# 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. Solutions of NCERT 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. CBSE 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.

## 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  ? 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

. It is given that :-

(ii)

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

So, by SAS congruence, we can write : In the previous part, we have proved that .

Thus by  c.p.c.t. , we can write : In the first part we have proved that   .

Thus by  c.p.c.t. , we can conclude : In the given figure consider AOD  and  BOC.

(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. 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 : 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 :

Q5 (ii) Line is the bisector of an angle and   is any point on . and are perpendiculars from to the arms of (see Fig. ). Show that:  or is equidistant from the arms of  . 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. 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 : 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 : In the previous part we have proved that .

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

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.

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 . (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. 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. Consider   and   ,

(i)      is common in both the triangles.

(ii)                  (Right angles)

(iii)                                (Given)

Thus by  AAS congruence, we can say that : 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. 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 : Consider ABC,
It is given that   AB = AC

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

Similarly in ACD,

and
So,

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

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

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

or

and

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

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 Consider   and    ,

(i)               (Common)

(ii)               (Isosceles triangle)

(iii)             (Isosceles triangle)

Thus by SSS congruency we can conclude that :

Q1 (ii)  Triangles ABC and Triangle DBC are two isosceles triangles on the same base BC and vertices A  and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that 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 : 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 . 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.

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.

In the previous part of the question we have proved that

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

Hence   bisects   .

(i)

(ii) (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,

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.

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

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. 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.) 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.  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. 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 :

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

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.

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.

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? (Hint : The parlour should be equidistant from A, B and C) 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 . 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

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## 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!