Geometry Chapter 4 Congruence Exercise 4 Solved Example Problems
Congruence
We say that the two objects are congruent when they are of the same size and shape.
In our daily life we come across many objects which are of the same size and shape.
For example, fifty paise coins, sheets of paper of a particular exercise book, keys of the same lock, shaving blades of the same brand etc.
These objects are called congruent objects. The relation of two objects being congruent is called congruence.
We shall, however, confine our attention to the congruence relation among plane figures only. In other words, we shall study those Figures which lie in the same plane and which have the same size and shape.
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Congruence of triangles
Two triangles are said to be congruent or equal in all respects if all the six parts namely three sides and three angles of one triangle are respectively equal to the corresponding six parts of other triangles.
If two triangles are congruent and we place one of them on the other then they will coincide.
In the two triangles ABC and DEF,
\(\overline{A B} \cong \overline{D E}, \quad \overline{B C} \cong \overline{E F}, \quad \overline{C A} \cong \overline{F D}\)Wbbse Class 7 Maths Solutions
∠BAC ≅ ∠EDF, ∠ABC ≅ ∠DEF, ∠BCA ≅ ∠EFD
Hence, the two triangles ABC and DEF are congruent and we write ΔABC ≅ ΔDEF.
Now place triangle ABC on triangle DEF so that \(\overline{B C}\) falls on \(\overline{E F}\) and the points B and C coincide respectively with the points E and F.
Then you will find that point A will coincide with point D and the two triangles will coincide entirely.
Corresponding sides and corresponding angles of two congruent triangles
If two triangles are congruent, then the sides opposite to the equal angles are called corresponding sides and the angles opposite to the equal sides are called corresponding angles.
Thus, in the two congruent triangles, ABC and DEF discussed in the previous article the corresponding sides are: AB and DE, BC and EF, and CA and FD.
Also, the corresponding angles are ∠BAC and ∠EDF, ∠ABC and ∠DEF, ∠BCA and ∠EFD.
Conditions of congruence of two triangles
The conditions under which the two triangles become congruent are :
- Of the two triangles, if the two sides and their included angle of one, are respectively equal to the two sides and their included angle of the other, then the triangles are congruent. (It is called Side-Angle- Side congruence or SAS congruence.)
- Of the two triangles, if the two angles and one side of one, are respectively equal to the two angles and one side of the other, then the triangles are congruent. (It is called Angle-Angle-Side congruence or AAS congruence.)
- Of the two triangles, if the three sides of one are respectively equal to the three sides of the other, then the triangles are congruent. (It is called Side-Side-Side congruence or SSS congruence.)
- Of two right triangles, if the hypotenuse and one side of the one are respectively equal to the hypotenuse and one side of the other, then the two right triangles are congruent. (It is called Right Angle-Hypotenuse — Side congruence or RHS congruence).
Verification of SAS congruence
Let us suppose that between two triangles ABC and DEF, AB = DE, AC = DF and the included ∠BAC = ∠EDF.
It is required to verify that ΔABC = ΔDEF.
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Cut off the triangle DEF along its border and place it on the triangle ABC such that A D EF coincides with BC.
You will find that D will coincide with A and the triangles DEF and ABC will also coincide.
Hence, it is verified that the two triangles are congruent.
Alternative method
Let ABC be a given triangle. Take a line segment DF congruent to AC. Now construct at the point D an angle ∠FDP congruent to ∠CAB. From DP cut off DE congruent to AB. Join EF.
Now measuring the angles of ΔDEF by a protractor it is found that ∠DEF = ∠ABC and ∠DFE =∠ACB.
Also measuring with a scale it is found that BC = EF.
Now if we set a correspondence between the two triangles ABC and DEF, such that A → D, B → E and C → F then all the six parts of ΔABC are congruent to the corresponding parts of ΔDEF. Hence, the two triangles are congruent.
But according to the construction two sides and their included angle of the triangle DEF were made congruent to the corresponding parts of the triangle ABC.
By actual measurement, the triangles are found to be congruent. Hence, the axiom is verified.
Verification of the AAS congruence
Let us suppose that between two triangles ABC and DEF, ∠ABC = ∠DEF, ∠BCA = ∠EFD and BC = EF.
It is required to verify that ΔABC ≅ ΔDEF.
Cut off the triangle DEF along its border and place it on the triangle ABC such that EF coincides with BC.
You will find that D will coincide with A and the triangles DEF and ABC will also coincide. Hence, it is verified that the two triangles are congruent.
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Alternative method
Let ABC be a given triangle. Take a line segment EF congruent to BC.
At the point E draw ∠DEF = ∠ABC and at point F draw ∠DFE = ∠ACB.
Now measuring with a protractor it will be found that ∠EDF = ∠BAC.
Also measuring with a scale it will be found that DE = AB and DF = AC.
Now, if we set a correspondence between the two triangles ABC and DEF, such that A → D, B → E and C → F then all the six parts of ΔABC are congruent to the corresponding parts of ΔDEF. Hence the two triangles are congruent.
But according to the construction two angles and one side of ΔDEF were made congruent to the corresponding parts of the ΔABC.
By actual measurement, the triangles are found to be congruent. Hence the axiom is verified.
Verification of SSS congruence
Let us suppose that between two triangles ABC and DEF, AB = DE, BC = EF and AC = DF.
It is required to verify that ΔABC = ΔDEF.
In this case, also we may verify the axiom by the two methods namely
- Paper cutting and
- Constructing on the basis of given parts as has already been discussed in two other cases.
Congruence of circles
If one of the two circles can be made to coincide with the other by translation, rotation, reflection or a combination of them, then the two circles are called congruent circles. It is obvious that the radii of two congruent circles are equal.
Axioms on congruent circles
- Incongruent circles (or in the same circle) equal chords cut off equal arcs and they subtend equal angles at the centre.
- Incongruent circles (or in the same circle) are those chords which cut off equal arcs or subtend equal angles at the centre at equal.
Verification of the above axioms
Let us consider two circles having centres O1 and O2. Let the circles be of the same radius r. Hence, the circles are congruent.
Consider the equal chords AB, CD and EF of the circles.
Let AGB, CHD and EIF be the minor arcs cut off by the chords AB, CD and EF respectively.
By measuring the lengths of these minor arcs with threads it is found that these minor arcs are equal in length.
Also, the circumferences of the two circles will be found to be equal by measuring with threads.
Also since the length of a major arc = length of the circumference – length of the corresponding minor arc
It is found that all the major arcs are also of equal length.
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Now join AO1,BO2,CO1,DO1,EO2 and FO2.
By measuring the angles ∠AO1B, ∠CO1D, ∠EO2F it is found that they are equal.
Repeating this process for different pair of circles and obtaining the same result you may conclude that :
Incongruent circles (or in the same circle) equal chords cut off equal arcs and they subtend equal angles at the centre.
To verify the second axiom i.e., the converse of the first take two congruent circles with centres O1 and O2 and construct at the centres the angles ∠AO1B, ∠CO1D and ∠EO2F of equal measure.
You may verify by a thread that the arcs AGB, CHD and EIF are equal in length and also measured by a scale you will find that the chords AB, CD and EF are equal in length.
Some examples
Example 1. Explain with reasons whether the following triangles are congruent or not.
Solution: Of the two triangles ΔABC and ΔDEF, we have AB = DF = 5 cm, BC = DE = 6 cm and AC = EF = 7 cm.
∴ ΔABC is congruent to ΔDEF as per SSS congruence condition.
Example 2. It is given that ΔABC: AB = 17 cm, BC = 15 cm, AC = 18 cm ΔPQR: PQ = 18 cm, QR = 17 cm, PR= 15 cm Verify if the two triangles are congruent. If they are congruent, write which angles are same among the two triangles.
Solution:
Given:
It is given that ΔABC: AB = 17 cm, BC = 15 cm, AC = 18 cm ΔPQR: PQ = 18 cm, QR = 17 cm, PR= 15 cm Verify if the two triangles are congruent.
Among the two triangles,
BC = PR = 15 cm, AB=QR= 17 cm and AC=PQ=18 cm.
The triangles are congruent as per SSS congruence condition.
According to the condition of congruency, the opposite angles of equal sides shall be equal to each other among the two triangles.
Hence, ∠A = ∠Q, ∠C = ∠P and ∠B = ∠R.
∠A = ∠Q, ∠C = ∠P and ∠B = ∠R angles are same among the two triangles
Example 3. Justify with reasons whether the two triangles are congruent or not.
Solution: Between the two triangles, ∠B = ∠F = 60°
∠C = ∠D = 45°
But BC ≠ FD
∴ The two triangles are not congruent.
Example 4. In given AB∥DC and AB=DC. Explain with reasons whether the triangles ΔACD and ΔCAB are congruent or not. Name the angle which is equal to ∠CAD.
Solution:
Given:
In given image AB∥DC and AB=DC.
Since ABIICD and AB=DC, hence quadrilateral ABCD is a parallelogram.
Among the two triangles ΔACD and ΔCAB, AB=DC (Given)
AC is the common side.
∠BAC = ∠ACD (ABIICDC, AC is the transversal, alternate internal angles are equal).
∴ The triangles are congruent as per SAS congruence condition.
According to the condition of congruency, the respective opposite angles of CD and AB shall be equal to each other.
∠CAD = ∠ACB.
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Example 5. Justify whether the given triangles are congruent or not.
Solution: The given triangles are right triangles each of which has PR, the opposite side of the right angle, as the hypotenuse.
Between the two triangles ΔPQR and ΔPSR, QR = SR = 4 cm
PR is the common hypotenuse
∠Q = ∠S = 90°.
ΔPQR and ΔPSR are congruent as per RHS congruence condition.
