Congruency : Congruent Triangles

ICSE Class 7 Mathematics · Selina Solutions Concise Mathematics Class 7 ICSE · 32 solved questions

Exercise

Marks

1Exercise 21

State, whether the pairs of triangles given in the following figures are congruent or not:

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(i) In the first triangle, 4 cm is a side containing the right angle, whereas in the second triangle, 4 cm is the hypotenuse.

Here, the hypotenuse of one right angled triangle is not equal to the hypotenuse of the other right angled triangle.

Hence, the two triangles are not congruent.

(ii) In the first triangle, the included angle = 180° − (40° + 30°) = 110°.

Here, two sides of one triangle are equal to the two sides of the other triangle and the included angles (each 110°) are equal.

Hence, the two triangles are congruent by S.A.S.

(iii) In the first triangle, the marked angle is included between the sides 5 cm and 4 cm, whereas in the second triangle, the marked angle is not included between the sides 5 cm and 4 cm.

Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal.

Hence, the two triangles are not congruent.

(iv) Here, three sides of one triangle (4 cm, 5 cm and 6 cm) are equal to three sides of the other triangle (4 cm, 5 cm and 6 cm), each to each.

Hence, the two triangles are congruent by S.S.S.

(v) Here, both the triangles are right-angled triangles, hypotenuse of one triangle = hypotenuse of the other triangle = 8 cm and one side of one triangle = corresponding side of the other triangle = 4 cm.

Hence, the two triangles are congruent by R.H.S.

(vi) Here, two angles of one triangle are equal to the two angles of the other triangle and the side 6 cm (not the included side) of one triangle is equal to the corresponding side of the other triangle.

∴ The triangles are congruent by A.A.S.

Also, since two angles of the triangles are equal, the third angles are also equal. So, the angles at the ends of the 6 cm sides are equal and 6 cm is the included side.

∴ The triangles are also congruent by A.S.A.

Hence, the two triangles are congruent by A.A.S. or A.S.A.

(vii) In ΔABC, AB = 2 cm, BC = 3.5 cm and ∠C = 80°.

Here, ∠C is not the angle included between the sides AB and BC (the included angle is ∠B).

In ΔDEF, DE = 2 cm, DF = 3.5 cm and ∠D = 80°.

Here, ∠D is the angle included between the sides DE and DF.

Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal.

Hence, the two triangles are not congruent.

2Exercise 21

In the given figure, prove that:

ΔABD ≅ ΔACD

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In ΔABD and ΔACD,

⇒ AB = AC (Given)

⇒ BD = CD (Given)

⇒ AD = AD (Common side)

∴ ΔABD ≅ ΔACD (By S.S.S. congruence rule)

Hence, proved that ΔABD ≅ ΔACD.

3Exercise 21

Prove that:

(i) ΔABC ≅ ΔADC

(ii) ∠B = ∠D

(iii) AC bisects angle DCB

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(i) In ΔABC and ΔADC,

⇒ AB = AD (Given)

⇒ BC = DC (Given)

⇒ AC = AC (Common side)

∴ ΔABC ≅ ΔADC (By S.S.S. congruence rule)

Hence, proved that ΔABC ≅ ΔADC.

(ii) Since, ΔABC ≅ ΔADC (Proved above)

∴ ∠B = ∠D (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠B = ∠D.

(iii) Since, ΔABC ≅ ΔADC (Proved above)

∴ ∠BCA = ∠DCA (Corresponding angles of congruent triangles are equal)

So, AC divides ∠DCB into two equal parts.

Hence, proved that AC bisects angle DCB.

4Exercise 21

Prove that:

(i) ΔABD ≅ ΔACD

(ii) ∠B = ∠C

(iii) ∠ADB = ∠ADC

(iv) ∠ADB = 90°

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(i) In ΔABD and ΔACD,

⇒ AB = AC (Given)

⇒ BD = CD (Given)

⇒ AD = AD (Common side)

∴ ΔABD ≅ ΔACD (By S.S.S. congruence rule)

Hence, proved that ΔABD ≅ ΔACD.

(ii) Since, ΔABD ≅ ΔACD (Proved above)

∴ ∠B = ∠C (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠B = ∠C.

(iii) Since, ΔABD ≅ ΔACD (Proved above)

∴ ∠ADB = ∠ADC (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠ADB = ∠ADC.

(iv) Since, BDC is a straight line,

⇒ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + ∠ADB = 180° (As ∠ADB = ∠ADC)

⇒ 2∠ADB = 180°

⇒ ∠ADB = $\frac{180°}{2}$ = 90°

Hence, proved that ∠ADB = 90°.

5Exercise 21

In the given figure, prove that:

(i) ΔACB ≅ ΔECD

(ii) AB = ED

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(i) In ΔACB and ΔECD,

⇒ AC = EC (Given)

⇒ ∠ACB = ∠ECD (Vertically opposite angles are equal)

⇒ CB = CD (Given)

∴ ΔACB ≅ ΔECD (By S.A.S. congruence rule)

Hence, proved that ΔACB ≅ ΔECD.

(ii) Since, ΔACB ≅ ΔECD (Proved above)

∴ AB = ED (Corresponding sides of congruent triangles are equal)

Hence, proved that AB = ED.

6Exercise 21

Prove that:

(i) ΔABC ≅ ΔADC

(ii) ∠B = ∠D

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(i) In ΔABC and ΔADC,

⇒ AB = AD (Given)

⇒ BC = DC (Given)

⇒ AC = AC (Common side)

∴ ΔABC ≅ ΔADC (By S.S.S. congruence rule)

Hence, proved that ΔABC ≅ ΔADC.

(ii) Since, ΔABC ≅ ΔADC (Proved above)

∴ ∠B = ∠D (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠B = ∠D.

7Exercise 21

In the given figure, prove that: BD = BC.

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In ΔABD and ΔABC,

⇒ ∠ABD = ∠ABC (Each 90°)

⇒ AD = AC (Given; hypotenuses)

⇒ AB = AB (Common side)

∴ ΔABD ≅ ΔABC (By R.H.S. congruence rule)

∴ BD = BC (Corresponding sides of congruent triangles are equal)

Hence, proved that BD = BC.

8Exercise 21

In the given figure, ∠1 = ∠2 and AB = AC.

Prove that:

(i) ∠B = ∠C

(ii) BD = DC

(iii) AD is perpendicular to BC.

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In ΔABD and ΔACD,

⇒ AB = AC (Given)

⇒ ∠1 = ∠2 (Given)

⇒ AD = AD (Common side)

∴ ΔABD ≅ ΔACD (By S.A.S. congruence rule)

(i) Since, ΔABD ≅ ΔACD,

∴ ∠B = ∠C (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠B = ∠C.

(ii) Since, ΔABD ≅ ΔACD,

∴ BD = DC (Corresponding sides of congruent triangles are equal)

Hence, proved that BD = DC.

(iii) Since, ΔABD ≅ ΔACD,

∴ ∠ADB = ∠ADC (Corresponding angles of congruent triangles are equal)

Since, BDC is a straight line,

⇒ ∠ADB + ∠ADC = 180°

⇒ ∠ADB + ∠ADB = 180°

⇒ 2∠ADB = 180°

⇒ ∠ADB = 90°

Hence, proved that AD is perpendicular to BC.

9Exercise 21

In the given figure, prove that:

(i) PQ = RS

(ii) PS = QR

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In ΔPSR and ΔRQP,

⇒ ∠PSR = ∠RQP (Given)

⇒ ∠SPR = ∠QRP (Given)

⇒ PR = RP (Common side)

∴ ΔPSR ≅ ΔRQP (By A.A.S. congruence rule)

(i) Since, ΔPSR ≅ ΔRQP,

∴ RS = PQ (Corresponding sides of congruent triangles are equal)

Hence, proved that PQ = RS.

(ii) Since, ΔPSR ≅ ΔRQP,

∴ PS = RQ (Corresponding sides of congruent triangles are equal)

Hence, proved that PS = QR.

10Exercise 21

In the given figure, prove that:

(i) ΔXYZ ≅ ΔXPZ

(ii) YZ = PZ

(iii) ∠YXZ = ∠PXZ

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(i) In ΔXYZ and ΔXPZ,

⇒ ∠XYZ = ∠XPZ (Each 90°)

⇒ XZ = XZ (Common hypotenuse)

⇒ XY = XP (Given)

∴ ΔXYZ ≅ ΔXPZ (By R.H.S. congruence rule)

Hence, proved that ΔXYZ ≅ ΔXPZ.

(ii) Since, ΔXYZ ≅ ΔXPZ (Proved above)

∴ YZ = PZ (Corresponding sides of congruent triangles are equal)

Hence, proved that YZ = PZ.

(iii) Since, ΔXYZ ≅ ΔXPZ (Proved above)

∴ ∠YXZ = ∠PXZ (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠YXZ = ∠PXZ.

11Exercise 21

In the given figure, prove that:

(i) ΔABC ≅ ΔDCB

(ii) AC = DB

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(i) In ΔABC and ΔDCB,

⇒ AB = DC (Given)

⇒ ∠ABC = ∠DCB (Each 90°)

⇒ BC = CB (Common side)

∴ ΔABC ≅ ΔDCB (By S.A.S. congruence rule)

Hence, proved that ΔABC ≅ ΔDCB.

(ii) Since, ΔABC ≅ ΔDCB (Proved above)

∴ AC = DB (Corresponding sides of congruent triangles are equal)

Hence, proved that AC = DB.

12Exercise 21

In the given figure, prove that:

(i) ΔAOD ≅ ΔBOC

(ii) AD = BC

(iii) ∠ADB = ∠ACB

(iv) ΔADB ≅ ΔBCA

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(i) In ΔAOD and ΔBOC,

⇒ AO = BO (Given)

⇒ ∠AOD = ∠BOC (Vertically opposite angles are equal)

⇒ DO = CO (Given)

∴ ΔAOD ≅ ΔBOC (By S.A.S. congruence rule)

Hence, proved that ΔAOD ≅ ΔBOC.

(ii) Since, ΔAOD ≅ ΔBOC (Proved above)

∴ AD = BC (Corresponding sides of congruent triangles are equal)

Hence, proved that AD = BC.

(iii) Since, ΔAOD ≅ ΔBOC (Proved above)

∴ ∠ADO = ∠BCO (Corresponding angles of congruent triangles are equal)

⇒ ∠ADB = ∠ADO and ∠ACB = ∠BCO

∴ ∠ADB = ∠ACB

Hence, proved that ∠ADB = ∠ACB.

(iv) In ΔADB and ΔBCA,

⇒ AD = BC (Proved above)

⇒ DB = DO + OB = CO + OA = CA (As DO = CO and OB = OA)

⇒ AB = BA (Common side)

∴ ΔADB ≅ ΔBCA (By S.S.S. congruence rule)

Hence, proved that ΔADB ≅ ΔBCA.

13Exercise 21

ABC is an equilateral triangle, AD and BE are perpendiculars to BC and AC respectively. Prove that:

(i) AD = BE

(ii) BD = CE

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In ΔADB and ΔBEC,

⇒ AB = BC (Sides of same equilateral triangle)

⇒ ∠ADB = ∠BEC (Each 90°)

⇒ ∠ABD = ∠BCE (Each 60°)

∴ ΔADB ≅ ΔBEC (By A.A.S. congruence rule)

(i) Since, ΔADB ≅ ΔBEC,

∴ AD = BE (Corresponding sides of congruent triangles are equal)

Hence, proved that AD = BE.

(ii) Since, ΔADB ≅ ΔBEC,

∴ BD = CE (Corresponding sides of congruent triangles are equal)

Hence, proved that BD = CE.

14Exercise 21

Use the information given in the following figure to prove that triangles ABD and CBD are congruent.

Also, find the values of x and y.

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In ΔABD and ΔCBD,

⇒ AB = CB (Given)

⇒ AD = CD (Given)

⇒ BD = BD (Common side)

∴ ΔABD ≅ ΔCBD (By S.S.S. congruence rule)

∴ ∠ABD = ∠CBD and ∠ADB = ∠CDB (Corresponding angles of congruent triangles are equal)

Since, ∠ABD = ∠CBD,

⇒ 50° = x + 5°

⇒ x = 50° − 5°

⇒ x = 45°

Since, ∠ADB = ∠CDB,

⇒ y − 7° = 38°

⇒ y = 38° + 7°

⇒ y = 45°

Hence, proved that ΔABD ≅ ΔCBD and x = y = 45°.

15Exercise 21

The given figure shows a triangle ABC in which AD is perpendicular to side BC and BD = CD. Prove that:

(i) ΔABD ≅ ΔACD

(ii) AB = AC

(iii) ∠B = ∠C.

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(i) In ΔABD and ΔACD,

⇒ BD = CD (Given)

⇒ ∠ADB = ∠ADC (Each 90°, as AD ⊥ BC)

⇒ AD = AD (Common side)

∴ ΔABD ≅ ΔACD (By S.A.S. congruence rule)

Hence, proved that ΔABD ≅ ΔACD.

(ii) Since, ΔABD ≅ ΔACD (Proved above)

∴ AB = AC (Corresponding sides of congruent triangles are equal)

Hence, proved that AB = AC.

(iii) Since, ΔABD ≅ ΔACD (Proved above)

∴ ∠B = ∠C (Corresponding angles of congruent triangles are equal)

Hence, proved that ∠B = ∠C.

11 Mark · Multiple Choice Questions

In the given figure:

(a)AB = AD
(b)BC = DC
(c)AC = AC
(d)all of these are true

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From the figure,

AB = AD (Given)

BC = DC (Given)

AC = AC (Common)

∴ All of these are true.

Hence, option 4 is the correct option.

21 Mark · Multiple Choice Questions

Triangles ABD and ACD are congruent by:

(a)SSS
(b)RHS
(c)SAS
(d)ASA

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In ΔABD and ΔACD,

AB = AC (Given)

∠BAD = ∠CAD (Given)

AD = AD (Common)

∴ ΔABD ≅ ΔACD by S.A.S.

Hence, option 3 is the correct option.

31 Mark · Multiple Choice Questions

Triangles ABD and ACD are congruent by:

(a)SSS
(b)RHS
(c)SAS
(d)ASA

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In ΔABD and ΔACD,

AB = AC (Given; hypotenuses)

AD = AD (Common)

∠ADB = ∠ADC (Each 90°)

∴ ΔABD ≅ ΔACD by R.H.S.

Hence, option 2 is the correct option.

41 Mark · Multiple Choice Questions

ΔAOB is congruent to ΔCOD by:

(a)SAS
(b)ASA
(c)SSS
(d)none of these

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∠AOB = ∠COD = 30° (Vertically opposite angles are equal)

In ΔAOB,

∠A = 180° − (30° + 70°) = 80°

In ΔCOD,

∠C = 180° − (30° + 70°) = 80°

In ΔAOB and ΔCOD,

∠A = ∠C = 80°

AB = CD (Given)

∠B = ∠D = 70°

AB and CD are the included sides.

∴ ΔAOB ≅ ΔCOD by A.S.A.

Hence, option 2 is the correct option.

51 Mark · Multiple Choice Questions

In the given figure, if AC = DB, triangles ABC and DCB are congruent to each by:

(a)SSS
(b)SAS
(c)RHS
(d)ASA

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In ΔABC and ΔDCB,

AC = DB (Given; hypotenuses)

BC = CB (Common)

∠ABC = ∠DCB (Each 90°)

∴ ΔABC ≅ ΔDCB by R.H.S.

Hence, option 3 is the correct option.

61 Mark · Multiple Choice Questions

Triangles BAC and CDB are congruent by:

(a)SSS
(b)SAS
(c)ASA
(d)RHS

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In ΔBAC and ΔCDB,

∠BAC = ∠CDB (Each 90°)

BC = CB (Common hypotenuse)

AB = DC (Given)

∴ ΔBAC ≅ ΔCDB by R.H.S.

Hence, option 4 is the correct option.

71 Mark · Multiple Choice Questions

Two circles are congruent to each other, if their:

(a)radii are equal
(b)circumferences are equal
(c)areas are equal
(d)all of the above are true

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Two circles coincide exactly when their radii are equal. If the radii are equal, their circumferences (2πr) and areas (πr²) are also equal, and conversely, equal circumferences or equal areas give equal radii.

∴ All of the above are true.

Hence, option 4 is the correct option.

81 Mark · Multiple Choice Questions

Two quadrilaterals are congruent to each other, if their:

(a)each angle is 90°
(b)perimeters are equal
(c)areas are equal
(d)none of these

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Congruent figures are exactly the same both in shape and size.

Two quadrilaterals with each angle 90°, or with equal perimeters, or with equal areas may not coincide exactly. For example, a rectangle 4 cm × 1 cm and a square 2 cm × 2 cm have equal areas (4 cm²) but are not congruent.

∴ None of these conditions ensures congruency.

Hence, option 4 is the correct option.

91 Mark · Multiple Choice Questions

In the given figure, AB = CD and AC = DB, then ΔABC is congruent to ΔDCB by:

(a)SSS
(b)SAS
(c)ASA
(d)RHS

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In ΔABC and ΔDCB,

AB = DC (Given)

AC = DB (Given)

BC = CB (Common)

∴ ΔABC ≅ ΔDCB by S.S.S.

Hence, option 1 is the correct option.

101 Mark · Multiple Choice Questions

In the given figure:

(a)ΔABC ≅ ΔDEF
(b)ΔABC ≅ ΔPQR
(c)ΔDEF ≅ ΔPQR
(d)none of these

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In ΔABC, AB = 10 cm, BC = 8 cm and the included angle ∠B = 55°.

In ΔDEF, DE = 10 cm, EF = 8 cm and ∠F = 55°, which is not the included angle.

In ΔPQR, PQ = 10 cm, QR = 8 cm and ∠P = 55°, which is not the included angle.

Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal. Also, in ΔDEF and ΔPQR, the 55° angles are not corresponding angles (∠F is opposite to the 10 cm side, while ∠P is opposite to the 8 cm side).

∴ None of the given pairs of triangles are congruent.

Hence, option 4 is the correct option.

111 Mark · Multiple Choice Questions

In the given figure, triangles ABD and ACD are congruent to each other, then:

(a)AB = AC
(b)∠B = ∠C
(c)BD = CD
(d)all of the above are true

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Since, ΔABD ≅ ΔACD,

AB = AC (Corresponding sides of congruent triangles)

∠B = ∠C (Corresponding angles of congruent triangles)

BD = CD (Corresponding sides of congruent triangles)

∴ All of the above are true.

Hence, option 4 is the correct option.

121 Mark · Multiple Choice Questions

The two triangles shown in the following figure are:

(a)congruent by AAA
(b)congruent by AAS
(c)congruent by SAS
(d)not congruent

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In ΔABC and ΔPQR,

∠A = ∠P (Given)

∠B = ∠Q (Given)

BC = QR (Given)

Here, BC and QR are not the included sides.

∴ ΔABC ≅ ΔPQR by A.A.S.

Hence, option 2 is the correct option.

13Statement I-II Type Questions

Statement 1: AAA (angle-angle-angle) axiom does not prove congruence.

Statement 2: SSA (side-side-angle) axiom does prove congruence.

Which of the following options is correct?

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

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If three angles of a triangle are equal to the three angles of another triangle, then the triangles are not necessarily congruent. For congruency at least one pair of corresponding sides must be equal.

So, statement 1 is true.

Triangles will be congruent by S.A.S., only when the angles included by the corresponding equal sides are equal. When the equal angle is not the included angle (SSA), the triangles are not necessarily congruent.

So, statement 2 is false.

Hence, option 3 is the correct option.

14Assertion-Reason Type Questions

Assertion (A): In the given figure ΔABC ≅ ΔPQR (RHS congruence axiom)

Reason (R): RHS axiom of congruence: One angle of both the triangles is a right angle, also the hypotenuse and one side of one triangle are equal to the hypotenuse and the corresponding side of the other triangle.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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In ΔABC and ΔPQR,

AC = PR (Given)

∠ABC = ∠PQR (= 90°)

∠ACB = ∠PRQ (Given)

⇒ ΔABC ≅ ΔPQR (AAS congruence axiom)

The given triangles are not congruent by RHS.

∴ Assertion (A) is false.

If AB = PQ, the given triangles are congruent by RHS.

∴ Reason (R) is true.

Hence, option 2 is the correct option.

15Assertion-Reason Type Questions

Assertion (A): In the given figure AC is the bisector of ∠BAD and ∠BCD.

Reason (R): SSS (Side-side-side) axiom of congruence: Two triangles are congruent if the sides of one triangle are correspondingly equal to sides of other triangle.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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In ΔABC and ΔADC,

AB = AD (Given)

BC = DC (Given)

AC = AC (Common)

⇒ ΔABC ≅ ΔADC (SSS congruence axiom)

⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (Corresponding angles of congruent triangles)

⇒ AC is the bisector of ∠BAD and ∠BCD.

∴ Assertion (A) is true.

The SSS axiom states that two triangles are congruent if the three sides of one triangle are correspondingly equal to the three sides of the other triangle.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

16Assertion-Reason Type Questions

Assertion (A): In the given figure ΔABD ≅ ΔADC (SAS congruence axiom)

⇒ AB = AC (c.p.c.t.)

⇒ ΔABC is an isosceles triangle.

Reason (R): SAS (side-angle-side) axiom of congruence: Two triangles are congruent if two sides and included angle of one triangle are correspondingly equal to two sides and included angle of the other triangle.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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In ΔABD and ΔACD,

BD = CD (Given)

∠ADB = ∠ADC (Each 90°)

AD = AD (Common)

⇒ ΔABD ≅ ΔACD (SAS congruence axiom)

Here, vertex B corresponds to vertex C and vertex D corresponds to vertex D. So, the correct congruence is ΔABD ≅ ΔACD and not ΔABD ≅ ΔADC.

∴ Assertion (A) is false.

The SAS axiom states that two triangles are congruent if two sides and the included angle of one triangle are correspondingly equal to two sides and the included angle of the other triangle.

∴ Reason (R) is true.

Hence, option 2 is the correct option.

17Assertion-Reason Type Questions

Assertion (A): In the given figure ΔAOC ≅ ΔBOD (ASA congruence axiom).

Reason (R): If any two angles and a side (not the included side) of one triangle are equal to two angles and the corresponding side of the other triangle then the two triangles are congruent.

(a)A is true, R is false.
(b)A is false, R is true.
(c)Both A and R are true.
(d)Both A and R are false.

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In ΔAOC and ΔBOD,

∠OAC = ∠OBD (Each 50°)

AC = BD (Each 4 cm)

∠OCA = ∠ODB (Each 70°)

Here, AC and BD are the included sides.

⇒ ΔAOC ≅ ΔBOD (ASA congruence axiom)

∴ Assertion (A) is true.

The reason states that if any two angles and a side (not the included side) of one triangle are equal to two angles and the corresponding side of the other triangle then the two triangles are congruent, which is AAS congruence criterion.

∴ Reason (R) is true.

Hence, option 3 is the correct option.

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