CBSE Class 9 Maths Revision Notes Chapter 7

Class 9 Mathematics Revision Notes for Triangles of Chapter 7

Class 9 Mathematics Chapter 7 Notes are available on Extramarks platform. Prepared according to the updated NCERT curriculum, these Class 9 Chapter 7 Mathematics Notes are written by experts to assist students in remembering all crucial points learned in this chapter. These Class 9 Mathematics Notes Chapter 7 also serve as a convenient resource for reviewing this important chapter during exams. This will not only allow them to save precious time, but it will also enable them to quickly review the chapter’s difficult topics.

Class 9 Mathematics Revision Notes for Triangles of Chapter 7

Access Class 9 Mathematics Chapter 7 – Congruent Triangles

Introduction:

Geometrical figures that are congruent share the same dimension and shape. Compare the results after superimposing one plane figure on top of the other to see if they are congruent.

Each figure’s shaded area can be moved over its uncovered area by dragging. If the shaded area matches (coincides) with the corresponding unshaded area, the pair of figures is said to be congruent.

Recall the axiom on real numbers, which states that if two real numbers a and b are present, they must adhere to one of the following three relations: a = b, or a b or a b.

Equality relation: a = b

The two magnitudes of the two line segments are the same, hence they are congruent and denoted as AB≅CD.

Congruent Circles:

Two circles are said to be congruent if they have the same radius.

In the figure given above, both circles have the same radius

OA = O ′ A ′ = r

Congruent Triangles:

Congruent triangles are those that share the same dimensions and shape. A triangle’s size and shape are determined by the triangle’s three sides and its three angles.

As a result, the congruence of triangles is determined by angles and sides.

Observe ΔKLM and ΔXYZ. If you superimpose one on top of the other, they will match, hence we can say that they are congruent. You’ll notice in both the triangles, that the three angles are congruent and the three sides, KL ≅ XY, LM ≅ YZ and KM ≅ XZ.

All six elements—three sides and three angles—of each triangle must match up with the corresponding six elements of the other for the two triangles to be said to be congruent.

Triangles’ Postulates of Congruency:

The previous observation serves as the foundation for the following definition of two triangles being congruent.

According to the definition, two triangles are said to be congruent if and only if all of their corresponding sides and angles are the same for both.

Sufficient Condition for Two Triangles to Be Congruent:

It is seen from the definition of congruence of two triangles that six requirements must be satisfied for two triangles to be congruent (3 sides and 3 angles). Through activities, it is noted that the remaining three requirements are met automatically if three of the six requirements are met.

Activity

In the two triangles, ΔABC and ΔDEF,

AB = DE = 4 cm

AC = DF = 4.5 cm and ∠BAC = ∠EDF = 95∘

Here, we can see that the two sides and their included angle of ΔABC are equal in magnitude to the corresponding sides and angle of ΔDEF.

By dragging ΔABC on ΔDEF, we can confirm that they match completely.

Therefore, ΔABC ≅ ΔDEF.

The SAS congruency postulate, where SAS stands for Side-Angle-Side, is the term used to describe this occurrence.

SAS Congruency Condition:

Two triangles are said to be congruent if their corresponding sides and included angles are the same on each side of the pair.

Theorem 9:

“Two triangles are congruent if any two sides and the included angle of one triangle is equal to the corresponding sides and the included angle of the other triangle”.

In the two given triangles,

ΔABC and ΔDEF

AB = DE 

AC = DF and ∠BAC = ∠EDF 

To prove:

ΔABC ≅ ΔDEF

Proof:

If we rotate and drag ΔABC on ΔDEF, such that the vertex B falls on the vertex of the other triangle E and places BC along EF, we will find that C falls on F, as AB = DE.

Also,

∠B = ∠E.

If AB falls on DE, C will coincide with the vertex F and A with D.

So, AC coincides with DF.

∴ΔABC coincides with ΔDEF

∴ΔABC ≅ ΔDEF

Application of SAS Congruency:

It is difficult to apply SAS in this figure because the provided data is insufficient. The SAS congruence criterion establishes the relationship between two triangles based on their angles and sides. Because of this, one cannot determine whether AC is the angle bisector of A and, consequently, whether the two triangles that may be displayed are congruent.

Theorem 10

“Angles opposite to equal sides are equal”.

In ΔABC

AB = AC

To prove:

∠ABC = ∠ACB

Construction: First, construct the angle bisector of A and name it AD.

Proof:

Comparing both the triangles, 

Given that

AB = AC

AD is the common side.

∠BAD = ∠DAC (AD being the angle bisector)

∴ΔBAD = ΔDAC (by SAS congruency rule)

∴∠ABD = ∠ACD (by CPCT)

∴∠B = ∠C

Hence, proved.

ASA Congruence Condition:

Two triangles are said to be congruent if their corresponding two angles and included sides are the same for both triangles.

Theorem 11:

Given:

In

ΔABC and ΔDEF

∠B = ∠E and ∠C = ∠F

BC = EF

To prove:

ΔABC ≅ ΔDEF

Proof:

There are three possibilities

Case I: AB = DE

Case II: AB  DE

Case III: AB  DE

Case I: In addition to data, if AB = DE

then,

ΔABC ≅ ΔDEF  (by the SAS congruence postulate)

Case II: If AB  DE 

Let K is any point on DE such that EK = AB

Join KF.

Now compare triangles ABC and KCF.

BC = EF (given)

∠B = ∠E(given)

Let,

AB = EK

ΔABC ≅ ΔKEF (SAS criterion)

Hence,

∠ABC = ∠KEF

But,

∠ABC = ∠DEF  (given)

Hence, K coincides with D.

Therefore, AB must be equal to DE.

Case III: If AB DE, then a similar justification can be applied. 

AB = DE.

Therefore, the only possibility is that AB must be equal to DE from SAS congruence condition

ΔABC ≅ ΔDEF

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 ΔABC, ∠B = ∠C

To prove: 

AB = AC

Construction:

Draw AD ⊥ BC

Proof:

Construct two right angle triangles, ADB and ADC, right-angled at D.

Here, ΔABC, ∠B = ∠C

∠ADB = ∠ADC = 90∘ (from the construction)

AD is common for both triangles.

∴ΔADB ≅ ΔADC    (by ASA postulate)

AB=AC   (corresponding sides)

AAS Congruence Condition:

The three sides of one triangle must match the corresponding three sides of the other triangle in order for two triangles to be said to be congruent.

Given: 

In triangles ABC and DEF,

BC = EF (non-included sides)

∠B = ∠E

∠A = ∠D

To prove:

ΔABC ≅ ΔDEF

Proof:

∠B = ∠E

  (given)

∠A = ∠D

  (given)

Now, adding both,

∠A +∠B = ∠E +∠D

…(1)

Since,

∠A +∠B +∠C = ∠E +∠D+∠F = 180∘

, considering (1) we can say that,

∠C = ∠F …(2)

Now in triangle ABC and DEF, 

∠B = ∠E

  (given)

∠C = ∠F

(proved in (2))

BC = EF (given)

ΔABC ≅ ΔDEF   (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, according to the SSS Congruence Condition.

Given: 

In triangles ABC and DEF,

AB = DE

BC = EF

AC = DF

Note:

Let BC and EF be the longest sides of triangles ABC and DEF respectively.

To prove:

ΔABC ≅ ΔDEF

Construction: If BC is the longest side, draw EG such that EG = AB and ∠GEF = ∠ABC.

Join GF and DG.

Proof:

In triangles ABC and GEF,

AB = GE (by construction)

BC = EF (given)

∠ABC = ∠GEF (by construction)

∴ΔABC = ΔGEF  (SAS congruence condition)

∠BAC = ∠EGF (by CPCT) 

and AC = GF (by construction)

But AB = DE (given)

∴DE = GE

Similarly, DF = GF

In ΔEDG,

DE = GE (Proved)

∴∠1=∠2…(1) (angles opposite equal sides)

In ΔEGF,

DF = GF (Proved)

∴∠3=∠4…(2) (angles opposite equal sides)

By adding (1) and (2), we get, 

∴∠1+∠3=∠4+∠2

∠EDF = ∠EGF but we have proved that ∠BAC = ∠EGF

Therefore, ∠EDF = ∠BAC 

Now in triangles ABC and DEF,

AB = DE

 (given)

AC = DF

 (given)

∠EDF = ∠BAC  (proved)

ΔABC ≅ ΔDEF (by SAS congruence)

Theorem of RHS (Right Angle Hypotenuse Side) Congruence

The two triangles are congruent if the hypotenuse and one side of one triangle match the hypotenuse and the corresponding side of the other triangle.

Given: 

ABC and DEF are two right-angled triangles such that

1. i) ∠B = ∠E = 90∘
2. ii) Hypotenuse AC = Hypotenuse DF and 

iii) Side BC  = Side EF

To prove:

ΔABC ≅ ΔDEF

Construction

Produce DE to M so that EM = AB. Join MF.

In triangles ABC and MEF,

EM = AB  (construction)

BC = EF

∠ABC = ∠MEF = 90∘ 

Therefore, ΔABC ≅ ΔMEF (SAS congruency)

Hence, ∠A = ∠M  …(1) (by CPCT)

AC=MF   …(2) (by CPCT)

Also, AC = DF   … (3) (given)

From (1) and (3), DF = MF

Therefore, ∠D = ∠M…(4) (Angles opposite to equal sides of ΔDFM)

From (2) and (4),

∠A = ∠D…(5)

Now compare triangles ABC and DEF,

∠A = ∠D (from 5)

∠B = ∠E = 90∘ (given)

∴∠C=∠F…(6)

Compare triangles ABC and DEF,

BC = EF (given)

AC = DF  (given)

∠C=∠F  (from 6)

Therefore, ΔABC ≅ ΔDEF (by SAS congruency)

Inequality of Angles:

Here, it is clear that the two angles are not equal: 135° < 160°

Inequality in a Triangle:

Construct a triangle ABC as shown in the figure.

Observe that in triangle ABC,

AC  is the smallest side (2 cm)

B is the angle opposite to AC and ∠B = 20∘

BC is the greatest side (6 cm)

A is the angle opposite to BCand ∠A = 100∘

From the measurements made above of the 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 it”.

Theorem on Inequalities:

If two sides of a triangle are unequal, the longer side has the greater angle opposite to it.

Draw a triangle according to the statement and data provided. 

Draw ΔABC, such that AC  AB

Data: AC  AB

To Prove: ∠ABC = ∠ACB

Construction: Mark a point D on AC at a distance that AB= AD. Join B and D.

Proof: 

In ΔABC, AB = AD  (by construction)

∴∠ABD = ∠ADB …(1)

but ∠ADB is the exterior angle with reference to ΔDBC. 

∠ADB ∠DCB … (2)

From relations (1) and (2) we can write ∠ABD ∠DCB

But ∠ABD is a part of ∠ACB.

∴∠ACB ∠DCB or ∴∠ABC ∠ACB

Hence, proved.

Angle Side Relation

Theorem 3

The longer side of a triangle is on the other side of the bigger angle. 

Given: In ΔABC, ∠ABC ∠ACB

To prove:  AC  AB

Proof: In ΔABC, AB and AC are two line segments. The three options, of which only one must be true, are as follows:

(I) either AB = AC, which means B = C, which defies the hypothesis.

∴AB ≠ AC

(II) AB  AC, therefore ∠B = ∠C, defying the prediction. then ∠B ∠C which is contrary to the hypothesis.

(III) AB  AC, this is the only condition we are left with, so AB  AC must be true.

Hence, proved.

Theorem 4Show that the length of the third side of any triangle is longer than the sum of the lengths of any two of its sides.

Draw Δ ABC

Data: ABC is a triangle.

To prove: 

AB+AC>BC

AB+BC>AC

AC+BC>BA

Construction:

Produce BC to D such that AC = CD. Join A to D.

Proof: 

AC= CD (by construction)

∠1=∠2 …(1)

From the figure, 

∠BAD = ∠2+∠3  …(2)

∠BAD ∠2

∴∠BAD ∠1   (proved from 1)

AB is opposite to ∠1 and BD is opposite to ∠BAD .

In ΔBAD , ∠1 ∠2+∠3

BD  AB 

(side opposite to greater angle is greater)

From figure BD = BC + CD

∴BC + CDAB

∴BC + ACAB

Since, CD = AC, hence, sum of two sides of a triangle is greater than the third side.

Similarly, AB+BC>AC and AC+BC>BA..

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.

Given:

P is a point that is not on line l. PM⊥l. Any point on l other than M is called N.

To prove: 

PM  PN

Proof:

In  Δ PMN, ∠M is the right angle.

∴According to the angle sum property, ∠N is an acute angle.

∵∠M ∠N

PN  PM (side opposite to greater angle)

∴PMPN.

Class 9 Mathematics Revision Notes for Triangles of Chapter 7

Properties of Triangles

Triangles are three-sided closed shapes. Depending on the dimensions of the sides and angles, triangles can take one of the following forms:

Equilateral Triangles: Equilateral Triangles have equal-sized sides and angles on each side. Because all of this triangle’s angles are 60 degrees, it is frequently referred to as an acute triangle.

Isosceles Triangle: An isosceles triangle has equal sides on each side, as well as equal angles on each side. 

Scalene Triangle: The scalene triangle has no sides or angles that are equivalent to one another.

Depending on the angle, the triangles can take one of the following shapes:

Acute Triangle: Acute triangles are triangles with all of their sides at acute angles to one another. The equilateral triangle is the most basic type of this triangle.

Obtuse Triangle: The triangle with an obtuse angle is known as an obtuse-angled triangle. This category of triangles includes scale and isosceles triangles.

Right-Angled Triangle: Triangle with an angle of 90 degrees is referred to as a “right-angled triangle.”

Congruent Triangles

If all three sides and all three angles are equal for two triangles, they are said to be congruent. However, all six dimensions must be identified. By knowing just three of the six values, it is possible to calculate the congruence of triangles. In Mathematics, two figures that are identical to one another based on their size and shape are said to be congruent.

Δ XYZ ≅ Δ LMN.

When placed over one another, each congruent object resembles the same figure. Similarly, congruent triangles are those triangles that are exact duplicates of one another in the sides and angles.

Similarity of Triangles

Two triangles are considered to be identical if their corresponding angles and sides are equal. In other words, similar triangles are nearly the same size but not exactly the same shape.

Inequalities of Triangles

The relationship between a triangle’s three sides is specified by the triangle inequality theorem. This theorem states that for any triangle, the number of two-side lengths is always greater than the third side. In other words, this theorem asserts that the shortest distance between any two points is always a straight line.

The triangle inequality theorem states that:

a < b + c,

b < a + c,

c < a + b

Pythagoras Theorem and its Applications

A right-angled triangle’s hypotenuse square is equal to the sum of the squares of its other two sides. 

(AC)²=(AB)²+(BC)²

AC is the hypotenuse.

AB is perpendicular.

BC is the base.

Basic Proportionality Theorem and Equal Intercept Theorem

A triangle’s other sides are proportionally divided by a line if it runs parallel to the side that splits the triangle into two different points. The fundamental proportionality theorem was first put forth by the Greek mathematician Thales. As a result, it is also referred to as the Thales Theorem.

DE is parallel to BC.

Then, according to the theorem,

AD/BD=AE/CE

Fun Facts about the Basic Proportionality Theorem

What are the two characteristics of the fundamental proportionality theorem? These two properties are:

• Property of an angle bisector.
• Property of Intercepts made by three parallel lines on a transversal.