# Congruence of Triangles Class-7 ML Aggarwal ICSE Maths

**Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 **Congruence of Triangles ** ICSE **Class-7th ML **Aggarwal Maths.

Our Solutions contain all type Questions with Exe-12.1 , Exe-12.2, Objective Type Questions ( including Mental Maths Multiple Choice Questions , HOTS ) and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-7 Mathematics.

**Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions Chapter-12

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**Objective Type Questions, **

**Multiple Choice Questions ,(MCQ)**

**Ex 12.1, Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions

Question 1.

If ΔABC and ΔDEF are congruent under the correspondence ABC ↔ FED, write all the corresponding congruent parts of the triangles.

#### Answer

ΔABC and ΔDEF are congruent

under the correspondence, ABC ↔ FED

∠A ↔ ∠F, ∠B ↔ ∠E, ∠C ↔ ∠D

AB ↔ FE, BC ↔ ED and AC ↔ FD

Question 2.

If ΔDEF = ΔBCA, then write the part(s) of ΔBCA that correspond to

(i) ∠E

(ii)

(iii) ∠F

(iv)

#### Answer

If ΔDEF = ΔBCA, then

(i) ∠E ↔ ∠C

(ii) EF ↔

(iii) ∠F ↔ ∠A

(iv) DF ↔

Question 3.

In the figure given below, the lengths of the sides of the triangles are indicated. By using SSS congruency rule, state which pairs of triangles are congruent. In the case of congruent triangles, write the result in symbolic form:

#### Answer

(i) In the given figure,

In ΔABC and ΔPQR

AB ↔ PQ, BC ↔ PR, and AC ↔ QR

Δs are congruent

ΔABC = ΔQPR

(ii) In the given figure,

In ΔABC and ΔPQR

AC ↔ PR, BC ↔ PQ

But AB ≠ QR

Δs are not congruent.

Question 4.

In the given figure, AB = 5 cm, AC = 5 cm, BD = 2.5 cm and CD = 2.5 cm

(i) State the three pairs of equal parts in ΔADB and ΔADC

(ii) Is ΔADB = ΔADC? Give reasons.

(iii) Is ∠B = ∠C? Why?

#### Answer

In the given figure,

AB = 5 cm, AC = 5 cm, BD = 2.5 cm

and CD = 2.5 cm

In ΔABD and ΔACD

(i) AB = AC = 5 cm

BD = CD = 2.5 cm

AD = AD (Common Side)

ΔABD = ΔACD

(ii) ΔADB = ΔADC (SSS axiom)

∠B = ∠C (c.p.c.t.)

Question 5.

In the given figure, AB = AC and D is the mid-point of .

(i) State the three pairs of equal parts in ΔADB and ΔADC.

(ii) Is ΔADB = ΔADC? Give reasons.

(iii) Is ∠B = ∠C? Why?

#### Answer

(i) In ΔABC,

AB = AC

D is the mid-point of BC

BD = DC

Now in ΔADB and ΔADC

AB = AC (Given)

AD = AD (Common)

BD = DC (D is mid-point of BC)

(ii) ΔADB = ΔADC (SSS axiom)

(iii) ∠B = ∠C (c.p.c.t.)

Question 6.

In the figure given below, the measures of some parts of the triangles are indicated. By using SAS rule of congruency, state which pairs of triangles are congruent. In the case of congruent triangles, write the result in symbolic form.

#### Answer

(i) In ΔABC and ΔDEF

AB = DE (Each = 2.5 cm)

AC = DF (Each = 2.8 cm)

∠A ≠ ∠D (Have different measure)

ΔABC is not congruent to ΔDEF

(ii) In ΔABC and ΔRPQ

AC = RP (Each = 2.5 cm)

CB = PQ (Each = 3 cm)

∠C = ∠P (Each = 35°)

ΔACB and ΔRPQ are congruent (SAS axiom)

(iii) In ΔDEF and ΔPQR

FD = QP (Each = 3.5 cm)

FE = QR (Each = 3 cm)

∠F = ∠Q (Each 40°)

ADEF and APQR are congruent

(iv) In ΔABC and ΔPRQ

AB = PQ (Each = 4 cm)

BC = QR (Each = 3 cm)

But included angles B and ∠Q are not equal

ΔABC and ΔPQR are not congruent.

Question 7.

By applying SAS congruence rule, you want to establish that ΔPQR = ΔFED. If is given that PQ = EF and RP = DF. What additional information is needed to establish the congruence?

#### Answer

In ΔPQR and ΔFED

PQ = FE

RP = DF

Their included angles ∠P must be equal to ∠F for congruency.

Hence, ∠P = ∠F

Question 8.

You want to show that ΔART = ΔPEN

(a) If you have to use SSS criterion, then you need to show

(i) AR = ……….

(ii) RT = ……….

(iii) AT = …………

(b) If it is given that ∠T = ∠N and you are to use the SAS criterion, you need to have

(i) RT = ………. and (ii) PN = ……….

#### Answer

(a) In ΔART and ΔPEN

For SSS criterion

AR = DE

RT = EN and AT = PN

ΔART = ΔPEN

(b) ∠T = ∠N (Given)

In ΔART and ΔPEN

If RT = EN

AT = PN and ∠T = ∠N

Then ΔART = ΔPEN (SAS criterion)

Question 9.

You have to show that ΔAMP = ΔAMQ.

In the following proof, supply the missing reasons.

Steps |
Reasons |

(i) PM = QM | (i) ……… |

(ii) ∠PMA = ∠QMA | (ii) ………. |

(iii) AM = AM | (iii) ………… |

(iv) ΔAMP = ΔAMQ | (iv) ………… |

#### Answer

In order to show that,

ΔAMP = ΔAMQ

PM = QM (Given)

∠PMA = ∠QMA (Given)

AM = AM (Common)

ΔAMP = ΔAMQ (SAS criterion)

Question 10.

In the given figure:

(i) State three pairs of equal parts in ΔPSR and ΔRQP.

(ii) Is ΔPSR = ΔRQP? Give reasons

(iii) Is PS = RQ? Why?

(iv) Is ∠S = ∠Q? Why?

#### Answer

In ΔPSR and ΔRQP

SR = PQ (each = 3.5 cm)

PR = PR (Common side)

∠SRP = ∠RPQ (Each = 30°)

ΔPSR = ΔRQP (SAS criterion)

PS = RQ (c.p.c.t.)

∠S = ∠Q (c.p.c.t.)

Question 11.

In the given figure, AB = DC and ∠ABC = ∠DCB.

(i) State three pairs of equal parts in AABC and ADCB.

(ii) Is ΔABC = ΔDCB? Give reasons.

(iii) Is AC = DB? Why?

#### Answer

In ΔABC and ΔDBC

AB = DC (Given)

∠ABC = ∠DCB (Given)

BC = BC (Common)

ΔABC = ΔDCB (SAS criterion)

AC = DB (c.p.c.t.)

Question 12.

In the quadrilateral, AC = AD, and AB bisect ∠CAD.

(i) State three pairs of equal parts in ΔABC and ΔABD.

(ii) Is ΔABC = ΔABD? Give reasons.

(iii) Is BC = BD? Why?

(iv) Is ∠C = ∠D? Why?

#### Answer

In quadrilateral ACBD,

AC = AD, AB bisects ∠CAD

Now in ΔABC and ΔABD

AC = AD (Given)

∠CAB = ∠DAB (Given)

(AB bisects ∠CAD)

AB = AB (Common)

ΔABC = ΔABD (SAS criterion)

BC = BD (c.p.c.t.)

∠C = ∠D (c.p.c.t.)

**Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions **Ex 12.2**

Question 1.

You want to establish ΔDEF = ΔMNP, using ASA rule of congruence. You are given that ∠D = ∠M and ∠F = ∠P. What additional information is needed to establish the congruence?

#### Answer

In ΔDEF and ΔMNP

∠D = ∠M (Given)

∠F = ∠P (Given)

For using ASA rule, we need

DF = MP (Included side)

Then ΔDEF = ΔMNP

Question 2.

In the given figure, two triangles are congruent. The corresponding parts are marked. We can write ΔRAT =?

#### Answer

In the given figure,

ΔART and ΔNOW

AT = NO (Given)

∠A = ∠O (Given)

∠T = ∠N (Given)

ΔART = ΔWON (ASA criterion)

Question 3.

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

#### Answer

In ΔABC and ΔPQR

∠B = ∠Q

∠C = ∠Q

Now we need BC = QR

ΔABC = ΔPQR (ASA criterion)

Question 4.

Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not, by ASA congruence rule. In the case of congruence, write its in symbolic form.

ΔDEF |
ΔPQR |

(i) ∠D = 60°, ∠F = 80°, DF = 5 cm | (i) ∠Q = 60°, ∠R = 80°, QR = 5 cm |

(ii) ∠D = 60°, ∠F = 80°, DF = 6 cm | (ii) ∠Q = 60°, ∠R = 80°, QR = 5 cm |

(iii) ∠E = 80°, ∠F = 30°, EF = 5 cm | (iii) ∠P = 80°, PQ = 5 cm, ∠R = 30° |

#### Answer

In ΔDEF and ΔPQR

(i) ∠D = 60°, ∠F = 80°, DF = 5 cm

∠Q = 60°, ∠R = 80°, QR = 5 cm

∠D = ∠Q (Each 60°)

∠F = ∠R (Each 80°)

Included side DF = QR

ΔDEF = ΔQPR (ASA criterion)

(ii) In ΔDEF and ΔPQR

∠D = 60°, ∠F = 80°, DF = 6 cm

∠Q = 60°, ∠R = 80° and QP = 6 cm

Here, ∠D = ∠Q (Each 80°)

∠F = ∠R (Each 80°)

But included side DF ≠ QR

ΔDEF and ΔPQR are not congruent.

(iii) In ΔDEF and ΔPQR

∠E = 80°, ∠F = 30°, EF = 5 cm

∠P = 80°, PQ = 5 cm, ∠R = 30°

Here, ∠E = ∠P (Each = 80°)

∠F = ∠R (Each = 30°)

But inlcuded sides are not equal.

ΔDEF and ΔPQR are not congruent.

Question 5.

In the adjoining figure, measures of some parts are indicated.

(i) State three pairs of equal parts in triangles ABC and ABD.

(ii) Is ΔABC = ΔBAD? Give reasons.

(iii) Is BC = AD? Why?

#### Answer

In the given figure,

∠DAC = 45°, ∠CAB = 30°, ∠CBD = 45° and ∠DBA = 30°

Now in ΔABC and ΔBAD,

∠DAC + ∠CAB = 45° + 30° = 75°

and ∠CBD + ∠DBA = 45° + 30° = 75°

∠DAB = ∠CBA

Now in ΔABC and ΔDAB

AB = AB (Common)

∠CBA = ∠DAB (Proved)

∠CAB = ∠DBA (Each = 30°)

ΔABC = ΔDAB (ASA criterion)

Yes, BC = AD (c.p.c.t.)

Question 6.

In the adjoining figure, ray AZ bisects ∠DAB as well as ∠DCB.

(i) State the three pairs of equal parts in triangles BAC and DAC.

(ii) Is ΔBAC = ΔDAC? Give reasons.

(iii) Is CD = CB? Give reasons.

#### Answer

In the given figure

∠DAC = ∠BAC

∠DCA = ∠BCA

Now in ΔBAC and ΔDAC

AC = AC (Common)

∠BAC = ∠DAC (Given)

∠BCA = ∠DCA (Given)

ΔBAC = ΔDAC (ASA criterion)

Yes,AB = AD (c.p.c.t.)

Yes, CD = CB (c.p.c.t.)

Question 7.

Explain why ΔABC = ΔFED?

#### Answer

In ΔABC and ΔFED

BC = DE

∠B = ∠E (Each = 90°)

∠A = ∠F

∠C = 90° – ∠A and ∠D = 90° – ∠F

But ∠A = ∠F (Given)

∠C = ∠D

Now in ΔABC and ΔDEF

BC = DE (Given)

∠B = ∠E (Given 90°)

∠C = ∠D (Proved)

ΔABC = ΔDEF (ASA criterion)

Question 8.

Given below are the measurements of some parts of triangles. Examine whether the two triangles are congruent or not, using RHS congruence rule. In the case of congruent triangles, write the result in symbolic form:

ΔABC |
ΔPQR |

(i) ∠B = 90°, AC = 8 cm, AB = 3 cm | (i) ∠P = 90°, PR = 3 cm, QR = 8 cm |

(ii) ∠A = 90°, AC = 5 cm, BC = 9 cm | (ii) ∠Q = 90°, PR = 8 cm, PQ = 5 cm |

Answer

We are given the measurement of some parts of the triangles.

We have to examine whether the two triangles are congruent

or not using RHS congruency rule.

In ΔABC and ΔPQR

(i) ∠B = 90°, AC = 8 cm, AB = 3 cm

∠P = 90°, PR = 3 cm, QR = 8 cm

We see that in two Δs ABC and RPQ

∠B = ∠P (Each = 90°)

Side AB = RP (Each = 3 cm)

Hypotenuse AC = RQ

ΔABC = ΔRPQ (RHS criterion)

(ii) In ΔABC and ΔPQR

∠A = ∠Q (Each = 90°)

Side AC = QP (Each = 5 cm)

But hypotenuse BC and PR are not equal to each other.

Triangles are not congruent.

Question 9.

In the given figure, measurements of some parts are given.

(i) State the three pairs of equal parts in ΔPQS and ΔPRS.

(ii) Is ΔPQS = ΔPRS? Give reasons.

(iii) Is S mid-point of ? Why?

#### Answer

In the given figure,

PQ = 3 cm, PR = 3 cm

PS ⊥ QR

(i) Now in right ΔPQS and ΔPRS right angles at S. (∵ PS ⊥ QR)

side PS = PS (Common)

Hypotenuse PQ = PR (Each = 3 cm)

(ii) ΔPQS = ΔPRS (RHS criterion)

(iii) QS = SR (c.p.c.t.)

S is the mid point of QR

Question 10.

In the given figure, O is mid-point of and ∠A = ∠B. Show that ΔAOC = ΔBOD.

#### Answer

In the given figure,

O is the mid-point of AB

AO = OB

Now in ΔAOC and ΔBOD

AO = OB (∵ O is mid-point of AB)

∠A = ∠B (Given)

∠AOC = ∠BOD (Vertically opposite angles)

ΔAOC = ΔBOD (ASA criterion)

### Objective Type Questions, **Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions Chapter-12

**Mental Maths**

Question 1.

Fill in the blanks:

(i) Two line segments are congruent if ……….

(ii) Among two congruent angles, one has a measure of 63°; the measure of the other angle is ……….

(iii) When we write ∠A = ∠B, we actually mean ………

(iv) The side included between ∠M and ∠N of ∆MNP is ……….

(v) The side QR of ∆PQR is included between angles ……….

(vi) If two triangles ABC and PQR are congruent under the correspondence A ↔ R, B ↔ P and C ↔ Q, then in symbolic form it can be written as ∆ABC = ………

(vii) If ∆DEF = ∆SRT, then the correspondence between vertices is ……….

#### Answer

(i) Two line segments are congruent if they are of the same length.

(ii) Among two congruent angles, one has a measure of 63°;

the measure of the other angle is 63°.

(iii) When we write ∠A = ∠B, we actually mean m∠A = m∠B.

(iv) The side included between ∠M and ∠N of ∆MNP is MN.

(v) The side QR of ∆PQR is included between angles ∠Q and ∠R.

(vi) If two triangles ABC and PQR are congruent

under the correspondence A ↔ R, B ↔ P and C ↔ Q,

then in symbolic form it can be written as ∆ABC = ∆RPQ.

(vii) If ∆DEF = ∆SRT, then the correspondence between vertices is

D ↔ S, E ↔ R and F ↔ T.

#### Question 2.

State whether the following statements are **true (T) or false (F):**

(i) All circles are congruent.

(ii) Circles having equal radii are congruent.

(iii) Two congruent triangles have equal areas and equal perimeters.

(iv) Two triangles having equal areas are congruent.

(v) Two squares having equal areas are congruent.

(vi) Two rectangles having equal areas are congruent.

(vii) All acute angles are congruent.

(vii)All right angles are congruent.

(ix) Two figures are congruent if they have the same shape.

(x) A two rupee coin is congruent to a five rupee coin.

(xi) All equilateral triangles are congruent.

(xii) Two equilateral triangles having equal perimeters are congruent.

(xii) If two legs of one right triangle are equal to two legs of another right angle triangle, then the two triangles are congruent by SAS rule.

(xiv) If three angles of two triangles are equal, then triangles are congruent.

(xv) If two sides and one angle of one triangle are equal to two sides and one angle of another triangle, then the triangle are congruent.

#### Answer

(i) All circles are congruent. (False)

Correct:

As if all circles have equal radii otherwise not.

(ii) Circles having equal radii are congruent. (True)

(iii) Two congruent triangles have equal areas

and equal perimeters. (True)

(iv) Two triangles having equal areas are congruent. (False)

Correct:

As they may have different sides and angles.

(v) Two squares having equal areas are congruent. (True)

(vi) Two rectangles having equal areas are congruent. (False)

Correct:

As their side can be different.

(vii) All acute angles are congruent. (False)

Correct:

As acute angles have different measures.

(viii) All right angles are congruent. (True)

(ix) Two figures are congruent if they have the same shape. (False)

##### Correct:

As the same shapes have different measures.

(x) A two rupee coin is congruent to a five rupee coin. (False)

Correct:

As they have different size.

(xi) All equilateral triangles are congruent. (False)

Correct:

As they have different sides in length.

(xii) Two equilateral triangles having equal perimeters are congruent. (True)

(xiii) If two legs of one right triangle are equal to

two legs of another right angle triangle,

then the two triangles are congruent by SAS rule. (True)

(xiv) If three angles of two triangles are equal,

then triangles are congruent. (False)

Correct:

They can be similar to each other.

(xv) If two sides and one angle of one triangle are equal to two sides

and one angle of another triangle, then the triangle is congruent. (False)

Correct:

If the angles are included, they can be congruent.

**Multiple Choice Questions, Congruence of Triangles Class-7 ML Aggarwal ICSE Maths Solutions Chapter-12**

**MCQs**

Choose the correct answer from the given four options (3 to 14):

Question 3.

Which one of the following is not a standard criterion of congruency of two triangles?

(a) SSS

(b) SSA

(c) SAS

(d) ASA

#### Answer

The axiom SSA is not a standard criterion

of congruency of triangles. (b)

Question 4.

If ∆ABC = ∆PQR and ∠CAB = 65°, then ∠RPQ is

(a) 65°

(b) 75°

(c) 90°

(d) 115°

#### Answer

∆ABC = ∆PQR

∠CAB = 65°

∠RPQ = 65° (corresponding angles) (a)

Question 5.

If ∆ABC = ∆EFD, then the correct statement is

(a) ∠A = ∠D

(b) ∠A = ∠F

(c) ∠A = ∠E

(d) ∠B = ∠E

#### Answer

∆ABC = ∆EFD

Then ∠A = ∠E (c)

Question 6.

If ∆ABC = ∆PQR, then the correct statement is

(a) AB = QR

(b) AB = PR

(c) BC = PR

(d) AC = PR

#### Answer

∆ABC = ∆PQR

Then AB = PQ

AC = PR (d)

Question 7.

If ∠D = ∠P, ∠E = ∠Q and DE = PQ, then ∆DEF = ∆PQR, by the congruence rule

(a) SAS

(b) ASA

(c) SSS

(d) RHS

#### Answer

In ∆DEF = ∆PQR

∠D = ∠P, ∠E = ∠Q

DE = PQ

∆DEF = ∆PQR (ASA axiom) (b)

Question 8.

In ∆ABC and ∆PQR, BC = QR and ∠C = ∠R. To establish ∆ABC = ∆PQR by SAS congruence rule, the additional information required is

(a) AC = PR

(b) AB = PR

(c) CA = PQ

(d) AB = PQ

#### Answer

If ∆ABC = ∆PQR by SAS

BC = QR and ∠C = ∠R, then AC = PR (a)

Question 9.

In the given figure, the lengths of the sides of two triangles are given. The correct statement is

(a) ∆ABC = ∆PQR

(b) ∆ABC = ∆QRP

(c) ∆ABC = ∆QPR

(d) ∆ABC = ∆RPQ

#### Answer

Correct statement is ∆ABC = ∆QRP. (b)

Question 10.

In the given figure, M is the mid-point of both AC and BD. Then

(a) ∠1 = ∠2

(b) ∠1 = ∠4

(c) ∠2 = ∠4

(d) ∠1 = ∠3

#### Answer

In the given figure,

M is mid-point of AC and BD both then ∠1 = ∠4. (b)

Question 11.

In the given figure, ∆PQR = ∆STU. What is the length of TU?

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) cannot be determined

#### Answer

In the given figure,

∆PQR = ∆STU

TU = QR = 6 cm (b)

Question 12.

In the given figure, ∆ABC and ∆DBC are on the same base BC. If AB = DC and AC = DB, then which of the following statement is correct?

(a) ∆ABC = ∆DBC

(b) ∆ABC = ∆CBD

(c) ∆ABC = ∆DCB

(d) ∆ABC = ∆BCD

#### Answer

In the given figure,

AB = DC, AC = DB

Then, ∆ABC = ∆DCB (c)

Question 13.

The two triangles shown in the given figure are:

(a) congruent by AAS rule

(b) congruent by ASA rule

(c) congruent by SAS rule

(d) not congruent.

#### Answer

In the given two triangles are not congruent.

In first triangle, AAS are given while in second ASA are given. (d)

Question 14.

In .the given figure, ∆ABC = ∆PQR. The values of x and y are:

(a) x = 63, y = 35

(b) x = 77, y = 35

(c) x = 35, y = 77

(d) x = 63, y = 40

#### Answer

In the given figure,

∆ABC = ∆PQR

∠A = ∠P and ∠B = ∠Q

Now x – 7 = 70°

⇒ x = 70° + 7 = 77°

and 2y + 5 = 75

⇒ 2y = 75° – 5 = 70°

⇒ y = 35°

x = 77°, y = 35° (b)

**Higher Order Thinking Skills, Congruence of Triangles Class-7 ML Aggarwal ICSE Maths Solutions Chapter-12**

** (HOTS)**

Question 1.

If all the three altitudes of a triangle are equal, then prove that it is an equilateral triangle.

#### Answer

Given: In ∆ABC,

AD, BE and CF are altitudes of the triangle

and AD = BE = CF.

To prove: ∆ABC is an equilateral.

Proof: In ∆ABD and ∆CFB

AD = CF (Given)

∠D = ∠F (Each = 90°)

∠B = ∠B (Common)

∆ABD = ∆CFB (AAS criterion)

AB = BC …….(i)

Similarly in ∆BEC and ∆ADC

BE = AD (Given)

∠C = ∠C (Common)

∠E = ∠D (Each = 90°)

∆BEC = ∆ADC (AAS criterion)

BC = AC ………(ii)

From (i) and (ii)

AB = BC = AC

∆ABC is an equilateral triangle.

Question 2.

In the given fig., if BA || RP, QP || BC and AQ = CR, then prove that ∆ABC = ∆RPQ.

#### Answer

In the given figure, BA || RP

QP || BC and AQ = CR

To prove : ∆ABC = ∆RPQ

Proof: AQ = CR

Adding CQ to both sides

AQ + CQ = CR + CQ

⇒ AC = RQ

Now in ∆ABC and ∆RPQ

∠A = ∠R (Alternate angles)

∠C = ∠Q (Alternate angles)

AC = RQ (Proved)

∆ABC = ∆RPQ (ASA criterion)

**Check Your Progress**

**Congruence of Triangles Class-7 ML** Aggarwal ICSE Maths Solutions Chapter-12

Question 1.

State, giving reasons, whether the following pairs of triangles are congruent or not:

#### Answer

(i) In the given figure, using the SSS criterion triangles one congruent.

(ii) Triangles are congruent for the criterion ASA criterion.

(iii) Triangles are congruent for the criterion RHS.

(iv) In the first triangle, third angle = 180° – (70° + 50°) = 180° – 120° = 60°

Now triangles are congruent for ASA criterion.

(v) Not congruent as included angles of the given two sides are not equal.

(vi) Not congruent as the included sides are different.

Question 2.

Given below are measurements of some parts of two triangles. Examine whether the two triangles are congruent or not. In case of congruence, give reasons and write in symbolic form:

∆ABC |
∆PQR |

(i) AB = 4 cm, BC = 5 cm, ∠B = 70° | (i) QR = 4 cm, RP = 5 cm, ∠R = 70° |

(ii) AB = 4 cm, BC = 5 cm, ∠B = 80° | (ii) PQ = 4 cm, RP = 5 cm, ∠R = 80° |

(iii) BC = 6 cm, ∠A = 90°, ∠C = 50° | (iii) QR = 6 cm, ∠R = 50°, ZQ = 40° |

(iv) AB = 5 cm, ∠A = 90°, BC = 8 cm | (iv) PR = 5 cm, ∠P = 90°, QR = 8 cm |

#### Answer

(i) In ∆ABC and ∆PQR

AB = QR = 4 cm

BC = RP = 5 cm

∠B = ∠R = 70°

∆ABC = ∆PQR (SAS criterion)

(ii) In ∆ABC and ∆PQR

AB = PQ = 4 cm

BC = RP = 5 cm not corresponding sides

∠B = ∠R = 80° not corresponding angles

Triangles are not congruent.

(iii) BC = QR = 6 cm

∠A = ∠P = 90° (Third angle)

∠C = ∠R = 50°

Triangles are congruent for ASA criterion.

(iv) AB = PR = 5 cm (Side)

∠A = ∠P = 90°

BC = QR = 8 cm (Hypotenuse)

Triangles are congruent for RHS criterion.

Question 3.

In the given figure, ABC is an isosceles triangle with AB = AC and AD is one of its altitudes.

(i) State the three pairs of equal parts in ∆ADB and ∆ADC.

(zz) Is ∆ADB = ∆ADC? Give reasons.

(iii) Is ∠B = ∠C? Why?

(iv) Is BD = DC? Why?

#### Answer

∆ABC is an isosceles triangle with AB = AC

and AD is one of the altitudes.

(i) In ∆ADB and ∆ADC

Side AD = AD (Common)

Hypotenuse, AB = AC (Given)

∠ADB = ∠ADC = 90° (∵ AB ⊥ BC)

∆ADB = ∆ADC

∠B = ∠C (c.p.c.t)

and BD = CD (c.p.c.t)

Question 4.

In the given figure, OA bisects ∠A and ∠ABO = ∠OCA. Prove that OB = OC.

#### Answer

In ∆OAB and ∆OAC

∠OAB = ∠OAC (∵ OA bisects ∠A)

∠ABO = ∠ACO (Given)

OA = OA (common)

∆OAB = ∆OAC (AAS congruence rule)

OB = OC (Corresponding parts of congruent As)

Question 5.

In the given figure , prove that

(i) AB = FC

(ii) AF = BC.

#### Answer

In ∆ABE and ∆DFC

∠B = ∠F (each 90°)

AE = DC (Given)

BE = DF (Given)

∆ABE = ∆DFC (RHS congruence rule)

(i) AB = FC (Corresponding parts of congruent ∆s)

(ii) As AB = FC (Proved above)

⇒ AF + FB = FB + BC

or AF + FB – FB = BC

hence AF = BC

— End of** Congruence of Triangles Class-7 ML Aggarwal ** Solutions :–

Return to **– ML Aggarwal Maths Solutions for ICSE Class -7**

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