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

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, ΔRATΔWON

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.

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

AC=CA   [common]

Hence, ΔADCΔABC  [By RHS congruence condition]

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

ΔABDΔACD #### SOLUTION

Solution : Steps: 1 Mark
Proof: 1 Mark

In ΔABD and ΔACD

AB=DC [Given]

BD=CA [Given]

ΔABDΔACD [SSS congruency criteria]

Which congruence criterion is used in the following?
Given: MLN=FGH,NML=HFG,ML=FG
So ΔLMNΔGFH  [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.

ΔLMNΔGFH    [By ASA congruence criterion]

In the given figure, ABCD and AB = CD.  [3 MARKS] Prove that: (i) ΔAOBΔDOC

(ii) AO = DO

(iii) BO = CO

#### SOLUTION

Solution : Each proof: 1 Mark In ΔAOB and ΔCOD

AB=CD [Given]

BAO=CDO [Alternate angles; as ABCD]

ABO=DCO [Alternate angles; as ABCD]

(i) ΔAOBΔDOC [A.S.A congruency criteria]

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

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

Prove that: [3 MARKS] (ii) B=D

#### SOLUTION

Solution : Each Proof: 1 Mark
Steps: 1 Marks

AB=DC [Given]

AC=AC [Common side]

B=D [Corresponding parts of congruent triangles]

In the given figure, prove that: [3 MARKS] (i) ΔACBΔECD

(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]

ΔACBΔECD (By SAS condition)

AB=ED [Corresponding parts of congruent triangles]

(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

(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.

(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]

ΔABCΔABD [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) ΔAOBΔAOC   (by SSS congruence)

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]

ΔAOEΔDOC   [By SAS condition]

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

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

1=2 [Given]

AB=AC [Given]

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

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

=90°

(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]

ΔPSRΔRQP [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.

In the given figure, prove that:  [4 MARKS] (i) ΔXYZΔXPZ

(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]

ΔXYZΔXPZ [RHS congruency criteria]

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

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

(a) In the given figure, prove that: (i) ΔABCΔDCB

(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]

ΔABCΔDCB [SAS congruency criteria]

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

(b) AB = AC (Given)
BD = DC  (D is mid point of BC)