Congruence of Triangles — Questions and Answers
Welcome to HSLC Guru. This page gives the complete solutions to ASSEB Class 7 New Mathematics Chapter 9, Congruence of Triangles — the idea of congruency, the five conditions under which two triangles are congruent, every “Work it Out” box, the in-text activities, and all of Exercise 9.
Summary
Two figures are said to be congruent when one is placed on the other and they coincide exactly — that is, they have the same shape and the same size. A blade on another blade, a stamp on an identical stamp, or a five-rupee note on another are everyday examples of congruence.
To decide whether two triangles are congruent we need not measure all three sides and all three angles. A triangle is congruent to another if any one condition holds — Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side (AAS), or, for right-angled triangles, Right angle-Hypotenuse-Side (RHS). Equal angles alone (Angle-Angle-Angle) or two sides with a non-included angle do not guarantee congruence.
The concept of congruence is also used to prove angle properties: in an isosceles triangle the angles opposite the equal sides are equal, and each angle of an equilateral triangle measures 60°.
Summary: This ASSEB Class 7 New Mathematics Chapter 9 (Congruence of Triangles) guide explains congruent figures and derives the five conditions of congruence — SSS, SAS, ASA, AAS and RHS — plus the ambiguous two-sides-and-a-non-included-angle case. It solves every “Work it Out” box (9.1–9.4), the in-text activities and all of Exercise 9, and applies congruence to the angle properties of isosceles and equilateral triangles.
The Five Conditions of Congruence
In the figure above $\triangle ABC \cong \triangle PQR$ because all three corresponding sides are equal (AB = PQ, BC = QR, CA = RP) — the Side-Side-Side (SSS) condition.
If two sides and the included angle (the angle between them) are equal, then $\triangle ABC \cong \triangle PQR$ — the Side-Angle-Side (SAS) condition.
If two angles and the common side between them are equal, then $\triangle ABC \cong \triangle PQR$ — the Angle-Side-Angle (ASA) condition. From this the Angle-Angle-Side (AAS) condition also follows, because the three angles of a triangle add up to 180°, so knowing two angles fixes the third.
If the hypotenuse (AC) and one side (BC) of a right-angled triangle equal the hypotenuse and the corresponding side of another right-angled triangle, then $\triangle ABC \cong \triangle XYZ$ — the Right angle-Hypotenuse-Side (RHS) condition.
Textbook Questions and Answers
Work it Out 9.1
1. Check if the two figures are congruent. (a) (b)
Answer: Place one figure on the other. A pair is congruent only if they coincide exactly — same shape and same size. So a pair whose shape and size match completely is congruent, while a pair whose sizes differ is not congruent.
2. Tick (✓) the congruent figures. (i) (ii) (iii) (iv)
Answer: Test each pair by placing (or folding) one figure onto the other. Tick only the pairs that superimpose completely — that is, the pairs equal in both shape and size. Do not tick a pair in which one figure is larger and the other smaller.
Work it Out 9.2
1. Find out the equal line segments. Lengths — (i) 3 cm, (ii) 5 cm, (iii) 3 cm, (iv) 5 cm, (v) 3 cm, (vi) 6 cm, (vii) 5 cm.
Answer: Two line segments are equal if their lengths are equal, whatever their position or direction. Hence —
- The 3 cm segments (i), (iii) and (v) are equal to one another.
- The 5 cm segments (ii), (iv) and (vii) are equal to one another.
- The 6 cm segment (vi) has no equal among the others.
2. Find out the equal angles. Measures — 35°, 35°, 90°, 35°, 90°, 45°, 35°, 90°.
Answer: Two angles are equal if their measures are equal, regardless of the length of the arms or their position. Hence —
- All the 35° angles are equal to one another.
- All the 90° angles (right angles) are equal to one another.
- The 45° angle has no equal among the others.
3. State True or False.
Answer: (i) For two line segments to be equal their positions must be the same — False (only equal length is needed). (ii) A line segment can have more than one equal line — True. (iii) For angles to be equal their measures must be the same irrespective of position — True. (iv) If the measures of two angles are equal though their arm lengths differ, the angles are equal — True.
4. Match column A with column B and choose the correct answer. A: [P] to draw a circle congruent to another, [Q] to draw a square congruent to another, [R] to draw a rectangle congruent to another, [S] two equal-sided figures to be congruent. B: (i) lengths of sides should be equal, (ii) corresponding length and breadth should be equal, (iii) the angle between them should be equal, (iv) the radius should be equal.
Answer: P → (iv) equal radius, Q → (i) equal sides, R → (ii) equal length and breadth, S → (iii) equal included angle. So the correct option is [D] P → (iv), Q → (i), R → (ii), S → (iii).
Activities
1. Are $\triangle ABC$ and $\triangle PRQ$ congruent? (AB = PR, BC = RQ, AC = PQ)
Answer: Draw and cut out both triangles and place one on the other; they coincide completely. Here AB = PR, BC = RQ, AC = PQ and $\angle A = \angle P$, $\angle B = \angle R$, $\angle C = \angle Q$. The corresponding vertices are A and P, B and R, C and Q, so by the Side-Side-Side condition $\triangle ABC \cong \triangle PRQ$. This may also be written $\triangle BCA \cong \triangle RQP$, $\triangle CAB \cong \triangle QPR$, etc. But we cannot write $\triangle ABC \cong \triangle PQR$, because $\angle B \ne \angle Q$ and AB ≠ PQ.
2. In parallelogram ABCD identify a pair of congruent triangles and write their corresponding sides, angles and vertices.
Answer: The diagonal AC divides the parallelogram into $\triangle ABC$ and $\triangle CDA$. Since opposite sides of a parallelogram are equal, AB = CD and BC = DA, and AC is the common side. Hence by the Side-Side-Side condition $\triangle ABC \cong \triangle CDA$. Corresponding vertices: A and C, B and D, C and A; corresponding sides: AB and CD, BC and DA, CA and AC; corresponding angles: $\angle BAC = \angle DCA$, $\angle ABC = \angle CDA$, $\angle BCA = \angle DAC$.
Work it Out 9.3
1. AB and CD are two equal chords of a circle with centre O. Prove that $\triangle AOB \cong \triangle COD$.
Answer: In $\triangle AOB$ and $\triangle COD$ — OA = OC (radii of the same circle), OB = OD (radii) and AB = CD (given equal chords). All three sides are correspondingly equal, so by the Side-Side-Side condition $\triangle AOB \cong \triangle COD$.
2. A kite ABCD is shown. Prove that (i) $\triangle ABC \cong \triangle ADC$; (ii) Does AC divide $\angle BAD$ and $\angle BCD$ into two equal parts? Give reasons.
Answer: (i) In a kite AB = AD and CB = CD, and AC is the common side. Hence by the Side-Side-Side condition $\triangle ABC \cong \triangle ADC$. (ii) Yes. Since corresponding angles of congruent triangles are equal, $\angle BAC = \angle DAC$ and $\angle BCA = \angle DCA$. So AC bisects both $\angle BAD$ and $\angle BCD$.
3. ABCD is a rectangle and AQD, BPC are two triangles (AQ = QD = BP = PC). Prove that $\triangle AQD \cong \triangle BPC$.
Answer: In $\triangle AQD$ and $\triangle BPC$ — AQ = BP and QD = PC (marked equal sides), and AD = BC (opposite sides of a rectangle are equal). All three sides are correspondingly equal, so by the Side-Side-Side condition $\triangle AQD \cong \triangle BPC$.
Work it Out 9.4
1. Identify the congruent triangles from the following and mention the condition of congruency. (a)–(e)
Answer:
- (a) In kite ABCD, AB = AD = 3 cm, $\angle BAC = \angle DAC$ (marked equal) and AC is common. So $\triangle ABC \cong \triangle ADC$ — Side-Angle-Side (SAS) condition.
- (b) $\triangle PQR$ has PQ = 5, QR = 4, PR = 5.5 cm and the other triangle has XY = 5, YZ = 4, XZ = 5.5 cm. All three sides are equal, so $\triangle PQR \cong \triangle XYZ$ — Side-Side-Side (SSS) condition.
- (c) The two triangles meet at O with $\angle ABO = \angle CDO = 70°$, BO = DO (marked equal) and $\angle AOB = \angle COD = 30°$ (vertically opposite). So $\triangle ABO \cong \triangle CDO$ — Angle-Side-Angle (ASA) condition.
- (d) Both triangles have angles 30°, 70° and 110°, but no side length is given. Angle-Angle-Angle is not a condition of congruence, so the two triangles need not be congruent.
- (e) Here AB = DC = 3 cm, $\angle A = \angle D = 80°$ and BC is common — that is, two sides and a non-included angle are given. This does not guarantee congruence, so the triangles cannot be said to be congruent.
2. ABC is an isosceles triangle with AB = AC and D is the midpoint of BC. (i) Is BD = CD? (ii) Is $\triangle ABD \cong \triangle ACD$? Give reasons.
Answer: (i) Yes. As D is the midpoint of BC, BD = CD. (ii) Yes. In $\triangle ABD$ and $\triangle ACD$, AB = AC (given), BD = CD (midpoint) and AD = AD (common side). All three sides are equal, so by the Side-Side-Side condition $\triangle ABD \cong \triangle ACD$.
3. The diagonals AC and BD of quadrilateral ABCD bisect each other at O. (i) Is AO = CO and BO = DO? (ii) Is $\angle AOB = \angle COD$? If yes, why? (iii) Is there a triangle congruent to $\triangle AOB$? (iv) Which is correct, $\triangle AOB \cong \triangle DOC$ or $\triangle AOB \cong \triangle COD$?
Answer: (i) Yes. As the diagonals bisect each other, AO = CO and BO = DO. (ii) Yes, $\angle AOB = \angle COD$ because they are vertically opposite angles. (iii) Yes, $\triangle COD$ is congruent to $\triangle AOB$ (AO = CO, BO = DO and the included angle $\angle AOB = \angle COD$ — SAS). (iv) The correct one is $\triangle AOB \cong \triangle COD$ (with A → C, O → O, B → D).
Angles of Isosceles and Equilateral Triangles
Isosceles triangle: In $\triangle ABC$, AB = AC and $\angle A = 70°$. Find $\angle B$ and $\angle C$.
Answer: Draw AD perpendicular to BC. In $\triangle ABD$ and $\triangle ACD$, AB = AC (given), $\angle ADB = \angle ADC = 90°$ (AD ⟂ BC) and AD = AD (common side). So by the RHS condition $\triangle ABD \cong \triangle ACD$, and therefore $\angle B = \angle C$ (corresponding parts). Thus in an isosceles triangle the angles opposite the equal sides are equal. Now $\angle A + \angle B + \angle C = 180°$, so $\angle B + \angle C = 180° − 70° = 110°$; since $\angle B = \angle C$, each is $110° ÷ 2 = 55°$.
Equilateral triangle: What are $\angle A$, $\angle B$ and $\angle C$ of equilateral triangle ABC?
Answer: Here AB = BC = CA. Since AB = CA, $\angle B = \angle C$; since AB = BC, $\angle A = \angle C$; hence $\angle A = \angle B = \angle C$. As $\angle A + \angle B + \angle C = 180°$, each angle $= 180° ÷ 3 = 60°$. So $\angle A = \angle B = \angle C = 60°$.
Exercise 9
1. Choose the correct answer.
(i) In $\triangle ABC$, AB = 4, BC = 5, AC = 6 cm and in $\triangle PQR$, PQ = 4, QR = 5, PR = 6 cm. Which is true?
Answer: B. $\triangle ABC \cong \triangle PQR$ (AB = PQ, BC = QR, AC = PR — SSS).
(ii) In $\triangle ABC$, $\angle A = 90°$, AB = AC, then —
Answer: A. $\angle B = \angle C = 45°$ (AB = AC gives $\angle B = \angle C$, and $\angle B + \angle C = 180° − 90° = 90°$).
(iii) Each angle of an equilateral triangle measures —
Answer: A. 60°.
(iv) In the figure AB = CD, AD = CB, $\angle DAB = \angle BCD$, then —
Answer: B. $\triangle BAD \cong \triangle DCB$ (BA = DC, AD = CB and the included angle $\angle BAD = \angle DCB$ — SAS).
(v) In $\triangle ABC$, AB = 7, BC = 5 cm, $\angle B = 50°$ and in $\triangle DEF$, DE = 5, EF = 7 cm, $\angle E = 50°$. Under which condition are they congruent?
Answer: A. S-A-S. AB = 7 = EF, BC = 5 = DE and the included angle $\angle B = \angle E = 50°$.
(vi) In $\triangle ABC$ and $\triangle PQR$, $\angle B = \angle P = 90°$ and AB = RP. The triangles are congruent if —
Answer: A. AC = RQ. AC and RQ are the hypotenuses of the two right-angled triangles and AB = RP is one side, so they are congruent by the R-H-S condition.
(vii) $\triangle ABC \cong \triangle DEF$ and $\angle A = 50°$, $\angle E = 85°$, then $\angle C = $?
Answer: B. 45°. Here $\angle A = \angle D = 50°$, $\angle B = \angle E = 85°$, so $\angle C = 180° − 50° − 85° = 45°$.
2. State True or False.
Answer: (i) The AAS condition can be derived from the ASA condition — True (the third angle is fixed). (ii) Two equilateral triangles are always congruent — False (equal angles but possibly different sizes). (iii) Angle-Angle-Angle is a condition for congruency — False. (iv) Corresponding angles of two congruent triangles are equal — True.
3. Assertion [A]: In $\triangle ABC$, if AB = AC and D is the midpoint of BC, then AD ⟂ BC. Reason [R]: If three sides of a triangle equal the three sides of another, the triangles are congruent.
Answer: C. Both A and R are true and R is the correct explanation of A. In $\triangle ABD$ and $\triangle ACD$, AB = AC, BD = CD and AD = AD, so by SSS they are congruent; hence $\angle ADB = \angle ADC$, and as they are supplementary each is 90°, i.e. AD ⟂ BC.
4. In $\triangle ABC$, $\angle B = \angle C$; BL and CM are the bisectors of $\angle B$ and $\angle C$. Prove that BL = CM.
Answer: Since $\angle B = \angle C$, their halves are equal: $\angle MBC = \tfrac{1}{2}\angle B = \tfrac{1}{2}\angle C = \angle LCB$. In $\triangle MBC$ and $\triangle LCB$ — $\angle MBC = \angle LCB$, BC = CB (common side) and $\angle MCB = \angle LBC$ (as $\angle B = \angle C$). So by the ASA condition $\triangle MBC \cong \triangle LCB$, and hence the corresponding sides give CM = BL, that is BL = CM.
5. BD and CE are two altitudes of $\triangle ABC$ with BD = CE. (i) Write three equal parts of $\triangle CBD$ and $\triangle BCE$. (ii) Is $\triangle CBD \cong \triangle BCE$? (iii) Is $\angle DCB = \angle EBC$? Give reasons.
Answer: (i) $\angle BDC = \angle CEB = 90°$ (altitudes), BD = CE (given) and BC = CB (common hypotenuse). (ii) Yes. From the three parts above, by the RHS condition $\triangle CBD \cong \triangle BCE$. (iii) Yes, $\angle DCB = \angle EBC$, being corresponding angles of congruent triangles.
6. AD bisects $\angle A$, and in $\triangle ABC$, AB = AC. Prove that the angles opposite the equal sides are equal.
Answer: In $\triangle ABD$ and $\triangle ACD$ — AB = AC (given), $\angle BAD = \angle CAD$ (AD bisects $\angle A$) and AD = AD (common side). So by the SAS condition $\triangle ABD \cong \triangle ACD$, and therefore $\angle B = \angle C$. Thus the angles opposite the equal sides AB and AC are equal.
7. In isosceles triangle ABC, AB = AC and AD is the altitude. (i) Is $\triangle ABD \cong \triangle ADC$? (ii) Is $\angle B = \angle C$? (iii) Is BD = CD? Give reasons.
Answer: (i) Yes. In $\triangle ABD$ and $\triangle ACD$, AB = AC (given), $\angle ADB = \angle ADC = 90°$ (AD is the altitude) and AD = AD (common side) — by the RHS condition $\triangle ABD \cong \triangle ACD$. (ii) Yes, $\angle B = \angle C$ (corresponding angles). (iii) Yes, BD = CD (corresponding sides).
8. In a circle with centre O, $\angle AOB = 70°$. If $\angle ODC = 55°$, find $\angle DOC$, $\angle DCO$, $\angle OBA$ and $\angle OAB$.
Answer: OA = OB = OC = OD (radii of the same circle). $\triangle OAB$ is isosceles (OA = OB), so $\angle OAB = \angle OBA = (180° − 70°) ÷ 2 = 55°$. $\triangle OCD$ is isosceles (OC = OD), so $\angle DCO = \angle ODC = 55°$ and $\angle DOC = 180° − 55° − 55° = 70°$. Hence $\angle DOC = 70°$, $\angle DCO = 55°$, $\angle OBA = 55°$, $\angle OAB = 55°$.
9. Fill in the blanks.
Answer: (i) If the shape and area of two triangles are equal, the triangles are congruent. (ii) If the corresponding hypotenuse and one side of two right-angled triangles are equal, the triangles are congruent under the Right angle-Hypotenuse-Side (RHS) criterion. (iii) If $\triangle PQR \cong \triangle NML$, then side PR and side NL are equal. (iv) If two triangles are congruent, their perimeters are equal.
Project
Find the unknown angles from the given figure.
Answer: This is an activity-project. In each triangle of the figure the sum of the three angles is 180°; use this to find an unknown angle (third angle = 180° − sum of the two known angles). Where needed, also use that angles on a straight line add up to 180° and that vertically opposite angles are equal, working out each unknown angle one after another.
Additional Questions and Answers
Multiple Choice Questions (MCQ)
1. The symbol used to denote congruence is — (a) = (b) ≈ (c) ≅ (d) ∼
Answer: (c) ≅.
2. Which of the following is NOT a condition of congruence of triangles? (a) SSS (b) ASA (c) AAA (d) RHS
Answer: (c) AAA (Angle-Angle-Angle).
3. The RHS condition applies only to which type of triangle? (a) equilateral (b) right-angled (c) isosceles (d) scalene
Answer: (b) right-angled triangle.
4. If $\triangle ABC \cong \triangle DEF$, the side corresponding to BC is — (a) DE (b) EF (c) DF (d) AC
Answer: (b) EF.
5. Each angle of an equilateral triangle measures — (a) 45° (b) 50° (c) 60° (d) 90°
Answer: (c) 60°.
6. In the SAS condition the angle must be — (a) any one angle (b) the angle included between the two sides (c) the angle opposite the hypotenuse (d) a right angle
Answer: (b) the angle included between the two sides.
7. If the vertex angle of an isosceles triangle is 40°, each base angle is — (a) 40° (b) 60° (c) 70° (d) 100°
Answer: (c) 70° [(180° − 40°) ÷ 2 = 70°].
8. If two sides and a non-included angle of two triangles are equal, the triangles are — (a) always congruent (b) not necessarily congruent (c) always equilateral (d) always right-angled
Answer: (b) not necessarily congruent.
9. In congruent triangles, “CPCT” stands for — (a) equal sides (b) Corresponding Parts of Congruent Triangles (c) equal angles (d) equal perimeters
Answer: (b) Corresponding Parts of Congruent Triangles.
10. Two equal chords of a circle form triangles at the centre that are congruent by which condition? (a) SSS (b) AAA (c) RHS (d) none
Answer: (a) SSS (two radii + the equal chord).
Fill in the Blanks
1. Congruent figures have the same shape and the same ______.
Answer: size.
2. The sum of the three angles of a triangle is ______ degrees.
Answer: 180.
3. The side opposite the right angle in a right-angled triangle is called the ______.
Answer: hypotenuse.
4. If $\triangle ABC \cong \triangle PQR$, the angle corresponding to $\angle A$ is ______.
Answer: $\angle P$.
5. In an isosceles triangle the angles opposite the equal sides are ______.
Answer: equal.
True or False
1. Two congruent figures have equal area.
Answer: True.
2. Two triangles are congruent merely because their three angles are equal.
Answer: False.
3. The RHS condition applies only to right-angled triangles.
Answer: True.
4. Congruent triangles may have different perimeters.
Answer: False (their perimeters are equal).
5. A diagonal of a parallelogram divides it into two congruent triangles.
Answer: True.
Short Answer Questions
1. What conditions must two figures satisfy to be congruent?
Answer: They must have the same shape and the same size; when one is placed on the other they must coincide exactly.
2. Name the five conditions of congruence of triangles.
Answer: Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), Angle-Angle-Side (AAS) and Right angle-Hypotenuse-Side (RHS).
3. Why is Angle-Angle-Angle (AAA) not a condition of congruence?
Answer: Two triangles can have all three angles equal yet be of different sizes (one larger, one smaller). Same shape without the same size does not make them congruent, so AAA is not a condition of congruence.
4. Show briefly why each angle of an equilateral triangle is 60°.
Answer: As all three sides of an equilateral triangle are equal, all three angles are equal, i.e. $\angle A = \angle B = \angle C$. Since $\angle A + \angle B + \angle C = 180°$, each angle $= 180° ÷ 3 = 60°$.
Key Terms
| Term | Meaning |
|---|---|
| Congruent | Exactly the same in shape and size |
| Congruence | The property of two figures being exactly the same |
| Corresponding side | A side of one triangle that matches a side of the congruent triangle |
| Included angle | The angle lying between two given sides |
| Common side | A side shared by both triangles |
| Hypotenuse | The longest side, opposite the right angle |
| Isosceles triangle | A triangle with two equal sides |
| Equilateral triangle | A triangle with all three sides equal |
| Bisector | A line that divides an angle or segment into two equal parts |
| Vertically opposite angles | The equal opposite angles formed where two lines cross |