RS Aggarwal Class 7 Mathematics Solutions Chapter 16 Congruence

Question 1. State the correspondence between the vertices, sides and angles of the following pairs of congruent triangles.
(i) \( \triangle ABC \cong \triangle EFD \)
(ii) \( \triangle CAB \cong \triangle QRP \)
(iii) \( \triangle XZY \cong \triangle QPR \)
(iv) \( \triangle MPN \cong \triangle SQR \)
Answer:
(i) \( \triangle ABC \cong \triangle EFD \)
Vertex correspondence: \( A \leftrightarrow E, B \leftrightarrow F, C \leftrightarrow D \)
Side correspondence: \( AB = EF, BC = FD, CA = DE \)
Angle correspondence: \( \angle A = \angle E, \angle B = \angle F, \angle C = \angle D \)

(ii) \( \triangle CAB \cong \triangle QRP \)
Vertex correspondence: \( C \leftrightarrow Q, A \leftrightarrow R, B \leftrightarrow P \)
Side correspondence: \( CA = QR, AB = RP, BC = PQ \)
Angle correspondence: \( \angle C = \angle Q, \angle A = \angle R, \angle B = \angle P \)

(iii) \( \triangle XZY \cong \triangle QPR \)
Vertex correspondence: \( X \leftrightarrow Q, Z \leftrightarrow P, Y \leftrightarrow R \)
Side correspondence: \( XZ = QP, ZY = PR, YX = RQ \)
Angle correspondence: \( \angle X = \angle Q, \angle Z = \angle P, \angle Y = \angle R \)

(iv) \( \triangle MPN \cong \triangle SQR \)
Vertex correspondence: \( M \leftrightarrow S, P \leftrightarrow Q, N \leftrightarrow R \)
Side correspondence: \( MP = SQ, PN = QR, NM = RS \)
Angle correspondence: \( \angle M = \angle S, \angle P = \angle Q, \angle N = \angle R \)
In simple words: When two triangles match exactly, you need to line up their vertices in the correct order. Once you know which vertex of one triangle goes with which vertex of the other, the matching sides and angles follow naturally - the side connecting two vertices in one triangle matches the side connecting the corresponding vertices in the other triangle, and the same rule applies to angles at those vertices.

Exam Tip: Always write the vertex correspondence first - this determines the entire answer. The order of letters in the congruence statement tells you exactly which vertices match.

Question 2. State the congruence property used in each of the following.
(i) \( \triangle ACB \cong \triangle DEF \) (SAS congruence property)
(ii) \( \triangle RPQ \cong \triangle LNM \) (RHS congruence property)
(iii) \( \triangle YXZ \cong \triangle TRS \) (SSS congruence property)
(iv) \( \triangle DEF \cong \triangle PNM \) (ASA congruence property)
(v) \( \triangle ACB \cong \triangle ACD \) (ASA congruence property)
Answer:
(i) \( \triangle ACB \cong \triangle DEF \) - SAS (Side - Angle - Side)
(ii) \( \triangle RPQ \cong \triangle LNM \) - RHS (Right angle - Hypotenuse - Side)
(iii) \( \triangle YXZ \cong \triangle TRS \) - SSS (Side - Side - Side)
(iv) \( \triangle DEF \cong \triangle PNM \) - ASA (Angle - Side - Angle)
(v) \( \triangle ACB \cong \triangle ACD \) - ASA (Angle - Side - Angle)
In simple words: Each congruence property is a shortcut. Instead of checking all three sides and all three angles, you only need to check which parts match - either the sides and one angle between them (SAS), or two angles with the side between them (ASA), or all three sides (SSS), or for right triangles, the longest side and one other side (RHS).

Exam Tip: Memorise which parts are checked in each rule - SAS means side, then angle, then side in that order; ASA means angle, then side, then angle. RHS only applies to right-angled triangles.

Question 3. Given: PL ⊥ OA, PM ⊥ OB, PL - PM. To prove: △PLO ≅ △PMO. Find whether OA = OB.
Answer: In \( \triangle PLO \) and \( \triangle PMO \):
\( \angle PLO = \angle PMO = 90° \) (each)
\( PO = PO \) (common)
\( PL = PM \) (given)
By RHS congruence property:
\( \triangle PLO \cong \triangle PMO \)

Therefore, \( OA = OB \) (corresponding parts of the congruent triangles).
In simple words: Since both triangles have right angles, share the same hypotenuse, and have one equal side, they must be identical. This means the distances OA and OB are the same.

Exam Tip: For RHS congruence, you must have a right angle and the hypotenuse must be the same side in both triangles - check this first.

Question 4. Given: AD = BC, AD ∥ BC. To prove: AB = DC.
Answer: Given:
\( AD = BC \)
\( AD \parallel BC \)
We must show that \( AB = DC \).

Proof:
\( AD \parallel BC \)
\( \therefore \angle BCA = \angle DAC \) (alternate angles)
In \( \triangle ABC \) and \( \triangle CDA \):
\( BC = DA \) (given)
\( \angle BCA = \angle DAC \) (proved above)
\( AC = AC \) (common)
By SAS congruence property:
\( \triangle ABC \cong \triangle CDA \)
\( \Rightarrow AB = CD \) (corresponding parts of the congruent triangles)
In simple words: The parallel lines create equal alternate angles. Combined with the equal sides and the common side between them, this gives us two triangles that match perfectly by the SAS rule, which guarantees the remaining sides are also equal.

Exam Tip: When parallel lines are given, always look for alternate angles or corresponding angles - these are equal and often form the key to the proof.

Question 5. Given: AB = AC, BD = DC. To prove: △ADB ≅ △ADC. Also find: (i) ∠ADB = ∠ADC = 90°, (ii) ∠BAD = ∠CAD.
Answer: Given:
\( AB = AC, BD = DC \)
To prove: \( \triangle ADB \cong \triangle ADC \)

Proof:
In \( \triangle ADB \) and \( \triangle ADC \):
\( AB = AC \) (given)
\( BD = DC \) (given)
\( DA = DA \) (common)
By SSS congruence property:
\( \triangle ADB \cong \triangle ADC \) \( \ldots(1) \)

\( \angle ADB \) and \( \angle ADC \) are on the straight line.
\( \therefore \angle ADB + \angle ADC = 180° \)
\( \angle ADB + \angle ADB = 180° \)
\( => 2\angle ADB = 180° \)
\( => \angle ADB = 90° \)

From (1):
\( \angle ADB = \angle ADC = 90° \)
(ii) \( \angle BAD = \angle CAD \) (corresponding parts of the congruent triangles)
In simple words: All three sides of one triangle match all three sides of the other, so they must be identical shapes. The angles at D must add up to 180° since they form a straight line, which means each one is 90°. The congruence also tells us the angles at A are equal.

Exam Tip: When angles lie on a straight line, they sum to 180°. If both angles are equal (from congruence), you can solve for each one directly.

Question 6. Given: AD is a bisector of ∠A, AD ⊥ BC. To prove: △ABC is isosceles.
Answer: Given:
\( AD \) is a bisector of \( \angle A \).
\( \Rightarrow \angle DAB = \angle DAC \) \( \ldots(1) \)
\( AD \perp BC \)
\( \Rightarrow \angle BDA = \angle CDA \) (90° each)

To prove:
\( \triangle ABC \) is isosceles.

Proof:
In \( \triangle DAB \) and \( \triangle DAC \):
\( \angle BDA = \angle CDA \) (90° each)
\( DA = DA \) (common)
\( \angle DAB = \angle DAC \) (from 1)
By ASA congruence property:
\( \triangle DAB \cong \triangle DAC \)
\( \Rightarrow AB = AC \) (corresponding parts of the congruent triangles)
Therefore, \( \triangle ABC \) is isosceles.
In simple words: The angle bisector divides angle A into two equal parts. When this bisector is also perpendicular to the base, the two resulting triangles match perfectly. This forces the two sides of the original triangle to be equal, making it isosceles.

Exam Tip: A line that both bisects an angle and is perpendicular to the opposite side creates two congruent triangles - this is a powerful tool for proving isosceles properties.

Question 7. Given: AB = AD, CB = CD. To prove: △ABC ≅ △ADC.
Answer: Given:
\( AB = AD \)
\( CB = CD \)

To prove:
\( \triangle ABC \cong \triangle ADC \)

Proof:
In \( \triangle ABC \) and \( \triangle ADC \):
\( AB = AD \) (given)
\( BC = DC \) (given)
\( AC = AC \) (common)
\( \therefore \triangle ABC \cong \triangle ADC \) (by SSS congruence property)
In simple words: Both triangles share the same side AC. Additionally, two sides of one triangle match two sides of the other triangle. Since all three sides match, the triangles must be identical.

Exam Tip: SSS congruence is often the quickest route when you have information about all three sides of both triangles.

Question 8. Given: PA ⊥ AB, QB ⊥ AB, PA = QB. To prove: △OAP ≅ △OBQ. Find whether OA = OB.
Answer: Given:
\( PA \perp AB \)
\( QB \perp AB \)
\( PA = QB \)

To prove: \( \triangle OAP \cong \triangle OBQ \)
Find whether \( OA = OB \).

Proof:
In \( \triangle OAP \) and \( \triangle OBQ \):
\( \angle POA = \angle QOB \) (vertically opposite angles)
\( \angle OAP = \angle OBQ \) (90° each)
\( PA = QB \) (given)
By AAS congruence property:
\( \triangle OAP \cong \triangle OBQ \)
\( \Rightarrow OA = OB \) (corresponding parts of the congruent triangles)
In simple words: The vertically opposite angles at O are equal. Both perpendiculars create 90° angles. Combined with the equal sides PA and QB, we get two matching triangles. The congruence tells us OA and OB must also be equal.

Exam Tip: Vertically opposite angles are always equal - use this whenever two lines cross. AAS (Angle - Angle - Side) is like ASA but with the side not between the two angles.

Question 9. Triangles ABC and DCB are right angled at A and D, respectively. AC = DB. To prove: △ABC ≅ △DCB.
Answer: Given:
Triangles \( ABC \) and \( DCB \) are right angled at \( A \) and \( D \), respectively.
\( AC = DB \)

To prove: \( \triangle ABC \cong \triangle DCB \)

In \( \triangle ABC \) and \( \triangle DCB \):
\( \angle CAB = \angle BDC \) (90° each)
\( BC = BC \) (common)
\( AC = DB \) (given)
By R.H.S. congruence property:
\( \triangle ABC \cong \triangle DCB \)
In simple words: Both triangles have a right angle. The side BC acts as the hypotenuse in both triangles and is shared. With one additional side (AC in one triangle, DB in the other) being equal, the RHS rule guarantees the triangles match perfectly.

Exam Tip: RHS stands for Right angle - Hypotenuse - Side. The hypotenuse must be the longest side in each triangle and must be the same in both.

Question 10. Given: △ABC is an isosceles triangle in which AB = AC. E and F are midpoints of AC and AB, respectively. To prove: BE = CF.
Answer: Given:
\( \triangle ABC \) is an isosceles triangle in which \( AB = AC \).
\( E \) and \( F \) are midpoints of \( AC \) and \( AB \), respectively.

To prove:
\( BE = CF \)

Proof:
\( E \) and \( F \) are midpoints of \( AC \) and \( AB \), respectively.
\( \Rightarrow AF = FB, AE = EC \)
\( AB = AC \)
\( \Rightarrow \frac{1}{2}AB = \frac{1}{2}AC \)
\( \Rightarrow FB = EC \)
\( \angle ABC = \angle ACB \) (angle opposite to equal sides are equal)
\( \Rightarrow \angle FBC = \angle ECB \)
Consider \( \triangle BCF \) and \( \triangle CBE \):
\( BC = BC \) (common)
\( \Rightarrow \angle FBC = \angle ECB \)
Consider \( \triangle BCF \) and \( \triangle CBE \):
\( BC = BC \) (common)
\( \angle FBC = \angle ECB \)
\( FB = EC \) (proved above)
By SAS congruence property:
\( \triangle BCF \cong \triangle CBE \)
\( BE = CF \) (corresponding parts of the congruent triangles)
In simple words: In an isosceles triangle, the base angles are equal. Since the equal sides are bisected, the two halves are equal too. The common base BC, combined with the equal half-sides and equal base angles, creates two matching triangles. Therefore the remaining sides must match as well.

Exam Tip: In isosceles triangles, remember that angles opposite equal sides are equal - this often provides the angles needed for congruence.

Question 11. Given: AB = AC, △ABC is an isosceles triangle, AP = AQ. To prove: BQ = CP.
Answer: Given:
\( AB = AC \)
\( \triangle ABC \) is an isosceles triangle.
\( AP = AQ \)

To prove:
\( BQ = CP \)

Proof:
\( AB = AC \) (given)
\( AP = AQ \) (given)
\( AB - AP = AC - AQ \)
\( \Rightarrow BP = CQ \)
\( \angle ABC = \angle ACB \) (angle opposite to the equal sides are equal)
\( \Rightarrow \angle PBC = \angle QCB \)
In \( \triangle PBC \) and \( \triangle QCB \):
\( PB = QC \) (proved above)
\( \angle PBC = \angle QCB \) (proved above)
\( BC = BC \) (common)
By SAS congruence property:
\( \triangle PBC \cong \triangle QCB \)
\( BQ = CP \) (corresponding parts of the congruent triangles)
In simple words: Because the triangle is isosceles, the base angles are equal. When equal lengths are subtracted from the equal sides, the remaining segments are equal. The base BC is common to both new triangles, and with two equal sides and equal angles, the triangles match perfectly by SAS.

Exam Tip: When you have equal lengths on equal sides and need to work with the remaining segments, subtract explicitly - this often reveals hidden equal sides in new triangles.

Question 12. Given: ABC is an isosceles triangle. AB = AC, BD = CE. To prove: BE = CD.
Answer: Given:
\( ABC \) is an isosceles triangle.
\( AB = AC \)
\( BD = CE \)

To prove:
\( BE = CD \)

Proof:
\( AB + BD = AC + CE \) (As, \( AB = AC, BD = CE \))
\( \Rightarrow AD = AE \)
Consider \( \triangle ACD \) and \( \triangle ABE \):
\( AC = AB \) (given)
\( \angle CAD = \angle BAE \) (common)
\( AD = AE \) (proved above)
By SAS congruence property:
\( \triangle ACD \cong \triangle ABE \)
\( \Rightarrow CD = BE \) (corresponding parts of the congruent triangles)
In simple words: Since the triangle starts isosceles and equal segments are removed from the equal sides, what remains must also be equal. The angle at A is shared by both new triangles. With two equal sides and the included angle matching, the SAS rule applies perfectly.

Exam Tip: "What's left over" - when equal amounts are removed from equal starting lengths, the remainders are equal. Use algebra to show this clearly.

Question 13. Given: △ABC is an isosceles triangle. AB = AC, BD = CD. To prove: AD bisects ∠A and ∠D.
Answer: Given:
\( \triangle ABC \) is an isosceles triangle.
\( AB = AC \)
\( BD = CD \)

To prove:
\( AD \) bisects \( \angle A \) and \( \angle D \).

Proof:
Consider \( \triangle ABD \) and \( \triangle ACD \):
\( AB = AC \) (given)
\( BD = CD \) (given)
\( AD = AD \) (common)
By SSS congruence property:
\( \triangle ABD \cong \triangle ACD \)
\( \Rightarrow \angle BAD = \angle CAD \) (by cpct)
\( \Rightarrow \angle BDA = \angle CDA \) (by cpct)
In simple words: All three sides of one triangle match all three sides of the other, so the triangles are identical. This means the angles at A are equal (so AD bisects angle A) and the angles at D are equal (so AD bisects angle D).

Exam Tip: When you prove two triangles congruent by SSS, all corresponding angles automatically become equal - use this to show angle bisection.

Question 14. Are two triangles congruent if the corresponding angles are equal?
Answer: No, it is not necessary. If the corresponding angles of two triangles are equal, then they may or may not be congruent. They may have proportional sides as shown in the following figure:

In simple words: Having the same three angles just means the triangles have the same shape - like two photos of the same scene, one enlarged and one smaller. The shapes match, but the sizes might be different, so the triangles are not necessarily the same size (congruent).

Exam Tip: Remember that similar triangles have equal corresponding angles but are NOT necessarily congruent. Congruence requires equal sides as well as equal angles.

Question 15. Are two triangles congruent if the corresponding sides and one angle are equal?
Answer: No, two triangles are not congruent if their two corresponding sides and one angle are equal. They will be congruent only if the said angle is the included angle between the sides.
In simple words: If you know two sides are equal and one angle is equal, you need to know whether that angle is sitting between the two equal sides. If it is (SAS), the triangles match. If the angle is not between the two sides, they might still be different shapes.

Exam Tip: The position of the angle matters in congruence rules. SAS means side - angle - side in that exact order. An angle not between the two equal sides does not guarantee congruence.

Question 17. List the properties that describe congruent figures.
Answer:
(i) the same length
(ii) the same measure
(iii) the same side length
(iv) the same radius
(v) the same length and the same breadth
(vi) equal parts
In simple words: Congruent figures are identical copies of each other - they have matching sizes in every way. All distances, angles, and dimensions line up exactly. They could be rotated or flipped, but every measurement will be the same.

Exam Tip: Congruence is about exact matching, not just similarity. Every corresponding measurement must be identical.

Question 18. State whether each statement is true or false. If false, explain why.
Answer:
(i) False. This is because they can be equal only if they have equal sides.

(ii) True. This is because if squares have equal areas, then their sides must be of equal length.

(iii) False. For example, if a triangle and a square have equal area, they cannot be congruent.

(iv) False. For example, an isosceles triangle and an equilateral triangle having equal area cannot be congruent.

(v) False. They can be congruent if two sides and the included angle of a triangle are equal to the corresponding two sides and the included corresponding angle of another triangle.

(vi) True. This is because of the AAS criterion of congruency.

(vii) False. Their sides are not necessarily equal.

(viii) True. This is because of the AAS criterion of congruency.

(ix) False. This is because two right triangles are congruent if the hypotenuse and one side of the first triangle are respectively equal to the hypotenuse and the corresponding side of the second triangle.

(x) True
In simple words: Equal areas do not guarantee congruence - the shapes have to be identical in every way. For congruence, you need specific matching patterns of sides and angles, like SAS, ASA, RHS, or SSS. Just because two shapes feel the same size does not mean they are the same shape.

Exam Tip: Distinguish carefully between "equal in area" and "congruent." Also remember the RHS rule applies only to right-angled triangles - it requires the hypotenuse plus one side to match.