Searching for accurate, teacher-verified solutions for DAV Class 7 Maths Chapter 9 Brain Teasers, HOTS & Value Based Questions (Congruent Triangles)? You’ve come to the right place! At NCERTBooksPDF, our detailed solutions strictly follow the DAV Secondary Mathematics textbook, helping students grasp triangle congruence criteria (SSS, SAS, ASA, RHS), CPCTC applications, geometric proofs, and higher-order analytical problems.
DAV Class 7 Maths Chapter 9: all worksheets All Class 7 Maths chapters →
Criteria for Congruence of Triangles (Quick Reference)
| Criterion | Full Name | Conditions for Congruence |
|---|---|---|
| SSS | Side-Side-Side | All three corresponding sides of two triangles are equal. |
| SAS | Side-Angle-Side | Two sides and the included angle of one triangle equal those of the other. |
| ASA / AAS | Angle-Side-Angle | Two angles and the included side (or any side) are respectively equal. |
| RHS | Right angle-Hypotenuse-Side | Hypotenuse and one leg of a right triangle equal those of another right triangle. |
Chapter 9 Brain Teasers Solutions
Q.1 (A). Tick the correct option:
(a) If ΔABC ≅ ΔPQR, then side BC corresponds to:
(i) PQ
(ii) QR
(iii) PR
(iv) None of these
Answer: (ii) QR
Explanation: In correspondence ABC ↔ PQR, vertex B corresponds to Q and C corresponds to R. Thus segment BC corresponds to segment QR.
(b) Which of the following is NOT a criterion for congruence of triangles?
(i) SSS
(ii) SAS
(iii) AAA
(iv) RHS
Answer: (iii) AAA
Explanation: AAA (Angle-Angle-Angle) guarantees that two triangles have the same shape (similar), but not necessarily the same size. Therefore AAA is not a congruence criterion.
(c) In ΔABC and ΔDEF, AB = DE, BC = EF, and ∠B = ∠E. Then ΔABC ≅ ΔDEF by:
(i) SSS
(ii) SAS
(iii) ASA
(iv) RHS
Answer: (ii) SAS
Explanation: ∠B is the included angle between sides AB and BC, and ∠E is the included angle between DE and EF. By SAS criterion, the two triangles are congruent.
(d) Two line segments are congruent if and only if they have:
(i) Equal lengths
(ii) Same endpoints
(iii) Parallel direction
(iv) None of these
Answer: (i) Equal lengths
Explanation: Two geometric figures are congruent if they coincide when superimposed. Two line segments coincide if and only if their lengths are equal.
(e) In ΔPQR, if PQ = PR and PM ⊥ QR, then ΔPMQ ≅ ΔPMR by which criterion?
(i) SSS
(ii) SAS
(iii) RHS
(iv) AAA
Answer: (iii) RHS
Explanation: In right triangles ΔPMQ and ΔPMR: ∠PMQ = ∠PMR = 90°, Hypotenuse PQ = Hypotenuse PR (given), and common side PM = PM. Hence congruent by RHS.
Q.1 (B). Answer the following questions:
(a) What is the full form of CPCTC?
Solution:
CPCTC stands for “Corresponding Parts of Congruent Triangles are Congruent”.
(b) If ΔABC ≅ ΔFED under the correspondence ABC ↔ FED, write the parts of ΔABC that correspond to:
(i) ∠E
(ii) Side ED
(iii) ∠D
Solution:
From ABC ↔ FED:
(i) ∠E corresponds to ∠B.
(ii) Side ED corresponds to Side BA (or AB).
(iii) ∠D corresponds to ∠C.
(c) Can two equilateral triangles of different perimeters be congruent? Why or why not?
Solution:
No. Congruent figures must have identical shape and size. Different perimeters mean their side lengths are unequal, violating SSS congruence condition.
(d) In ΔABC, AD is the perpendicular bisector of side BC. Prove that AB = AC.
Solution:
In ΔADB and ΔADC:
1. BD = CD (since AD bisects BC)
2. ∠ADB = ∠ADC = 90° (since AD ⊥ BC)
3. AD = AD (common side)
By SAS criterion: ΔADB ≅ ΔADC.
Therefore, AB = AC (by CPCTC). Hence ΔABC is isosceles.
(e) State whether the following statement is True or False: “Two squares with equal areas are always congruent.”
Solution:
True. Area of square = s2. If areas are equal, side lengths s are equal. Since all four angles of a square are 90° and side lengths are equal, the two squares coincide exactly.
Q.2. Geometric Proofs & Congruence Applications:
Problem 1: In a triangle ABC, altitudes BD and CE are drawn to opposite sides AC and AB respectively such that BD = CE. Prove that ΔBDC ≅ ΔCEB and deduce that ΔABC is an isosceles triangle.
Proof:
In right-angled triangles ΔBDC and ΔCEB:
1. ∠BDC = ∠CEB = 90° (altitudes BD ⊥ AC, CE ⊥ AB)
2. BC = BC (common hypotenuse)
3. BD = CE (given altitudes are equal)
By RHS criterion:
ΔBDC ≅ ΔCEB.
By CPCTC:
∠DCB = ∠EBC, which means ∠ACB = ∠ABC.
Since the two base angles of ΔABC are equal, the sides opposite to equal angles are equal:
AB = AC.
Therefore, ΔABC is an isosceles triangle. (Hence Proved!)
Problem 2: In ΔABC, AB = AC and D is a point on BC such that AD ⊥ BC. Prove that:
(a) ΔABD ≅ ΔACD
(b) BD = CD
(c) ∠BAD = ∠CAD
Proof:
In right triangles ΔABD and ΔACD:
1. ∠ADB = ∠ADC = 90° (since AD ⊥ BC)
2. AB = AC (hypotenuse, given)
3. AD = AD (common side)
By RHS criterion:
(a) ΔABD ≅ ΔACD.
From this congruence, by CPCTC:
(b) BD = CD (AD bisects BC)
(c) ∠BAD = ∠CAD (AD bisects vertical angle ∠A). (Hence Proved!)
HOTS (Higher Order Thinking Skills) Questions
HOTS 1: The diagonals AC and BD of a quadrilateral ABCD bisect each other at right angles at point O. Prove that ABCD is a rhombus.
Proof:
Given: OA = OC, OB = OD, and AC ⊥ BD at O.
Consider triangles ΔAOB and ΔCOB:
1. OA = OC (given)
2. ∠AOB = ∠COB = 90° (given)
3. OB = OB (common side)
By SAS criterion: ΔAOB ≅ ΔCOB → AB = BC (CPCTC).
Similarly, taking ΔCOB and ΔCOD gives BC = CD.
Taking ΔCOD and ΔAOD gives CD = DA.
Thus, AB = BC = CD = DA.
A quadrilateral having all four sides equal is a rhombus. (Hence Proved!)
HOTS 2: If the medians BE and CF of a triangle ABC are equal, prove that ΔABC is isosceles with AB = AC.
Proof:
Let G be the centroid of ΔABC. The centroid divides each median in the ratio 2 : 1.
Since BE = CF, BG = (2/3)BE = (2/3)CF = CG, and GE = (1/3)BE = (1/3)CF = GF.
In ΔBGF and ΔCGE:
1. BG = CG
2. ∠BGF = ∠CGE (vertically opposite angles)
3. GF = GE
By SAS criterion: ΔBGF ≅ ΔCGE → BF = CE (CPCTC).
Since F is midpoint of AB → AB = 2 BF.
Since E is midpoint of AC → AC = 2 CE.
Since BF = CE → 2 BF = 2 CE → AB = AC.
Therefore, ΔABC is an isosceles triangle. (Hence Proved!)
Value Based Questions (VBQs)
Question: During a school craft competition, Neha designed a symmetrical kite ABCD where AB = AD and BC = CD. The central bamboo spine AC connects the top and bottom corners.
(a) Prove that ΔABC ≅ ΔADC using a suitable congruence criterion.
Solution:
In ΔABC and ΔADC:
1. AB = AD (given)
2. BC = CD (given)
3. AC = AC (common side)
By SSS criterion: ΔABC ≅ ΔADC.
(b) Prove that diagonal AC bisects the angles ∠BAD and ∠BCD.
Solution:
Since ΔABC ≅ ΔADC, by CPCTC:
• ∠BAC = ∠DAC → AC bisects ∠BAD.
• ∠BCA = ∠DCA → AC bisects ∠BCD.
(c) Why is mathematical symmetry important in flight and kite aerodynamics?
Solution:
Symmetry ensures equal lift distribution, stability, and aerodynamic balance. Asymmetric weight or surface area would cause the kite or aircraft to spin uncontrollably and crash.
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