# 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.

### Answer:

In the given triangles we are given that:-

(i)  $\small AC=AD$

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

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

Hence by SAS congruence, we can say that  :   $\small \Delta ABC\cong \Delta ABD$

By c.p.c.t. (corresponding parts of congruent triangles are equal) we can say that$BC\ =\ BD$

$\small \Delta ABD\cong \Delta BAC$.

### Answer:

It is given that :-

(i)  AD  =  BC

(ii)  $\small \angle DAB= \angle CBA$

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

So, by SAS congruence, we can write :

$\small \Delta ABD\cong \Delta BAC$

### Answer:

In the previous part, we have proved that $\small \Delta ABD\cong \Delta BAC$.

Thus by  c.p.c.t. , we can write :  $\small BD=AC$

### Answer:

In the first part we have proved that   $\small \Delta ABD\cong \Delta BAC$.

Thus by  c.p.c.t. , we can conclude :

$\small \angle ABD= \angle BAC$

### Answer:

In the given figure consider $\Delta$AOD  and  $\Delta$BOC.

(i)      AD  =  BC           (given)

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

(iii)    $\angle$ AOD  =  $\angle$ BOC       (vertically opposite angles).

Thus by AAS Postulate, we have

$\Delta AOD\ \cong \ \Delta BOC$

Hence by c.p.c.t. we can write :                  $AO\ =\ OB$

And thus CD bisects AB.

### Answer:

In the given figure, consider $\Delta$ ABC and $\Delta$ CDA :

(i)    $\angle\ BCA\ =\ \angle DAC$

(ii)    $\angle\ BAC\ =\ \angle DCA$

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

Thus by ASA congruence, we have :

$\Delta ABC\ \cong \ \Delta CDA$

### Answer:

In the given figure consider $\small \Delta APB$   and   $\small \Delta AQB$,

(i)      $\angle P\ =\ \angle Q$                      (Right angle)

(ii)      $\angle BAP\ =\ \angle BAQ$       (Since it is given that I is bisector)

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

Thus AAS congruence, we can write :

$\small \Delta APB\cong \Delta AQB$

### Answer:

In the previous part we have proved that    $\small \Delta APB\cong \Delta AQB$.

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

$BP\ =\ BQ$

Thus B is equidistant from arms of angle A.

### Answer:

From the given figure following result can be drawn :-

$\angle BAD\ =\ \angle EAC$

Adding $\angle DAC$ to the both sides, we get :

$\angle BAD\ +\ \angle DAC\ =\ \angle EAC\ +\ \angle DAC$

$\angle BAC\ =\ \angle EAD$

Now consider $\Delta ABC$  and   $\Delta ADE$ ,    :-

(i)               $AC\ =\ AE$                          (Given)

(ii)   $\angle BAC\ =\ \angle EAD$                      (proved above)

(iii)          $AB\ =\ AD$                              (Given)

Thus by SAS congruence we can say that :

$\Delta ABC\ \cong \ \Delta ADE$

Hence by c.p.c.t., we can say that :                $BC\ =\ DE$

### Answer:

From the figure, it is clear that :
$\angle EPA\ =\ \angle DPB$

Adding $\angle DPE$ both sides, we get :

$\angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE$

or                                                            $\angle DPA =\ \angle EPB$

Now, consider $\Delta DAP$ and   $\Delta EBP$ :

(i)   $\angle DPA =\ \angle EPB$                 (Proved above)

(ii)    $AP\ =\ BP$                             (Since P is the midpoint of line AB)

(iii)   $\small \angle BAD=\angle ABE$                    (Given)

Hence by ASA congruence, we can say that :

$\small \Delta DAP\cong \Delta EBP$

### Answer:

In the previous part we have proved that $\small \Delta DAP\cong \Delta EBP$.

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

$\small AD=BE$

### Answer:

Consider $\Delta AMC$   and   $\Delta BMD$  ,

(i)    $AM\ =\ BM$          (Since M is the mid-point)

(ii)    $\angle CMA\ =\ \angle DMB$             (Vertically opposite angles are equal)

(iii)    $CM\ =\ DM$           (Given)

Thus by SAS congruency, we can conclude that :

$\small \Delta AMC\cong \Delta BMD$

### Answer:

In the previous part, we have proved that   $\small \Delta AMC\cong \Delta BMD$.

By c.p.c.t. we can say that    :            $\angle ACM\ =\ \angle BDM$

This implies side AC is parallel to BD.

Thus we can write :            $\angle ACB\ +\ \angle DBC\ =\ 180^{\circ}$          (Co-interior angles)

and,                                            $90^{\circ}\ +\ \angle DBC\ =\ 180^{\circ}$

or                                                                 $\angle DBC\ =\ 90^{\circ}$

Hence $\small \angle DBC$ is a right angle.

### Answer:

Consider $\Delta DBC$  and  $\Delta ACB$,

(i)          $BC\ =\ BC$          (Common in both the triangles)

(ii)   $\angle ACB\ =\ \angle DBC$                    (Right angle)

(iii)         $DB\ =\ AC$            (By c.p.c.t. from the part (a) of the question.)

Thus SAS congruence we can conclude that :

$\small \Delta DBC\cong \Delta ACB$

### Answer:

In the previous part we have proved that  $\Delta DBC\ \cong \ \Delta ACB$.

Thus by c.p.c.t., we can write :        $DC\ =\ AB$

$DM\ +\ CM\ =\ AM\ +\ BM$

or                                  $CM\ +\ CM\ =\ AB$                                              (Since M is midpoint.)

or                                                      $\small CM=\frac{1}{2}AB$.

Hence proved.

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

### Answer:

In the triangle ABC,

Since AB  =  AC, thus        $\angle B\ =\ \angle C$

or                                        $\frac{1}{2}\angle B\ =\ \frac{1}{2}\angle C$

or                                      $\angle OBC\ =\ \angle OCB$      (Angles bisectors are equal)

Thus       $\small OB=OC$   as sides opposite to equal are angles are also equal.

### Answer:

Consider $\Delta AOB$  and  $\Delta AOC$,

(i)   $AB\ =\ AC$                  (Given)

(ii)   $AO\ =\ AO$            (Common in both the triangles)

(iii)   $OB\ =\ OC$          (Proved in previous part)

Thus by  SSS congruence rule, we can conclude that  :

$\Delta AOB\ \cong \ \Delta AOC$

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

$\angle BAO\ =\ \angle CAO$

Hence  AO bisects $\angle A$.

### Answer:

Consider $\Delta$ABD  and  $\Delta$ADC,

(i)  $AD\ =\ AD$                       (Common in both the triangles)

(ii)   $\angle ADB\ =\ \angle ADC$       (Right angle)

(iii)  $BD\ =\ CD$                      (Since AD is the bisector of BC)

Thus by SAS congruence axiom, we can state :

$\Delta ADB\ \cong \ \Delta ADC$

Hence by c.p.c.t., we can say that :           $\small AB=AC$

Thus   $\Delta ABC$ is an isosceles triangle with AB and AC as equal sides.

### Answer:

Consider $\Delta AEB$   and  $\Delta AFC$,

(i)     $\angle A$    is common in both the triangles.

(ii)   $\angle AEB\ =\ \angle AFC$           (Right angles)

(iii)    $AB\ =\ AC$              (Given)

Thus by AAS congruence axiom, we can conclude that :

$\Delta AEB\ \cong \Delta AFC$

Now, by c.p.c.t. we can say :     $BE\ =\ CF$

Hence these altitudes are equal.

### Answer:

Consider $\Delta ABE$  and  $\Delta ACF$ ,

(i)    $\angle A$  is common in both the triangles.

(ii)    $\angle AEB\ =\ \angle AFC$              (Right angles)

(iii)   $BE\ =\ CF$                             (Given)

Thus by  AAS congruence, we can say that :

$\small \Delta ABE \cong \Delta ACF$

### Answer:

From the prevoius part of the question we found out that  :        $\Delta ABE\ \cong \Delta ACF$

Now, by c.p.c.t. we can say that :           $AB\ =\ AC$

Hence $\Delta \ ABC$  is an isosceles triangle.

### Answer:

Consider $\Delta ABD$  and   $\Delta ACD$,

(i)   $AD\ =\ AD$    (Common in both the triangles)

(ii)   $AB\ =\ AC$         (Sides of isosceles triangle)

(iii)   $BD\ =\ CD$       (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that :

$\small \angle ABD\ \cong \ \angle ACD$

### Answer:

Consider $\Delta$ABC,
It is given that   AB = AC

So, $\angle ACB = \angle ABC$    (Since angles opposite to the equal sides are equal.)

Similarly in $\Delta$ACD,

We have       AD = AB
and       $\angle ADC = \angle ACD$
So,

$\angle CAB + \angle ACB + \angle ABC = 180^{\circ}$

$\angle CAB\ +\ 2\angle ACB = 180^{\circ}$
or                                                                 $\angle CAB\ = 180^{\circ}\ -\ 2\angle ACB$                  ...........................(i)

And in $\Delta$ADC,
$\angle CAD\ = 180^{\circ}\ -\ 2\angle ACD$              ..............................(ii)

Adding (i) and (ii), we get :
$\angle CAB\ +\ \angle CAD\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

or                                                                              $180^{\circ}\ = 360^{\circ}\ -\ 2\angle ACD\ -\ 2\angle ACB$

and                                                                              $\angle BCD\ =\ 90^{\circ}$

### Answer:

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

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

Thus,            $\angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180^{\circ}$.

So, we have :

$\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

or                                                   $90^{\circ} +\ 2\angle B\ =\ 180^{\circ}$

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

Hence                        $\angle B\ =\ \angle C\ =\ 45^{\circ}$

### Answer:

Consider a triangle ABC which has all sides equal.

We know that angles opposite to equal sides are equal.

Thus we can write :                    $\angle A\ =\ \angle B\ =\ \angle C$

Also, the sum of the interior angles of a triangle is $180 ^{\circ}$.

Hence,                                        $\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

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

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

So, all the angles of the equilateral triangle are equal ($60^{\circ}$).

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

### Answer:

Consider $\Delta ABD$  and   $\Delta ACD$ ,

(i)  $AD\ =\ AD$             (Common)

(ii)   $AB\ =\ AC$            (Isosceles triangle)

(iii)   $BD\ =\ CD$          (Isosceles triangle)

Thus by SSS congruency we can conclude that :

$\small \Delta ABD\cong \Delta ACD$

### Answer:

Consider $\Delta ABP$  and   $\Delta ACP$,

(i)  $AP$  is common side in both the triangles.

(ii)  $\angle PAB\ =\ \angle PAC$                       (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii)   $AB\ =\ AC$                                    (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

$\small \Delta ABP \cong \Delta ACP$

### Answer:

In the first part, we have proved that $\small \Delta ABD\cong \Delta ACD$.

So, by c.p.c.t.   $\angle PAB\ =\ \angle PAC$.

Hence AP bisects $\angle A$.

Now consider $\Delta BPD$  and  $\Delta CPD$,

(i)  $PD\ =\ PD$                         (Common)

(ii)  $BD\ =\ CD$                         (Isosceles triangle)

(iii)  $BP\ =\ CP$                         (by c.p.c.t. from the part (b))

Thus by SSS congruency we have   :

$\Delta BPD\ \cong \ \Delta CPD$

Hence by c.p.c.t. we have :      $\angle BDP\ =\ \angle CDP$

or  AP bisects $\angle D$.

### Answer:

In the previous part we have proved that   $\Delta BPD\ \cong \ \Delta CPD$.

Thus by c.p.c.t. we can say that  :       $\angle BPD\ =\ \angle CPD$

Also,                                                         $BP\ =\ CP$

SInce BC is a straight line, thus  :           $\angle BPD\ +\ \angle CPD\ =\ 180^{\circ}$

or                                                                                    $2\angle BPD\ =\ 180^{\circ}$

or                                                                                       $\angle BPD\ =\ 90^{\circ}$

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

### Answer:

Consider $\Delta ABD$   and   $\Delta ACD$ ,

(i)  $AB\ =\ AC$               (Given)

(ii)  $AD\ =\ AD$              (Common in both triangles)

(iii)  $\angle ADB\ =\ \angle ADC\ =\ 90^{\circ}$

Thus by RHS axiom we can conclude that :

$\Delta ABD\ \cong \ \Delta ACD$

Hence by c.p.c.t. we can say that  :         $BD\ =\ CD$      or        AD bisects BC.

### Answer:

In the previous part of the question we have proved that    $\Delta ABD\ \cong \ \Delta ACD$

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

$\angle BAD\ =\ \angle CAD$

Hence $AD$  bisects  $\angle A$ .

(i) $\small \Delta ABM \cong \Delta PQN$

(ii) $\small \Delta ABC \cong \Delta PQR$

### Answer:

(i)  From the figure we can say that :

$BC\ =\ QR$

or                                          $\frac{1}{2}BC\ =\ \frac{1}{2}QR$

or                                            $BM\ =\ QN$

Now, consider $\Delta ABM$   and   $\Delta PQN$,

(a)  $AM\ =\ PN$                        (Given)

(b)   $AB\ =\ PQ$                          (Given)

(c)   $BM\ =\ QN$                        (Prove above)

Thus by SSS congruence rule, we can conclude that :

$\small \Delta ABM \cong \Delta PQN$

(ii)   Consider $\Delta ABC$  and   $\Delta PQR$ :

(a)  $AB\ =\ PQ$                  (Given)

(b)   $\angle ABC\ =\ \angle PQR$   (by c.p.c.t. from the above proof)

(c)   $BC\ =\ QR$                 (Given)

Thus by SAS congruence rule,

$\small \Delta ABC \cong \Delta PQR$

### Answer:

Using the given conditions, consider $\Delta BEC$   and     $\Delta CFB$,

(i)  $\angle BEC\ =\ \angle CFB$                 (Right angle)

(ii)    $BC\ =\ BC$                              (Common in both the triangles)

(iii)   $BE\ =\ CF$                             (Given that altitudes are of the same length. )

Thus by RHS axiom, we can say that  :                $\Delta BEC\ \cong \Delta CFB$

Hence by c.p.c.t.,     $\angle B\ =\ \angle C$

And thus  $AB\ =\ AC$      (sides opposite to equal angles are also equal). Thus ABC is an isosceles triangle.

### Answer:

Consider $\Delta ABP$  and  $\Delta ACP$,

(i)   $\angle APB\ =\ \angle APC\ =\ 90^{\circ}$       (Since it is given that AP is altitude.)

(ii)   $AB\ =\ AC$                                      (Isosceles triangle)

(iii)   $AP\ =\ AP$                                     (Common in both triangles)

Thus by RHS axiom we can conclude that :

$\Delta ABP\ \cong \Delta ACP$

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

$\angle B\ =\ \angle C$

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

### 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,                                    $\angle A\ +\ \angle B\ +\ \angle C\ =\ 180^{\circ}$

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

or                                                       $\angle B\ +\ \angle C\ =\ 90^{\circ}$

Hence $\angle B$ and $\angle C$ are less than $\angle A$   ($90^{\circ}$).

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

Hence the side BC is largest is the hypotenuse of the $\Delta ABC$.

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

### Answer:

We are given that,

$\small \angle PBC < \angle QCB$                                                                                        ......................(i)

Also,                            $\angle ABC\ +\ \angle PBC\ =\ 180^{\circ}$                    (Linear pair of angles)          .....................(ii)

and                              $\angle ACB\ +\ \angle QCB\ =\ 180^{\circ}$                    (Linear pair of angles)          .....................(iii)

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

$\angle ABC\ > \ \angle ACB$

Thus   $AC\ > AB$   ( 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 $\small \angle B <\angle A$  and $\small \angle C <\angle D$ .

Thus,                     $BO\ > AO$              ..............(i)

and                        $OC\ > OD$             ...............(ii)

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

$AO\ +\ OD\ <\ BO\ +\ OC$

or                                        $AD\ <\ BC$

Hence proved.

### Answer:

Consider $\Delta ADC$ in the above figure :

$AD\ <\ CD$          (Given)

Thus         $\angle CAD\ > \angle ACD$                   (as angle opposite to smaller side is smaller)

Now consider $\Delta ABC$,

We have :                $BC\ > AB$

and                            $\angle BAC\ > \angle ACB$

Adding the above result we get,

$\angle BAC\ +\ \angle CAD > \angle ACB\ +\ \angle ACD$

or                                                     $\small \angle A>\angle C$

Similarly, consider $\Delta ABD$,

we have               $AB\ <\ AD$

Therefore               $\angle ABD\ > \angle ADB$

and in  $\Delta BDC$  we have,

$CD\ >\ BC$

and                                $\angle CBD\ >\ \angle CDB$

from the above result we have,

$\angle ABD\ +\ \angle CBD\ >\ \angle ADB\ +\ \angle CDB$

or                                                                $\small \angle B>\angle D$

Hence proved.

### Answer:

We are given that     $\small PR>PQ$.

Thus                 $\angle PQR\ =\ \angle PRQ$

Also,  PS bisects $\small \angle QPR$ , thus :

$\angle QPS\ =\ \angle RPS$

Now, consider $\Delta QPS$

$\angle PSR\ =\ \angle PQR\ +\ \angle QPS$                (Exterior angle)

Now, consider $\Delta PSR$

$\angle PSQ\ =\ \angle PRQ\ +\ \angle RPS$

Thus from the above the result we can conclude that :

$\small \angle PSR>\angle PSQ$

### Answer:

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

Then              $\angle B\ >\ \angle A\ or\ \angle C$            (Since $\angle B\ =\ 90^{\circ}$)

Thus the side opposite to largest angle is also largest.        $AC\ >\ BC\ or\ AB$

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

### 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.

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)

### 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 $\Delta ABC$

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 $\Delta ABC$.

### Answer:

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

Consider the hexagonal structure :

Area of hexagon  =   6  $\times$   Area of 1 equilateral

Thus area of the equilateral triangle :

$=\ \frac{\sqrt{3}}{4}\times a^2$

or        $=\ \frac{\sqrt{3}}{4}\times 5^2$

or        $=\ \frac{25\sqrt{3}}{4}\ cm^2$

So, the area of the hexagon is  :

$=\6\times \frac{25\sqrt{3}}{4}\ =\ \frac{75\sqrt{3}}{2}\ cm^2$

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

$=\ \frac{\sqrt{3}}{4}\times 1^2$

or    $=\ \frac{\sqrt{3}}{4}\ cm^2$

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

$=\ \frac{\frac{75\sqrt{3}}{2}}{\frac{\sqrt{3}}{4}}\ =\ 150$

Similarly for star-shaped rangoli :

Area :

$=\12\times \frac{\sqrt{3}}{4}\times 5^2 \ =\ 75\sqrt{3}\ cm^2$

Thus the number of equilateral triangles are :

$=\ \frac{75\sqrt{3}}{\frac{\sqrt{3}}{4}}\ =\ 300$

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!

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