Congruence

(i)

AB = QP = 2.5 cm

BC = QR = 4 cm

AC = PR = 3.5 cm

Condition: SSS

(ii)

∠B = ∠Q = 90° $\quad$[Right angle]

AC = PR $\quad$[Hypotenuse]

AB = PQ $\quad$[Side]

Condition: RHS

(iii)

∠A = ∠P = 35° $\quad$[Angle]

AB = PQ $\quad$[Side]

∠B = ∠Q = 90° $\quad$[Angle]

Condition: ASA

(iv)

AC = QR $\quad$[Side]

∠C = ∠R = 90° $\quad$[Angle]

BC = RP $\quad$[Side]

Condition: SAS

(i)

Consider △ABC,

∠A = 180° - (60° + 50°) = 70°.

Consider △FDE,

∠F = 180° - (60° + 50°) = 70°.

In △ABC and △FDE:

∠A = ∠F = 70° $\quad$[Angle]

AB = FD $\quad$[Side]

∠B = ∠D = 60° $\quad$[Angle]

Since two angles and the included side match, they are congruent by ASA condition of congruence.

△ABC ≅ △FDE

(ii)

In △ABC and △PQR:

AB = PR $\quad$[Side]

∠B = ∠R = 35° $\quad$[Angle]

BC = QR $\quad$[Side]

Since two sides and the included angle match, they are congruent by SAS condition of congruence.

△ABC ≅ △PRQ

(iii)

In △DEH and △PQR:

DH = RP $\quad$[Side]

∠H = ∠P = 90° $\quad$[Right-angle]

HE = PQ $\quad$[Side]

Since two sides and the included right angle match, they are congruent by SAS condition of congruence.

△DEH ≅ △RQP

(iv)

In △ABC and △PQR:

AB = RQ $\quad$[Side]

∠B = ∠Q = 90° $\quad$[Right-angle]

CB = PQ $\quad$[Side]

Since two sides and the included right angle match, they are congruent by SAS condition of congruence.

△ABC ≅ △RQP

(v)

Consider △XYZ,

∠X = 180° - (30° + 80°) = 70°.

In △LMN and △XYZ:

LM = ZX $\quad$[Side]

∠M = ∠X = 70° $\quad$[Angle]

MN = XY $\quad$[Side]

Since two sides and the included angle match, they are congruent by SAS condition of congruence.

△LMN ≅ △ZXY

△ABC in which ∠A = 50°, ∠B = 60°, BC = 4.5 cm and △DEF in which ∠E = 60°, EF = 4.5 cm, ∠F = 70°.

Consider △ABC,

∠C = 180° - (50° + 60°) = 70°.

In △ABC and △DEF:

∠C = ∠F = 70° $\quad$[Angle]

BC = EF = 4.5 cm $\quad$[Side]

∠B = ∠E = 60° $\quad$[Angle]

Since two angles and the included side match,

They are congruent by ASA condition of congruence.

△DEF in which ∠E = 48°, DE = 6 cm, EF = 8 cm and △MNR in which ∠R = 48°, MN = 6 cm, MR = 8 cm.

In △DEF, the angle 48° is between the two given sides (SAS).

In △MNR, the angle 48° is at vertex R, which is not the included angle between side MN and MR (the included angle would be ∠M).

The given angle is not in the corresponding "included" position for both triangles.

∴ They are not congruent.

△KLM in which KM = 4 cm, ∠K = 75°, ∠M = 40° and △PQR in which PR = 4 cm, ∠Q = 65°, ∠R = 40°.

Consider △PQR,

∠P = 180° - (65° + 40°) = 75°.

In △KLM and △PQR:

∠K = ∠P = 75° $\quad$[Angle]

KM = PR = 4 cm $\quad$[Side]

∠M = ∠R = 40° $\quad$[Angle]

Since two angles and the included side match,

△KLM ≅ △PRQ by ASA condition of congruence.

△ABC in which AB = 3 cm, ∠A = 90°, BC = 5 cm and △KLM in which KM = 3 cm, ∠K = 90°, LM = 5 cm.

In △ABC and △KML:

AB = KM = 3 cm $\quad$[Side]

∠A = ∠K = 90° $\quad$[Right angle]

BC = LM = 5 cm $\quad$[Hypotenuse]

Since one side, the hypotenuse and the right angle match,

△ABC ≅ △KML by RHS condition of congruence.

(i) △ABD ≅ △CDB

In △ABD and △CDB:

AB = CD $\quad$[Given]

AD = CB $\quad$[Given]

BD = DB $\quad$[Common side]

∴ △ABD ≅ △CDB $\quad$[By SSS condition of congruence]

(ii) ∠A = ∠C

Since △ABD ≅ △CDB, corresponding angles are equal.

∴ ∠A = ∠C $\quad$[CPCT]

⇒ Hence proved

To prove that the sides are equal, we first show that the two triangles formed by diagonal PR are congruent.

In △SPR and △QRP:

∠SPR = ∠QRP $\quad$[Given]

∠RSP = ∠PQR $\quad$[Given]

∴ ∠SRP = ∠QPR $\quad$[Third angles of triangles]

PR = RP $\quad$[Common side]

△SPR ≅ △QRP $\quad$[By ASA condition of congruence]

Therefore, by CPCT:

(i) PQ = RS

(ii) PS = QR

⇒ Hence proved

(i) To prove △AOC ≅ △BOD

In △AOC and △BOD:

AO = BO $\quad$[Given]

∠AOC = ∠BOD $\quad$[Vertically opposite angles are always equal]

CO = DO $\quad$[Given]

△AOC ≅ △BOD $\quad$[By SAS condition of congruence]

(ii) To prove AC = BD

Since △AOC ≅ △BOD, we have:

AC = BD $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

(i) To prove △ABD ≅ △CDB

In △ABD and △CDB:

AB = CD $\quad$[Given]

∠ABD = ∠CDB = 90° $\quad$[Right-angle]

BD = BD $\quad$[Common side to both triangles]

△ABD ≅ △CDB $\quad$[By SAS condition of congruence]

(ii) To prove AD = CB

Since △ABD ≅ △CDB, we have

AD = CB $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

(i) To prove △OLP ≅ △OMP

In △OLP and △OMP:

OL = OM $\quad$[Given]

∠OLP = ∠OMP = 90° $\quad$[Given PL ⊥ OA and PM ⊥ OB]

OP = OP $\quad$[Common side to both triangles, acting as hypotenuse]

△OLP ≅ △OMP $\quad$[By RHS condition of congruence]

(ii) To prove PL = PM

Since △OLP ≅ △OMP, we have:

PL = PM $\quad$[Corresponding Parts of Congruent Triangles]

(iii) To prove ∠LOP = ∠MOP

Again, using the property of CPCT from congruence proof in part (i):

∠LOP = ∠MOP

⇒ Hence proved

(i) To prove △PQR ≅ △SRQ

In △PQR and △SRQ:

PQ = SR $\quad$[Given]

PR = SQ $\quad$[Given]

QR = QR $\quad$[Common base for both triangles]

△PQR ≅ △SRQ $\quad$[By SSS condition of congruence]

(ii) To prove ∠PQR = ∠SRQ

Since △PQR ≅ △SRQ, we can conclude that all their corresponding parts are equal.

By the property of CPCT (Corresponding Parts of Congruent Triangles):

∠PQR = ∠SRQ

⇒ Hence proved

To prove CA = CB, first prove △DCA ≅ △DCB

In △DCA and △DCB:

DA = DB $\quad$[Given]

∠DCA = ∠DCB = 90° $\quad$[Given AC ⊥ CD and BC ⊥ CD]

CD = CD $\quad$[Common side]

△DCA ≅ △DCB $\quad$[By RHS condition of congruence]

Since △DCA ≅ △DCB, we have:

CA = CB $\quad$[Corresponding Parts of Congruent Triangles]

⇒ Hence proved

(i) To prove △ADB ≅ △ADC

In △ADB and △ADC:

AB = AC $\quad$[Given]

BD = CD $\quad$[Given that AD is a median, which by definition bisects the base BC]

AD = AD $\quad$[Common side to both triangles]

△ADB ≅ △ADC $\quad$[By SSS condition of congruence]

(ii) To prove ∠BAD = ∠CAD

Since △ADB ≅ △ADC, we have:

∠BAD = ∠CAD $\quad$[Corresponding angles of congruent triangles]

⇒ Hence proved

(i) To prove △ADB ≅ △ADC

In △ADB and △ADC:

AB = AC $\quad$[Given]

∠BAD = ∠CAD $\quad$[Given, as AD is the bisector of ∠A]

AD = AD $\quad$[Common side to both triangles]

∴ △ADB ≅ △ADC $\quad$[By SAS condition of congruence]

(ii) To prove ∠B = ∠C

Since △ADB ≅ △ADC, we have:

∠B = ∠C $\quad$[Corresponding angles of congruent triangles]

(iii) To prove BD = DC

Since △ADB ≅ △ADC, we have:

BD = DC $\quad$[Corresponding parts of congruent triangles]

(iv) To prove AD ⊥ BC

From CPCT, we know that:

∠ADB = ∠ADC

Since BC is a straight line, these two angles form a linear pair:

∴ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + ∠ADB = 180° $\quad$[∵∠ADB = ∠ADC]

⇒ 2∠ADB = 180°

⇒ ∠ADB = $\frac{180^\circ}{2}$

⇒ ∠ADB = 90°

Since the angle is 90°,

AD ⊥ BC

⇒ Hence proved

(i) To prove △BMC ≅ △CNB

Since AB = AC, the base angles of isosceles triangle ABC are equal.

∴ ∠ABC = ∠ACB

As N lies on AB and M lies on AC,

∠CBN = ∠BCM

Also,

∠BMC = ∠CNB = 90° $\quad$[Given, BM ⊥ AC and CN ⊥ AB]

∴ ∠MBC = ∠BCN $\quad$[Third angles of triangles]

In △BMC and △CNB:

∠MBC = ∠BCN $\quad$[Angle]

BC = CB $\quad$[Common side]

∠BCM = ∠CBN $\quad$[Angle]

∴ △BMC ≅ △CNB $\quad$[By ASA condition of congruence]

(ii) To prove BM = CN

Since △BMC ≅ △CNB, we have:

BM = CN $\quad$[Corresponding parts of congruent triangles]

⇒ Hence proved

Congruence implies that one figure can be placed exactly over the other. To achieve this, both the shape (angles and proportions) and the size (actual measurements) must be identical.

Hence, Option 3 is the correct option.

= → means equal in value or area.

≡ → denotes identity or equivalence.

≈ → means approximately equal.

≅ → is the specific mathematical symbol for congruence.

Hence, Option 4 is the correct option.

For 1D objects like line segments, the only "size" they have is length. If two segments have the same length, they are congruent, regardless of where they are positioned or if they are parallel.

Hence, Option 2 is the correct option.

Two angles are congruent if they have the same degree measure. The length of the arms does not matter because arms are rays that extend infinitely.

Hence, Option 3 is the correct option.

Since all circles have the same shape, the only factor that determines their size is the radius.

If two circles have equal radii, they are congruent. They do not need to share the same center (that would make them concentric, not just congruent).

Hence, Option 2 is the correct option.

SSS, SAS, and ASA are valid congruence criteria.

AAA (Angle-Angle-Angle) only proves that triangles are the same shape (similar), but not necessarily the same size. A small equilateral triangle and a giant equilateral triangle have the same angles but are not congruent.

Hence, Option 4 is the correct option.

In a congruence statement, the order of letters matters.

AB corresponds to DE

BC corresponds to EF

AC corresponds to DF

Therefore, only BC = EF is necessarily true based on the corresponding parts of congruent triangles (CPCT).

Hence, Option 2 is the correct option.

Congruent triangles must have the same area, perimeter, and shape. However, one triangle can be rotated, flipped, or moved to a different position (orientation) and still remain congruent to the original.

Hence, Option 4 is the correct option.

(i) Two triangles having corresponding angles equal are always similar.

(ii) Two squares are congruent if they have the same side length.

(iii) If two triangles are similar, then they have corresponding angles equal.

(iv) If two triangles are congruent, then they have corresponding sides and angles equal.

(v) Two congruent triangles are always similar.

Explanation

(i) Two triangles have the same shape, but they might be different sizes unless a corresponding side is also equal.

(ii) All squares have 90° angles, so the only factor that determines congruence is the length of the side.

(iii) The corresponding sides of two similar triangles are proportional, but not necessarily equal.

(iv) If two triangles are congruent, then they have corresponding sides and angles equal. This is often referred to as CPCT — Corresponding Parts of Congruent Triangles.

(v) Congruence is a special case of similarity where the ratio of corresponding sides is 1:1.

(i) False
Reason — While all squares have the same shape (all angles are 90°), they can be different sizes. One square might have a side of 2 cm and another 5 cm; these are similar but not congruent.

(ii) False
Reason — Concentric circles share the same center but must have different radii to be distinct. Since their radii are different, their sizes are different, making them non-congruent.

(iii) True
Reason — The area of a square is side². If the areas are equal, the side lengths must also be equal ($s = \sqrt{Area}$). Since they have the same side lengths and same shape, they are congruent.

(iv) False
Reason — Figures can have the same area but completely different shapes. For example, a rectangle with sides 4 cm and 9 cm has an area of 36 cm², and a square with side 6 cm also has an area of 36 cm², but they are not congruent.

(v) True
Reason — For an equilateral triangle, the area is fixed by the side length ($Area = \frac{\sqrt{3}}{4}s^2$). If the areas are equal, the sides must be equal. Since all equilateral triangles have 60° angles, equal sides guarantee congruence by SSS or SAS.

(vi) True
Reason — If one acute angle is equal, and we know the right angle (90°) is equal, then the third angle must also be equal. Since the hypotenuse (a side) is also equal, the triangles are identical in shape and size. This is the ASA (Angle-Side-Angle) criterion.

Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Explanation

The sides of both triangles are equal (3.5 cm, 5 cm, 6 cm), so triangles are congruent by SSS.

So, Assertion is true.

The statement in reason is the formal definition of the SSS (Side-Side-Side) congruence criterion.

So, Reason is true and it correctly explains the assertion.

Hence, option 1 is the correct option.

Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

Explanation

Perimeter of a square = 4 × side. If the perimeters are equal, the sides are equal. Since the sides are equal and all angles are 90°, the squares are congruent.

So, Assertion is true.

Every square has the same internal angles 90° and proportional sides, so all squares are indeed similar.

So, Reason is true.

The reason the squares are congruent is because their side lengths are equal. The similarity rule in Reason only explains why they have the same shape, not why they are the same size.

Hence, option 2 is the correct option.

Statement 1 — All equilateral triangles have three equal sides and all angles equal to 60°, so they have the same shape. However, they can have different side lengths.

For example, an equilateral triangle of side 2 cm and another of side 5 cm have the same shape but different sizes. So they are similar but not necessarily congruent.

So, Statement 1 is false.

Statement 2 — All right triangles have one angle equal to 90°, but the lengths of the other two sides can vary. For example, a right triangle with legs 3 cm and 4 cm is not the same size as a right triangle with legs 5 cm and 12 cm.

So, Statement 2 is false.

Hence, Option 4 is the correct option.

The sides of the four triangles are:

△ABC : AB = 6 cm, BC = 4.5 cm, AC = 4.5 cm

△XYZ : XY = 4 cm, YZ = 6 cm, XZ = 4.5 cm

△PQR : PQ = 6 cm, QR = 4.5 cm, PR = 4.5 cm

△MNO : MN = 5.5 cm, NO = 4 cm, MO = 5.5 cm

Since the sides of △ABC and △PQR are the same, let's compare △ABC and △PQR :

AB = PQ = 6 cm $\quad$[Side]

BC = QR = 4.5 cm $\quad$[Side]

AC = PR = 4.5 cm $\quad$[Side]

Since all three pairs of corresponding sides are equal, the triangles are congruent by the SSS condition of congruence.

∴ △ABC ≅ △PQR

The remaining pairs of triangles do not have all three sides equal, so they are not congruent.

Hence, Option 4 is the correct option.

In △ABC :

∠A = 40°, ∠B = 60° and AB = 4 cm.

The side AB = 4 cm lies between the two given angles ∠A and ∠B.

Checking each option :

Option 1 (△XYZ) : ∠X = 61°, ∠Y = 41° and XY = 4 cm. The angles do not match those of △ABC, so it is not congruent to △ABC.

Option 2 (△PQR) : ∠P = 40°, ∠R = 60° and PR = 4 cm. The side PR = 4 cm lies between the two given angles ∠P and ∠R.

Comparing △ABC and △PRQ :

∠A = ∠P = 40° $\quad$ [Angle]

AB = PR = 4 cm $\quad$ [Side]

∠B = ∠R = 60° $\quad$ [Angle]

Since two angles and the included side are equal, the triangles are congruent by the ASA condition of congruence.

∴ △ABC ≅ △PRQ

Option 3 (△MNO) : ∠M = 90°, ∠N = 60° and MN = 2.7 cm. The angles and the side length do not match those of △ABC, so it is not congruent to △ABC.

Option 4 (△UVW) : ∠V = 61°, ∠W = 40° and VW = 4 cm. The angles do not match those of △ABC, so it is not congruent to △ABC.

Hence, Option 2 is the correct option.

From the figure:

• A, F and G are collinear.
• B, G and H are collinear.
• C, H and E are collinear.
• D, E and F are collinear.

Since EFGH is a square, its adjacent sides are perpendicular. Therefore,

∠AGB = ∠BHC = ∠CED = ∠DFA = 90°

Also, since ABCD is a square,

AB = BC = CD = DA

Consider △AGB and △BHC:

∠AGB = ∠BHC = 90°

AB = BC

Also, ∠GAB = ∠HBC

because AG ⊥ BH and AB ⊥ BC.

∴ △AGB ≅ △BHC

by the AAS condition of congruence.

Similarly,

△BHC ≅ △CED

and

△CED ≅ △DFA.

Thus, options 1, 2 and 3 are true.

However, D, E and F are collinear. Hence, they do not form a triangle, so △DEF cannot be congruent to △DFA.

Hence, Option 4 is the correct option.