NCERT Solutions for Class 9 Maths Chapter 7 Triangles

NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Komal MiglaniUpdated on 01 Sep 2025, 08:23 AM IST

A triangle is a geometrical shape having three sides and three angles. We can see different shapes around us that resemble a triangle, such as a slice of pizza, a clothes hanger, etc. Triangles are one of the important topics in the Class 9 NCERT syllabus. In this chapter, students learn about the special types of triangles, such as equilateral triangles, isosceles triangles, and right-angled triangles, etc. It introduces the concept of congruency of triangles, wherein triangles are said to be congruent if they have identical form and length. The chapter explains important congruence rules like SSS, SAS, AAS, and RHS. It also covers properties of triangles, inequalities in a triangle, and basic theorems such as the angle sum property. The main benefit of NCERT Solutions for Class 9 Maths is that they explain answers clearly, making learning simpler and more effective.

This Story also Contains

  1. Triangles Class 9 Questions And Answers PDF Free Download
  2. NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise Questions
  3. Triangles Class 9 NCERT Solutions: Exercise-wise
  4. Class 9 Maths NCERT Chapter 7: Extra Question
  5. Triangles Class 9 Chapter 7: Topics
  6. NCERT Triangles Class 9 Solutions - Important Formulae And Points
  7. Approach to Solve Questions of Triangles Class 9
  8. NCERT Solutions for Class 9 - Chapter Wise
NCERT Solutions for Class 9 Maths Chapter 7 Triangles
NCERT Solutions for Class 9 Maths Chapter 7 Triangles

These NCERT Solutions for Class 9 are trustworthy and reliable, as they are created by subject matter experts at Careers360, making them an essential resource for exam preparation. By using the solutions, you can prepare well for your exams at the school level as well as create a strong base for the competitive examinations. Many toppers rely on NCERT Solutions since they are designed as per the latest syllabus. For complete syllabus coverage, solved exercises, and downloadable PDFs, check out this NCERT article

Triangles Class 9 Questions And Answers PDF Free Download

To make maths learning easier, Careers360 experts have created these NCERT Solutions for Class 9 Maths Chapter 7 Triangles. Students can also access these solutions in PDF form.

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles: Exercise Questions

Class 9 Maths Chapter 7 Question Answers: Exercise: 7.1
Total Questions: 8
Page number: 92-94

Question 1: In quadrilateral $\small ABCD$ , $\small AC=AD$ and $\small AB$ bisects $\small \angle A$ (see Fig.). Show that $\small \Delta ABC\cong \Delta ABD$ . What can you say about $\small BC$ and $\small BD$?

1640157021241

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 $BC\ =\ BD$

Question 2: (i) $ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB= \angle CBA$ (see Fig. ). Prove that $\small \Delta ABD\cong \Delta BAC$ .

1640157040178

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 :

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

Question 2: (ii) $\small ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB=\angle CBA$ (see Fig.). Prove that $\small BD=AC$

1640157066047

Answer:

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

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

Question 2: (iii) $\small ABCD$ is a quadrilateral in which $\small AD=BC$ and $\small \angle DAB= \angle BAC$ (see Fig.). Prove that $\small \angle ABD= \angle BAC$ .

1640157095232

Answer:

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

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

$\small \angle ABD= \angle BAC$

Question 3: $\small AD$ and $\small BC$ are equal perpendiculars to a line segment $\small AB$ (see Fig.). Show that $\small CD$ bisects $\small AB$ .

1640157122932

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.

Question 4: $\small l$ and $\small m$ are two parallel lines intersected by another pair of parallel lines $\small p$ and $\small q$ (see Fig. ). Show that $\small \Delta ABC\cong \Delta CDA$

1640157162694

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 :

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

Question 5: (i) Line $\small l$ is the bisector of an angle $\small \angle A$ and B is any point on $\small l$ .$\small BP$ and $\small BQ$ are perpendiculars from $\small B$ to the arms of $\small \angle A$ (see Fig.). Show that: $\small \Delta APB\cong \Delta AQB$

1640157183354

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 triangles.

Thus, AAS congruence, we can write :

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

Question 6: In Fig, $\small AC=AE,AB=AD$ and $\small \angle BAD= \angle EAC$ . Show that $\small BC=DE$.

1640157243651

Answer:

From the given figure following result can be drawn:-

$\angle BAD\ =\ \angle EAC$

Adding $\angle DAC$ to 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 : $BC\ =\ DE$

Question 7: (i) $\small AB$ is a line segment and $\small P$ is its mid-point. $\small D$ and $\small E$ are points on the same side of $\small AB$ such that $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPB$ (see Fig). Show that $\small \Delta DAP\cong \Delta EBP$

1640157265740

Answer:

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

Adding $\angle DPE$ on 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 :

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

Question 7: (ii) AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that $\small \angle BAD=\angle ABE$ and $\small \angle EPA=\angle DPA$ (see Fig). Show that $\small AD=BE$

1640157291575

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 :

$\small AD=BE$

Question 8: (i) 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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \Delta AMC\cong \Delta BMD$

1640157315976

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 :

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

Question 8: (ii) 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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \angle DBC$ is a right angle.

1640157364987

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.

Question 8: (iii) 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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small \Delta DBC\cong \Delta ACB$

1640157384655

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 :

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

Question 8: (iv) 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 $\small DM=CM$. Point D is joined to point B (see Fig.). Show that: $\small CM=\frac{1}{2}AB$

1640165513559

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.

Class 9 Maths Chapter 7 Question Answers: Exercise: 7.2
Total Questions: 8
Page number: 97-98

Question 1: (i) In an isosceles triangle ABC, with $\small AB=AC$, the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that: $\small OB=OC$

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.

Question 1: (ii) In an isosceles triangle ABC, with $\small AB=AC$, the bisectors of $\small \angle B$ and $\small \angle C$ intersect each other at O. Join A to O. Show that: AO bisects $\small \angle A$

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 the 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 $\angle A$.

Question 2: In $\Delta ABC$ AD is the perpendicular bisector of BC (see Fig). Show that $\small \Delta ABC$ is an isosceles triangle in which $\small AB=AC$ .

1640157417015

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 SASthe 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.

Question 3: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig). Show that these altitudes are equal.

1640165544901

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 the 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.

Question 4: (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\small \Delta ABE \cong \Delta ACF$

1640165562618

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 :

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

Question 4: (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $\small AB=AC$, i.e., ABC is an isosceles triangle.

1640157444769

Answer:

From the previous 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.

Question 5: ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\small \angle ABD\ \cong \ \angle ACD$.

1640165593723

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 :

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

Question 6: $\Delta ABC$ is an isosceles triangle in which $AB=AC$. Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\angle BCD$ is a right angle.

1640157465047

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

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

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

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

Answer:

Consider a triangle ABC that 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 ( $60^{\circ}$ ).

Class 9 Maths Chapter 7 Question Answers: Exercise: 7.3
Total Questions: 5
Page number: 102

Question 1: (i) $\small \Delta ABC$ and $\small \Delta DBC$ 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 $\small \Delta ABD\cong \Delta ACD$

1640157485851

Answer:

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

(i) $AD\ =\ AD$ (Common)

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

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

Thus by SS,S congruency, we can conclude that :

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

Question 1: (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 $\small \Delta ABP \cong \Delta ACP.$

1640157528618

Answer:

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

(i) $AP$ is the 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 the SAS axiom, we can conclude that :

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

Question 1: (iii) $\small \Delta ABC$ and $\small \Delta DBC$ 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 AP bisects $\small \angle A$ as well as $\small \angle D$.

1640157546671

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 $\angle D$ .

Question 1: (iv) $\small \Delta ABC$ and $\small \Delta DBC$ 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 AP is the perpendicular bisector of BC.

1640157712234

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.

Question 2: (i) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that AD bisects 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 the 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.

Question 2: (ii) AD is an altitude of an isosceles triangle ABC in which $\small AB=AC$. Show that AD bisects $\small \angle A$

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 $\angle A$ .

Question 3: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\small \Delta PQR$ (see Fig). Show that:

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

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

1640157733376

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 the 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 the SAS congruence rule,

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

Question 4: BE and CF are two equal altitudes of a triangle ABC. Using the RHS congruence rule, prove that the triangle ABC is isosceles.

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.

Question 5: ABC is an isosceles triangle with $\small AB=AC$. Draw $\small AP\perp BC$ to show that $\small \angle B=\angle C$ .

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 the RHS axiom, we can conclude that :

$\Delta ABP\ \cong \Delta ACP$

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

$\angle B\ =\ \angle C$

Triangles Class 9 NCERT Solutions: Exercise-wise

Exercise-wise NCERT Solutions of Triangles Class 9 Maths Chapter 7 are provided in the link below.

Class 9 Maths NCERT Chapter 7: Extra Question

Question: In triangle ABC, AB = AC and $\angle B$ = 50°. Find $\angle A$.

Answer:

Given AB = AC, so it is an isosceles triangle

Therefore, $\angle B$ = $\angle C$ = 50°

Now, using the angle sum property:

$\angle A + \angle B + \angle C$ = 180°

$\Rightarrow \angle A$ + 50° + 50° = 180°

$\Rightarrow \angle A$ = 80°

Triangles Class 9 Chapter 7: Topics

The topics discussed in the NCERT Solutions for Class 9 Maths Chapter 7 Triangles are:

  • Introduction
  • Congruence of Triangles
  • Criteria for Congruence of Triangles
  • Some Properties of a Triangle
  • More Criteria for the Congruence of Triangles

NCERT Triangles Class 9 Solutions - Important Formulae And Points

Congruence:

  • Congruent refers to figures that are identical in all aspects, including their shapes and sizes. For example, two circles with the same radius or two squares with the same side lengths are considered congruent.

Congruent Triangles:

  • Two triangles are considered congruent if and only if one of them can be superimposed (placed or overlaid) over the other in such a way that they entirely cover each other.

Congruence Rules for Triangles:

  • Side-Angle-Side (SAS) Congruence:

  • Angle-Side-Angle (ASA) Congruence:

  • Angle-Angle-Side (AAS) Congruence:.

  • Side-Side-Side (SSS) Congruence:

  • Right-Angle Hypotenuse Side (RHS) Congruence:


Approach to Solve Questions of Triangles Class 9

Using these approaches, students can tackle the Triangles Class 9 Chapter 7 Question Answers with greater confidence.

  • Understand triangle classification: Master the classification of triangles using side and angle properties, such as scalene, isosceles, equilateral, in addition to acute, right and obtuse.
  • Apply the angle sum property: The total measurement of interior triangle angles always amounts to 180° to find missing angles.
  • Use congruence rules: Apply the SSS, SAS, ASA, RHS and AAS master criteria to prove triangle congruence, together with solving related problems.
  • Relate properties of isosceles and equilateral triangles: Theorems demonstrate that when two triangle sides are equal, the opposing angles match in size, and when two angles have matching measures, the connected sides are equivalent in length.
  • Explore inequalities in triangles: Explain the relationship between angles and side lengths, as well as the principle that the two side lengths combined always measure more than the third side.
  • Practice reasoning-based proofs: Students need to complete statement-reason questions involving triangle problems by implementing logical sequences and theorem usage.

Frequently Asked Questions (FAQs)

Q: What are the important topics in chapter 7 maths class 9?
A:

Topics in Maths Chapter 7 Class 9 include:

  • Congruence of Triangles
  • Criteria for Congruence of Triangles
  • Some Properties of a Triangle
  • More Criteria for the Congruence of Triangles
Q: Where can I find free NCERT Solutions for Class 9 Maths Chapter 7?
A:

Free NCERT Triangles Class 9 solutions are available online on many educational platforms, such as Careers360, in PDF and web formats, covering all exercises with clear, step-by-step explanations.

Q: What is triangle congruence?
A:

Two triangles are congruent if all their corresponding sides and angles are equal.
Congruence can be proved using the criteria:

  • SSS (Side-Side-Side)

  • SAS (Side-Angle-Side)

  • ASA (Angle-Side-Angle)

  • RHS (Right-angle-Hypotenuse-Side)

Q: What is the Pythagoras theorem?
A:

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides: 

c² = a² + b²

where,

c = hypotenuse,

a = base and

b = perpendicular

Q: How many exercises are there in NCERT Class 9 Maths Chapter 7?
A:

There are 3 exercises, covering conceptual questions, numerical problems, and proofs based on congruence and triangle properties.

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