Class 7 Maths Congruence of Triangles | Criteria For Congruence of Triangles |

__Criteria For Congruence of Triangles__

There are four standard congruence tests for two triangles to be congruent. These are:

**SSS congruence ****criterion (Side-Side-Side)**

If the three sides of one triangle are respectively equal to the three sides of another, then the two triangles are congruent.

If we are given the lengths of the three sides of a triangle, then only one such triangle can be constructed.

In the given figure,

AB = PR (= 3.5 cm)

BC = PQ (= 7.1 cm)

AC = QR (= 5 cm)

Since all three sides of triangle ABC is equal to triangle PQR.

So, triangle ABC is congruent to triangle PQR.

i.e. ΔABC ≅ ΔRPQ

**SAS congruence ****criterion (Side-Angle-Side)**

If two sides and an include angle of one triangle are respectively equal to two sides and the included angle of another triangle, then the triangles are congruent.

If we are given the lengths of two sides of a triangle and the size of the included angle, then only one such triangle can be constructed (up to congruence).

In the given figure,

AB = EF (= 7 cm)

BC = DE (= 5 cm)

∠B = ∠E (= 50°)

Since length of two sides and an include angle of triangle ABC is equal to triangle DEF

So, triangle ABC is congruent to triangle DEF.

i.e. ΔABC ≅ ΔDEF

**ASA congruence ****criterion (Angle-Side-Angle)**

If two angles and one side of a triangle are respectively equal to two angles and the matching side of another triangle, then the two triangles are congruent.

If we are given the angles sizes of a triangle and the length of a specific side, then only one such triangle can be constructed.

In the two triangles AOC and BOD,

∠C = ∠D (each 70°)

Also, ∠AOC = ∠BOD = 30° (vertically opposite angles)

So, ∠A of ΔAOC = 180° – (70° + 30°) = 80° (using angle sum property of a triangle)

Similarly, ∠B of ΔBOD = 180° – (70° + 30°) = 80°

Thus, we have ∠A = ∠B, AC = BD and ∠C = ∠D

Now, side AC is between ∠A and ∠C and side BD is between ∠B and ∠D.

So, by ASA congruence rule, ΔAOC ≅ ΔBOD.

**RHS congruence ****criterion (Right angle- Hypotenuse-Side)**

The hypotenuse and one side of one right-angled triangle are respectively equal to the hypotenuse and one other side of another right-angled triangle then the two triangles are congruent.

If we are given the length of the hypotenuse and one other side of a right-angled triangle, then only one such triangle can be constructed.

In the given figure,

∠B = ∠P = 90^{0}

hypotenuse AC = hypotenuse RQ (= 8 cm)

and side AB = side RP (= 3 cm)

So, ΔABC ≅ ΔRPQ

__Problem:__ You want to show that Δ ART ≅ Δ PEN:

(a) If you have to use SSS criterion, then you need to show:

(i) AR = (ii) RT = (iii) AT =

(b)If it is given that Ð T = Ð N and you are to use SAS criterion, you need to have:

(i) RT = and (ii) PN =

(c) If it is given that AT = PN and you are to use ASA criterion, you need to have:

(i) ? (ii) ?

Answer:

(a) Using SSS criterion, Δ ART ≅ Δ PEN

(i) AR = PE (ii) RT = EN (iii) AT = PN

(b) Given, Ð T = Ð N

Using SAS criterion, Δ ART ≅ Δ PEN

(i) RT = EN (ii) PN = AT

(c) Given, AT = PN

Using SAS criterion, Δ ART @ Δ PE

(i) <RAT = <EPN (ii) < RTA = < ENP

__Problem:__ Explain, why Δ ABC ≅ Δ FED

Answer:

Given, <A = < F, BC = ED, < B = < E

In Δ ABC and Δ FED,

< B = < E = 90^{0}

< A = < F

BC = ED

Therefore, Δ ABC ≅ Δ FED [By RHS congruence rule]

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