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Congruence of Triangles - Criteria for Congruence of Triangles (SSS, SAS, ASA, RHS)

Grade 7CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Meaning of Congruence: Two plane figures are congruent if they have the exact same shape and size. For triangles, this means that if one triangle is placed over the other, they cover each other perfectly. The symbol for congruence is ≅\cong. In congruent triangles ΔABC≅ΔPQR\Delta ABC \cong \Delta PQR, the vertices AA corresponds to PP, BB to QQ, and CC to RR.

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Side-Side-Side (SSS) Criterion: Two triangles are congruent if the three sides of one triangle are equal to the three corresponding sides of the other triangle. Visually, if you have two triangles where every side length in the first triangle has a matching partner of the same length in the second, they are identical in every way.

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Side-Angle-Side (SAS) Criterion: Two triangles are congruent if two sides and the angle included between them of one triangle are equal to the corresponding two sides and the included angle of the other triangle. The word 'included' is critical; the angle must be the one formed by the meeting point of the two known equal sides, looking like a V-shape where the corner angle and both arms are equal.

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Angle-Side-Angle (ASA) Criterion: Two triangles are congruent if two angles and the side included between them of one triangle are equal to the corresponding two angles and the included side of the other triangle. In a diagram, this looks like a single side with two specific angles 'sitting' on its endpoints.

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Right Angle-Hypotenuse-Side (RHS) Criterion: This specific rule applies only to right-angled triangles. Two right-angled triangles are congruent if the hypotenuse (the side opposite the 90∘90^{\circ} angle) and one side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle.

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Corresponding Parts of Congruent Triangles (CPCT): Once two triangles are proved to be congruent using any of the criteria (SSS, SAS, ASA, RHS), all their other corresponding parts (the remaining sides and angles) are automatically equal. This is used as a reason to equate parts of triangles after proving congruence.

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Angle-Angle-Side (AAS) Variation: While ASA focuses on the included side, if any two pairs of angles and one pair of corresponding sides are equal, the triangles are congruent. This is because the third pair of angles must also be equal due to the angle sum property (180∘180^{\circ}) of a triangle.

📐Formulae

ΔABC≅ΔPQR  ⟹  AB=PQ,BC=QR,AC=PRΔ ABC \cong Δ PQR \implies AB=PQ, BC=QR, AC=PR

ΔABC≅ΔPQR  ⟹  ∠A=∠P,∠B=∠Q,∠C=∠RΔ ABC \cong Δ PQR \implies \angle A=\angle P, \angle B=\angle Q, \angle C=\angle R

SSS Condition: Side1=Side1,Side2=Side2,Side3=Side3Side_1 = Side_1, Side_2 = Side_2, Side_3 = Side_3

SAS Condition: Side1=Side1,∠Included=∠Included,Side2=Side2Side_1 = Side_1, \angle Included = \angle Included, Side_2 = Side_2

ASA Condition: ∠1=∠1,SideIncluded=SideIncluded,∠2=∠2\angle 1 = \angle 1, Side_{Included} = Side_{Included}, \angle 2 = \angle 2

RHS Condition: ∠90∘=∠90∘,Hypotenuse1=Hypotenuse2,Side1=Side2\angle 90^{\circ} = \angle 90^{\circ}, Hypotenuse_1 = Hypotenuse_2, Side_1 = Side_2

💡Examples

Problem 1:

In ΔABCΔ ABC and ΔADCΔ ADC, it is given that AB=ADAB = AD and CB=CDCB = CD. Prove that ΔABC≅ΔADCΔ ABC \cong Δ ADC.

Solution:

In ΔABCΔ ABC and ΔADCΔ ADC:

  1. AB=ADAB = AD (Given)
  2. CB=CDCB = CD (Given)
  3. AC=ACAC = AC (Common side to both triangles) Therefore, by the SSS congruence criterion, ΔABC≅ΔADCΔ ABC \cong Δ ADC.

Explanation:

We identified three pairs of corresponding sides that are equal. Since all three sides of ΔABCΔ ABC match the three sides of ΔADCΔ ADC, we use the Side-Side-Side (SSS) rule.

Problem 2:

In ΔPQRΔ PQR, PSPS is the perpendicular bisector of QRQR (where SS lies on QRQR). Show that ΔPQS≅ΔPRSΔ PQS \cong Δ PRS.

Solution:

In ΔPQSΔ PQS and ΔPRSΔ PRS:

  1. QS=RSQS = RS (Since PSPS bisects QRQR)
  2. ∠PSQ=∠PSR=90∘\angle PSQ = \angle PSR = 90^{\circ} (Since PSPS is perpendicular to QRQR)
  3. PS=PSPS = PS (Common side) Therefore, by the SAS congruence criterion, ΔPQS≅ΔPRSΔ PQS \cong Δ PRS.

Explanation:

We have two sides and the angle included between them equal. Side QS=RSQS=RS, side PSPS is common, and the angle formed at SS is 90∘90^{\circ} for both triangles. This satisfies the Side-Angle-Side (SAS) rule.