CBSE Class 9 Maths Notes Chapter 7 Triangles
Ananya Gupta21 May, 2024
CBSE Class 9 Maths Notes Chapter 7: In CBSE Class 9 Maths, Chapter 7 is all about triangles. Triangles are basic shapes with three sides, and this chapter teaches us about their different types and properties. We learn about triangles with equal sides, called equilateral triangles, and those with two equal sides, known as isosceles triangles.
The chapter also covers the Pythagorean theorem, which helps us find the sides of right-angled triangles. We learn about triangle congruence and similarity, which are important for comparing and analyzing different triangles. By studying this chapter, we can understand triangles better and solve problems involving them more easily.CBSE Class 9 Maths Notes Chapter 7 Triangles Overview
The Class 9 Maths Notes for Chapter 7, "Triangles," are made by subject experts of Physics Wallah. They're designed to make learning easy and clear for students. This chapter teaches about triangles, explaining their properties and features step by step. It covers everything from the basic definition of triangles to more advanced ideas like the Pythagoras' theorem. With these notes, students can understand different types of triangles and learn how to solve problems using theorems. It's a great resource to help students improve their math skills and succeed in their studies.CBSE Class 9 Maths Notes Chapter 7 PDF
You can access the CBSE Class 9 Maths Notes for Chapter 7 "Triangles" in PDF format using the provided link. These notes provide comprehensive explanations and examples to help you grasp the concepts of triangles effectively.CBSE Class 9 Maths Notes Chapter 7 PDF
CBSE Class 9 Maths Notes Chapter 7 Triangles
Introduction
Understanding the concepts covered in the "Triangles" chapter of Class 9 Mathematics is crucial as they form the foundation for higher education levels. Students are encouraged to grasp these concepts thoroughly as they play a significant role in various mathematical applications. The chapter aims to teach students the following key concepts:- Triangle congruence: Understanding when two triangles are congruent, meaning they have the same shape and size.
- Congruence rules: Exploring different criteria for triangle congruence, such as Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Side-Side-Side (SSS), and Right Angle-Hypotenuse-Side (RHS).
- Triangle properties: Learning about the properties of triangles, including angles, sides, and interior angles' sum.
- Triangle inequalities: Understanding the conditions that determine whether a set of given side lengths can form a triangle or not.
Congruent Triangles
For example, in Figure line segments AB and CD are congruent because they have the same length.
In Figure┬а angles a and b, as well as angles m and n, are congruent because they share the same measure.Angles m and n are considered congruent as they are vertically opposite angles.
Congruent Circles
Two circles are said to be congruent if they have the same radius.
In the above figure, both the circles have the same radius
Congruent Triangles
Consider triangles KLM and XYZ. When you align one on top of the other, they match perfectly, indicating congruency. Both triangles have congruent angles and sides: KL тЙЕ XY, LM тЙЕ YZ, and KM тЙЕ XZ.
For two triangles to be congruent, all three angles and three sides of one triangle must match the corresponding angles and sides of the other triangle.
Triangles' Postulates of Congruency
Sufficient Condition for Triangle Congruence
In the activity comparing triangles ╬ФABC and ╬ФDEF, we observed that when two sides and the included angle of one triangle match those of another, the triangles are congruent. This principle is known as the Side-Angle-Side (SAS) congruency condition. It simplifies the process of determining triangle congruence by requiring only three conditions to be met instead of all six.Theorem 9
"Two triangles are congruent if any two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other triangle".
In the two given triangles, ╬Ф ABC ╬ФABC and ╬Ф DEF ╬ФDEF , AB = DE AB = DE
AC = DF AC = DF and
тИа BAC = тИа EDF тИаBAC =┬атИаEDF .
To prove: ╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Proof:
If we rotate and drag ╬Ф ABC ╬ФABC on ╬Ф DEF ╬ФDEF , such the vertices B falls on the vertex of the other triangle E and place BC along EF, we will find that, since AB = DE AB = DE , C falls on F.
Also, тИа B = тИа E тИаB =┬атИаE .
AB falls on DE, A will coincide with the vertex D and C with F.
So, AC coincides with DF.
тИ┤ ╬Ф ABC тИ┤╬ФABC coincides with ╬Ф DEF ╬ФDEF .
тИ┤ ╬Ф ABC тЙЕ ╬Ф DEF
Application of SAS Congruency
Theorem 10
"Angles opposite to equal sides are equal".
In ╬Ф ABC ╬ФABC , AB = AC AB = AC
To prove:
тИа ABC = тИа ACB тИаABC =┬атИаACB
Construction: Draw the angle bisector of A, AD.
Proof:
Comparing both the triangles,
Given that AB = AC AB = AC
AD is the common side.
тИа BAD = тИа DAC тИаBAD =┬атИаDAC (as AD is the angle bisector)
тИ┤ ╬Ф BAD = ╬Ф DAC тИ┤╬ФBAD =┬а╬ФDAC (by SAS congruency rule)
тИ┤ тИа ABD = тИа ACD тИ┤тИаABD =┬атИаACD (by CPCT)
тИ┤ тИа B = тИа C тИ┤тИаB =┬атИаC
Hence, proved.
ASA Congruence Condition
The Angle-Side-Angle (ASA) congruence condition states that two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle. This condition simplifies the process of determining triangle congruence by requiring the equality of two angles and the side between them, thereby ensuring that the triangles have the same shape and size.
Theorem 11
Given:
In ╬Ф ABC ╬ФABC and ╬Ф DEF ╬ФDEF ,
тИа B = тИа E тИаB =┬атИаE and тИа C = тИа F тИаC =┬атИаF
BC = EF BC = EF .
To prove:
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Proof:
There are three possibilities
Case I:
Case II:
Case III:
Case I: In addition to data, if
then
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF (by SAS congruence postulate)
Case II: If
and let K is any point on DE such that
.
Join KF.
Now compare triangles ABC and KCF.
(given)
тИа B = тИа E тИаB =┬атИаE (given)
Let
(SAS criterion)
Hence,
But,
(given)
Hence, K coincides with D.
Therefore, AB must be equal to DE.
Case III: If
, then a similar argument applies.
AB must be equal to DE.
Hence the only possibility is that AB must be equal to DE and from SAS congruence condition
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Hence the theorem is proved.
Theorem 12
"In a triangle the sides opposite to equal angles are equal".
This theorem can also be stated as
"The sides opposite to equal angles of a triangle are equal".
Given:
In
To prove:
AB ┬п┬п┬п┬п┬п┬п┬п = AC ┬п┬п┬п┬п┬п┬п┬п AB┬п┬а=┬аAC┬п
Construction:
Draw AD ┬п┬п┬п┬п┬п┬п┬п┬п тКе BC ┬п┬п┬п┬п┬п┬п┬п AD┬п┬атКе┬аBC┬п
Proof:
Construct two right angle triangles, ADB and ADC, right angled at D.
Here,
тИа ADB = тИа ADC = 90 тИШ тИаADB =┬атИаADC = 90тИШ (from the construction)
AD ┬п┬п┬п┬п┬п┬п┬п┬п AD┬п is common for both the triangles.
тИ┤ ╬Ф ADB тЙЕ ╬Ф ADC тИ┤╬ФADB┬атЙЕ┬а╬ФADC (by ASA postulate)
AB ┬п┬п┬п┬п┬п┬п┬п = AC ┬п┬п┬п┬п┬п┬п┬п AB┬п┬а=┬аAC┬п (corresponding sides)
AAS Congruence Condition
The Angle-Angle-Side (AAS) congruence condition states that two triangles are congruent if two angles and a non-included side of one triangle are equal to the corresponding two angles and the side of the other triangle. This condition provides another method for determining triangle congruence by requiring the equality of two angles and a side not included between them, ensuring that the triangles have the same shape and size.
Given:
In triangles ABC and DEF,
(non-included sides)
To prove:
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Proof:
(given)
(given)
Now, adding both,
тАж(1)
Since,
, considering (1) we can say that,
тАж(2)
Now in triangle ABC and DEF,
(given)
(proved in (2))
BC = EF BC = EF (given)
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF (by SAS congruency)
SSS Congruence Condition
"Two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle".
In triangles ABC and DEF,
Note:
Let BC and EF be the longest sides of triangles ABC and DEF respectively.
To prove:
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Construction: If BC is the longest side, draw EG such that EG = AB EG = AB and тИа GEF = тИа ABC тИаGEF =┬атИаABC .
Join GF and DG.
Proof:
In triangles ABC and GEF,
(by construction)
(given)
тИа ABC = тИа GEF тИаABC =┬атИаGEF (by construction)
тИ┤ ╬Ф ABC = ╬Ф GEF тИ┤╬ФABC =┬а╬ФGEF (SAS congruence condition)
тИа BAC = тИа EGF тИаBAC =┬атИаEGF (by CPCT)
and AC = GF AC = GF (by construction)
But AB = DE AB = DE (given)
тИ┤ DE = GE тИ┤DE = GE
Similarly, DF = GF DF = GF
In ╬Ф EDG ╬ФEDG ,
DE = GE DE = GE (Proved)
тИ┤ тИа 1 = тИа 2 тИ┤тИа1=тИа2 тАж(1) (angles opposite equal sides)
In ╬Ф EGF ╬ФEGF ,
DF = GF DF = GF (Proved)
тИ┤ тИа 3 = тИа 4 тИ┤тИа3=тИа4 тАж(2) (angles opposite equal sides)
By adding (1) and (2), we get,
тИ┤ тИа 1 + тИа 3 = тИа 4 + тИа 2 тИ┤тИа1+тИа3=тИа4+тИа2
тИа EDF = тИа EGF тИаEDF =┬атИаEGF but we have proved that тИа BAC = тИа EGF тИаBAC =┬атИаEGF
Therefore, тИа EDF = тИа BAC тИаEDF =┬атИаBAC
Now in triangles ABC and DEF,
(given)
(given)
тИа EDF = тИа BAC тИаEDF =┬атИаBAC (proved)
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF (by SAS congruency)
Theorem of RHS (Right Angle Hypotenuse Side) Congruence
If the hypotenuse and one side of one triangle are equal to the hypotenuse and corresponding side of the other triangle, the two triangles are congruent.
ABC and DEF are two right-angled triangles such that
-
тИа B = тИа E = 90 тИШ тИаB =┬атИаE = 90тИШ
-
Hypotenuse AC ┬п┬п┬п┬п┬п┬п┬п┬п┬п = AC┬а┬п┬а= Hypotenuse DF ┬п┬п┬п┬п┬п┬п┬п DF┬п and
-
Side BC ┬п┬п┬п┬п┬п┬п┬п┬п = BC┬а┬п= Side EF ┬п┬п┬п┬п┬п┬п┬п EF┬п
To prove:
╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF
Construction
Produce DE to M so that
. Join MF.
Proof:
In triangles ABC and MEF,
(construction)
(given) тИа ABC = тИа MEF = 90 тИШ тИаABC =┬атИаMEF = 90тИШ
Therefore, ╬Ф ABC тЙЕ ╬Ф MEF ╬ФABC┬атЙЕ┬а╬ФMEF (SAS congruency)
Hence, тИа A = тИа M тИаA =┬атИаM тАж(1) (by CPCT)
AC ┬п┬п┬п┬п┬п┬п┬п = MF ┬п┬п┬п┬п┬п┬п┬п┬п AC┬п=MF┬п тАж(2) (by CPCT)
Also, AC = DF AC = DF тАж (3) (given)
From (1) and (3), DF = MF DF = MF
Therefore, тИа D = тИа M тИаD =┬атИаM ...(4) (Angles opposite to equal sides of
)
From (2) and (4),
тИа A = тИа D тИаA =┬атИаD тАж(5)
Now compare triangles ABC and DEF,
тИа A = тИа D тИаA =┬атИаD (from 5)
тИа B = тИа E = 90 тИШ тИаB =┬атИаE = 90тИШ (given)
тИ┤ тИа C= тИа F тИ┤тИаC=тИаF тАж(6)
Compare triangles ABC and DEF,
BC ┬п┬п┬п┬п┬п┬п┬п = EF ┬п┬п┬п┬п┬п┬п┬п BC┬п=EF┬п (given)
AC ┬п┬п┬п┬п┬п┬п┬п = DF ┬п┬п┬п┬п┬п┬п┬п AC┬п=DF┬п (given)
тИа C= тИа F тИаC=тИаF (from 6)
Therefore, ╬Ф ABC тЙЕ ╬Ф DEF ╬ФABC┬атЙЕ┬а╬ФDEF (by SAS congruency)
Inequality of Angles
Here we can see that, both these angles are not equal, as 135 тИШ < 160 тИШ 135тИШ<160тИШ
Inequality in a Triangle
Construct a triangle ABC as shown in the figure.
Observe that in triangle ABC, AC ┬п┬п┬п┬п┬п┬п┬п AC┬п is the smallest side (2 cm)
B is the angle opposite to AC ┬п┬п┬п┬п┬п┬п┬п AC┬п and тИа B = 20 тИШ тИаB = 20тИШ
BC ┬п┬п┬п┬п┬п┬п┬п BC┬п is the greatest side (6 cm)
A is the angle opposite to BC ┬п┬п┬п┬п┬п┬п┬п BC┬п and тИа A = 100 тИШ тИаA = 100тИШ
From the measurements made above of side and angle opposite to it, we can write the relation in the form of a statement.
"If two sides of a triangle are unequal then the longer side has the greater angle opposite to itтАЭ.
Theorem on Inequalities
Theorem 1:
If two sides of a triangle are unequal, the longer side has the greater angle opposite to it.
Read the statement and draw a triangle as per data.
Draw
, such that
.
Data:
To Prove:
тИа ABC = тИа ACB тИаABC =┬атИаACB
Construction:
Take a point D on AC ┬п┬п┬п┬п┬п┬п┬п AC┬п such that AB ┬п┬п┬п┬п┬п┬п┬п = AD ┬п┬п┬п┬п┬п┬п┬п┬п AB┬п┬а=┬аAD┬п . Join B to D.
Proof:
In ╬Ф ABC ╬ФABC , AB ┬п┬п┬п┬п┬п┬п┬п = AD ┬п┬п┬п┬п┬п┬п┬п┬п AB┬п┬а=┬аAD┬п (by construction)
тИ┤ тИа ABD = тИа ADB тИ┤тИаABD =┬атИаADB тАж(1)
but тИа ADB тИаADB is the exterior angle with reference to ╬Ф DBC ╬ФDBC .
тИа ADB тИа DCB тИаADB┬атИаDCB тАж (2)
From relation (1) and (2) we can write тИа ABD тИа DCB тИаABD┬атИаDCB
But тИа ABD тИаABD is a part of тИа ACB тИаACB .
тИ┤ тИа ACB тИа DCB тИ┤тИаACB┬атИаDCB or тИ┤ тИа ABC тИа ACB тИ┤тИаABC┬атИаACB
Hence, proved.
Angle Side Relation
Theorem 2:
In a triangle, if two angles are unequal, the side opposite to greater angle is longer than the side opposite to the smaller angle.
Theorem 3
In a triangle, the greater angle has the longer side opposite to it.
In ╬Ф ABC ╬ФABC , тИа ABC тИа ACB тИаABC┬атИаACB
To prove:
Proof:
In
, AB and AC are two line segments. So the following are the three possibilities of which
exactly one must be true.
-
either
AB = AC AB = AC, then тИа B = тИа C тИаB =┬атИаC which is contrary to the hypothesis.
-
AB┬а AC AB┬а AC
, then тИа B тИа C тИаB┬атИаC which is contrary to the hypothesis.
-
AB┬а AC AB┬а AC
, this is the only condition we are left with, so
AB┬а AC AB┬а ACmust be true.
Hence, proved.
Theorem 4
Prove that in any triangle the sum of the lengths of any two sides of a triangle is greater than the length of its third side. Draw ╬Ф ABC ╬Ф┬аABC .
Data:
ABC is a triangle.
To prove:
AB ┬п┬п┬п┬п┬п┬п┬п + AC ┬п┬п┬п┬п┬п┬п┬п > BC ┬п┬п┬п┬п┬п┬п┬п AB┬п+AC┬п>BC┬п
AB ┬п┬п┬п┬п┬п┬п┬п + BC ┬п┬п┬п┬п┬п┬п┬п > AC ┬п┬п┬п┬п┬п┬п┬п AB┬п+BC┬п>AC┬п
AC ┬п┬п┬п┬п┬п┬п┬п + BC ┬п┬п┬п┬п┬п┬п┬п > BA ┬п┬п┬п┬п┬п┬п┬п AC┬п+BC┬п>BA┬п
Construction:
Produce BC ┬п┬п┬п┬п┬п┬п┬п BC┬п to D such that
. Join A to D.
Proof:
(by construction)
тИа 1 = тИа 2 тИа1=тИа2 тАж(1)
From the figure,
тИа BAD = тИа 2+ тИа 3 тИаBAD =┬атИа2+тИа3 тАж(2)
тИа BAD тИа 2 тИаBAD┬атИа2
тИ┤ тИа BAD тИа 1 тИ┤тИаBAD┬атИа1 (proved from 1)
AB ┬п┬п┬п┬п┬п┬п┬п AB┬п is opposite to тИа 1 тИа1 and BD ┬п┬п┬п┬п┬п┬п┬п BD┬п is opposite to тИа BAD тИаBAD .
In ╬Ф BAD ╬ФBAD , тИа 1 тИа 2+ тИа 3 тИа1┬атИа2+тИа3
(side opposite to greater angle is greater)
From figure
Since, CD = AC CD = AC , hence, sum of two sides of a triangle is greater than the third side.
Similarly, AB ┬п┬п┬п┬п┬п┬п┬п + BC ┬п┬п┬п┬п┬п┬п┬п > AC ┬п┬п┬п┬п┬п┬п┬п AB┬п+BC┬п>AC┬п and AC ┬п┬п┬п┬п┬п┬п┬п + BC ┬п┬п┬п┬п┬п┬п┬п > BA ┬п┬п┬п┬п┬п┬п┬п AC┬п+BC┬п>BA┬п .
Hence, proved.
Theorem 5
Of all the line segments that can be drawn to a given line, from a point not lying on it, the perpendicular line segment is the shortest.
l ЁЭСЩ is a line and P is a point not lying on it.
. N is any point on l other than M.
To prove:
Proof:
In ╬Ф PMN ╬Ф┬аPMN , тИа M тИаM is the right angle.
тИ┤ тИа N тИ┤тИаN is an acute angle, from angle sum property.
тИ╡ тИа M тИа N тИ╡тИаM┬атИаN
PN┬а PM PN┬а PM (side opposite to greater angle)
тИ┤ PMPN тИ┤PMPN .
Benefits of CBSE Class 9 Maths Notes Chapter 7 Triangles
- Concept Clarity : These notes provide clear explanations of various concepts related to triangles, making it easier for students to understand the fundamentals.
- Structured Learning : The notes follow a structured format, covering each topic comprehensively, ensuring that students don't miss out on any essential concepts.
- Exam Preparation : The notes include important formulas, theorems, and properties related to triangles, which are crucial for exam preparation. Students can use these notes for quick revision before exams.
- Visual Aid: Diagrams and illustrations included in the notes help students visualize geometric concepts and understand them better.
CBSE Class 9 Maths Notes Chapter 7 FAQs
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