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.
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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
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.
Class 9 Maths Chapter 7 Question Answers: Exercise: 7.1 Total Questions: 8 Page number: 92-94 |
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 $\small \Delta ABD\cong \Delta BAC$ .
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 $\small BD=AC$
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 $\small \angle ABD= \angle BAC$ .
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 triangles.
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 the arms of angle A.
Question 6: In Fig, $\small AC=AE,AB=AD$ and $\small \angle BAD= \angle EAC$ . Show that $\small BC=DE$.
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 : $BC\ =\ DE$
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 :
$\small \Delta DAP\cong \Delta EBP$
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.
Class 9 Maths Chapter 7 Question Answers: Exercise: 7.2
Total Questions: 8
Page number: 97-98
|
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 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 $\angle A$.
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.
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 $\small \Delta ABE \cong \Delta ACF$
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.
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.
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, $\angle B\ =\ \angle C\ =\ 45^{\circ}$
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
|
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 :
$\small \Delta ABD\cong \Delta ACD$
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 :
$\small \Delta ABP \cong \Delta ACP$
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.
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 $\angle A$ .
(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 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,
$\small \Delta ABC \cong \Delta PQR$
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 the RHS axiom, we can conclude that :
$\Delta ABP\ \cong \Delta ACP$
Now, by c.p.c.t. we can say that :
$\angle B\ =\ \angle C$
Exercise-wise NCERT Solutions of Triangles Class 9 Maths Chapter 7 are provided in the link below.
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°
The topics discussed in the NCERT Solutions for Class 9 Maths Chapter 7 Triangles are:
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:
Using these approaches, students can tackle the Triangles Class 9 Chapter 7 Question Answers with greater confidence.
We at Careers360 compiled all the NCERT class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.
Also, read,
Given below are direct links to the NCERT Books and the latest NCERT Syllabus for your convenience.
Frequently Asked Questions (FAQs)
Topics in Maths Chapter 7 Class 9 include:
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.
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)
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
There are 3 exercises, covering conceptual questions, numerical problems, and proofs based on congruence and triangle properties.
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