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# NCERT Solutions for Class 9 Maths Chapter 7 - Exercise LIVE
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## NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.2) Exercise 7.2

Free PDF download of NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 (Ex 7.2) and all chapter exercises at one place prepared by an expert teacher as per NCERT (CBSE) books guidelines. Class 9 Maths Chapter 7 Triangles Exercise 7.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Download Maths NCERT Solutions Class 9 here at Vedantu. Students can also avail of NCERT Class 9 Science from our website. Besides, find NCERT Solutions to get more understanding of various subjects. The solutions are up-to-date and are sure to help in your academic journey.

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NCERT Solutions for Class 9 Maths Chapter 7 - Exercise TRIANGLES L-1 (𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 𝐨𝐟 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬 & 𝐂𝐫𝐢𝐭𝐞𝐫𝐢𝐚 𝐟𝐨𝐫 𝐂𝐨𝐧𝐠𝐫𝐮𝐞𝐧𝐜𝐞 𝐨𝐟 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬) CBSE 9 Maths -Term 1
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## Access NCERT Solutions for Class 9 Maths Chapter 7 – Triangles

Exercise 7.2

1. In an isosceles triangle $\text{ABC}$, with $\text{AB}=\text{AC}$, the bisectors of $\angle \text{B}$ and $\angle \text{C}$ intersect each other at $O$. Join $\text{A}$ to $\text{O}$. Show that:

(i) $\text{OB}=\text{OC}$

(ii) $AO$ bisects $\angle \text{A}$

Ans:

(i) It is given that in triangle $\text{ABC}$ is an isosceles triangle with $\text{AB}=\text{AC}$.

$\therefore \,\,\angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are equal)

Multiplying $\frac{1}{2}$ on both sides, we get

$\frac{1}{2}\angle \text{ACB}=\frac{1}{2}\angle \text{ABC}$

$\angle \text{OCB}=\angle \text{OBC}$

$\therefore \text{OB}=\text{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\Delta \text{OAB}$ and $\Delta \text{OAC}$,

Side AO is present in both triangle. Therefore,

$\text{AO}=\text{AO}$ (Common)

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\text{OB}=\text{OC}$ (Proved above)

Therefore,

$\Delta \text{OAB}\cong \Delta \text{OAC}$ (By SSS axiom of congruency)

Therefore, $\angle \text{BAO}=\angle \text{CAO}$.

As they are corresponding parts of the corresponding triangle.

$\therefore$ AO bisects $\angle \text{A}$.     ($\angle \text{BAO}=\angle \text{CAO}$)

2. In $\Delta \text{ABC},\text{AD}$ is the perpendicular bisector of $\text{BC}$ (see the given figure). Show that $\vartriangle \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$.

Ans:

In $\Delta \text{ADC}$ and $\Delta \text{ADB}$,

Side AD is present in both triangle.

$\therefore \,\,\text{AD}=\text{AD}$ (Common)

Since AD is the perpendicular bisector of BC.

$\angle ADC=\angle ADB={{90}^{{}^\circ }}$ and

$\text{CD}=\text{BD}$

$\therefore \Delta \text{ADC}\cong \Delta \text{ADB}$ (By SAS axiom of congruency)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

Hence, $\Delta ABC$ is an isosceles triangle as two of it’s sides are equal, i.e., $AB=AC$.

3. $\text{ABC}$ is an isosceles triangle in which altitudes $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$ respectively (see the given figure). Show that these altitudes are equal.

Ans:

In $\Delta \text{AEB}$ and $\Delta \text{AFC}$,

Since $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$. Therefore,

$\angle \text{AEB}$ and $\angle \text{AFC}$ (Each ${{90}^{{}^\circ }}$)

As $\angle A$ lies in both the triangles. Therefore,

$\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side AB is equal to AC, i.e.,$\text{AB}=\text{AC}$ (Given)

$\therefore \Delta \text{AEB}\cong \Delta \text{AFC}$ (By AAS axiom of congruency)

Therefore, $\text{BE}=\text{CF}$.

As they are corresponding parts of the corresponding triangle.

4.$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$ are equal (see the given figure). Show that

(i) $\Delta \text{ABE}\cong \Delta \text{ACF}$

(ii) $\text{AB}=\text{AC}$, ie., $\text{ABC}$ is an isosceles triangle.

Ans:

(i) In $\Delta \text{ABE}$ and $\Delta \text{ACF}$,

$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$.

Therefore, $\angle \text{ABE}$ and $\angle ACF={{90}^{{}^\circ }}$

Since $\angle A$ is present in both triangles,

$\therefore \,\,\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side BE is equal to CF, i.e., $\text{BE}=\text{CF}$ (Given)

$\therefore \Delta \text{ABE}\cong \Delta \text{ACF}$ (By AAS axiom of congruency)

(ii) Since $\Delta \text{ABE}\cong \Delta \text{ACF}$   (proved above)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

5. $\text{ABC}$ and $\text{DBC}$ are two isosceles triangles on the same base $\text{BC}$ (see the given figure). Show that $\angle \text{ABD}=\angle \text{ACD}$.

Ans:

Construction: join $\text{AD}$.

In $\Delta \text{ABD}$ and $\Delta \text{ACD}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

Length of side BD is equal to CD, i.e., $\text{BD}=\text{CD}$ (Given)

As side AD is present in both triangles. Therefore,

$\text{AD}=\text{AD}$ (Common side)

$\therefore \Delta \text{ABD}\cong \Delta \text{ACD}$ (By SSS axiom of congruency)

Therefore, $\angle \text{ABD}=\angle \text{ACD}$.

As they are corresponding parts of the corresponding triangle.

6. $\Delta \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$. Side $\text{BA}$ is produced to $\text{D}$ such that $\text{AD}=\text{AB}$ (see the given figure). Show that $\angle \text{BCD}$ is a right angle.

Ans:

In $\Delta \text{ABC}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\therefore \angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{ACD}$,

Length of side AC is equal to AD, i.e., $\text{AC}=\text{AD}$  (Given)

$\therefore \angle \text{ADC}=\angle \text{ACD}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{BCD}$,

$\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{ACB}+\angle \text{ACB}+\angle \text{ACD}+\angle \text{ACD}={{180}^{{}^\circ }}$  ($\angle BCD=\angle BCA+\angle ACD$)

$2(\angle \text{ACB}+\angle \text{ACD})={{180}^{{}^\circ }}$

$2(\angle \text{BCD})={{180}^{{}^\circ }}$

$\therefore \angle \text{BCD}={{90}^{{}^\circ }}$

7. $\text{ABC}$ is a right-angled triangle in which $\angle \text{A}={{90}^{{}^\circ }}$ and $\text{AB}=\text{AC}$. Find $\angle \text{B}$ and $\angle \text{C}$.

Ans:

It is given that the side $\text{AB}=\text{AC}$

Angles opposite to equal sides are also equal

$\therefore \angle \text{C}=\angle \text{B}$  …(i)

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$  …from equation (i)

${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{B}={{180}^{{}^\circ }}$

$2\angle \text{B}={{90}^{{}^\circ }}$

$\angle \text{B}={{45}^{{}^\circ }}$

\begin{align} & \because \angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,\,\,\,...\text{from equation}\,\left( i \right) \\ & \therefore \angle \text{C}={{45}^{{}^\circ }} \\ \end{align}

8. Show that the angles of an equilateral triangle are ${{60}^{{}^\circ }}$ each.

Ans:

Consider that $\text{ABC}$ is an equilateral triangle in which all sides are equal.

Therefore, $\text{AB}=\text{BC}=\text{AC}$

$\text{AB}=\text{AC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{C}=\angle \text{B}\,\,\,\,\,\,...\left( i \right)$

Also, $\text{AC}=\text{BC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{B}=\angle \text{A}\,\,\,\,\,\,\,\,\,...\left( ii \right)$

Therefore, from equation (i) and (ii), we get

$\angle \text{A}=\angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,...\left( iii \right)$

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$   (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{A}+\angle \text{A}+\angle \text{A}={{180}^{{}^\circ }}$     …from equation (iii)

$3\angle \text{A}={{180}^{{}^\circ }}$

$\angle \text{A}={{60}^{{}^\circ }}$

$\therefore \angle \text{A}=\angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}$

From above we can conclude that all the interior angles of an equilateral triangle has an angle of ${{60}^{{}^\circ }}$.

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Opting for the NCERT solutions for Ex 7.2 Class 9 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 7.2 Class 9 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 9 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 9 Maths Chapter 7 Exercise 7.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 9 Maths Chapter 7 Exercise 7.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it.

Do not delay any more. Download the NCERT solutions for Class 9 Maths Chapter 7 Exercise 7.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

## FAQs on NCERT Solutions for Class 9 Maths Chapter 7 - Exercise

1. How many exercises are made available for class 9 Maths Chapter 7?

Triangles is one of the most important chapters in the Class 9 syllabus. These form the basis for further chapters regarding trigonometry and related chapters. Therefore students must ensure that they know all the basic concepts of triangles in detail.

Following are the exercises provided in this chapter for the benefit of the students.

• Exercise 7.1 Solution -  8 Questions (6 Short Answer Questions, 2 Long Answer Question)

• Exercise 7.3 Solution - 5 Questions (3 Short Answer Questions, 2 Long Answer Question)

• Exercise 7.4 Solution - 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

• Exercise 7.5 (Optional) Solution - 4 Questions

Students must ensure that they solve all the above in order to perform better in the examination.

2. What are the important concepts covered in the Triangle chapters?

Students need to learn a lot of concepts in order to be able to solve and understand concepts in higher classes. Following are the concepts that you will learn in this chapter:

• Congruence of triangles.

• Criteria for congruence of triangles (SAS congruence rule, ASA congruence rule, SSS congruence rule, RHS congruence rule).

• Properties of triangles.

• Inequalities of the triangle.

These topics help students understand the concept of the triangle better. Understanding triangles and its properties are very important as it carries high marks percentage in the exam paper as well. Therefore students must ensure that they solve all the exercises in order to perform better in the exam.

3. What are the marks distribution for the Triangle chapter in class 9 syllabus?

The marks distribution for the triangles chapter in class 9 examination is as follows:

• Topics: Introduction to Euclid’s Geometry, Lines and Angles, Triangles, Quadrilaterals, Areas, Circles, Constructions

• Multiple Choice Questions - 4 Qs of 1 mark each

• Short Answer Questions - 2 Qs of 3 marks each

• Long Answer Questions - 2 Qs of 6 marks each

You must not just mug up the theorems and equations related to triangles. You must also be able to apply the concepts and use them to solve all kinds of triangle problems. Also, students must learn to prove the theorems as they will also be asked to prove the same in the exam paper.

4. What are the theorems covered in the class 9 maths chapter 7 exercise 7.2?

The exercise of class 9 maths chapter 7 exercise 7.2 covers the properties of Triangles along with the following theorems:

• Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.

• Theorem 7.3: The sides opposite to equal angles of a triangle are equal.

These theorems can be proved by the congruence rule that students have learnt in the previous exercise.  Therefore students must ensure that they learn all the theorems in detail. Also, they need to learn problems regarding these theorems. In case they are unable to understand anything, they can always refer to the NCERT solutions provided here, by Vedantu. Vedantu is the top online education platform in India. These solutions are prepared by subject matter experts in a simplified and easy manner in order to make the process of studying easy for the students. Apart from NCERT solutions, they also provide solved previous years question papers, sample papers and related syllabus to aid students in their study process.