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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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Congruence of triangles means that two triangles are identical in shape and size. When ΔABC≅ΔPQRΔ ABC \cong Δ PQR, their corresponding parts (sides and angles) match exactly. This relationship is called Corresponding Parts of Congruent Triangles (CPCT).

Two congruent triangles ABC and PQR illustrating corresponding vertices.
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The SSS (Side-Side-Side) Criterion states that two triangles are congruent if all three sides of one triangle are equal to the corresponding three sides of the other triangle.

Two triangles with sides of equal lengths demonstrating SSS criterion.
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The SAS (Side-Angle-Side) Criterion requires two sides and the included angle (the angle between the sides) to be equal in both triangles.

Triangles showing SAS congruence with equal sides and included angle.
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The ASA (Angle-Side-Angle) Criterion applies when two angles and the included side (the side between the angles) are equal.

Triangle illustrating ASA with two angles and the side between them.
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The RHS (Right angle-Hypotenuse-Side) Criterion is specific to right-angled triangles. It states that if the hypotenuse and one side of a right-angled triangle are equal to the hypotenuse and one side of another right-angled triangle, they are congruent.

📐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.

Problem 3:

In the given figure, line segment ABAB is parallel to another line segment CDCD. OO is the midpoint of ADAD. Prove that ΔAOB≅ΔDOCΔ AOB \cong Δ DOC.

Figure showing two triangles AOB and DOC meeting at point O with parallel bases AB and CD.

Solution:

  1. In ΔAOBΔ AOB and ΔDOCΔ DOC:
  2. ∠BAO=∠CDO\angle BAO = \angle CDO (Alternate interior angles, as AB∥CDAB \parallel CD and ADAD is the transversal).
  3. AO=DOAO = DO (Given that OO is the midpoint of ADAD).
  4. ∠AOB=∠DOC\angle AOB = \angle DOC (Vertically opposite angles).
  5. Therefore, ΔAOB≅ΔDOCΔ AOB \cong Δ DOC by the ASA congruence criterion.

Explanation:

We use the properties of parallel lines and midpoints to identify two equal angles and one equal included side.

Problem 4:

In ΔABCΔ ABC, ADAD is the altitude to BCBC such that AD⊥BCAD \perp BC and AB=ACAB = AC. Show that ΔABD≅ΔACDΔ ABD \cong Δ ACD.

An isosceles triangle ABC with altitude AD from A to BC.

Solution:

  1. In right-angled triangles ΔABDΔ ABD and ΔACDΔ ACD:
  2. ∠ADB=∠ADC=90∘\angle ADB = \angle ADC = 90^{\circ} (Given AD⊥BCAD \perp BC).
  3. AB=ACAB = AC (Given Hypotenuse are equal).
  4. AD=ADAD = AD (Common side).
  5. Hence, ΔABD≅ΔACDΔ ABD \cong Δ ACD by the RHS congruence criterion.

Explanation:

Since the triangles are right-angled and have equal hypotenuses and a common side, the RHS criterion is satisfied.