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NCERT Solutions for Class 9 Maths Chapter 7 Triangles

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Complete Resource of NCERT Solutions for Maths Chapter 7 Triangles Class 9 - Free PDF Download

NCERT Solutions for Triangles Class 9 focuses on triangles, an essential topic in geometry. This chapter helps students understand different kinds of triangles like isosceles, equilateral, and right-angled, and explores their properties and  an overview of triangle congruence and congruence principles. Learning about triangles is important because they are a basic shape used in many aspects of mathematics and everyday life. Also study characteristics and inequalities of triangles. The key areas covered in this chapter include understanding the angle sum property, which is that the angles inside any triangle add up to 180 degrees, and theorems like the Pythagorean theorem. Vedantu has provided clear explanations and step-by-step guidance to help students master these concepts. Concentrating on these areas is crucial for building a strong foundation in geometry.

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Table of Content
1. Complete Resource of NCERT Solutions for Maths Chapter 7 Triangles Class 9 - Free PDF Download
2. Glance of NCERT Solutions for Class 9 Maths Trianglesย  Chapter 7 | Vedantu
3. Access Exercise Wise NCERT Solutions for Chapter 7 Maths Class 9
4. Exercises Under NCERT Solutions for Class 9 Maths Triangles Chapter 7
5. Access NCERT Answers for Class-9 Maths Chapter 7 โ€“ Triangles
    5.1Exercise- 7.1
    5.2Exercise- 7.2
    5.3Exercise- 7.3
6. Overview of Deleted Syllabus for CBSE Class 9 Maths Chapter 7 Triangles
7. Class 9 Maths Chapter 7: Exercises Breakdown
8. Conclusion
9. Other Study Material for CBSE Class 9 Maths Chapter 7
10. Chapter-Specific NCERT Solutions for Class 9 Maths
11. Important Study Materials for CBSE Class 9 Maths
FAQs


Glance of NCERT Solutions for Class 9 Maths Triangles  Chapter 7 | Vedantu

  • Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.

  • Basic Proportionality Theorem (Thales Theorem) in a triangle, a line that is parallel to one side divides the other two sides proportionally.

  • Angle Sum Property is the sum of angles in any triangle is always 180ยฐ.

  • Congruence Criteria includes SAS, ASA, SSS, and RHS, which are essential for determining the congruence of triangles.

  • This article contains detailed notes explaining the properties of triangles, different types of triangles, and criteria for congruence and similarity of triangles.

  • Exercises are provided with solutions for applying congruence and similarity tests, proving properties using Pythagorasโ€™ Theorem, and working with special types of triangles like equilateral and isosceles.

  • This article contains links of chapter notes, important questions, exemplar solutions, exercises and video tutorials for better understanding.

  • There are three exercises (21 fully solved questions) in Class 9th Maths Chapter 7 Fractions and Decimal.


Access Exercise Wise NCERT Solutions for Chapter 7 Maths Class 9

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles
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Exercises Under NCERT Solutions for Class 9 Maths Triangles Chapter 7

Exercise 7.1:

This exercise comprises eight questions, and it mainly deals with the basic properties of triangles such as interior angles, exterior angles, and inequalities. The questions are mostly objective and require students to apply simple formulas to find the values of angles and sides of triangles.


Exercise 7.2:

This exercise consists of eight questions and mainly focuses on the congruence of triangles. The questions in this exercise deal with the criteria for congruence of triangles such as SSS, SAS, ASA, RHS, and AAS. The questions require students to apply these criteria to determine if the given triangles are congruent or not.


Exercise 7.3:

This exercise comprises five questions, and it deals with the properties of isosceles and equilateral triangles. The questions in this exercise require students to apply the properties of these triangles to find the values of angles and sides.


Access NCERT Answers for Class-9 Maths Chapter 7 โ€“ Triangles

Exercise- 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects \[\angle A\] (See the given figure). Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\]. What can you say about BC and BD?

 

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Quadrilateral ABCD

 

Ans: Given: In quadrilateral ACBD, AC = AD and AB is bisected by \[\angle A\]

To find: Show that \[\Delta ABC{\text{ }} \cong \Delta ABD\]. 

In \[\Delta ABC{\text{ , }}\Delta ABD\]

\[AC{\text{ }} = {\text{ }}AD\] (Given)

\[\angle CAB{\text{ }} = \angle DAB\]     (AB bisects โˆ A)

\[AB{\text{ }} = {\text{ }}AB\]        (Common)

\[\therefore \Delta ABC{\text{ }} \cong \Delta ABD\]            (By SAS congruence rule)

\[\therefore BC{\text{ }} = {\text{ }}BD\]     (By CPCT)

Therefore, BC and BD are of equal lengths.

 

2. ABCD is a quadrilateral in which AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]  (See the given figure). Prove that

(i)  \[\Delta ABD{\text{ }} \cong \Delta BAC\]

(ii) BD = AC

(iii)\[\angle ABD{\text{ }} = \angle BAC\].

 

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Quadrilateral ABCD with AD = BC

 

Ans : Given: ABCD is a quadrilateral where AD = BC and \[\angle DAB{\text{ }} = \angle CBA\]

To prove: 

(i)  \[\Delta ABD{\text{ }} \cong \Delta BAC\]

(ii) BD = AC

(iii)\[\angle ABD{\text{ }} = \angle BAC\].

In \[\Delta ABD{\text{ , }}\Delta BAC\],

AD = BC    (Given)

\[\angle DAB{\text{ }} = \angle CBA\]   (Given)

AB = BA     (Common)

\[\therefore \Delta ABD{\text{ }} \cong \Delta BAC\]              (By SAS congruence rule)

\[\therefore BD{\text{ }} = {\text{ }}AC\]          (By CPCT)

And, \[\angle ABD{\text{ }} = \angle BAC\]           (By CPCT)

 

3. AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.

 

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Figure showing AD and BC equal perpendiculars to a line segment AB

 

Ans: Given: AD and BC are equal perpendiculars to a line segment AB 

To prove: CD bisects AB.

In \[\Delta BOC{\text{ , }}\Delta AOD\],

\[\angle BOC{\text{ }} = \angle AOD\]     (Vertically opposite angles)

\[\angle CBO{\text{ }} = \angle DAO\]        (Each right angle )

BC = AD     (Given)

\[\therefore \Delta BOC{\text{ }} \cong \Delta AOD\]     (AAS congruence rule)

\[\therefore BO{\text{ }} = {\text{ }}AO\]     (By CPCT)

\[ \Rightarrow \]CD bisects AB.

 

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that \[\Delta ABC \cong \Delta CDA\].

 

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Two  Parallel Lines L and M

 

Given: l and m are two parallel lines intersected by another pair of parallel lines p and q To prove:  \[\Delta ABC \cong \Delta CDA\].

In \[\Delta ABC{\text{ , }}\Delta CDA\],

\[\angle BAC{\text{ }} = \angle DCA\]       (Alternate interior angles, as\[p{\text{ }}||{\text{ }}q\])

2AC = CA     (Common)

\[\angle BCA{\text{ }} = \angle DAC\]       (Alternate interior angles, as \[l{\text{ }}||{\text{ }}m\])

\[\therefore \Delta ABC \cong \Delta CDA\]          (By ASA congruence rule)

 

5. Line l is the bisector of an angle \[\angle A\] and B is any point on l. BP and BQ are $$perpendiculars from B to the arms of \[\angle A\] (see the given figure). Show that:

 

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Line l the Bisector of an Angle A

 

(i) \[\Delta APB \cong \Delta AQB\]

(ii) BP = BQ or B is equidistant from the arms of โˆ A.

Ans: Given: Line l is the bisector of an angle \[\angle A\] and B is any point on l.

To prove: (i) \[\Delta APB \cong \Delta AQB\]

(ii) \[BP{\text{ }} = {\text{ }}BQ\] or B is equidistant from the arms of \[\angle A\].

In \[\Delta APB{\text{ , }}\Delta AQB\],

\[\angle APB{\text{ }} = \angle AQB\]        (Each right angle )

\[\angle PAB{\text{ }} = \angle QAB\]      (l is the angle bisector of \[\angle A\])

AB = AB (Common)

\[\therefore \Delta APB \cong \Delta AQB\]       (By AAS congruence rule)

\[\therefore BP{\text{ }} = {\text{ }}BQ\]        (By CPCT)

Or, it can be said that B is equidistant from the arms of \[\angle A\].

 

6. In the given figure, \[AC{\text{ }} = {\text{ }}AE,{\text{ }}AB{\text{ }} = {\text{ }}AD\] and \[\angle BAD{\text{ }} = \angle EAC\]. Show that BC = DE.

Ans:

 

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Angle BAD = angle EAC

 

Given: \[\angle BAD{\text{ }} = \angle EAC\]

To prove: BC = DE

It is given that \[\angle BAD{\text{ }} = \angle EAC\]

\[\angle BAD{\text{ }} + \angle DAC{\text{ }} = \angle EAC{\text{ }} + \angle DAC\]

\[\angle BAC{\text{ }} = \angle DAE\]

In \[\Delta BAC{\text{ , }}\Delta DAE\],

AB = AD     (Given)

\[\angle BAC{\text{ }} = \angle DAE\]      (Proved above)

AC = AE    (Given)

\[\therefore \Delta BAC{\text{ }} \cong \Delta DAE\]       (By SAS congruence rule)

\[\therefore BC{\text{ }} = {\text{ }}DE\]           (By CPCT)

 

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\] (See the given figure). Show that

 

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Line segment AB with its mid point P


(i) \[\Delta DAP \cong \Delta EBP\]

(ii) AD = BE

Ans: Given: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that \[\angle BAD{\text{ }} = \angle ABE\] and \[\angle EPA{\text{ }} = \angle DPB\]

To prove: (i) \[\Delta DAP \cong \Delta EBP\]

(ii) AD = BE

It is given that \[\angle EPA{\text{ }} = \angle DPB\]

\[\angle EPA{\text{ }} + \angle DPE{\text{ }} = \angle DPB{\text{ }} + \angle DPE\]

\[\therefore \angle DPA{\text{ }} = \angle EPB\]

In \[\Delta DAP{\text{ , }}\Delta EBP\],

\[\angle DAP{\text{ }} = \angle EBP\] (Given)

AP = BP        (P is mid-point of AB)

\[\angle DPA{\text{ }} = \angle EPB\]   (From above)

\[\therefore \Delta DAP \cong \Delta EBP\]    (ASA congruence rule)

\[\therefore AD{\text{ }} = {\text{ }}BE\]          (By CPCT)

 

8. 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 DM = CM. Point D is joined to point B (see the given figure). Show that:

 

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Triangle ABC right - angled at C and midpoint M of AB

 

i) \[\Delta AMC \cong \Delta BMD\]

ii) \[\angle DBC\] is a right angle.

iii) \[\Delta DBC \cong \Delta ACB\]

iv) \[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

Ans: Given: M is the mid-point of hypotenuse AB. DM = CM

(i) In \[\Delta AMC{\text{ }},{\text{ }}\Delta BMD\],

AM = BM (M is the mid-point of AB)

\[\angle AMC{\text{ }} = \angle BMD\]        (Vertically opposite angles)

CM = DM          (Given)

\[\therefore \Delta AMC \cong \Delta BMD\]          (By SAS congruence rule)

\[\therefore AC{\text{ }} = {\text{ }}BD\] (By CPCT)

And, \[\angle ACM{\text{ }} = \angle BDM\](By CPCT)

(ii) \[\angle ACM{\text{ }} = \angle BDM\]

However,  \[\angle ACM{\text{ , }}\angle BDM\] are alternate interior angles.

Since alternate angles are equal,

It can be said that \[DB{\text{ }}||{\text{ }}AC\]

(Co-interior angles)

(iii) In \[\Delta DBC{\text{ , }}\Delta ACB\],

DB = AC      (Already proved)

\[\angle DBC{\text{ }} = \angle ACB\]          (Each \[{90^o}\])

BC = CB (Common)

\[\therefore \Delta DBC \cong \Delta ACB\]           (SAS congruence rule)

(iv) \[\therefore \Delta DBC \cong \Delta ACB\]

AB = DC (By CPCT)

AB = 2 CM

\[CM{\text{ }} = {\text{ }}\dfrac{1}{2}{\text{ }}AB\]

 

Exercise- 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of โˆ B and โˆ C intersect each other at O. Join A to O. Show that: 

(i) OB = OC (ii) AO bisects โˆ A

 

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Isosceles triangle with AB = AC

 

Ans: We know, $\angle ABC = \angle ACB$ ${\text{[Equal angles of isosceles triangle]}}$

$ \Rightarrow \dfrac{1}{2}\angle ABC = \dfrac{1}{2}ACB$

$ \Rightarrow \angle OBC = \angle OCB$

$\therefore OB = OC$ ${\text{[Sides opposite to equal angles of Isosceles triangle are equal]}}$

Now, In $\Delta ABO$ and $\Delta ACO$,

$AB = AC$ ${\text{[Equal sides of Isosceles triangle]}}$

$OB = OC$ ${\text{[Proved above]}}$

AO=AO [Common]

$\therefore \Delta ABO \cong \Delta ACO$ ${\text{[By SSS]}}$

$\therefore \angle BAO = \angle CAO$ ${\text{[CPCT]}}$

$\therefore $ AO bisects angle A.

 

2. In ฮ”ABC, AD is the perpendicular bisector of BC (see the given figure). Show that ฮ”ABC is an isosceles triangle in which AB = AC.

 

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Triangle ABC , AD being the bisector of BC

 

Ans: $\because $ AD is perpendicular bisector of BC,

$\therefore $$BD = DC$

& $\angle ADB = \angle ADC$ ${\text{[Each 90ยฐ]}}$

Now, in$\Delta ABD$ and $\Delta ACD$,

$AD = AD$ (Common)

$\angle ADB = \angle ADC$ ${\text{[Proved above]}}$ $BD = CD$ ${\text{[Proved above]}}$

$\therefore \Delta ADB \cong \Delta ACD$ ${\text{[by SAS]}}$

$\therefore AB = AC$ (CPCT)

So, $\Delta ABC$ is an isosceles triangle with \[AB = AC\]

 

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

 

seo images


Isosceles triangle ABC

 

Ans: In $\Delta ABE$and $\Delta ACF$,

$\angle AEB = \angle AFE$ ${\text{[Each 90ยฐ]}}$

$\angle BAE = \angle CAF$ (Common)

$AB = AC$ [Given]

$\therefore \Delta ABE \cong \Delta ACF$ ${\text{[By AAS]}}$

$\therefore BE = CF$ (CPCT)

 

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that

(i) \[\Delta ABE \cong \Delta ACF\]

(ii) AB = AC i.e., ABC is an isosceles triangle

 

seo images


Triangle ABC with equal altitudes

 

Ans: In $\Delta ABE$and $\Delta ACF$,

\[\angle BAE = \angle CAF\] (Common)

$\angle BEA = \angle CFA$               ${\text{[Each 90ยฐ]}}$

$BE = CF$ [Given]

$\therefore \Delta ABE \cong \Delta ACF$ ${\text{[by AAS]}}$

$\therefore AB = AC$ (CPCT)

And therefore, $\Delta ABC$ is an isosceles triangle with AB=AC.

 

5. ABC and DBC are two isosceles triangles on the same base BC (see the given figure). Show that โˆ ABD = โˆ ACD.

 

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Two Isosceles Triangles

 

Ans: Since AB=AC, and DB=DC,

$\therefore $ABDC is a quadrilateral with adjacent sides being equal.

$ \Rightarrow $ABDC is a kite.

We know that one pair of opposite (obtuse) angles of kite are equal.

Hence, $\angle ABD = \angle ACD$

 

6. ฮ”ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see the given figure). Show that โˆ BCD is a right angle.

 

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Isosceles Triangle in which AB = AC

 

Ans: In $\Delta ABC$

$\angle ABC = \angle ACB$ ${\text{[angles opposite to equal sides of a triangle]}}$ โ€ฆ (1)

Similarly, in $\Delta ADC$

$\angle ADC = \angle ACD$ ${\text{[angles opposite to equal sides of a triangle]}}$ โ€ฆ (2)

Now since BD is a straight line,

$\angle BAC + \angle CAD = 180^\circ $ โ€ฆ (3)

And we know,

$\angle BAC = \angle ADC + \angle ACD$ ${\text{[Exterior angle of a triangle = sum of opposite interior angles]}}$

$ \Rightarrow \angle BAC = 2\angle ACD$ ${\text{[From 2]}}$ โ€ฆ (4)

Similarly,

$\angle CAD = \angle ABC + \angle ACB$

$ \Rightarrow \angle CAD = 2\angle ACB$ ${\text{[From 1]}}$ โ€ฆ (5)

$\therefore 2\angle ACB + 2\angle ACD = 180^\circ $         ${\text{[From 3, 4 and 5]}}$

$ \Rightarrow \angle ACB + \angle ACD = 90^\circ  $

$ \Rightarrow \angle BCD = 90^\circ  $

Hence proved.

 

7. ABC is a right-angled triangle in which โˆ A = 90ยฐ and AB = AC. Find โˆ B and โˆ C.

 

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Right - Angled Triangle ABC

 

Ans: Since AB = AC,

$\angle B = \angle C$ ${\text{[Angles opposite to equal sides of a triangle]}}$

Now, we know

$\angle A + \angle B + \angle C = 180^\circ $ ${\text{[Angle sum property]}}$

$ \Rightarrow \angle A + 2\angle C = 180^\circ $

$ \Rightarrow 90^\circ  + 2\angle C = 180^\circ  $

$ \Rightarrow 2\angle C = 90^\circ  $

$  \Rightarrow \angle C = 45^\circ   $

$\therefore \angle B = 45^\circ $

Hence $\angle B$ and $\angle C$ are each 45$^\circ $

 

8. Show that the angles of an equilateral triangle are 60ยบ each.

Ans:

 

seo images


Equilateral Triangle

 

In $\angle B = \angle C$ ${\text{[Angles opposite to equal sides of a triangle]}}$

Similarly,

$\angle A = \angle B$ ${\text{[Angles opposite to equal sides of a triangle]}}$

$ \Rightarrow \angle A = \angle B = \angle C$

Now, we know that

$\angle A + \angle B + \angle C = 180^\circ $ ${\text{[Angle sum property]}}$

$ \Rightarrow \angle A + \angle A + \angle A = 180^\circ  $

$\Rightarrow 3\angle A = 180^\circ  $

$ \Rightarrow \angle A = 60^\circ $ 

$\therefore \angle A = \angle B = \angle C = 60^\circ $

Hence proved.

 

Exercise- 7.3

1. ฮ”ABC and ฮ”DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that

(i) \[\Delta ABD \cong \Delta ACD\]

(ii) \[\Delta ABP \cong \Delta ACP\]

(iii) AP bisects $\angle A$ as well as $\angle D$

(iv)AP is the bisector of BC

 

seo images


Two Isosceles Triangle on the same Base

 

Ans: (i) In $\Delta ABD$ and $\Delta ACD$,

$AB = AC$ ${\text{[Equal sides of isosceles triangle]}}$

$DB = DC$ ${\text{[Equal sides of isosceles triangle]}}$

AD=AD (Common)

$\therefore \Delta ABD \cong \Delta ACD$ ${\text{[By SSS]}}$

$ \Rightarrow \angle BAD = \angle CAD$ [CPCT]

$ \Rightarrow \angle BAP = \angle CAP$ โ€ฆ (1)

And $\angle ADB = \angle ADC$ (CPCT) โ€ฆ (2)

(ii) In $\Delta ABP$ and $\Delta ACP$

$AB = AC$ ${\text{[Equal sides of isosceles triangle]}}$

$\angle BAP = \angle CAP$ ${\text{[From 1]}}$

$AP = AP$ (Common)

$\therefore \Delta ABP \cong \Delta ACP$ ${\text{[By SAS]}}$

$\therefore BP = CP$ (CPCT) โ€ฆ (3)

Similarly, $\angle APB = \angle APC$ (CPCT) โ€ฆ (4)

(iii) AP is bisector of $\angle A$ ${\text{[From 1]}}$

Now, since AP is a line segment

$\therefore \angle ADB + \angle BDP = 180^\circ $ โ€ฆ (5)

Similarly, $\angle ADC + \angle CDP = 180^\circ $ โ€ฆ (6)

Comparing equations 2, 5 and 6 we can say that

$\angle BDP = \angle CDP$

$\therefore $AP bisects $\angle D$

Hence AP bisects both $\angle A$ and $\angle D$

(iv)We know,

$\angle APB + \angle APC = 180^\circ $

$ \Rightarrow \angle APB + \angle APB = 180^\circ $ ${\text{[From 4]}}$

$ \Rightarrow \angle APB = 90^\circ $ โ€ฆ (7)

From equations 3 and 7 we can say that,

AP is perpendicular bisector of BC.

 

2. AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects $\angle A$

Ans:

 

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Isosceles Triangle BC with AB = AC

 

In $\Delta ADB$ and $\Delta ADC$

$AB = AC$ (Given)

$AD = AD$ (Common)

$\angle ADB = \angle ADC$ ${\text{[Each 90ยฐ]}}$

$\therefore \Delta ADB \cong \Delta ADC$ ${\text{[By RHS]}}$

$ \Rightarrow BD = DC$ (CPCT)

Therefore, AD is bisector of BC

Similarly, $\angle BAD = \angle CAD$ (CPCT)

Therefore, AD bisects $\angle A$as well.

 

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 ฮ”PQR (see the given figure). Show that:

(i)\[\Delta ABM \cong \Delta PQN\]

(ii)\[\Delta ABC \cong \Delta PQR\]

 

seo images


Two Triangles with Their Corresponding Parts

 

Ans: (i) We know that $BC = QR$ โ€ฆ (1)

Now, Since AM is median of ฮ”ABC,

$ \Rightarrow BM = \dfrac{1}{2}BC$ โ€ฆ (2)

Similarly, PN is median of ฮ”PQR,

$ \Rightarrow QN = \dfrac{1}{2}QR$ โ€ฆ (3)

From equations 1, 2 and 3, we can say that,

$BM = QN$ โ€ฆ (4)

Now in $\Delta ABM$ and $\Delta PQN$

\[AB = PQ\] (Given)

\[BM = QN\] ${\text{[From 4]}}$

AM=PN (Given)

$\therefore \Delta ABM \cong \Delta PQN$ ${\text{[By SSS]}}$

$ \Rightarrow \angle ABM = \angle AQN$ (CPCT)

$ \Rightarrow \angle ABC = \angle PQR$ โ€ฆ (5)

(ii) Now in $\Delta ABC$ and $\Delta PQR$,

$AB = PQ$ (Given)

$\angle ABC = \angle PQR$ ${\text{[From 5]}}$

$BC = QR$ (Given)

\[\Delta ABC \cong \Delta PQR\]         ${\text{[By SAS]}}$

 

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

Ans:

 

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Two Equal Altitudes BE and CF of Triangle ABC

 

In $\Delta BEC$ and $\Delta CFB$

$BE = CF$ (Given)

$\angle BEC = \angle CFB$ ${\text{[Each 90ยฐ]}}$

BC=CB (Common)

$\therefore \Delta BEC \cong \Delta CFB$ ${\text{[By RHS Congruency]}}$

$ \Rightarrow \angle BCE = \angle CBF$               (CPCT)

$\therefore AB = AC$ ${\text{[Sides opposite to equal angles of a triangle are equal]}}$

Therefore, ฮ”ABC is an isosceles triangle.

 

5. ABC is an isosceles triangle with AB = AC. Draw AP โŠฅ BC to show that โˆ B= โˆ C.

 

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Isosceles Triangle ABC with AB = AC

 

Ans: In $\Delta ABP$ and $\Delta ACP$

$AB = AC$ (Given)

$AP = AP$ (Common)

$\angle APB = \angle APC$ ${\text{[Each 90ยฐ]}}$

$\therefore \Delta ABP \cong \Delta ACP$ ${\text{[By RHS]}}$

$ \Rightarrow \angle ABP = \angle ACP$         (CPCT)


Overview of Deleted Syllabus for CBSE Class 9 Maths Chapter 7 Triangles

Chapter

Dropped Topics

Triangles

Exercise 7.6: Inequalities in triangles.



Class 9 Maths Chapter 7: Exercises Breakdown

Exercise

Number of Questions

Exercise 7.1

8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.2

8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.3

5 Questions (3 Short Answer Questions, 2 Long Answer Question)



Conclusion

Triangle-related NCERT Solutions for Class 9 Chapter 7 from Vedantu are an excellent resource for understanding important geometric ideas. Because it provides the foundation for understanding more difficult geometrical structures that students will encounter in higher grades, this chapter is essential. Along with several congruence criteria including SAS, SSS, and ASA, it concentrates on key theorems such as the Pythagorean Theorem, Angle Sum Property, and Triangle Inequality. The importance of understanding both the theoretical and practical elements of these geometric principles is made clear by the three to five problems from this chapter that have frequently appeared in earlier year's question papers.


Other Study Material for CBSE Class 9 Maths Chapter 7



Chapter-Specific NCERT Solutions for Class 9 Maths

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.



Important Study Materials for CBSE Class 9 Maths

FAQs on NCERT Solutions for Class 9 Maths Chapter 7 Triangles

1. What does class 9th maths chapter 7 include?

In class 9th maths chapter 7, students get to learn about triangles. In this chapter, the different properties of triangles are discussed, along with the meaning of triangles. The criteria for inequalities of triangles are also mentioned. The chapter ends with a discussion on the rules of congruence of triangles.

2. What are some important topics of Class 9 maths chapter 7 solutions?

Some important topics of class 9 maths chapter 7 are:

  • Congruence of triangle

  • Properties of triangle

  • Inequalities of triangle

  • Rules of congruence of triangle

3. From where can I download NCERT solutions class 9 maths chapter 7?

Students can download NCERT solutions from Vedantu for free. To download the solutions, all you have to do is simply install the Vedantu app.

4. What are the properties of triangles?

There are six main properties of triangles. And these properties are:

  • In a triangle, the angle opposite to the longer side is greater

  • In a triangle, the side that is opposite to the greater angle is longer

  • The sum of any two sides of a triangle is greater than the third side

  • Every angle of an equilateral triangle measures to be 60 degrees

  • The sides opposite to equal angles of a triangle are equal

  • The angles opposite to equal sides of a triangle are equal

5. Are there any practice exercises available in NCERT Solutions for Chapter 7?

Yes, NCERT Solutions typically include practice exercises with step-by-step solutions. These exercises help you reinforce your understanding and improve your problem-solving skills.

6. Is it essential to memorize the theorems and properties mentioned in this chapter?

While memorization can be helpful, it's more important to understand the underlying principles and how to apply them. Focus on grasping the concepts, and you'll find it easier to remember the theorems and properties.

7. Can I use NCERT Solutions as a primary study resource for Class 9 Mathematics Chapter 7?

NCERT Solutions are an excellent primary study resource, but it's also beneficial to refer to other textbooks, practice additional problems, and seek clarifications from teachers when needed to gain a more comprehensive understanding of the chapter.

8. Is triangles Class 9 difficult?

Class 9 triangles may seem difficult since they need an understanding of multiple fundamental geometry principles and theorems. However, the principles become easier to understand with thorough study and application. By providing clear explanations and visual aids, Vedantu's solutions and video tutorials may help in the simplification of these difficult concepts, making the learning experience easier and more enjoyable.

9. What is the basic concept of a triangle Class 9?

In Class 9 triangles, the basic concept of a triangle is to understand its characteristics, varieties, and angles and sides theorems. It covers the study of triangle congruence and similarity, the connection between a triangle's sides and angles, and the use of Pythagoras' Theorem in right-angled triangles.

10. What is the theory of triangles Class 9?

Triangle congruence and similarity criteria (like SSS, SAS, ASA, and RHS), Pythagoras' Theorem, properties like the idea that the sum of a triangle's angles is 180 degrees, and the triangle inequality theoremโ€”which claims that the sum of any two of a triangle's sides is greater than the length of the third sideโ€”are the main topics covered in Class 9's theory of triangles.

11. What is the median of a triangle in class 9 triangles?

A line segment from a vertex to the opposite side's midway is called a triangle's median. There are three medians in every triangle, and they are significant because they meet at a single location known as the centroid, which is the center of gravity of the triangle.

12. What are the 7 properties of a Triangles Class 9?

  • The sum of the angles in a triangle is always 180 degrees.

  • The exterior angle of a triangle is equal to the sum of the opposite interior angles.

  • The sum of the lengths of any two sides of a triangle is greater than the length of the third side.

  • In a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

  • Each angle in an equilateral triangle is 60 degrees.

  • The medians of a triangle intersect at a single point (centroid), which is the triangle's center of gravity.

  • The perpendicular bisectors of a triangle's sides intersect at a point (circumcenter), which is equidistant from the triangleโ€™s vertices.