ML Aggarwal Solution Class 9 Chapter 10 Triangles Multiple Choice Question

 Multiple Choice Question

Choose the correct answer from the given four options (1 to 18):

Question 1

Which of the following is not a criterion for congruency of triangles?

(a) SAS

(b) ASA

(c) SSA

(d) SSS

Sol :

Criteria of congruency of two triangles ‘SSA’ is not the criterion. (c)


Question 2

In the adjoining figure, AB = FC, EF=BD and ∠AFE = ∠CBD. Then the rule by which ∆AFE = ∆CBD is

(a) SAS

(b) ASA

(c) SSS

(d) AAS









Sol :

In the figure given,

⇒∆AFE≅∆CBD by SAS axiom

⇒AB+BF=BF+FC  (∵AB=FC)

⇒AF=BC
⇒EF=BD
∠AFE=∠CBD 
Ans (b)

Question 3

In the adjoining figure, AB ⊥ BE and FE ⊥ BE. If AB = FE and BC = DE, then

(a) ∆ABD ≅ ∆EFC

(b) ∆ABD ≅ ∆FEC

(c) ∆ABD ≅ ∆ECF

(d) ∆ABD ≅ ∆CEF

Sol :

In the figure given,










⇒AB⊥BE and FE⊥BE

⇒AB=FE, BC+CD=CD+DE  (∵BC=DE)

⇒AB=FE and BD=CE, ∠B=∠E (Each 90°)

∴∆ABD ≅ ∆FEC

Ans (b)


Question 4

In the adjoining figure, AB=AC and AD is median of ∆ABC, then ∆ADC is equal to

(a) 60°

(b) 120°

(c) 90°

(d) 75°

Sol :

In the given figure , AB=AC

⇒AD is median of ΔABC








∴D is mid-point 

⇒BD=DC

∴AD⊥BC

∴∠ADC=90

Ans (c)


Question 5

In the adjoining figure, O is mid point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to

(a) ∠OCA

(b) ∠ODB

(c) ∠OBD

(d) ∠BOD

Sol :
In the given figure, O is mid-point of AB
⇒∠ACO=∠BDO
⇒∠AOC=∠BOD  (Vertically Opposite Angles)





∴ ΔOAC≅ΔOBD (AAS)

∴∠OAC=∠OBD

Ans (c)


Question 6

In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to

(a) ∠BAD

(b) ∠ABC

(c) ∠ABD

(d) ∠BDA










Sol :

In the figure , AC=BD

⇒∠CAB=∠DBA

⇒AB=AB (Common)

∴ΔABC≅ΔABD (SAS axiom)

∴∠ACB=∠BDA  (c.p.c.t)

Ans (d)


Question 7

In the adjoining figure, ABCD is a quadrilateral in which BN and DM are drawn perpendiculars to AC such that BN = DM. If OB = 4 cm, then BD is

(a) 6 cm

(b) 8 cm

(c) 10 cm

(d) 12 cm

Sol :
In the given figure,
⇒ABCD is a quadrilateral








⇒BN⊥AC, DM⊥AC

⇒BN=DM, OB=4 cm

In ΔONB and ΔOMD

⇒BN=DM

⇒∠N=∠M (each 90°)

⇒∠BON=∠DOM (Vertically Opposite Angle)

∴ΔONB≅ΔOMD

∴OB=OD

But OB=4 cm

∴BD=BO+OD

=4+4=8 cm 

Ans (b)


Question 8

In ∆ABC, AB = AC and ∠B = 50°. Then ∠C is equal to

(a) 40°

(b) 50°

(c) 80°

(d) 130°

Sol :
In ΔABC, AB=AC










∴∠C=∠B°  (Angles opposite to equal sides)

∠B=50°

∴∠C=50°


Question 9

In ∆ABC, BC = AB and ∠B = 80°. Then ∠A is equal to

(a) 80°

(b) 40°

(c) 50°

(d) 100°

Sol :
In ΔABC=BC=AB












∴∠A=∠C (Angles opposite to equal sides)

⇒∠B=80°

∴∠A+∠C=180°-80°=100°

But ∠A=∠C=100°

and 2∠A

⇒∠A$=\frac{100^{\circ}}{2}$=50°

Ans (c)


Question 10

In ∆PQR, ∠R = ∠P, QR = 4 cm and PR = 5 cm. Then the length of PQ is

(a) 4 cm

(b) 5 cm

(c) 2 cm

(d) 2.5 cm

Sol :
In ∆PQR
⇒∠R=∠P, QR=4 cm
⇒PR=5 cm












∴∠P=∠R

⇒PQ=QR

∴(Sides opposite to equal angles)

∴PQ=4 cm

Ans (a)


Question 11

In ∆ABC and ∆PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are

(a) isosceles but not congruent

(b) isosceles and congruent

(c) congruent but isosceles

(d) neither congruent nor isosceles

Sol :
In ∆ABC and ∆PQR
⇒AB=AC, ∠C=∠P
⇒∠B=∠Q






∵In ∆ABC, AB=AC

⇒∠C=∠B  (Opposite to equal sides)

But ∠C=∠P and ∠B=∠Q

∴∠P=∠Q

∴RQ=PR

∴ΔRPQ is an isosceles triangle but not congruent.

Ans (a)


Question 12

Two sides of a triangle are of length 5 cm and 1.5 cm. The length of the third side of the triangle can not be

(a) 3.6 cm

(b) 4.1 cm

(c) 3.8 cm

(d) 3.4 cm

Sol :
In a triangle, two sides are 5 and 1.5 cm

∵Sum of any two sides of a triangle is greater than its third side

∴Third side<(5+1.5)

⇒Third side<(6.5 cm)

or third side +1.5>5 cm

or third side >5-1.5=3.5 cm

∴Third side can not be equal to 3.4 cm


Question 13

If a, b, c are the lengths of the sides of a triangle, then

(a) a – b > c

(b) c > a + b

(c) c = a + b

(d) c < A + B

Sol :

a, b, c are the lengths of the sides of a triangle than a + b> c or c < a + b

(Sum of any two sides is greater than its third side) (d)


Question 14

It is not possible to construct a triangle when the lengths of its sides are

(a) 6 cm, 7 cm, 8 cm

(b) 4 cm, 6 cm, 6 cm

(c) 5.3 cm, 2.2 cm, 3.1 cm

(d) 9.3 cm, 5.2 cm, 7.4 cm

Sol :

We know that sum of any two sides of a triangle is greater than its third side 2.2 + 3.1 = 5.3 ⇒ 5.3 = 5.3 is not possible (c)


Question 15

In ∆PQR, if ∠R> ∠Q, then

(a) QR > PR

(b) PQ > PR

(c) PQ < PR

(d) QR < PR

Sol :

In ∆PQR, ∠R> ∠Q

∴ PQ > PR (b)


Question 16

If triangle PQR is right angled at Q, then

(a) PR = PQ

(b) PR < PQ

(c) PR < QR

(d) PR > PQ

Sol :

In right angled ΔPQR,

∠Q=90°

Side opposite to greater angle is greater

∴PR>PQ










Question 17

If triangle ABC is obtuse angled and ∠C is obtuse, then

(a) AB > BC

(b) AB = BC

(c) AB < BC

(d) AC > AB

Sol :
In ΔABC, ∠C is obtuse angle

AB>BC

(Side opposite to greater angle is greater)









Ans (a)


Question P.Q.

A triangle can be constructed when the lengths of its three sides are

(a) 7 cm, 3 cm, 4 cm

(b) 3.6 cm, 11.5 cm, 6.9 cm

(c) 5.2 cm, 7.6 cm, 4.7 cm

(d) 33 mm, 8.5 cm, 49 mm

Sol :

We know that in a triangle, if sum of any two sides is greater than its third side, it is possible to construct it 5.2 cm, 7.6 cm, 4.7 cm is only possible.

Ans (c)


Question P.Q.

A unique triangle cannot be constructed if its

(a) three angles are given

(b) two angles and one side is given

(c) three sides are given

(d) two sides and the included angle is given

Sol :

A unique triangle cannot be constructed if its three angle are given, (a)


Question 18

If the lengths of two sides of an isosceles are 4 cm and 10 cm, then the length of the third side is

(a) 4 cm

(b) 10 cm

(c) 7 cm

(d) 14 cm

Sol :

Lengths of two sides of an isosceles triangle are 4 cm and 10 cm, then length of the third side is 10 cm

(Sum of any two sides of a triangle is greater than its third side and 4 cm is not possible as 4 + 4 > 10 cm.

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