Triangles Test

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Triangles Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

PQR is an isosceles triangle with PQ = PR. The bisectors of and meet at point O. By which of the following conditions is POQ POR?

A
SSS

B
SSA

C
AAS

D
ASA

Solution

PQR is an isosceles .
PQ = PR
( OQ and OR are bisectors)

In QOR,

Then, OQ = OR (sides opposite to equal angles are equal)
Now, in POQ and POR,
PQ = PR (given)
OQ = OR (proved above)
OP = OP (common)
POQ POR by SSS criterion.
Question 2
1 / -0

In the given triangle, ADBC and B > C.

Which of the following relations is definitely correct?

A
AB > BC

B
BD < AB

C
AC < AB

D
None of these

Solution

B > C
In ABD,

ADB = 90° [ ADBC]
BAD < ADB [ADB = 90°]
BD < AB [Side opposite to the larger angle is always larger.]
Question 3
1 / -0

PQR is an isosceles triangle such that PQ = PR and perpendiculars from vertices Q and R meet their opposite sides at points M and N, respectively. Which of the following conditions will make ΔNQR ΔMRQ?

A
RHS

B
SAS

C
ASA

D
AAS

Solution

PQR is an isosceles triangle.
=
Now, in ΔNQR and MRQ,
(each 90°)
= (proved above)
QR = RQ (common side)
ΔNQR ΔMRQ by AAS congruency criterion.
Question 4
1 / -0

In the given figure, ABCD, AD || BC, AD = BC and OC = OD.

Which of the following options is the correct relationship for the given figure?

A
AOD COB

B
AOD BOC

C
AOD BCO

D
ADO COB

Solution

In AOD and BOC,

AD = BC [given]
1 = 2 [alternate interior angles]
OD = OC [given]
AOD BOC
Question 5
1 / -0

In the given figure, if AC bisects BCD and BAD, which of the following criteria can be used to prove that ABC ADC?

A
SAS

B
ASA

C
AAS

D
Data insufficient

Solution

In ABC and ADC,

1 = 2 (Given)
3 = 4
AC = AC (Common)
ABC ADC by ASA criterion (A pair of corresponding angles and the included side are equal.)
Question 6
1 / -0

Based on the figure, which of the following options gives the correct comparison between MN and MP?

A
MN < MP

B
MN > MP

C
MN = MP

D
Cannot be determined

Solution

In MNO:

MN = OM
MON = MNO [Angles opposite to equal sides are also equal.]
Also, MON > MPO
MNO > MPO
MPO < MNO
MN < MP
Question 7
1 / -0

If in a PQR, PQ < QR, then which of the following is definitely true?

A
QR < PR

B
PR – QR > PQ

C
QP + PR < QR

D
QR – PQ < PR

Solution

The difference between two sides of a triangle must be less than the third side.
QR – PQ < PR, which is always true for a triangle.
The sum of two sides of a triangle must be greater than the third side.
QR < PR + PQ, which is always true for a triangle.
Question 8
1 / -0

In the given figures, if ABC DEF, find the value of x.

A
7

B
8

C
9

D
6

Solution

Triangles are congruent means corresponding sides are equal.
So, AB = DE,
5x - 4 = 36
5x = 40
x = 8
Question 9
1 / -0

What will be the ratio of the perimeters of two congruent triangles?

A
1 : 1

B
1 : 3

C
2 : 1

D
2 : 3

Solution

Triangles are congruent means corresponding sides are equal.
So, perimeters will be equal.
So, ratio = 1 : 1
Question 10
1 / -0

In the given figure, D is the midpoint of BE, E is the midpoint of CD, AB > BD and AD > CE. Which of the following relations is definitely true?

A
AE < AD

B
AB > AC

C
ED < AB

D
AD > AC

Solution

ED = DB [D is the midpoint of BE]

And, ED = EC [E is the midpoint of CD]
ED = DB = EC
Also, AB > BD
AB > ED [ BD = ED]
Question 11
1 / -0

If MN is a line segment and PO is its perpendicular bisector, bisecting it at O, then which of the following is true?

A
MO = PN

B
PM = PN

C
PM = PO

D
PN = PO

Solution

Consider POM and PON:
We have: MO = ON (O is the midpoint of MN.)
POM = PON = 90° (Given)
PO = PO (Common)
So, POM PON (by SAS criterion)
Hence, PM = PN, as they are corresponding sides of congruent triangles.
Question 12
1 / -0

Which of the following options is correct?

A
ABE CDE

B
ABE DEC

C
ABE ECD

D
BAE EDC

Solution

In ΔABE and ΔCDE, we have
B = D = 45°
E = E (Common)
BE = DE = 4 cm
So, by ASA congruency rule,
Question 13
1 / -0

Which of the following criteria is most appropriate to prove that BAC is congruent to ABD?

A
SAS

B
AAS

C
RHS

D
SSS

Solution

We can see in the given figure that there are two right-angled triangles.

BAC and ABD
In BAC, In ABD,
Base = AC = 2 cm Base = BD = 2 cm

Perpendicular = BC Perpendicular = AD

Hypotenuse = AB = 3.5 cm Hypotenuse = AB = 3.5 cm

Right-angled at C Right-angled at D

Hence, AB = AB, C = D = 90°and base = AC = BD = 2 cm

BAC ABD (by RHS criterion)
Question 14
1 / -0

In the given figure, >

Which of the following relations is correct?

A

B

C

D

Solution

In SPR,
= 180° (angle sum property of )
Or = 90° (∵ ∠P = 90°) … (i)
InQSR,
180°
Or + = 90° (∵ ∠Q = 90°) ... (ii)
From (i) and (ii),
=
∵
Question 15
1 / -0

Given PQ = FE and RP = DF, what additional information is needed to prove that PQR and FED are congruent by SAS congruency criterion?

A
QR = DE

B
P = F

C
P = D

D
P = E

Solution

We know that in PQR and FED, the corresponding two sides are equal.
i.e.
PQ = FE
RP = DF
To make the triangles PQR and FED congruent by SAS criterion, we need two corresponding sides and an angle between them equal.
So, the angle between PQ and RP is P.
The angle between FE and DF is F.
If P = F, then PQR FED by SAS criterion.
Question 16
1 / -0

Which of the following relations is true, if A, B and C of ΔABC are in the ratio 2 : 3 : 4, respectively?

A
BC > AC

B
AB > AC

C
BC > AB

D
AC > AB

Solution

The angles of the triangle are in the ratio 2 : 3 : 4.
Let the angles A, B, C be 2x, 3x and 4x respectively.
Opposite side of greatest angle is greatest. So, AB is the greatest.
Opposite side of smallest angle is smallest. So, BC is the smallest.
AB > CA > BC
Therefore, option 2 is correct.
Question 17
1 / -0

According to which rule is ?

A
SAS

B
ASA

C
RHS

D
SSS

Solution

B = P
C = R
and, BC = PR = 6 cm
The rule used here is ASA as two corresponding angles are equal and corresponding side is also equal.
Question 18
1 / -0

In the following figure, AB = BC and AC = CD. Which of the following is true?

A
BAD : ADB = 3 : 1

B
BAD : ADB = 2 : 1

C
BAD : ADB = 4 : 1

D
BAD : ADB = 1 : 1

Solution

ADB = 30°
BAD = BAC + CAD = 60° + 30° = 90°
And, = =
Question 19
1 / -0

Which of the following options is true for the two given congruent triangles?

A
ABC DEF

B
ABC EFD

C
BCA DEF

D
ABC DFE

Solution

We just need to put both the triangles into the same orientation so that we can compare them. We find that ABC DFE.
Question 20
1 / -0

IfABC A'C'B', C'= 3x - 40° and B = 2x - 10°, then what is the measure of B?

A
30°

B
50°

C
60°

D
40°

Solution

As ABC A'C'B'So,B = C'
2x - 10° = 3x - 40°
x = 30°
B = 2x - 10° = 50°

Question 21

1 / -0

Fill in the blanks.

(1) In a triangle ABC, AB = 5 units and AC = 3 units. Then, side BC W 8.
(2) Two line segments AB and PQ bisect each other at O. Then, △AOQ △BOP by X criterion.
(3) If PQR MNO, then NO correspond to Y .

	W	X	Y
(A)	>	SAS	PR
(B)	≤	ASA	RQ
(C)	<	SAS	QR
(D)	≥	AAA	PQ

(A)

(B)

(C)

(D)

Solution

(1) The sum of two sides of a triangle is always greater than the third side.
So, AB + AC > BC
5 units + 3 units > BC
BC < 8 units

(2)

In AOQ and BOP,
AO = OB [O bisects AB and PQ.]
QO = OP
AOQ = BOP [Vertically opposite angles]
AOQ BOP by SAS criterion

(3)

Given: PQR MNO.
So, the side NO will correspond to the corresponding side of PQR, i.e. QR.

Question 22
1 / -0

In the figure below (not drawn to scale), if AD = CD = BC and BCE = 96°, then what is the measure of DBC?

A
32°

B
84°

C
64°

D
48°

Solution

Given, AD = CD = BC and BCE = 96°

CAD = ACD ...(i)
CDB = CBD ...(ii)
Now, CDB = CAD + ACD = 2CAD ...(iii)
Also, CAD + CBD = 96°
CAD + 2CAD = 96° ...from (iii)
CAD = 32°
CDB = 64° ...from (iii)
DBC = 64° ...from (ii)
Question 23
1 / -0

Which of the following statement is/are correct?

(A) In PQR, PQ + PR > QR is always true.
(B) If two angles and one side of a triangle are equal to two angles and the corresponding side of another triangle, then the two triangles are congruent (AAS Congruence Rule).
(C) Angles opposite to equal sides of an isosceles triangle are equal.
(D) In a triangle, the side opposite to the larger (greater) angle is longer.

A
All of them

B
Only (B)

C
Only (A), (C) and (D)

D
Only (B) and (D)

Solution

(A) In PQR, PQ + PR > QR is always true.
This is true as this is a condition for the formation of a triangle.

(B) If two angles and one side of a triangle are equal to two angles and the corresponding side of another triangle, then the two triangles are congruent (AAS Congruence Rule).
This is a true statement.

(C) Angles opposite to equal sides of an isosceles triangle are equal.
This is true for every isosceles triangle.

(D) In a triangle, the side opposite to the larger (greater) angle is longer.
This is true for every triangle.

Question 24

1 / -0

Match the following:

Column A	Column B
(i) ABC DCB by	(A) ASA Rule
(ii) Given, AD = DC and ∠ADB = ∠CDB. Then, ADB CDB by	(B) RHS Rule
(iii) EAB DCB by	(C) SAS Rule
(iv) Given, PM \|\| NQ and O is the midpoint of line segment PQ. Then, POM QON by	(D) SSS Rule

(i) - A, (ii) - B, (iii) - C, (iv) - D

(i) - B, (ii) - A, (iii) - D, (iv) - C

(i) - A, (ii) - D, (iii) - C, (iv) - B

(i) - D, (ii) - C, (iii) - B, (iv) - A

Solution

(i) AB = CD (Given)
AC = BD (Given)
BC = BC (Common)
BAC CDB (By SSS criterion)

(ii) In ADB and CDB:
AD = DC (Given)
ADB = CDB (Given)
DB = DB (Common)
ADB CDB (By SAS criterion)

(iii) Two corresponding sides of both the triangles are equal and one angle of both the triangles is a right angle. So, both the triangles are congruent by RHS criterion.

(iv) PM || NQ and PQ is transversal.
MPO =NQO (Alternate interior angles)
In POM and QON:
MPO = NQO (Proved above)
PO = OQ (O is the midpoint of PQ.)
POM = QON (Vertically opposite angles)
POM QON (By ASA criterion)

Question 25
1 / -0

In the figure above, the area of triangular region PQR is 36 sq. units. What is the area of triangular region SQR?

A
12 sq. units

B
16 sq. units

C
24 sq. units

D
30 sq. units

Solution

In triangle PQS, PQ² + PS² = QS²4² + PS² = 5²PS² = 9 units
PS = 3 units
Area of triangle PQS = × 4 × 3 = 6 sq. units
Now, area of triangle SQR = ar(PQR) - ar(PQS)
= 36 - 6 = 30 sq. units

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Triangles Test - 6

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Triangles Test - 6

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SHARING IS CARING

Question 1

SSS

SSA

AAS

ASA

Solution

Question 2

AB > BC

BD < AB

AC < AB

None of these

Solution

Question 3

RHS

SAS

ASA

AAS

Solution

Question 4

AOD COB

AOD BOC

AOD BCO

ADO COB

Solution

Question 5

SAS

ASA

AAS

Data insufficient

Solution

Question 6

MN < MP

MN > MP

MN = MP

Cannot be determined

Solution

Question 7

QR < PR

PR – QR > PQ

QP + PR < QR

QR – PQ < PR

Solution

Question 8

7

8

9

6

Solution

Question 9

1 : 1

1 : 3

2 : 1

2 : 3

Solution

Question 10

AE < AD

AB > AC

ED < AB

AD > AC

Solution

Question 11

MO = PN

PM = PN

PM = PO

PN = PO

Solution

Question 12

ABE CDE

ABE DEC

ABE ECD

BAE EDC

Solution