, CM = DM. \\ To Prove: (i) ΔAMC ≅ ΔBMD \\ (ii) ∠ DBC is a right angle. \\ (iii) ΔDBC ≅ ΔACB \\ (iv) CM = \frac{1}{2} AB \\ Proof : (i) In ΔAMC and ΔBMD \\ AM = BM [Given] \\ ∠ AMC = ∠ BMD [Vertica l opposite angles] \\ CM = DM [Given] \end{array}$ $\begin{array}{l} ∴ ΔAMC ≅ ΔBMD [S . A . S .] \\ AC = DB [By C.P.C.T.] \\ (ii) ∠ ACM = ∠ BDM [By C.P.C.T.] \\ \Rightarrow ∠ ACD = ∠ BDC \\ \Rightarrow AC ∥ DB [Alternate angles are equal.] \\ ∠ ACB + ∠ DBC = 180 ° [Co-interior angles] \\ 90 ° + ∠ DBC = 180 ° \\ ∠ DBC = 180 ° - 90 ° \\ = 90 ° \\ So, ∠ DBC is a right angle. \\ (iii) In Δ DCB and ΔACB \\ DB = AC [Proved] \\ ∠ ACB = ∠ DBC [Each 90°] \\ BC = BC [Common] \\ ∴ Δ DCB ≅ ΔACB [S . A . S .] \\ (iv) DC = AB [By C.P.C.T.] \\ 2 CM = AB [CM = DM] \\ CM = \frac{1}{2} AB \end{array}$

Q.9 In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A.

Ans.

$\begin{array}{l} Given: In Δ ABC, AB = AC and bisectors of ∠ B and ∠ C intersect \\ at O. \\ To Prove: (i) OB = OC (ii) AO bisects ∠ A \\ Proof: (i) In Δ ABC, AB = AC \\ So, ∠ ACB = ∠ ABC [\begin{array}{l} Opposite angles of equal \\ sides are equal. \end{array}] \\ \Rightarrow \frac{1}{2} ∠ ACB = \frac{1}{2} ∠ ABC \\ \Rightarrow ∠ OCB = ∠ OBC [\begin{array}{l} Given, as OB and OC are angle \\ bisectors. \end{array}] \\ \Rightarrow OB = OC [\begin{array}{l} Opposite angles of equal \\ sides are equal. \end{array}] \end{array}$ $\begin{array}{l} (ii) In ΔO AB and ΔO AC, \\ OA = OA [Common] \\ AB = AC [Given] \\ OB = OC [Proved above] \\ ∴ ΔO AB ≅ ΔO AC [By S.S.S.] \\ Hence, ∠ OAB = ∠ OAC [By C.P.C.T.] \\ Thus, OA bisects ∠ A. \end{array}$

Q.10 In Δ ABC, AD is the perpendicular bisector of BC (see figure below). Show that Δ ABC is an isosceles triangle in which AB = AC.

Ans.

$\begin{array}{l} Given : In Δ ABC, BD = DC and AD ⊥ BC. \\ To Prove: AB = AC \\ Proof: In Δ ABD and Δ ACD \\ AD = AD [Common] \\ ∠ ADB = ∠ ADC [Given] \\ B D = DC [Given] \\ ∴ Δ ABD ≅ Δ ACD [By S.A.S.] \\ ∴ AB = AC [By C.P.C.T.] \\ Therefore, Δ ABC is isosceles triangle. \end{array}$

Q.11 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure below). Show that these altitudes are equal.

Ans.

$\begin{array}{l} Given:In ΔABC, AC = AB, BE ⊥ AC and CF ⊥ AB. \\ To prove: BE = CF \\ Proof: In ΔABE and ΔACF \\ ∠ BAE = ∠ CAF [Common angle] \\ ∠ BEA = ∠ CFA [Each 90°] \\ AB = AC [Given] \\ ∴ ΔABE ≅ ΔACF [By A.A.S.] \\ ∴ BE = CF [By C.P.C.T.] \\ Hence proved. \end{array}$

Q.12

$\begin{array}{l} ABC is a triangle in which altitudes BE and CF to sides AC \\ and AB are equal (see figure below) . Show that \\ (i) ΔABE ≅ ΔACF \\ (ii) AB = AC, i . e ., ABC is an isosceles triangle . \end{array}$

Ans.

$\begin{array}{l} Given : In ΔABC, BE = CF, BE ⊥ AC and CF ⊥ AB . \\ To prove : (i) ΔABE ≅ ΔACF \\ (ii) AB = AC, i . e ., ABC is an isosceles triangle . \\ Proof : (i) In ΔABE and ΔACF \\ ∠ BAE = ∠ CAF [Common angle] \\ ∠ BEA = ∠ CFA [Each 90 °] \\ BE = CF [Given] \\ ∴ ΔABE ≅ ΔACF [By A . A . S .] \\ (ii) AB = AC [By C . P . C . T .] \\ i . e ., ABC is an isosceles triangle . \\ Hence proved . \end{array}$

Q.13 ABC and DBC are two isosceles triangles on the same base BC (see figure below). Show that ∠ABD = ∠ACD.

Ans.

$\begin{array}{l} Given : In ΔABC, AB = AC and in ΔDBC, DB = DC . \\ To prove: ∠ ABD = ∠ ACD \\ Proof: In ΔABC \\ AB = AC [Given] \\ ∴ ∠ ACB = ∠ ABC . .. (i) [\begin{array}{l} Opposite angles of equal sides \\ are equal. \end{array}] \\ In ΔDBC \\ DB = DC [Given] \\ ∴ ∠ DCB = ∠ DBC . .. (ii) [\begin{array}{l} Opposite angles of equal sides \\ are equal. \end{array}] \\ Adding equation (i) and equation (ii), we get \\ ∠ ACB + ∠ DCB = ∠ ABC + ∠ DBC \\ ∴ ∠ ACD = ∠ ABD \\ \Rightarrow ∠ ABD = ∠ ACD \\ Hence proved. \end{array}$

Q.14 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure below). Show that ∠BCD is a right angle.

Ans.

$\begin{array}{l} Given:In Δ ABC, AB = AC, side BA is produced to D such that \\ AD = AB. \\ To prove: ∠ BCD is a right angle . \\ Proof:In Δ ABC, \\ AB = AC [Given] \\ ∠ ACB = ∠ ABC [Opposite angles of equal sides are equal.] \\ = x (let) \\ In Δ ACD, \\ AC = AD [Given] \\ ∠ ADC = ∠ ACD [Opposite angles of equal sides are equal.] \\ = y (let) \\ Now, In Δ BCD \\ ∠ BCD + ∠ C BD + ∠ BDC = 180 ° \\ (x + y) + x + y = 180 ° \\ 2 (x + y) = 180 ° \\ (x + y) = \frac{180 °}{2} \\ ∠ BCD = 90 ° \\ \Rightarrow ∆ BCD is right triangle. \end{array}$

Q.15 ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans.

$\begin{array}{l} Given :In Δ ABC, ∠ A = 90° and AB = AC. \\ To find : ∠ B and ∠ C \\ Sol : In Δ ABC, \\ AB = AC [Given] \\ So, ∠ ACB = ∠ ABC = x (let) \\ Since, ∠ A + ∠ B + ∠ C = 180 ° \\ 90 ° + x + x = 180 ° \\ 2 x = 180 ° - 90 ° \\ x = \frac{90 °}{2} = 45 ° \\ So, ∠ B = 45 ° and ∠ C = 45 ° . \end{array}$

Q.16 Show that the angles of an equilateral triangle are 60° each.

Ans.

$\begin{array}{l} Given : ΔABC is an equilateral triangle. \\ To prove : ∠ A = ∠ B = ∠ C = 60 ° \\ Proof : In ΔABC, \\ AB = AC [Given] \end{array}$

$\begin{array}{l} ∠ C = ∠ B . .. (i) [\begin{array}{l} Angles opposite to equal sides \\ are equal. \end{array}] \\ AB = BC [Given] \\ ∠ C = ∠ A . .. (ii) [\begin{array}{l} Angles opposite to equal sides \\ are equal. \end{array}] \\ By angle sum property in a triangle, \\ ∠ A + ∠ B + ∠ C = 180 ° \\ ∠ C + ∠ C + ∠ C = 180 ° [From equation (i) and equation (ii)] \\ 3 ∠ C = 180 ° \\ ∠ C = \frac{180 °}{3} = 60 ° \\ So, ∠ A = 60 ° and ∠ B = 60 ° \\ Therefore, each angle of an equilateral triangle is 60° each. \end{array}$

Q.17

$\begin{array}{l} ΔABC and ΔDBC are two isosceles triangle son the same \\ base BC and vertices A and D are on the same side of BC \\ (see the figure below) . If AD is extended to intersect BC at P, show \\ that \end{array}$ $\begin{array}{l} (i) ΔABD ≅ ΔACD \\ (ii) ΔABP ≅ ΔACP \\ (iii) AP bisects ∠ A as well as ∠ D . \\ (iv) AP is the perpendicular bisector of BC . \end{array}$

Ans.

$\begin{array}{l} Given : ΔABC and ΔDBC are two isosceles triangles on the same \\ base BC and vertices A and D are on the same side of BC . \\ To prove : (i) ΔABD ≅ ΔACD \\ (ii) ΔABP ≅ ΔACP \\ (iii) AP bisects ∠ A as well as ∠ D . \\ (iv) AP is the perpendicular bisector of BC . \\ Proof : (i) In ΔABD and ΔACD \\ AB = AC [Given] \\ DB = DC [Given] \\ AD = AD [Common] \\ ∴ ΔABD ≅ ΔACD [By S . S . S .] \\ ∠ BAD = ∠ CAD [By C . P . C . T .] \end{array}$ $\begin{array}{l} (ii) In ΔABP andΔACP \\ AB = AC [Given] \\ ∠ BAP = ∠ CAP [Proved above] \\ AP = AP [Common] \\ ∴ ΔABP ≅ ΔACP [By S.A.S.] \\ ∠ BAP = ∠ CAP [By C.P.C.T.] \\ BP = CP [By C.P.C.T.] \\ (iii) ∵ ∠ BAP = ∠ CAP [By C.P.C.T.] \\ \Rightarrow AP bisects ∠ A . \\ In ΔBDP andΔ CDP \\ BD = CD [Given] \\ DP = DP [Common] \\ BP = CP [By C.P.C.T.] \\ ∴ ΔBDP ≅ Δ CDP [By S.S.S.] \\ ∠ BDP = ∠ CDP [By C.P.C.T.] \\ \Rightarrow DP bisects ∠ BDC i.e., ∠ D . \\ Thus, AP bisects ∠ A as well as ∠ D . \\ (iv) ∠ APB = ∠ APC [By C.P.C.T.] \\ and ∠ APB + ∠ APC = 180° \\ \Rightarrow ∠ APB + ∠ APB = 180° \\ \Rightarrow 2 ∠ APB = 180 ° \\ \Rightarrow ∠ APB = \frac{180 °}{2} = 90 ° \\ And BP = C P \\ So, AP is perpendicular bisector of BC. \end{array}$

Q.18 AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A.

Ans.

$\begin{array}{l} (i) Given : AD is altitude in isosceles Δ ABC in which AB = AC. \\ To prove : (i) AD bisects BC (ii) AD bisects ∠ A \\ Proof : (i) In Δ ABD and Δ ACD \\ AB = AC [Given] \\ ∠ ADB = ∠ ADC [Each 90°] \\ AD = AD [Common] \\ Δ ABD ≅ Δ ACD [By R.H.S.] \\ So, BD = CD [By C.P.C.T.] \\ Therefore, AD bisects BC. \\ (ii) ∠ BAD = ∠ CAD [By C.P.C.T.] \\ Therefore, AD bisects ∠ A . Hence proved. \end{array}$

Q.19

$\begin{array}{l} Two sides AB and BC and median AM of one triangle ABC are \\ respectively equal to sides PQ and QR and median PN of \\ ΔPQR (see figure below) . Show that : \\ (i) ΔABM ≅ ΔPQN \\ (ii) ΔABC ≅ ΔPQR \end{array}$

Ans.

$\begin{array}{l} Given:In Δ ABC and Δ PQR, AB = PQ, BC = QR and AM = P N . \\ To prove: (i) Δ A B M ≅ Δ P Q N \\ (i i) Δ A B C ≅ Δ P Q R \\ Proof : (i)In Δ ABM and Δ PQN \\ AB = PQ [Given] \\ BM = QN [\frac{1}{2} B C = \frac{1}{2} Q R] \\ AM = P N [G i v e n] \\ ∴ Δ ABC ≅ Δ PQR [B y S .S .S .] \\ ∠ A B M = ∠ P Q N [B y C .P .C .T .] \\ \Rightarrow ∠ B = ∠ Q \\ (i i) I n Δ ABC and Δ PQR \\ AB = PQ [Given] \\ ∠ B = ∠ Q [Proved above] \\ BC = QR [Given] \\ ∴ Δ A B C ≅ Δ P Q R [B y S .A .S .] \\ Hence Proved . \end{array}$

Q.20 BE and CF are two equal altitudes of a triangle ABC.
Show that:
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e ABC is an isosceles triangle.

Ans.

$\begin{array}{l} Given : In Δ ABC, BE ⊥ AC and CF ⊥ AB and BE = CF \\ To prove: Δ ABC is isosceles triangle. \\ Proof:(i) In Δ BCE and Δ CBF \\ ∠ BEC = ∠ CFB [Each 90°] \\ BC = BC [Common] \\ BE = CF [Given] \\ ∴ Δ BCE ≅ Δ CBF [By R.H.S.] \\ (ii) AB = AC [By CPCT] \\ i.e ABC is an isosceles triangle \end{array}$

Q.21 ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Ans.

$\begin{array}{l} Given:In Δ ABC, AB = AC and AP ⊥ BC \\ To prove: Δ ABC is isosceles triangle. \\ Proof:In Δ ABP and ΔA CP \\ ∠ APB = ∠ APC [Each 90°] \\ AB = AC [Given] \\ AP = AP [Common] \\ Δ ABP ≅ ΔA CP [By R.H.S.] \\ ∴ ∠ ABP = ∠ ACP [By C.P.C.T.] \\ \Rightarrow ∠ B = ∠ C \\ Hence Proved. \end{array}$

Q.22 Show that in a right angled triangle, the hypotenuse is the longest side.

Ans.

$\begin{array}{l} Given : In Δ ABC, ∠ B = 90° \\ To prove: AC is the longest side of Δ ABC . \\ Proof:In Δ ABC, \\ ∠ A + ∠ B + ∠ C = 180 ° \\ ∠ A + 90 ° + ∠ C = 180 ° \\ ∠ A + ∠ C = 180 ° - 90 ° \\ ∠ A + ∠ C = 90 ° \\ Hence, the other two angles have to be acute (i . e ., less than 9 0 º) . \\ ∴ ∠ B is the largest angle of the triangle. \\ ∠ B > ∠ A and ∠ B > ∠ C \\ ∴ AC > BC and AC > A B \\ [\begin{array}{l} In any triangle, \\ the side opposite to the larger (greater) angle is longer . \end{array}] \\ Therefore, AC is the largest side of the triangle. \end{array}$

Q.23 In Fig. shown below, sides AB and AC of Δ ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

Ans.

$\begin{array}{l} Given: In Δ ABC, AB and AC are produced upto P and Q respectively. ∠ PBC < ∠ QCB. \\ To prove: AC > AB \\ Proof: Since, ∠ PBC + ∠ ABC = 180° [Linear Pair of angles] \\ ∠ PBC = 180° - ∠ ABC . .. (i) \\ and ∠ QCB + ∠ ACB = 180° [Linear Pair of angles] \\ ∠ QCB = 180° - ∠ ACB . .. (ii) \\ It is given that, ∠ PBC < ∠ QCB \\ 180° - ∠ ABC < 180° - ∠ ACB \\ \Rightarrow - ∠ ABC < - ∠ ACB \\ \Rightarrow ∠ ACB < ∠ ABC \\ \Rightarrow AB < AC [\begin{array}{l} Side opposite to smaller \\ angle is smaller. \end{array}] \\ \Rightarrow AC > AB \\ Hence proved. \end{array}$

Q.24

In Fig. shown, ∠B<∠A and ∠C<∠D. Show that AD
<bc< p=””></bc<>

Ans.

$\begin{array}{l} Given : In Δ OAB, ∠ B < ∠ A and in Δ OCD, ∠ C < ∠ D . \\ To prove: AD < BC \\ Proof : In Δ OAB, \\ ∠ B < ∠ A [Given] \\ \Rightarrow OA < OB . .. (i) [\begin{array}{l} Side opposite to smaller angle is \\ smaller. \end{array}] \\ In Δ OCD, \\ ∠ C < ∠ D [Given] \\ \Rightarrow OD < OC . .. (ii) [\begin{array}{l} Side opposite to smaller angle is \\ smaller. \end{array}] \\ Adding relation (i) and (ii), we get \\ OA + OD < OB + OC \\ \Rightarrow AD < BC Hence Proved. \end{array}$

Q.25 AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see figure below). Show that ∠A > ∠C and ∠B > ∠D.

Ans.

$\begin{array}{l} Given : In quadrilateral ABCD, A B is the smallest side and CD is the largest side. \\ To prove: ∠ A > ∠ C and ∠ B > ∠ D . \end{array}$

$\begin{array}{l} Proof : In Δ ABC, \\ BC > AB [Given] \\ ∠ BAC > ∠ ACB . .. (i) [\begin{array}{l} Angle opposite to greater side is \\ greater. \end{array}] \end{array}$ $\begin{array}{l} In Δ ACD, \\ CD > AD [Given] \\ ∠ CAD > ∠ ACD . .. (ii) [\begin{array}{l} Angle opposite to greater side is \\ greater. \end{array}] \\ Adding relation (i) and relation (ii), we get \\ ∠ BAC + ∠ CAD > ∠ ACB + ∠ ACD \\ ∠ BAD > ∠ BCD \\ \Rightarrow ∠ A > ∠ C . .. (iii) \\ In Δ ABD, \\ AD > AB [Given] \\ ∠ ABD > ∠ ADB . .. (iv) [\begin{array}{l} Angle opposite to greater side is \\ greater. \end{array}] \\ In Δ BCD, \\ CD > BC [Given] \\ ∠ CBD > ∠ BDC . .. (v) [\begin{array}{l} Angle opposite to greater side is \\ greater. \end{array}] \\ Adding relation (iv) and relation (v), we get \\ ∠ ABD + ∠ CBD > ∠ ADB + ∠ BDC \\ ∠ ABC > ∠ ADC \\ \Rightarrow ∠ B > ∠ D . .. (vi) \\ From relation (iii) and relation (vi), we have \\ ∠ A > ∠ C and ∠ B > ∠ D. Hence proved. \end{array}$

Q.26 In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

Ans.

$\begin{array}{l} Given : In Δ PQR, PR > PQ and PS bisects ∠ QPR. \\ To prove: ∠ PSR > ∠ PSQ \\ Proof: In Δ PQR, PS bisects ∠ QPR \\ i . e ., ∠ QPS = ∠ RPS . .. (i) \\ In Δ PQS, \\ Ex. ∠ PSR = ∠ PQS + ∠ QPS \\ \Rightarrow Ex. ∠ PSR = ∠ PQR + ∠ QPS \\ \Rightarrow ∠ PQR = ∠ PSR - ∠ QPS . .. (ii) \\ In Δ PRS, \\ Ex. ∠ PSQ = ∠ PRS + ∠ SPR \\ \Rightarrow Ex. ∠ PSQ = ∠ PRQ + ∠ RPS \\ \Rightarrow ∠ PRQ = ∠ PSQ - ∠ RPS \\ \Rightarrow ∠ PRQ = ∠ PSQ - ∠ QPS . .. (iii) [From equation (i)] \\ and PR > PQ \\ ∠ PQR > ∠ PRQ [\begin{array}{l} Opposite angle of greater side \\ is greater. \end{array}] \\ ∠ PSR - ∠ QPS > ∠ PSQ - ∠ QPS [From equation (ii) and (iii)] \\ ∠ PSR > ∠ PSQ Hence proved. \end{array}$

Q.27 Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Ans.

$\begin{array}{l} Given : A line l and a point A which is away from line l. \\ To prove: Perpendicular line segment from A to line l is the shortest. \\ Contruction: Draw AB ⊥ \overset{\leftrightarrow}{l}, AC and AD. \\ Proof: In ΔABC, ∠ B=90° \\ Then, ∠ A + ∠ C = 90° \\ \Rightarrow ∠ B > ∠ C \\ \Rightarrow AC > AB [Side opposite to greater angle is greater.] \\ Similarly, ∠ B > ∠ D \\ \Rightarrow AD > AB [Side opposite to greater angle is greater.] \\ Therefore, it can be observed that of all line segments drawn \\ from a given point not on it, the perpendicular line segment is \\ the shortest . \end{array}$

Q.28 ABC is a triangle. Locate a point in the interior of Δ ABC which is equidistant from all the vertices of ΔABC.

Ans.

Circumcentre of a triangle is always equidistant from all the vertices of that triangle. Circumcentre is the point where perpendicular bisectors of all the sides of the triangle meet together.

In ∆ABC, we can find the circumcentre by drawing the perpendicular bisectors of sides AB, BC, and CA. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of ∆ABC.

Q.29 In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

Ans.

A point in the interior of a triangle which is equidistant from all the sides is in centre. In centre is a point which is obtained by the intersection of angle bisectors.

Here, in ∆ABC, we can find the in centre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of ∆ABC.

Q.30 In a huge park, people are concentrated at three points (see figure below):

A: where there are different slides and swings for children,
B: near which a man-made lake is situated,
C: which is near to a large parking and exit. Where should an icecream parlour be set up so that maximum number of persons can approach it?

Ans.

Icecream parlour should be set up at equidistant point from three points A, B and C so that maximum number of people will approach there.

When we join three points A, B and C, we get a triangle. A point which is equidistant from three vertex is called Circumcentre.

In figure, Circumcentre O is obtained by intersection of perpendicular bisectors of sides of triangle ABC.

At this point of icecream parlour, maximum number of people will approach.

Q.31 Complete the hexagonal and star shaped Rangolies [see the figure below (i) and (ii)] by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

Ans.

$\begin{array}{l} (i) A r e a (Δ A O B) = \frac{\sqrt{3}}{4} {(s i d e)}^{2} \\ = \frac{\sqrt{3}}{4} {(5)}^{2} \\ = \frac{25 \sqrt{3}}{4} {cm}^{2} \end{array}$

$\begin{array}{l} Area of hexagonal – shaped \\ rangoli = 6 \times A r e a (Δ A O B) \\ = 6 \times \frac{25 \sqrt{3}}{4} {cm}^{2} \\ = \frac{75 \sqrt{3}}{2} {cm}^{2} \\ A r e a o f equilateral triangle having its side as 1 cm \\ = \frac{\sqrt{3}}{4} {(s i d e)}^{2} \\ = \frac{\sqrt{3}}{4} {(1)}^{2} \\ = \frac{\sqrt{3}}{4} {cm}^{2} \\ N u m b e r of equilateral triangles of 1 cm side that can be filled \\ in hexagonal-shaped rangoli \\ = \frac{(\frac{75 \sqrt{3}}{2})}{(\frac{\sqrt{3}}{4})} = 150 \end{array}$

$\begin{array}{l} \end{array}$ $\begin{array}{l} Star - shaped rangoli has 12 equilateral triangles of side 5 cm in it . \\ Area of equilateral triangle \\ = \frac{\sqrt{3}}{4} {(s i d e)}^{2} \\ = \frac{\sqrt{3}}{4} {(5)}^{2} \\ = \frac{25 \sqrt{3}}{4} {cm}^{2} \\ Area of star-shaped rangoli \\ = 12 \times \frac{25 \sqrt{3}}{4} {cm}^{2} \\ = 75 \sqrt{5} {cm}^{2} \\ Area o f equilateral triangle having its side as 1 cm \\ = \frac{\sqrt{3}}{4} {(s i d e)}^{2} \\ = \frac{\sqrt{3}}{4} {(1)}^{2} \\ = \frac{\sqrt{3}}{4} {cm}^{2} \\ Number of equilateral triangles of 1 cm side that can be filled \\ in hexagonal-shaped rangoli \\ = \frac{(75 \sqrt{3} {cm}^{2})}{(\frac{\sqrt{3}}{4})} = 300 \\ Therefore, star-shaped rangoli has more equilateral \\ triangles in it . \end{array}$

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1. What are the important topics covered in NCERT Class 9 Mathematics Chapter 7?

Properties of triangles, congruence of triangles, rules of congruency, equalities of triangles, inequalities of triangles, basic proportionality theorem, similarities of triangles etc are some of the important topics covered in NCERT Class 9 Mathematics Chapter 7.

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You can find chapter related as well as extra information in NCERT Solutions for Class 9 Mathematics Chapter 7 available on Extramarks’ website. Once you refer to it, you will develop stronger conceptual understanding.Extramarks which tries to do away with rote learning and supplements their studies with experiential learning and other innovative educational materials.

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