7th Grade > Mathematics
CONGRUENCE OF TRIANGLES MCQs
Total Questions : 103
| Page 3 of 11 pages
Answer: Option A. -> True
:
A
If all the side lengths of one triangle are equal to the side lengthsofanother triangle, then the triangles are congruent. This is calledSSS criterion.
:
A
If all the side lengths of one triangle are equal to the side lengthsofanother triangle, then the triangles are congruent. This is calledSSS criterion.
:
In the given figure:
In ΔABCandΔPQR
AB = PR = 5 cm
BC = PQ
AC = QR = 7 cm
Therefore,ΔABC≅ΔPQR by SSS criterion.
Since perimeter = 18cm
⇒AB + AC + BC = 18 cm,
⇒ BC = 18 - 5 - 7 =6 cm.
Therefore,
AB = PR = 5 cm
BC = PQ = 6 cm
AC = QR = 7 cm
So, the answer is 7 cm.
Answer: Option D. -> AAA
:
D
The criteria for congruence of triangles are SSS criterion, SAScriterion, ASAcriterion and RHS criterion.
AAA is not a criterion for congruence as it does not ensurethe equality of sides of the two triangles.
Note: AAA is acriterion for 'Similarity'of triangles.
:
D
The criteria for congruence of triangles are SSS criterion, SAScriterion, ASAcriterion and RHS criterion.
AAA is not a criterion for congruence as it does not ensurethe equality of sides of the two triangles.
Note: AAA is acriterion for 'Similarity'of triangles.
Answer: Option C. -> PQ
:
C
SinceΔABC≅ΔPQR,
corresponding sides of congruent triangles will be equal.
Hence, AB = PQ.
:
C
SinceΔABC≅ΔPQR,
corresponding sides of congruent triangles will be equal.
Hence, AB = PQ.
Answer: Option C. -> ∠C, AB
:
C
We know that if two triangles are congruent, then their corresponding parts are equal.
Since ΔDEF≅ΔBCA, therefore∠E=∠C and FD=AB.
:
C
We know that if two triangles are congruent, then their corresponding parts are equal.
Since ΔDEF≅ΔBCA, therefore∠E=∠C and FD=AB.
:
Naming: 1 Mark
Criterion: 1 Mark
Given, ΔABC andΔPQR 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 andΔPQR are congruent by ASA congruence rule.
Then one additional pair is BC = QR.
:
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
:
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
:
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)
InΔAOEandΔDOC,
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]
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