CBSE Class 7 Mathematics Congruence Of Triangles Notes

CBSE Class 7 Congruence of Triangles Concepts. Learning the important concepts is very important for every student to get better marks in examinations. The concepts should be clear which will help in faster learning. The attached concepts made as per NCERT and CBSE pattern will help the student to understand the chapter and score better marks in the examinations.

Congruence of Triangles

Congruence of Plane Figures

If two objects are of exactly the same shape and size, they are said to be congruent. The relation between two congruent objects is called congruence.  The method of superposition examines the congruence of plane figures, line segments and angles.  Two plane figures are congruent if each, when superimposed on the other, covers it exactly.

Example: Two plane figures, say, P₁ and P₂ are congruent if the trace copy of  P₁ fits exactly on that of P₂.  We  write 

If two line segments have the same or equal length, they are congruent. Also, if two line segments are congruent, then they have the same length.

If two angles have the same measure, they are congruent. Also, if two angles are congruent, their measures are the same.

A plane figure is any shape that can be drawn in two dimensions.

Congruence of Triangles

Consider triangles ABC and XYZ. Cut triangle ABC and place it over XYZ. The two triangles cover each other exactly, and they are of the same shape and size. Also notice that A falls on X, B on Y, and C on Z. Also, side AB falls along XY, side BC along YZ, and side AC along XZ. So, we can say that triangle ABC is congruent to triangle XYZ. Symbolically, it is represented as ΔABC = ΔXYZ

So, in general, we can say that two triangles are congruent if all the sides and all the angles of one triangle are equal to the corresponding sides and angles of the other triangle.

In two congruent triangles ABC and XYZ, the corresponding vertices are A and X, B and Y, and C and Z, that is, A corresponds to X, B to Y, and C to Z. Similarly, the corresponding sides are AB ,XY, BC and YZ, and AC and XZ. Also, angle A corresponds to X, B to Y, and C to Z. So, we write ABC corresponds to XYZ.

We can tell if two triangles are congruent using 4 axioms: SAS axiom, ASA axiom, SSS axiom and RHS axiom.

SSS congruence criterion: Two triangles are congruent if three sides of one triangle are equal to the three corresponding sides of the other triangle.

SAS congruence criterion: Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.

RHS congruence criterion: Two right-angled triangles are congruent if the hypotenuse and a side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle.

ASA congruence criterion: 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.

SSS Congruence Criterion: Two triangles are congruent, if three sides of one triangle are equal to the three corresponding sides of another triangle.

We can tell if two triangles are congruent using 4 axioms: SAS axiom, ASA axiom, SSS axiom and RHS axiom.

SAS Congruence Criterion: Two triangles are congruent if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle. 

ASA Congruence Criterion: 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.

RHS Congruence Criterion: Two right angled triangles are congruent, if the hypotenuse and a side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle.

Example 1: In the given figure, OA = OB and OD = OC. Show that AOD BOC.

Solution: In ∠AOD and ∠BOC,

OA = OB(given)

OD = OC (given)

Also, since ∠AOD and ∠BOC form a pair of vertically opposite angles, we have

∠AOD = ∠BOC.

So, ∠AOD ≅ ∠BOC (by the SAS congruence rule)

Example 2: AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.

Solution: Line l AB and passes through C which is the mid-point of AB .

To show that PA = PB.

In Δ PCA and Δ PCB.

AC = BC (C is the mid-point of AB)

∠PCA = ∠PCB = 90° (Given)

PC = PC (Common)

So, Δ PCA ≅ Δ PCB (SAS rule)

and so, PA = PB, as they are corresponding sides of congruent triangles.

Example 3: Line-segment AB is parallel to another line-segment CD. O is the mid-point of AD. Show that ΔAOB ≅ ΔDOC

Solution: In Δ AOB and Δ DOC.

∠BAO = ∠CDO (Alternate angles as AB || CD and BC is the transversal)

∠AOB = ∠DOC (Vertically opposite angles)

OA = OD (Given)

Therefore, ΔAOB ≅ ΔDOC (ASA rule)

Example 4: In Δ ABC, the bisector AD of ∠A is perpendicular to side BC

Show that AB = AC and Δ ABC is isosceles.

Solution: In ΔABD and ΔACD,

∠BAD = ∠CAD (Given)

AD = AD (Common)

∠ADB = ∠ADC = 90° (Given)

So, Δ ABD ≅ Δ ACD (ASA rule)

So, AB = AC (CPCT) or, Δ ABC is an isosceles triangle.

Example 7: AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B Show that Δ PAQ ≅ Δ PBQ

Solution: In Δ PAQ and Δ PBQ

PA = PB (given)

QA = QB (given)

PQ = PQ (Common)

So, Δ PAQ ≅ Δ PBQ (SSS rule)

Example 8: P is a point equidistant from two lines l and m intersecting at point A. Show that Δ PAB ≅ Δ PAC

Solution: Given that lines l and m intersect each other at A.

Let PB l, PC m.

In Δ PAB and Δ PAC,

PB = PC (Given)

∠PBA = ∠PCA = 90° (Given)

PA = PA (Common)

So, Δ PAB Δ PAC (RHS rule)

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