Chapter 11: Congruence of Triangles - Mock Test
Instructions
- •This test has 20 questions covering all topics from Chapter 11
- •Questions are arranged as per CBSE paper pattern
- •Section A: 8 questions × 1 mark = 8 marks
- •Section B: 8 questions × 2 marks = 16 marks
- •Section C: 4 questions × 4 marks = 16 marks
- •No negative marking
- •Show all steps in your proofs and solutions
- •You can take this test multiple times
Mock Test Questions
Section A: 1 Mark Questions (8 × 1 = 8 marks)
1. What does the symbol ≅ represent?
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Answer: ≅ represents congruence. △ABC ≅ △DEF means triangle ABC is congruent to triangle DEF.
2. If △PQR ≅ △STU, which side of △STU corresponds to PQ?
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Answer: ST. Since vertices are in order P↔S, Q↔T, R↔U, side PQ corresponds to ST.
3. Name the criterion that requires two sides and the included angle.
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Answer: SAS (Side-Angle-Side) criterion. The angle must be between the two sides.
4. Can SSA (Side-Side-Angle) prove triangle congruence?
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Answer: No. SSA does not prove congruence. It can produce two different triangles (ambiguous case).
5. For which type of triangles is RHS criterion applicable?
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Answer: Right triangles only. RHS = Right angle-Hypotenuse-Side criterion.
6. If △ABC ≅ △XYZ by SSS, what can you conclude about angles?
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Answer: Corresponding angles are equal by CPCT. ∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z.
7. In triangle XYZ, XY = 5 cm, YZ = 6 cm, ZX = 7 cm. In triangle ABC, AB = 5 cm, BC = 6 cm, CA = 7 cm. By which criterion are they congruent?
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Answer: SSS (Side-Side-Side) criterion. All three sides of both triangles are equal.
8. What is CPCT used for?
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Answer: CPCT (Corresponding Parts of Congruent Triangles) is used to find unknown sides and angles after proving congruence.
Section B: 2 Mark Questions (8 × 2 = 16 marks)
1. △ABC has AB = 7 cm, BC = 5 cm, CA = 6 cm. △PQR has PQ = 7 cm, QR = 5 cm, RP = 6 cm. State whether they are congruent and give the criterion.
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Answer: Yes, △ABC ≅ △PQR by SSS criterion. Reason: AB = PQ = 7 cm, BC = QR = 5 cm, CA = RP = 6 cm. All three sides are equal.
2. In right triangle ABC with ∠B = 90°, hypotenuse AC = 13 cm, AB = 5 cm. In right triangle PQR with ∠Q = 90°, hypotenuse PR = 13 cm, PQ = 5 cm. Prove congruence using RHS.
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Answer: △ABC ≅ △PQR by RHS criterion. Reason: ∠B = ∠Q = 90°, AC = PR = 13 cm (hypotenuses), AB = PQ = 5 cm (one leg).
3. △DEF and △MNO have ∠D = ∠M = 60°, DE = MN = 6 cm, ∠E = ∠N = 70°. State the congruence criterion.
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Answer: △DEF ≅ △MNO by ASA criterion. Reason: ∠D = ∠M, DE = MN (included side), ∠E = ∠N. Side DE is between angles D and E.
4. △XYZ ≅ △ABC. If XY = 8 cm, ∠Z = 45°, find the corresponding measurements in △ABC.
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Answer: AB = 8 cm (by CPCT, XY corresponds to AB), ∠C = 45° (by CPCT, ∠Z corresponds to ∠C).
5. In △PQR, PQ = PR. If ∠Q = 55°, find ∠R. Justify your answer.
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Answer: ∠R = 55°. Justification: Since PQ = PR, triangle PQR is isosceles. In isosceles triangles, base angles are equal. Therefore, ∠Q = ∠R = 55°.
6. Can two triangles with sides 3 cm, 4 cm, and 5 cm be non-congruent? Why or why not?
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Answer: No, they must be congruent. Reason: By SSS criterion, if all three sides are equal, the triangles are definitely congruent.
7. In △ABC, AB = 5 cm, ∠A = 50°, AC = 6 cm. In △PQR, PQ = 5 cm, ∠P = 50°, PR = 6 cm. Are they congruent? State the criterion.
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Answer: Yes, △ABC ≅ △PQR by SAS criterion. Reason: AB = PQ = 5 cm, ∠A = ∠P = 50° (included), AC = PR = 6 cm.
8. If △MNO ≅ △RST, and ∠M = 35°, ∠N = 65°, find ∠O and corresponding angles in △RST.
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Answer: ∠O = 80° (since sum of angles = 180°: 35° + 65° + 80° = 180°) By CPCT: ∠R = 35°, ∠S = 65°, ∠T = 80° (corresponding to ∠M, ∠N, ∠O respectively)
Section C: 4 Mark Questions (4 × 4 = 16 marks)
1. In quadrilateral ABCD, AB = CD and AB ∥ CD. Diagonal AC divides it. Prove that △ABC ≅ △CDA using SAS criterion and state what we can conclude using CPCT.
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Answer: Given: AB = CD, AB ∥ CD, AC is diagonal To Prove: △ABC ≅ △CDA Proof: AB = CD (given) ∠BAC = ∠DCA (alternate angles, since AB ∥ CD) AC = AC (common side) By SAS criterion: △ABC ≅ △CDA By CPCT: - BC = AD - ∠ABC = ∠CDA - ∠BCA = ∠DAC
2. △ABC is isosceles with AB = AC. If D is a point on BC such that AD ⊥ BC, prove that △ABD ≅ △ACD. What does this tell us about D?
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Answer: Given: AB = AC, D on BC, AD ⊥ BC To Prove: △ABD ≅ △ACD Proof: AB = AC (given) ∠ADB = ∠ADC = 90° (given AD ⊥ BC) AD = AD (common side) By RHS criterion: △ABD ≅ △ACD (both are right triangles) By CPCT: - BD = DC - ∠BAD = ∠CAD Conclusion: D is the midpoint of BC, and AD bisects ∠BAC.
3. In two right triangles, the hypotenuses are equal to 10 cm. One triangle has a leg of 6 cm, and the other has a leg of 6 cm. Are they congruent? Prove using RHS criterion and find the other leg in both triangles.
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Answer: Given: Right △ABC: ∠B = 90°, AC (hypotenuse) = 10 cm, AB = 6 cm Right △PQR: ∠Q = 90°, PR (hypotenuse) = 10 cm, PQ = 6 cm Proof of Congruence: ∠B = ∠Q = 90° (right angles) AC = PR = 10 cm (hypotenuses equal) AB = PQ = 6 cm (one leg equal) By RHS criterion: △ABC ≅ △PQR Finding other leg: Using Pythagoras theorem: AC² = AB² + BC² 10² = 6² + BC² 100 = 36 + BC² BC² = 64 BC = 8 cm Similarly, QR = 8 cm (by CPCT) Answer: Yes, they are congruent. Other legs are 8 cm each.
4. Prove that if two triangles have two sides and the angle opposite the longer side equal, then they are congruent (this extends understanding beyond standard criteria). Explain why SSA usually fails but this variant works.
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Answer: This is an extension of RHS principle. When the angle is opposite the longer side: Let △ABC and △PQR have: AB = PQ (side) AC = PR (side, let's say AC is longer) ∠B = ∠Q (angle opposite to AC and PR respectively) Since AC is the longest side (hypotenuse in right case), and ∠B is opposite to it, the angle is uniquely determined. Why standard SSA fails: When angle is opposite the shorter side, two positions of the third vertex are possible. Why this variant works: When angle is opposite the longer side, there's only one possible position for the third vertex, making triangles unique and congruent. This principle is important in understanding the conditions for uniqueness of triangles.
Test Summary
- Total Marks: 40 marks
- Time Limit: 60 minutes
- Expected Duration per Question: Section A: 2-3 min each, Section B: 3-4 min each, Section C: 8-10 min each
- Key Topics Covered: Congruence criteria (SSS, SAS, ASA, RHS), CPCT, Identifying corresponding parts, Solving geometry problems
- Passing Score: 24/40 (60%)