# NCERT Solutions for Class 9 Maths Chapter 7 Triangles

**NCERT solutions for class 9 maths chapter 7 Triangles- **A geometrical figure is made up of three straight lines is known as Triangles. A triangle has three vertices, three sides, and three angles. In the previous classes, you have studied the construction and types of triangles. You have studied briefly about triangles and their properties in the previous classes. In Class 9 Triangles, you will learn something of a higher level. NCERT solutions for class 9 maths chapter 7 Triangles can be a good tool whenever you are stuck in any of the problems. This chapter will be covering the properties of triangles like congruence of triangles, isosceles triangle, etc in detail.

In this chapter, there are a total of 5 exercises with 31 questions in them. NCERT solutions for class 9 maths chapter 7 Triangles is covering the entire chapter including the optional exercises. The chapter is full of properties and theorems that's why the examples and theorems are as important as the practice exercises. There is another aspect to look at the importance of this chapter, apart from school exams this is an essential part of competitive examinations like- CAT, SSC, NTSE, INO, etc. By using NCERT solutions for class 9 maths chapter 7 Triangles, you can prepare 360 degree for your school as well as for the competitive examinations. If you need NCERT Solutions for other classes and subjects then you can download all these for free by clicking on the given link. Here you will get NCERT solutions for class 9 also.

## ** NCERT solutions for class 10 maths chapter 7 Triangles Excercise: 7.1 **

** Q1 ** In quadrilateral , and bisects (see Fig.). Show that . What can you say about and ?

** Answer: **

In the given triangles we are given that:-

(i)

(ii) Further, it is given that AB bisects angle A. Thus BAC BAD.

(iii) Side AB is common in both the triangles.

Hence by SAS congruence, we can say that :

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that

** Q2 (i) ** is a quadrilateral in which and (see Fig. ). Prove that

** Answer: **

It is given that :-

(i) AD = BC

(ii)

(iii) Side AB is common in both the triangles.

So, by SAS congruence, we can write :

** Q2 (ii) ** is a quadrilateral in which and (see Fig.). Prove that

** Answer: **

In the previous part, we have proved that .

Thus by ** c.p.c.t. ** , we can write :

** Q2 (iii) ** is a quadrilateral in which and (see Fig.). Prove that .

** Answer: **

In the first part we have proved that .

Thus by ** c.p.c.t. ** , we can conclude :

** Q3 ** and are equal perpendiculars to a line segment (see Fig.). Show that bisects .

** Answer: **

In the given figure consider AOD and BOC.

(i) AD = BC (given)

(ii) A = B (given that the line AB is perpendicular to AD and BC)

(iii) AOD = BOC (vertically opposite angles).

Thus by AAS Postulate, we have

Hence by c.p.c.t. we can write :

And thus CD bisects AB.

** Answer: **

In the given figure, consider ABC and CDA :

(i)

(ii)

(iii) Side AC is common in both the triangles.

Thus by ASA congruence, we have :

** Answer: **

In the given figure consider and _{} ,

(i) (Right angle)

(ii) (Since it is given that I is bisector)

(iii) Side AB is common in both the triangle.

Thus AAS congruence, we can write :

** Answer: **

In the previous part we have proved that _{} .

Thus by c.p.c.t. we can write :

Thus B is equidistant from arms of angle A.

** Q6 ** In Fig, _{} and . Show that .

** Answer: **

From the given figure following result can be drawn :-

Adding to the both sides, we get :

Now consider and , :-

(i) (Given)

(ii) (proved above)

(iii) (Given)

Thus by SAS congruence we can say that :

Hence by c.p.c.t., we can say that :

** Answer: **

From the figure, it is clear that :

Adding both sides, we get :

or

Now, consider and :

(i) (Proved above)

(ii) (Since P is the midpoint of line AB)

(iii) (Given)

Hence by ASA congruence, we can say that :

** Answer: **

In the previous part we have proved that .

Thus by c.p.c.t., we can say that :

** Answer: **

Consider and ,

(i) (Since M is the mid-point)

(ii) (Vertically opposite angles are equal)

(iii) (Given)

Thus by SAS congruency, we can conclude that :

** Answer: **

In the previous part, we have proved that .

By c.p.c.t. we can say that :

This implies side AC is parallel to BD.

Thus we can write : (Co-interior angles)

and,

or

Hence is a right angle.

** Answer: **

Consider and ,

(i) (Common in both the triangles)

(ii) (Right angle)

(iii) (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :

** Answer: **

In the previous part we have proved that .

Thus by c.p.c.t., we can write :

or (Since M is midpoint.)

or .

Hence proved.

## ** NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.2 **

** Answer: **

In the triangle ABC,

Since AB = AC, thus

or

or (Angles bisectors are equal)

Thus as sides opposite to equal are angles are also equal.

** Answer: **

Consider and ,

(i) (Given)

(ii) (Common in both the triangles)

(iii) (Proved in previous part)

Thus by SSS congruence rule, we can conclude that :

Now, by c.p.c.t.,

Hence AO bisects .

** Q2 ** In , AD is the perpendicular bisector of BC (see Fig). Show that is an isosceles triangle in which .

** Answer: **

Consider ABD and ADC,

(i) (Common in both the triangles)

(ii) (Right angle)

(iii) (Since AD is the bisector of BC)

Thus by SAS congruence axiom, we can state :

Hence by c.p.c.t., we can say that :

Thus is an isosceles triangle with AB and AC as equal sides.

** Answer: **

Consider and ,

(i) is common in both the triangles.

(ii) (Right angles)

(iii) (Given)

Thus by AAS congruence axiom, we can conclude that :

Now, by c.p.c.t. we can say :

Hence these altitudes are equal.

** Q4 (i) ** ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that

** Answer: **

Consider and ,

(i) is common in both the triangles.

(ii) (Right angles)

(iii) (Given)

Thus by AAS congruence, we can say that :

** Answer: **

From the prevoius part of the question we found out that :

Now, by c.p.c.t. we can say that :

Hence is an isosceles triangle.

** Q5 ** ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that .

** Answer: **

Consider and ,

(i) (Common in both the triangles)

(ii) (Sides of isosceles triangle)

(iii) (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

** Answer: **

Consider ABC,

It is given that AB = AC

So, (Since angles opposite to the equal sides are equal.)

Similarly in ACD,

We have AD = AB

and

So,

or ...........................(i)

And in ADC,

..............................(ii)

Adding (i) and (ii), we get :

or

and

** Q7 ** ABC is a right angled triangle in which and . Find and .

** Answer: **

In the triangle ABC, sides AB and AC are equal.

We know that angles opposite to equal sides are also equal.

Thus,

Also, the sum of the interior angles of a triangle is .

So, we have :

or

or

Hence

** Q8 ** Show that the angles of an equilateral triangle are each.

** Answer: **

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write :

Also, the sum of the interior angles of a triangle is .

Hence,

or

or

So, all the angles of the equilateral triangle are equal ( ).

## ** NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.3 **

** Answer: **

Consider and ,

(i) (Common)

(ii) (Isosceles triangle)

(iii) (Isosceles triangle)

Thus by SSS congruency we can conclude that :

** Answer: **

Consider and ,

(i) is common side in both the triangles.

(ii) (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii) (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

** Answer: **

In the first part, we have proved that .

So, by c.p.c.t. .

Hence AP bisects .

Now consider and ,

(i) (Common)

(ii) (Isosceles triangle)

(iii) (by c.p.c.t. from the part (b))

Thus by SSS congruency we have :

Hence by c.p.c.t. we have :

or AP bisects .

** Answer: **

In the previous part we have proved that .

Thus by c.p.c.t. we can say that :

Also,

SInce BC is a straight line, thus :

or

or

Hence it is clear that AP is a perpendicular bisector of line BC.

** Q2 (i) ** AD is an altitude of an isosceles triangle ABC in which . Show that AD bisects BC

** Answer: **

Consider and ,

(i) (Given)

(ii) (Common in both triangles)

(iii)

Thus by RHS axiom we can conclude that :

Hence by c.p.c.t. we can say that : or AD bisects BC.

** Q2 (ii) ** AD is an altitude of an isosceles triangle ABC in which . Show that AD bisects .

** Answer: **

In the previous part of the question we have proved that

Thus by c.p.c.t., we can write :

Hence bisects .

** Answer: **

(i) From the figure we can say that :

or

or

Now, consider and ,

(a) (Given)

(b) (Given)

(c) (Prove above)

Thus by SSS congruence rule, we can conclude that :

(ii) Consider and :

(a) (Given)

(b) (by c.p.c.t. from the above proof)

(c) (Given)

Thus by SAS congruence rule,

** Answer: **

Using the given conditions, consider and ,

(i) (Right angle)

(ii) (Common in both the triangles)

(iii) (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that :

Hence by c.p.c.t.,

And thus (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

** Q5 ** ABC is an isosceles triangle with . Draw to show that .

** Answer: **

Consider and ,

(i) (Since it is given that AP is altitude.)

(ii) (Isosceles triangle)

(iii) (Common in both triangles)

Thus by ** RHS axiom ** we can conclude that :

Now, by c.p.c.t.we can say that :

## ** NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.4 **

** Q1 ** Show that in a right-angled triangle, the hypotenuse is the longest side.

** Answer: **

Consider a right-angled triangle ABC with right angle at A.

We know that the sum of the interior angles of a triangle is 180.

So,

or

or

Hence and are less than ( ).

Also, the side opposite to the largest angle is also the largest.

Hence the side BC is largest is the hypotenuse of the .

Hence it is proved that in a right-angled triangle, the hypotenuse is the longest side.

** Q2 ** In Fig, sides AB and AC of are extended to points P and Q respectively. Also, . Show that .

** Answer: **

We are given that,

......................(i)

Also, (Linear pair of angles) .....................(ii)

and (Linear pair of angles) .....................(iii)

From (i), (ii) and (iii) we can say that :

Thus ( Sides opposite to the larger angle is larger.)

** Answer: **

In this question, we will use the property that sides opposite to larger angle are larger.

We are given and .

Thus, ..............(i)

and ...............(ii)

Adding (i) and (ii), we get :

or

Hence proved.

** Answer: **

Consider in the above figure :

(Given)

Thus (as angle opposite to smaller side is smaller)

Now consider ,

We have :

and

Adding the above result we get,

or

Similarly, consider ,

we have

Therefore

and in we have,

and

from the above result we have,

or

Hence proved.

** Q5 ** In Fig , and PS bisects . Prove that .

** Answer: **

We are given that .

Thus

Also, PS bisects , thus :

Now, consider ,

(Exterior angle)

Now, consider ,

Thus from the above the result we can conclude that :

** Answer: **

Consider a right-angled triangle ABC with right angle at B.

Then (Since )

Thus the side opposite to largest angle is also largest.

Hence the given statement is proved that all line segments are drawn from a given point, not on it, the perpendicular line segment is the shortest.

** NCERT solutions for class 9 maths chapter 7 Triangles Excercise: 7.5 **

** Q1 ** ABC is a triangle. Locate a point in the interior of which is equidistant from all the vertices of .

** Answer: **

We know that ** circumcenter ** of a triangle is equidistant from all the vertices. Also, circumcenter is the point of intersection of the perpendicular bisectors of the sides of a triangle.

Thus, draw perpendicular bisectors of each side of the triangle ABC. And let them meet at a point, say O.

Hence O is the required point which is equidistant from all the vertices.

** Answer: **

The required point is called ** in-centre ** of the triangle. This point is the intersection of the angle bisectors of the interior angles of a triangle.

Hence the point can be found out in this case just by drawing angle bisectors of all the angles of the triangle.

** Q3 ** In a huge park, people are concentrated at three points (see Fig.):

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.

** Answer: **

The three main points form a triangle ABC. Now we have to find a point which is equidistant from all the three points.

Thus we need to find the circumcenter of the .

We know that circumcenter is defined as the point as the intersection point of the perpendicular bisectors of the sides of the triangle.

Hence the required point can be found out by drawing perpendicular bisectors of .

** Answer: **

For finding the number of triangles we need to find the area of the figure.

Consider the hexagonal structure :

Area of hexagon = 6 Area of 1 equilateral

Thus area of the equilateral triangle :

or

or

So, the area of the hexagon is :

And the area of an equilateral triangle having 1cm as its side is :

or

Hence a number of equilateral triangles that can be filled in hexagon are :

Similarly for star-shaped rangoli :

Area :

Thus the number of equilateral triangles are :

Hence star-shaped rangoli has more equilateral triangles.

** NCERT solutions for class 9 maths chapter wise **

** NCERT solutions for class 9 subject wise **

** **

** How to use NCERT solutions for class 9 maths chapter 7 Triangles **

- Learn some basic properties of triangles using some books of previous classes.
- Go through all the theorems and examples given in the chapter.
- Once you have memorized all the theorems and properties, then you can practice the questions from practice exercises
- During the practice, you can use NCERT solutions for class 9 maths chapter 7 Triangles as a helpmate.
- After doing all the exercises you can do some questions from past papers.

* Keep Working Hard and Happy Learning! *

## Frequently Asked Question (FAQs) - NCERT Solutions for Class 9 Maths Chapter 7 Triangles

**Question: **What are the important topics in chapter Triangles ?

**Answer: **

Congruence of triangles, Criteria for congruence of triangles, Properties of triangles, Inequalities of triangles are the important topics covered in this chapter.

**Question: **Does CBSE provides the solutions of NCERT for class 9 maths ?

**Answer: **

No, CBSE doesn’t provide NCERT solutions for any class or subject.

**Question: **Where can I find the complete solutions of NCERT for class 9 maths ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 maths by clicking on the link.

**Question: **Does CBSE class 9 maths is tougher than CBSE class 10 maths ?

**Answer: **

CBSE class 9 maths is base for the class 10 maths. Some topics are new in this class, so some students find it tougher. Overall class 9 maths is not tougher, it's just that we study maths in the previous class was very basic, so may find some difficulties initially.

**Question: **Where can I find the complete solutions of NCERT for class 9 ?

**Answer: **

Here you will get the detailed NCERT solutions for class 9 by clicking on the link.

**Question: **How many chapters are there in the CBSE class 9 maths ?

**Answer: **

There are 15 chapters starting from the number system to probability in the CBSE class 9 maths.

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