Q1. What does \( \triangle ABC \cong \triangle PQR\) mean?

Ans: Triangles are congruent with correspondence \(A\leftrightarrow P, B\leftrightarrow Q, C\leftrightarrow R\).

Q2. If two segments are congruent, what does it mean?

Ans: They have equal length.

Q3. State one congruence test with its name.

Ans: SSS — three corresponding sides equal.

Q4. State SAS test in words.

Ans: Two sides and the included angle equal → triangles congruent.

Q5. State ASA test.

Ans: Two angles and included side equal → triangles congruent.

Q6. State AAS (SAA) test.

Ans: Two angles and any corresponding side → triangles congruent.

Q7. What is RHS (hypotenuse-side) test?

Ans: For right triangles, if hypotenuse and one side are equal, triangles congruent.

Q8. Are two triangles with corresponding parts equal necessarily congruent?

Ans: Yes — if three corresponding parts meet a congruence test (SSS, SAS, ASA, AAS, RHS).

Q9. If \( \triangle ABC \cong \triangle PQR\), is \( \angle A = \angle R\)?

Ans: Not necessarily; vertex order matters. Only \(A\leftrightarrow P\), so \( \angle A = \angle P\).

Q10. If two triangles are congruent, are corresponding sides proportional?

Ans: Yes — corresponding sides are equal (special case of proportional with ratio 1).

Q11. Give a quick reason: Why AAS implies ASA?

Ans: Because sum of triangle angles is \(180^\circ\), two angles determine the third; AAS gives included side via ASA equivalence.

Q12. If two triangles share a side and other two sides are equal respectively, which test could you try?

Ans: SSS (if third sides are equal) or SAS if one included angle is equal.

Q13. Are SSA (two sides and non-included angle) sufficient for congruence?

Ans: No — SSA is not a valid general congruence test (ambiguous case).

Q14. Name the six corresponding equalities implied by \( \triangle ABC \cong \triangle PQR\).

Ans: \(AB=PQ, BC=QR, CA=RP, \angle A=\angle P, \angle B=\angle Q, \angle C=\angle R.\)

Q15. If \(AB=PQ, BC=QR, CA=RP\), which test is used?

Ans: SSS test.

Q16. If triangles are congruent, are their areas equal?

Ans: Yes — congruent triangles have equal area.

Q17. In right triangles congruence, why hypotenuse matters?

Ans: Hypotenuse is the side opposite \(90^\circ\); RHS uses its equality with another side to fix triangle shape.

Q18. If \( \angle A = \angle P\) and \(AB=PQ\) and \(AC=PR\), which test?

Ans: SAS (two sides and included angle) — note angle must be included between those sides.

Q19. If two triangles are congruent by ASA, what else can you immediately conclude?

Ans: The third angle and all three sides correspondingly equal.

Q20. If triangle \( \triangle ABD\) and \( \triangle CBD\) share BD and AB=CB and AD=CD, are they congruent?

Ans: Yes — SSS or by noting common side BD and two pairs of equal sides → SSS.

Q1. Write a one-to-one correspondence implied by \( \triangle ABC \cong \triangle PQR\).

Ans: \(A\leftrightarrow P,\; B\leftrightarrow Q,\; C\leftrightarrow R.\)

Q2. Using SSS: If \(AB=DE,\; BC=EF,\; CA=FD\), show triangles congruent.

Ans: By SSS, \( \triangle ABC \cong \triangle DEF\) with correspondence \(A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F.\)

Q3. Using SAS: Given \(AB=PQ,\; \angle B = \angle Q,\; BC=QR\), conclude.

Ans: Two sides and included angle equal → \( \triangle ABC \cong \triangle PQR\).

Q4. Explain why ASA implies congruence using angle-sum property.

Ans: ASA gives two pairs of equal angles and included side; the third angle equals by angle sum, thus ASA forms unique triangle congruence.

Q5. Right triangle: hypotenuse \(=13\), one leg \(=5\). Another right triangle has same; are they congruent?

Ans: Yes by RHS test (hypotenuse and a corresponding side equal) — triangles congruent.

Q6. Show a counterexample why SSA fails (short explanation).

Ans: Two different triangles can be formed with same two sides and non-included angle (ambiguous case) — so SSA not sufficient.

Q7. If \( \triangle ABC \cong \triangle PQR\), find \( \angle B + \angle Q\).

Ans: Since \( \angle B = \angle Q\), sum \(=2\angle B\) (numerical value depends on angle measure).

Q8. In figure, AD=EC and \( \angle A = \angle C\) and AC common — which test?

Ans: AAS (two angles and non-included side) → triangles congruent.

Q9. If AB || DE and AC || DF, and \( \triangle ABC\) and \( \triangle DEF\) share vertex order, what can you say?

Ans: Corresponding angles equal (parallel lines), often giving ASA or AAS → congruence depending on additional equal side.

Q10. Triangles have equal perimeters and two equal corresponding sides — is congruence guaranteed?

Ans: No — equal perimeters do not force congruence; need side-side-side or other tests.

Q11. If \( \triangle ABD \cong \triangle CBD\) (common BD), which angles equal?

Ans: Corresponding angles: \( \angle ABD = \angle CBD\), \( \angle BAD=\angle BCD\), etc.

Q12. Given SSS congruence, show corresponding medians are equal.

Ans: Congruent triangles have all corresponding parts equal; median length from corresponding vertices equal.

Q13. If two triangles are congruent, are their corresponding altitudes equal?

Ans: Yes — corresponding altitudes (perpendicular distances to corresponding sides) are equal.

Q14. If triangles congruent by ASA, can you deduce SSS as well?

Ans: Yes — once triangles are congruent, the three corresponding sides are equal (SSS holds after proving congruence).

Q15. If \( \angle A = \angle P\) and \(AB=PQ\) but \(BC\ne QR\), can triangles be congruent?

Ans: Not necessarily; need three appropriate matching parts satisfying a congruence test.

Q16. If two triangles are congruent, show corresponding circumradii are equal.

Ans: Congruent triangles are identical in size — their circumcircles have equal radii.

Q17. Provide the correspondence for congruence if identical marks show AB = DE and AC = DF and \( \angle A\) common.

Ans: Correspondence \(A\leftrightarrow D,\ B\leftrightarrow E,\ C\leftrightarrow F\) → SAS if angle is included.

Q18. In a parallelogram, two triangles formed by a diagonal are they congruent?

Ans: Yes — diagonal divides parallelogram into two congruent triangles by SSS or SAS.

Q19. If two triangles are congruent, show corresponding bisected angles correspond.

Ans: Angle bisectors correspond — each half-angle matches the other triangle's corresponding half-angle.

Q20. If \(\triangle ABC\) congruent to \(\triangle PQR\), what is the relation between perimeters?

Ans: Equal perimeters: \(AB+BC+CA = PQ+QR+RP.\)

Q1. Prove: If two triangles are SSS congruent, corresponding angles are equal.

Ans: By SSS, triangles match exactly. Construct congruence mapping; since side lengths equal, triangle correspondence fixes shape, so corresponding angles equal. (Alternatively use rigidity — triangle uniquely determined by 3 side lengths.)

Q2. Given \(AB=DE, BC=EF, \angle B=\angle E\). Are triangles congruent? If yes, which test?

Ans: If \(AB\) and \(BC\) are the two sides adjacent to angle \(B\), this is SAS: two sides and included angle. So \( \triangle ABC \cong \triangle DEF.\)

Q3. In diagram, triangles \(ABD\) and \(CBD\) share BD, and AB=CB, AD=CD. Prove congruence and state corresponding angles.

Ans: By SSS (AB=CB, AD=CD, BD common) → \( \triangle ABD \cong \triangle CBD\). Hence \( \angle BAD=\angle BCD\) and \( \angle ABD=\angle CBD\) etc.

Q4. Right triangles: Show if hypotenuse and an acute side equal, triangles congruent (RHS).

Ans: Let right triangles have hypotenuses equal and one corresponding leg equal. Reflect one triangle onto the other so right angle matches; remaining sides forced equal → congruent by RHS.

Q5. Given \( \triangle ABC\) and \( \triangle PQR\) with \(AB=PQ, BC=QR\) and \( \angle C=\angle R\). Are they congruent?

Ans: Need to check whether angle \(C\) is included between \(BC\) and \(CA\). If \( \angle C\) is included between the given equal sides (i.e., between \(BC\) and \(CA\) corresponding to \(QR\) and \(RP\)), then SAS holds. If not, more info is needed.

Q6. Triangles ABC and DEF: AB=DE, AC=DF, and \( \angle A = \angle D\). Show congruence.

Ans: Sides AB and AC with included angle \(A\) equal → SAS → triangles congruent.

Q7. Provide full solution: In \( \triangle ABC\) and \( \triangle DEF\), if \( \angle A=\angle D\), \( \angle B=\angle E\), and \(AB=DE\), are they congruent?

Ans: Two angles equal, and a side equal. If the side is included between the two equal angles, it's ASA; if not included (side opposite one of these angles), it's AAS. In either case triangles are congruent.

Q8. Show why SSA fails with a numeric example.

Ans: Let side \(a=10\), other side \(b=6\), angle opposite \(b\) be \(30^\circ\). Two different heights can be constructed giving two different triangles (ambiguous case). Thus SSA not sufficient.

Q9. If diagonals of rhombus bisect each other at right angles, show triangles formed are congruent.

Ans: Diagonals bisect, so two triangles share a side (half-diagonal), and adjacent sides equal (rhombus sides), and include right angle → SAS or RHS depending on configuration → congruent.

Q10. Prove: In a rectangle, diagonals are equal and they bisect each other — use congruence.

Ans: In rectangle \(ABCD\), consider \( \triangle AOB\) and \( \triangle COD\) where O is intersection of diagonals. \(AO=CO\) and \(BO=DO\) and \( \angle AOB = \angle COD\) vertically opposite. So SSS/SAS → triangles congruent; hence diagonals equal and bisect each other.

Q11. Given two triangles with sides \(5,12,13\) and \(13,12,5\) (order changed), are they congruent?

Ans: Yes — SSS (all corresponding side lengths equal) — triangles congruent although vertex order should preserve correspondence.

Q12. Show that a triangle is uniquely determined by SAS — provide short reasoning.

Ans: Two sides and included angle fix the relative positions of the sides; rotating and translating changes nothing — unique triangle up to congruence → SAS valid.

Q13. Solve textbook example: In figure identical marks show congruent parts; write correspondence and test. (Refers to example in chapter).

Ans: See chapter example: if marks show AB=PQ, AC=PR and angle A= P, test SAS → \( \triangle ABC \cong \triangle PQR\) (correspondence \(A\leftrightarrow P\)).

Q14. Prove: If two triangles are congruent, corresponding altitudes to corresponding sides are equal.

Ans: Congruent triangles have equal corresponding sides; altitude length depends only on side length and area (equal), so altitudes equal.

Q15. Using congruence, show opposite angles of kite with equal adjacent sides are equal.

Ans: In kite with AB=BC and AD=DC, triangles formed by diagonal are congruent (SAS) → corresponding angles equal → one pair of opposite angles equal.

Q16. Give steps to verify whether \( \triangle ABC \cong \triangle CAB\) is correct (vertex order).

Ans: Check correspondence: \(A\leftrightarrow C,\ B\leftrightarrow A,\ C\leftrightarrow B\). Compare marked equal parts; if marking matches this mapping, congruence holds; otherwise wrong mapping.

Q17. If triangles are congruent, do medians correspond? Provide reason.

Ans: Yes — medians (segments from a vertex to midpoint of opposite side) are corresponding parts; congruence preserves midpoints and distances → medians equal.

Q18. In example, triangles STU and XZY are congruent; write two different congruences (re-orderings) that are valid.

Ans: If correspondence STU \( \leftrightarrow \) XZY then \( \triangle STU \cong \triangle XZY\). Another valid re-ordering is \( \triangle UST \cong \triangle YXZ\) rotating vertex order but preserving correspondence.

Q19. If triangles have equal area and one equal side, are they congruent?

Ans: No — equal area and one equal side not sufficient; need full congruence test (SSS/SAS/ASA/AAS/RHS).

Q20. Full solution (textbook): In figure, parts marked equal — state correspondence & test. (Refer to solved Ex (1) and (2) in chapter.)

Ans: Follow the chapter examples: identify which sides/angles are equal, check which congruence test applies (SSS/SAS/ASA/AAS/RHS), then write correspondence preserving vertex order. (E.g. SSS → \(PQR\leftrightarrow UTS\)).

Ex: In figure where identical marks show corresponding parts — identify correspondence and write congruence in two ways; check if \( \triangle XYZ \cong \triangle STU\) is correct.

Solution (brief): From the figure the matching is \(S\leftrightarrow X,\ T\leftrightarrow Z,\ U\leftrightarrow Y\) so one congruence is \( \triangle STU \cong \triangle XZY\). Another valid cyclic re-ordering is \( \triangle UST \cong \triangle YXZ\). The statement \( \triangle XYZ \cong \triangle STU\) is wrong because vertex correspondence doesn't match the marked equal parts (it would imply wrong side matches).

Ex: In triangles ABD and ACD with AD common and given marks AB=CB and AD common; state correspondence and congruence.

Solution: Correspondence \(A\leftrightarrow A,\ B\leftrightarrow C,\ D\leftrightarrow D\). Triangles \( \triangle ABD \cong \triangle ACD\) by SSS (if third side equal is given or deduced). Then corresponding angles \( \angle ABD = \angle ACD\), etc.

Q1.(i) For each pair of triangles with equal marks, state test & correspondence.

Solution approach: Identify three marked congruent parts — check which test (SSS/SAS/ASA/AAS/RHS) fits. Then list correspondence preserving vertex order: e.g. if AB=PQ, BC=QR, AC=PR → SSS: \( \triangle ABC\cong\triangle PQR\) (A↔P,B↔Q,C↔R).

Q2. In triangle pairs where one side is common (say CA), and other equal parts marked — what additional info needed to use AAS?

Ans: Need one more angle equality (besides the two) or that the given side is adjacent to the two equal angles — then AAS or ASA applies. For example, if CA common and \( \angle C = \angle A\) (in other triangle) and another angle equal, use AAS.

Q3. Prove: If in quadrilateral two triangles having matching three sides are congruent, then the third pair of corresponding angles are equal.

Ans: By SSS congruence, triangles are identical; hence all corresponding angles equal, including the third pair.

Q1. For each pair of triangles (with given equal marks) find congruence test, vertex correspondence & remaining congruent parts.

Solution style: (1) Identify SSS/SAS/ASA etc; (2) Write mapping e.g. \(M\leftrightarrow S\), etc.; (3) State remaining equal angles/sides using correspondence. (Because the chapter's diagrams vary, follow the marking to produce exact mapping.)

Q2. In the adjacent figure seg AD = EC. What else is needed to prove \( \triangle ABD \cong \triangle EBC\) by AAS?

Ans: Need two angles: e.g. \( \angle BAD = \angle CEB\) and \( \angle ABD = \angle EBC\). With AD = EC and two angle equalities → AAS → congruence.