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°