📊 Mark Distribution
Section A: MCQ (8 questions) - 1 mark each
8 marks
Section B: Short Answer (10 questions) - 2 marks each
20 marks
Section C: Short Answer (10 questions) - 3 marks each
30 marks
Section D: Long Answer (4 questions) - 5 marks each
20 marks
TOTAL
80 marks
Choose the correct option for each question.
Q1. Two figures are congruent if they have the same _____ and _____.
(A) Size, color
(B) Shape, size
(C) Color, position
(D) Position, shape
1 mark
Q2. In the SAS rule, the angle must be the _____ angle.
(A) Exterior
(B) Obtuse
(C) Included
(D) Right
1 mark
Q3. Which is NOT a valid congruence rule?
(A) SSS
(B) SAS
(C) ASS
(D) RHS
1 mark
Q4. CPCT stands for _____
(A) Corresponding Parts of Congruent Triangles
(B) Corresponding Pairs of Congruent Triangles
(C) Common Parts of Congruent Triangles
(D) Corresponding Perpendiculars of Congruent Triangles
1 mark
Q5. In isosceles △ABC with AB = AC, which angles are equal?
(A) ∠A = ∠B
(B) ∠B = ∠C
(C) ∠A = ∠C
(D) None of these
1 mark
Q6. Each angle of an equilateral triangle is _____
(A) 30°
(B) 45°
(C) 60°
(D) 90°
1 mark
Q7. The RHS rule applies only to _____ triangles.
(A) Isosceles
(B) Right-angled
(C) Equilateral
(D) Scalene
1 mark
Q8. If △ABC ≅ △PQR, then AB corresponds to _____
(A) PQ
(B) PR
(C) QR
(D) QP
1 mark
Answer the following questions briefly.
Q9. State the SAS congruence rule and give one example.
2 marks
Q10. Explain why the ASS rule is NOT valid for triangle congruence.
2 marks
Q11. In △ABC, if AB = 5 cm, BC = 6 cm, and AC = 7 cm. Which congruence rule would you use to prove it congruent to another triangle with the same side lengths? Why?
2 marks
Q12. What is the relationship between angles in an isosceles triangle? Explain with an example.
2 marks
Q13. Define CPCT and explain its importance in proving geometric statements.
2 marks
Q14. State the ASA congruence rule. Is ASA different from AAS? Explain.
2 marks
Q15. In a right triangle, if the hypotenuse and one side of one triangle equal the hypotenuse and one side of another, are the triangles congruent? Which rule applies?
2 marks
Q16. Write the correspondence between vertices if △ABC ≅ △DEF. What does this correspondence mean?
2 marks
Q17. If a triangle has ∠A = ∠B, what can you conclude about the sides opposite to these angles?
2 marks
Q18. Real-life example: Ice cube trays have congruent molds. Explain why this is important using the concept of congruence.
2 marks
Answer the following questions with detailed explanations.
Q19. In △ABC, if AB = AC and ∠A = 80°, find ∠B and ∠C. Also state which theorem you used.
3 marks
Q20. Explain with an example why equality of three angles (AAA) is NOT sufficient to prove triangle congruence.
3 marks
Q21. State Theorem 7.2 and prove it using SAS congruence rule.
3 marks
Q22. In right triangle ABC with ∠C = 90°, AB = 10 cm, and AC = 6 cm. Find BC and state the relationship.
3 marks
Q23. Compare and contrast SAS and ASA rules with one example each.
3 marks
Q24. If △PQR ≅ △STU by SSS rule, list all corresponding equal angles without using the congruence statement.
3 marks
Q25. In an equilateral triangle, prove that all three angles are 60° each. Also prove that it's congruent to another equilateral triangle with the same side length.
3 marks
Q26. E and F are the midpoints of equal sides AB and AC of isosceles triangle ABC (AB = AC). Prove that BF = CE.
3 marks
Q27. State and prove Theorem 7.3 (converse of the isosceles triangle theorem).
3 marks
Q28. In △ABC, the perpendicular bisector of BC passes through A. Prove that AB = AC.
3 marks
Answer the following questions with complete proofs and explanations.
Q29. In △ABC and △PQR, if AB = PQ, ∠A = ∠P, and ∠B = ∠Q, prove that △ABC ≅ △PQR. State which congruence rule you used and explain why the other sides and angles are also equal.
5 marks
Q30. AB is a line segment. P and Q are points on opposite sides of AB such that PA = PB and QA = QB. Prove that the line PQ is the perpendicular bisector of AB. (Use SSS and SAS rules in your proof.)
5 marks
Q31. In an isosceles triangle ABC with AB = AC, if D and E are points on BC such that BE = CD, prove that AD = AE. Explain each step of your proof clearly.
5 marks
Q32. In right triangle ABC (right angle at C), M is the midpoint of hypotenuse AB. If C is joined to M and extended to point D such that MD = CM, prove that:
(i) △AMC ≅ △BMD
(ii) △DBC ≅ △ACB
(iii) ∠DBC = 90°
5 marks
SECTION A: ANSWERS & EXPLANATIONS (MCQ)
Question 1
Correct Answer: (B) Shape, size
Explanation:
Two figures are congruent if they have:
- The same shape (same sides and angles)
- The same size (identical measurements)
Congruence is NOT about color or position. Two identical photographs printed on different colored papers are still congruent to each other.
Question 2
Correct Answer: (C) Included
Explanation:
In the SAS (Side-Angle-Side) rule, the angle must be the included angle – the angle that lies between the two equal sides.
Important: If the angle is not between the sides (like in ASS or SSA), the triangles may NOT be congruent. This is a critical distinction!
Question 3
Correct Answer: (C) ASS
Explanation:
The four VALID congruence rules are:
- SSS - Side-Side-Side ✓
- SAS - Side-Angle-Side ✓
- ASA - Angle-Side-Angle ✓
- RHS - Right angle-Hypotenuse-Side ✓
ASS is INVALID because it doesn't guarantee congruence. Two triangles with two equal sides and one equal angle (not included) may not be congruent.
Question 4
Correct Answer: (A) Corresponding Parts of Congruent Triangles
Explanation:
CPCT is one of the most important concepts in triangle congruence:
Once you prove two triangles are congruent, you can immediately conclude that ALL their corresponding sides and angles are equal.
This is the power of CPCT – you don't need to prove each part separately; they're automatically equal once congruence is established!
Question 5
Correct Answer: (B) ∠B = ∠C
Explanation (Theorem 7.2):
In an isosceles triangle, if two sides are equal (AB = AC), then the angles opposite to those equal sides are also equal (∠B = ∠C).
Key Rule: In an isosceles triangle, if AB = AC, then the angles at the base (∠B and ∠C) are equal. These angles are called the "base angles."
Question 6
Correct Answer: (C) 60°
Explanation:
In an equilateral triangle, all three sides are equal, so all three angles must be equal.
Since the sum of angles in a triangle = 180°:
Each angle = 180° ÷ 3 = 60°
This is true for ANY equilateral triangle, regardless of its side length.
Question 7
Correct Answer: (B) Right-angled
Explanation:
RHS (Right angle-Hypotenuse-Side) congruence rule applies ONLY to right triangles because:
- It specifically mentions a right angle (90°)
- It uses the hypotenuse (the longest side opposite the right angle)
- It's a special case of the SSA rule that works only for right triangles
Question 8
Correct Answer: (A) PQ
Explanation:
When we write △ABC ≅ △PQR, the correspondence is:
A ↔ P, B ↔ Q, C ↔ R
Therefore, the sides correspond as:
Important: The order of vertices in the congruence statement is crucial! Always write the vertices in corresponding order.
SECTION B: ANSWERS & EXPLANATIONS (2 Marks Each)
Question 9
SAS Congruence Rule:
Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.
Example:
In △AOD and △BOC:
Given: OA = OB (Side 1)
∠AOD = ∠BOC (Vertically opposite angles - Included Angle)
OD = OC (Side 2)
Therefore, △AOD ≅ △BOC (by SAS rule)
Question 10
Why ASS Rule is Not Valid:
The ASS (or SSA) rule is not a valid congruence rule because two triangles with two equal sides and one non-included angle may NOT be congruent.
Explanation:
When the angle is not between (included) the two equal sides, multiple different triangles can be formed with the same measurements. This is called the "ambiguous case."
Example: If you have two sides of 4 cm and 5 cm with an angle of 50° (not between them), you can construct two different triangles with these measurements.
The SAS rule specifically requires the angle to BE INCLUDED (between the sides) to ensure only one unique triangle can be formed.
Question 11
Answer: SSS (Side-Side-Side) Rule
Explanation:
Since all three sides are given (5 cm, 6 cm, 7 cm) and they match with another triangle's sides, we use the SSS rule.
Why SSS?
- No angles are mentioned
- All three sides are known and equal
- SSS directly states: three equal sides → congruent triangles
This is the simplest and most direct approach when all three sides are known.
Question 12
Relationship in Isosceles Triangle:
In an isosceles triangle, the angles opposite to equal sides are equal. Conversely, if two angles are equal, the sides opposite to them are equal.
Example:
In △ABC with AB = AC:
Two sides are equal: AB = AC
Therefore, angles opposite to them are equal: ∠C = ∠B
Theorem 7.2: Equal sides → Equal opposite angles
Theorem 7.3: Equal angles → Equal opposite sides
Question 13
Definition of CPCT:
CPCT = Corresponding Parts of Congruent Triangles. It means that if two triangles are congruent, then all their corresponding sides and angles are equal.
Importance:
Once congruence is established, CPCT automatically proves all corresponding parts are equal. This saves time and effort in geometric proofs.
Example: If △ABC ≅ △PQR, then by CPCT:
- AB = PQ, BC = QR, CA = RP
- ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
Without CPCT, you'd need to prove each equality separately, which would be very tedious.
Question 14
ASA Rule:
Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of the other.
Difference from AAS:
ASA: The side MUST be between the two angles (included)
AAS: Two angles and ONE corresponding side (not necessarily between them)
Despite the difference, both rules prove congruence. However, they're used in different situations. AAS is more flexible since the side doesn't have to be included.
Question 15
Answer: YES, the triangles are congruent. RHS Rule applies.
Explanation:
This is exactly what the RHS (Right angle-Hypotenuse-Side) rule states:
RHS Congruence Rule:
In two right triangles, if the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other, then the triangles are congruent.
Why this works: The right angle guarantees a fixed structure, making this the only case where SSA (Side-Side-Angle) type rule is valid.
Question 16
Correspondence:
If △ABC ≅ △DEF, then:
Vertex Correspondence:
A ↔ D (A corresponds to D)
B ↔ E (B corresponds to E)
C ↔ F (C corresponds to F)
What This Means:
- Side AB corresponds to side DE
- Side BC corresponds to side EF
- Side CA corresponds to side FD
- ∠A corresponds to ∠D, ∠B to ∠E, ∠C to ∠F
Critical: The order matters! Writing △CBA ≅ △DEF would be incorrect.
Question 17
Conclusion: The sides opposite to these angles are equal.
If ∠A = ∠B, then the sides opposite to them are equal: AC = BC
Theorem 7.3 (Converse of Isosceles Triangle Theorem):
If two angles of a triangle are equal, then the sides opposite to those angles are also equal.
Why? Equal angles in a triangle imply equal opposite sides. This is the converse of the basic isosceles triangle property.
Example: In △ABC, if ∠A = ∠B, then BC = AC (the triangle is isosceles).
Question 18
Explanation Using Congruence Concept:
Ice cube trays have congruent molds because identical ice cubes are needed.
Why This Matters:
When two molds are congruent:
- They produce identical ice cubes with the same shape and size
- It ensures consistency across the tray
- Frozen ice cubes from different molds fit together properly
- They freeze at the same rate because they have equal volumes
This is a practical application of the concept: Congruent figures are "equal in all respects" and can be used interchangeably.
SECTION C: ANSWERS & EXPLANATIONS (3 Marks Each)
Question 19: Finding Angles in Isosceles Triangle
Given: △ABC with AB = AC, ∠A = 80°
Theorem Used: Theorem 7.2
Solution:
1 By Theorem 7.2, since AB = AC, angles opposite to them are equal: ∠B = ∠C
2 By angle sum property: ∠A + ∠B + ∠C = 180°
3 80° + ∠B + ∠C = 180°, so ∠B + ∠C = 100°
4 Since ∠B = ∠C: 2∠B = 100°, so ∠B = 50°
5 Therefore: ∠B = 50° and ∠C = 50°
Question 20: Why AAA is Not Sufficient
Why AAA is Not Sufficient:
Two triangles can have equal angles (AAA) but still have different sizes, making them similar but not congruent.
Example:
Consider two triangles:
Triangle 1: Angles = 60°, 60°, 60° (and sides 3 cm, 3 cm, 3 cm - equilateral)
Triangle 2: Angles = 60°, 60°, 60° (and sides 5 cm, 5 cm, 5 cm - equilateral)
Analysis:
- Both have the same angles (AAA) ✓
- But they have DIFFERENT SIZES
- Therefore, they are NOT congruent ✗
- They are only SIMILAR
Key Difference: Congruence requires both SAME SHAPE AND SAME SIZE. Angles alone only guarantee same shape, not same size.
Question 21: Theorem 7.2 & Proof
Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.
Proof Using SAS:
Given: △ABC with AB = AC
To Prove: ∠B = ∠C
Construction: Draw the bisector of ∠A, meeting BC at point D.
Proof:
In △ABD and △ACD:
AB = AC (Given)
∠BAD = ∠CAD (AD bisects ∠A by construction)
AD = AD (Common side)
Therefore, △ABD ≅ △ACD (by SAS rule)
By CPCT: ∠ABD = ∠ACD
Hence, ∠B = ∠C (Proved) ✓
Question 22: Finding Side in Right Triangle
Given: Right △ABC with ∠C = 90°, AB = 10 cm (hypotenuse), AC = 6 cm
Find: BC
Solution:
1 Using Pythagorean Theorem: In a right triangle: (Hypotenuse)² = (Side 1)² + (Side 2)²
AB² = AC² + BC²
2 Substitute values:
10² = 6² + BC²
100 = 36 + BC²
3 Solve for BC:
BC² = 100 - 36
BC² = 64
BC = 8 cm
Verification: 6² + 8² = 36 + 64 = 100 = 10² ✓ (Pythagorean triplet: 6-8-10)
Question 23: Comparing SAS and ASA
Comparison of SAS and ASA Rules
| Feature |
SAS Rule |
ASA Rule |
| Full Name |
Side-Angle-Side |
Angle-Side-Angle |
| What You Need |
2 Sides + 1 Angle |
2 Angles + 1 Side |
| Position |
Angle MUST be INCLUDED (between sides) |
Side MUST be INCLUDED (between angles) |
| If Condition Not Met |
Congruence NOT guaranteed (ASS/SSA invalid) |
Still valid (AAS also works) |
| Type |
Axiom (cannot be proved) |
Theorem (can be proved using SAS) |
SAS Example:
In △ABC: AB = 5 cm, ∠B = 60°, BC = 6 cm
In △PQR: PQ = 5 cm, ∠Q = 60°, QR = 6 cm
Then △ABC ≅ △PQR (by SAS)
ASA Example:
In △ABC: ∠A = 50°, AB = 7 cm, ∠B = 60°
In △PQR: ∠P = 50°, PQ = 7 cm, ∠Q = 60°
Then △ABC ≅ △PQR (by ASA)
Question 24: Equal Angles in Congruent Triangles
Given: △PQR ≅ △STU by SSS rule
Find: All corresponding equal angles
Vertex Correspondence:
From △PQR ≅ △STU, the correspondence is:
P ↔ S, Q ↔ T, R ↔ U
Equal Angles (by CPCT):
∠P = ∠S
∠Q = ∠T
∠R = ∠U
Key Point: Although we proved congruence using SSS (three sides), by CPCT, all corresponding angles are automatically equal without needing to measure them!
Question 25: Angles of Equilateral Triangle
Prove: All angles of an equilateral triangle are 60° each, and it's congruent to another equilateral triangle with the same side length.
Part 1: All Angles are 60°
Given: Equilateral triangle ABC (AB = BC = CA)
To Prove: ∠A = ∠B = ∠C = 60°
Step 1: Since AB = BC, by Theorem 7.2, ∠A = ∠C (angles opposite equal sides)
Step 2: Since BC = CA, by Theorem 7.2, ∠A = ∠B
Step 3: From Steps 1 and 2: ∠A = ∠B = ∠C
Step 4: By angle sum property: ∠A + ∠B + ∠C = 180°
Step 5: Let ∠A = ∠B = ∠C = x
Then: x + x + x = 180°
3x = 180°
x = 60°
Therefore, ∠A = ∠B = ∠C = 60° ✓
Part 2: Congruence with Another Equilateral Triangle
Given: △ABC and △PQR both equilateral with side length = a cm
To Prove: △ABC ≅ △PQR
In △ABC: AB = BC = CA = a cm
In △PQR: PQ = QR = RP = a cm
Comparison:
AB = PQ = a
BC = QR = a
CA = RP = a
By SSS Rule: △ABC ≅ △PQR ✓
Question 26: Proving BF = CE
Given: Isosceles △ABC with AB = AC. E and F are midpoints of AB and AC respectively
To Prove: BF = CE
Proof:
Step 1: Since AB = AC (Given), and E, F are midpoints:
AE = AF (halves of equal sides)
Step 2: In △ABF and △ACE:
• AB = AC (Given)
• ∠A = ∠A (Common angle)
• AF = AE (From Step 1)
Step 3: By SAS congruence rule:
△ABF ≅ △ACE
Step 4: By CPCT:
BF = CE ✓ (Proved)
Question 27: Theorem 7.3 & Proof
Theorem 7.3: The sides opposite to equal angles of a triangle are equal. (Converse of Theorem 7.2)
Proof Using ASA:
Given: △ABC with ∠B = ∠C
To Prove: AB = AC
Construction: Draw the angle bisector of ∠A, meeting BC at point D.
In △ABD and △ACD:
• ∠BAD = ∠CAD (AD bisects ∠A, by construction)
• ∠ABD = ∠ACD (Given: ∠B = ∠C)
• AD = AD (Common side)
By ASA rule: △ABD ≅ △ACD
By CPCT: AB = AC ✓ (Proved)
Question 28: Perpendicular Bisector Proof
Given: In △ABC, the perpendicular bisector of BC passes through A
To Prove: AB = AC
Proof:
Given Information:
• The perpendicular bisector of BC passes through A
• Let M be the midpoint of BC (where perpendicular bisector meets BC)
• Therefore, BM = MC and AM ⊥ BC, so ∠AMB = ∠AMC = 90°
Consider △ABM and △ACM:
• BM = MC (M is midpoint of BC)
• ∠AMB = ∠AMC = 90° (AM is perpendicular to BC)
• AM = AM (Common side)
By SAS rule: △ABM ≅ △ACM
By CPCT: AB = AC ✓ (Proved)
SECTION D: ANSWERS & EXPLANATIONS (5 Marks Each)
Question 29: Proving △ABC ≅ △PQR (ASA Rule)
Given: △ABC and △PQR with AB = PQ, ∠A = ∠P, ∠B = ∠Q
To Prove: △ABC ≅ △PQR
Proof Using ASA Rule:
In △ABC and △PQR:
• ∠A = ∠P (Given Angle 1)
• AB = PQ (Given Side - included between angles)
• ∠B = ∠Q (Given Angle 2)
By ASA Rule: △ABC ≅ △PQR ✓
Why Other Parts Are Also Equal (By CPCT):
Once congruence is established, by CPCT:
- BC = QR (corresponding sides of congruent triangles)
- CA = RP (corresponding sides of congruent triangles)
- ∠C = ∠R (corresponding angles of congruent triangles)
Significance: This demonstrates that once we establish congruence using ANY valid rule, all remaining sides and angles automatically become equal through CPCT, without needing separate proofs.
Question 30: PQ is Perpendicular Bisector of AB
Given: AB is a line segment. P and Q are points on opposite sides of AB such that PA = PB and QA = QB.
To Prove: Line PQ is the perpendicular bisector of AB (PQ ⊥ AB and PQ bisects AB)
Complete Proof:
Let C = intersection of PQ and AB
Part 1: Proving AC = BC (PQ bisects AB)
First, consider △PAQ and △PBQ (Using SSS rule):
• PA = PB (Given)
• QA = QB (Given)
• PQ = PQ (Common side)
By SSS: △PAQ ≅ △PBQ
By CPCT: ∠APQ = ∠BPQ
(This means PQ bisects ∠APB)
Now, consider △PAC and △PBC (Using SAS rule):
• AP = BP (Given)
• ∠APC = ∠BPC (Proved above: ∠APQ = ∠BPQ)
• PC = PC (Common side)
By SAS: △PAC ≅ △PBC
By CPCT: AC = BC
(Therefore, C is the midpoint of AB, i.e., PQ bisects AB ✓)
Also by CPCT: ∠ACP = ∠BCP
Part 2: Proving PQ ⊥ AB
Since ∠ACP = ∠BCP and they form a linear pair:
∠ACP + ∠BCP = 180°
∠ACP + ∠ACP = 180°
2∠ACP = 180°
∠ACP = 90°
Therefore, PQ ⊥ AB ✓
Conclusion: PQ is the perpendicular bisector of AB (proved both ⊥ and bisects) ✓
Question 31: Proving AD = AE
Given: Isosceles △ABC with AB = AC. D and E are points on BC such that BE = CD
To Prove: AD = AE
Detailed Proof with Explanation:
Step 1: Identify the given information
✓ AB = AC (isosceles triangle)
✓ BE = CD (given condition for points on BC)
✓ D and E are on side BC
Step 2: Use properties of isosceles triangle
Since AB = AC, by Theorem 7.2:
∠B = ∠C (base angles of isosceles triangle)
Step 3: Manipulate the given condition BE = CD
Given: BE = CD
Subtracting ED from both sides:
BE - ED = CD - ED
BD = CE
(This is the key step! We find another pair of equal lengths)
Step 4: Apply SAS rule to △ABD and △ACE
In △ABD and △ACE:
• AB = AC (From Step 1 - given) ... (1)
• ∠B = ∠C (From Step 2 - angles of isosceles triangle) ... (2)
• BD = CE (From Step 3 - derived) ... (3)
Step 5: Conclude using SAS
From (1), (2), and (3), using SAS congruence rule:
△ABD ≅ △ACE
Step 6: Use CPCT to get the final result
Since △ABD ≅ △ACE, by CPCT:
AD = AE ✓ (Proved)
Key Insight: The crucial step was recognizing that from "BE = CD," we could derive "BD = CE" by subtracting the common segment ED. This gave us the third piece of information needed for SAS congruence.
Question 32: Right Triangle with Extended Point
Given: Right triangle ABC with right angle at C, M is midpoint of hypotenuse AB, C is extended to D such that MD = CM
To Prove:
(i) △AMC ≅ △BMD
(ii) △DBC ≅ △ACB
(iii) ∠DBC = 90°
Complete Proof:
Proof of (i): △AMC ≅ △BMD
In △AMC and △BMD:
• AM = BM (M is midpoint of AB) ... (1)
• CM = MD (Given) ... (2)
• ∠AMC = ∠BMD (Vertically opposite angles) ... (3)
By SAS rule: △AMC ≅ △BMD ✓
Proof of (ii): △DBC ≅ △ACB
From (i), we have: △AMC ≅ △BMD
By CPCT from (i):
• AC = BD (corresponding sides)
• ∠ACM = ∠BDM (corresponding angles)
Now consider △DBC and △ACB:
• BC = BC (Common side)
• BD = AC (From above)
• ∠DBC and ∠ACB are corresponding angles
By SAS rule (or checking all three sides):
△DBC ≅ △ACB ✓
Proof of (iii): ∠DBC = 90°
From (ii), we have: △DBC ≅ △ACB
By CPCT:
∠DBC = ∠ACB (corresponding angles)
Given that: ∠ACB = 90° (right angle at C)
Therefore: ∠DBC = 90° ✓
Interesting Note: This problem also proves that CM = ½ AB (half the hypotenuse equals the distance from right angle to midpoint of hypotenuse), which is an important property of right triangles!
