7th Grade > Mathematics
CONGRUENCE OF TRIANGLES MCQs
Total Questions : 103
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Answer: Option B. -> OD = OC
:
B
Consider ΔBOCand ΔAOD
1)AD = BC ( Given )
2) ∠CBO=∠DAO= 90°
3) ∠BOC=∠AOD ......(vertically opposite angles)
∴ΔBOC≅ΔAOD ....[AAS Criterion]
⇒ OC = OD ....(congruent parts of congruent triangle)
:
B
Consider ΔBOCand ΔAOD
1)AD = BC ( Given )
2) ∠CBO=∠DAO= 90°
3) ∠BOC=∠AOD ......(vertically opposite angles)
∴ΔBOC≅ΔAOD ....[AAS Criterion]
⇒ OC = OD ....(congruent parts of congruent triangle)
Answer: Option C. -> Both Statement 1 and statement 2 are true.
:
C
Two line segments are congruent only if they superimpose which is possible only if they have equal length and the converse is also true.
Hence, both the given statementsare true.
:
C
Two line segments are congruent only if they superimpose which is possible only if they have equal length and the converse is also true.
Hence, both the given statementsare true.
Answer: Option A. -> True
:
A
In two triangles,
If a pair of corresponding angles and the included side are equal, then they are congruent [ASA congruence criterion].
If a pair of corresponding angles and a non-included side are equal,then they are congruent [AAS congruence criterion].
Therefore, given statement is true.
:
A
In two triangles,
If a pair of corresponding angles and the included side are equal, then they are congruent [ASA congruence criterion].
If a pair of corresponding angles and a non-included side are equal,then they are congruent [AAS congruence criterion].
Therefore, given statement is true.
Answer: Option A. -> ∠ABD = ∠BAC
:
A
InΔABDandΔBAC,
AD = BC(Given)
∠BAD= ∠CBA(Given)
AB = BA (Side common to both triangles)
HenceΔABD≅ΔBAC
(by SAS congruence condition).
Thus, ∠ABD = ∠BAC
(congruent parts of congruent triangles).
:
A
InΔABDandΔBAC,
AD = BC(Given)
∠BAD= ∠CBA(Given)
AB = BA (Side common to both triangles)
HenceΔABD≅ΔBAC
(by SAS congruence condition).
Thus, ∠ABD = ∠BAC
(congruent parts of congruent triangles).
:
(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 thatthe 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, bySSScongruence criterion ΔABCis congruent to ΔDEF.
:
Steps: 1 Mark
Proof: 2Marks
In Δ1andΔ2:
∠AOD=∠COD=90∘(Diagonals of square intersect at right angles)
AD=CD (Sides of a square; hypotenuse)
OD=DO (Common)
Hence, Δ1≅Δ2(By RHS congruence rule) ---------------------1
Similarly, Δ4≅Δ3(By RHS congruence rule) ---------------------2
In Δ1andΔ4
∠AOD=∠AOB=90∘(Diagonals of square intersect at right angles)
AD=AB(Sides of a square; hypotenuse)
OA=AO (Common)
Hence, Δ1≅Δ4 (By RHS congruence rule) -------------------3
Similarly, Δ2≅Δ3 (By RHS congruence rule) ----------------4
From 1, 2, 3 and 4 we can say that all the triangles i.e. Δ1, Δ2, Δ3 and Δ4are congruent to each other.
5, 6, 7 and 8 have relatively small area. Since they have same shape and size, they are also congruent. So, we can say that all the slices of the pizza are congruent to each other.
:
Each Part: 2 Marks
(a)
In ΔBDAandΔCDA
∠B=∠C [Given]
∠BAD=∠CAD [Given, DA is an angle bisector]
AD=AD [Common side]
⇒ΔBDA≅ΔCDA [ AAS criteria]
(b) we observethat in the given figures, there are no pairs of congruent sides. Since all of the congruency theorems call for at least one pair of congruent sides, there isn'tenough information to prove that the triangles are congruent. Two triangles cannot be proved congruent just by AAA because triangles with same angles can have different sizes.
:
If two angles have the same measure, they are congruent. Also, if two angles arecongruent, their measures are same.
Hence,the measure of the other angle is also60o.