# Free Congruence of Triangles Subjective Test 02 Practice Test - 7th grade

### Question 1

In the figure, the two triangles are congruent. The corresponding parts are marked. Complete the congruent statement ΔRAT≅ [1 MARK]

#### SOLUTION

Solution :Solution: 1 Mark

In the figure, the two triangles are congruent.

So, the corresponding congruent parts are:

∠A=∠O,∠R=∠W,∠T=∠N

Side AT = Side ON, Side AR = Side OW

∴ We can write, ΔRAT≅ΔWON

### Question 2

If ΔABC and ΔPQR are to be congruent, name one additional pair of corresponding parts. Which criterion did you use? [2 MARKS]

#### SOLUTION

Solution :Naming: 1 Mark

Criterion: 1 Mark

Given, ΔABC and ΔPQR are congruent with,

∠B=∠Q=90∘

∠C=∠R

For ΔABC and ΔPQR to be congruent, the side in between the equal angles needs to be equal.

¯¯¯¯¯¯¯¯BC=¯¯¯¯¯¯¯¯¯QR

⇒ΔABC and ΔPQR are congruent by ASA congruence rule.

Then one additional pair is BC = QR.

### Question 3

Prove that in the following kite, ΔADC is congruent to ΔABC. Given that AD = AB and ∠ADC = ∠ABC = 90∘

[2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Proof: 1 Mark

In ΔADC and ΔABC

AD=AB [Given]

∠ADC=∠ABC=90o [Given]

AC=CA [common]

Hence, ΔADC≅ΔABC [By RHS congruence condition]

### Question 4

In the given figure, prove that: [2 MARKS]

ΔABD≅ΔACD

#### SOLUTION

Solution :Steps: 1 Mark

Proof: 1 Mark

In ΔABD and ΔACD

AB=DC [Given]

BD=CA [Given]

AD=AD [Common]

⇒ΔABD≅ΔACD [SSS congruency criteria]

### Question 5

Which congruence criterion is used in the following?

Given: ∠MLN=∠FGH,∠NML=∠HFG,ML=FG

So ΔLMN≅ΔGFH [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Proof: 1 Mark

It is given that,

∠MLN=∠FGH,

∠NML=∠HFG,

ML=FG.

⇒ The two angles and an included side of one triangle are equal to the corresponding angles and an included side of other triangles.

∴ΔLMN≅ΔGFH [By ASA congruence criterion]

### Question 6

In the given figure, AB∥CD and AB = CD. [3 MARKS]

Prove that: (i) ΔAOB≅ΔDOC

(ii) AO = DO

(iii) BO = CO

#### SOLUTION

Solution :Each proof: 1 Mark

In ΔAOB and ΔCOD

AB=CD [Given]

∠BAO=∠CDO [Alternate angles; as AB∥CD]

∠ABO=∠DCO [Alternate angles; as AB∥CD]

(i) ∴ΔAOB≅ΔDOC [A.S.A congruency criteria]

(ii) AO=DO [Corresponding sides of congruent triangles]

also, (iii) BO=CO [Corresponding sides of congruent triangles]

### Question 7

Prove that: [3 MARKS]

(i) ΔABC≅ΔADC

(ii) ∠B=∠D

#### SOLUTION

Solution :Each Proof: 1 Mark

Steps: 1 Marks

In ΔABC and ΔADC

AB=DC [Given]

BC=AD [Given]

AC=AC [Common side]

⇒ΔABC≅ΔADC [SSS congruency criteria]

∴∠B=∠D [Corresponding parts of congruent triangles]

### Question 8

In the given figure, prove that: [3 MARKS]

(i) ΔACB≅ΔECD

(ii) AB=ED

#### SOLUTION

Solution :Concept : 1 Mark

Proof : 2 Marks

In ΔACB and ΔECD

AC=EC [Given]

BC=DC [Given]

∠ACB=∠ECD [Vertically opposite angles]

⇒ΔACB≅ΔECD (By SAS condition)

∴AB=ED [Corresponding parts of congruent triangles]

### Question 9

(a) Show with an example that two triangles can't be congruent using AAA criterion.

(b) Which congruence criterion will you use in the following?

Given: AC = DF

AB = DE

BC = EF

So, ΔABC ≅ ΔDEF

[3 MARKS]

#### SOLUTION

Solution :(a) Proof: 2 Marks

(b) Answer: 1 Mark

(a)Consider the two triangles :

In the two triangles,

∠ABC = ∠PQR

∠BAC = ∠QPR

∠ACB = ∠PRQ

But clearly, ΔABC is not congruent to ΔPQR.

As the sides of triangle are not equal.

Thus, AAA cannot be a congruence condition.

It actually tells that the two triangles are similar, but not congruent.

(b) Since, the three sides of the first triangle is equal to the corresponding three sides of the second triangle, by SSS congruence criterion ΔABC is congruent to ΔDEF.

### Question 10

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

(b) In the given figure, ΔABC and ΔOBC are both isosceles.

Prove that ΔAOC is congruent to ΔAOB.

[4 MARKS]

#### SOLUTION

Solution :Each Proof: 2 Marks

(a) In ΔABC and ΔABD

AB=AB [Common side]

∠ABC=∠ABD=90∘ [Given]

AC=AD [Given]

⇒ΔABC≅ΔABD [RHS congruency criteria]

∴BD=BC [Corresponding parts of congruent triangles]

(b)

In ΔAOB and ΔAOC,AB=AC (Sides of isosceles ΔABC).OB=OC (Sides of isosceles ΔOBC).AO=AO (Common side) ∴ΔAOB≅ΔAOC (by SSS congruence)

### Question 11

In the isosceles ΔABC with AB = AC. E and O are the midpoints of AB and AC respectively. 2OD = DE and ∠OCD = 70o. Prove using congruency that ∠EAO = 70o. [4 MARKS]

#### SOLUTION

Solution :Steps: 2 Marks

Proof: 2 Marks

In ΔABC,

OD + OE = DE

Multiplying both sides with 2:

2OD + 2OE = 2DE

DE + 2OE = 2DE (2OD = DE; given in question)

2OE = DE

So, OD = OE -------------- (1)

In ΔAOE and ΔDOC,

OD = OE [From (1)]

∠AOE = ∠DOC [Vertically Opposite Angles]

AO = OC [O is the mid-point of AC]ΔAOE≅ΔDOC [By SAS condition]

Hence, ∠EAO = ∠OCD = 70o [Corresponding parts of corresponding triangles]

### Question 12

In the given figure; ∠1=∠2 and AB = AC. Prove that: [4 MARKS]

(i) ∠B=∠C

(ii) BD = DC

(iii) AD is perpendicular to BC.

#### SOLUTION

Solution :Steps: 1 Mark

Each proof: 1 Mark

In ΔADB and ΔADC

∠1=∠2 [Given]

⇒∠BAD=∠CAD

AD=AD [Common side]

AB=AC [Given]

⇒ΔADB≅ΔADC [SAS congruency criteria]

(i) ∴∠B=∠C [Corresponding parts of congruent triangles]

(ii) BD=DC [Corresponding parts of congruent triangles]

(iii) ΔADB≅ΔADC [proved above]

∠ADB+∠ADC=180° [Linear pair]

⇒∠ADB=∠ADC [c.p.c.t]

∴∠ADB+∠ADB=180°

⇒2∠ADB=180°

⇒∠ADB=180°2

=90°

⇒AD⊥BC

### Question 13

(a) In the given figure, prove that:

(i) PQ = RS

(ii) PS = QR

(b) Which congruence criterion will you use in the following?

(i) Given: ∠ MLN = ∠ FGH

∠ NML = ∠ GFH

ML = FG

So, ΔLMN ≅ ΔGFH

(ii) Given: EB = DB

AE = BC

∠ A = ∠ C = 90°

So, ΔABE ≅ ΔCDB

[4 MARKS]

#### SOLUTION

Solution :Each part: 2 Marks

(a) In ΔPSR and ΔRQP

∠PSR=∠RQP [Given]

∠SPR=∠QRP [Given]

PR=PR [Common]

⇒ΔPSR≅ΔRQP [AAS congruency criteria]

(i) ∴PQ=RS [Corresponding parts of congruent triangles]

(ii) Also, PS=QR [Corresponding parts of congruent triangles]

(b) (i) ASA, as two angles and the side included between these angles of ΔLMN, are equal to two angles and the side included between these angles of ΔGFH.

(ii) RHS, as in the given two right-angled triangles, one side and the hypotenuse are respectively equal.

### Question 14

In the given figure, prove that: [4 MARKS]

(i) ΔXYZ≅ΔXPZ

(ii) YZ = PZ

(iii) ∠YXZ=∠PXZ

#### SOLUTION

Solution :Steps: 1 Mark

Each Proof: 1 Mark

(i) In ΔXYZ and ΔXPZ

XY=XP [Given]

∠XYZ=∠XPZ=90∘ [Given]

XZ=XZ [Common]

⇒ΔXYZ≅ΔXPZ [RHS congruency criteria]

(ii) YZ=PZ [Corresponding parts of congruent triangles]

(iii) ∠YXZ=∠PXZ [Corresponding parts of congruent triangles]

### Question 15

(a) In the given figure, prove that:

(i) ΔABC≅ΔDCB

(ii) AC = DB

(b) In the figure, it is given that ∠A = 90∘, AB = AC, and D is the mid point of BC. Find ∠ADC.

[4 MARKS]

#### SOLUTION

Solution :Each Part: 2 Marks

(a)

(i) In ΔABC and ΔDCB

AB=DC [Given]

∠ABC=∠DCB=90∘ [Given]

BC=BC [Common side]

⇒ΔABC≅ΔDCB [SAS congruency criteria]

(ii) ∴AC=DB [Corresponding parts of congruent triangles]

(b)

AB = AC (Given)

BD = DC (D is mid point of BC)

AD = AD (Common side)

Therefore, △ADB ≅ △ADC [SSS congruency criteria]

thus, ∠ADB = ∠ADC (c.p.c.t.) ...(1)

but, ∠ADB + ∠ADC = 180∘ [linear pair]

∴∠ADC + ∠ADC = 180∘

⇒2∠ADC = 180∘

⇒∠ADC=180°2

⇒∠ADC=90°