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Triangles - Aptitude MCQ Questions and Solutions with Explanations

Triangles MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Triangles with easy and logical explanations. Triangles MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-9 section-1

Each section contains 100 MCQs on Triangles .
81
If angle bisector of a triangle bisects the opposite side, then what type of triangle is it?
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
AB = AC
BD = DC
The triangle will be isosceles and equilateral triangle
82
If the sides of a right angled triangle are three consecutive integers, then the length of the smallest side is
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
ABC is a right angle triangle
By using Pythagoras theorem
AC2 = BC2 + AB2
52 = 32 + 42
25 = 9 + 16
25 = 25 (satisfied)
∴ Smallest length of right angle triangle is 3 units
83
In a triangle ABC, BC is produced to D so that CD = AC. If ∠BAD = 111° and ∠ACB = 80°, then the measure of ∠ABC is:
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
Triangles mcq solution image
Given :
AC = CD
∠BAD = 111°
∠ACB = 80°
∴ ∠ACD = 180° - 80°
∠ACD = 100°
In isosceles triangle ACD
   ∠ACD + ∠CAD + ∠ADC = 180°
   2∠CAD = 180° - 100°
   ∠CAD = 40°
∴ ∠CAB = 111° - 40° = 71°
∴ ∠ABC = 180° - 71° - 80°
   ∠ABC = 29°
84
In a ΔABC, AB = BC, ∠B = x° and ∠A = (2x - 20)°, Then ∠B is :
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
Triangles mcq solution image
Given :
AB = AC
∠C = ∠A = 2x - 20°
∠B = x°
As we know that
∠A + ∠B + ∠C = 180°
(2x - 20)° + x + (2x - 20)° = 180°
5x = 220°
x = 44°
∴ ∠B = 44°
85
If two angles of a triangle are 21° and 38°, then the triangle is :
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
Given :
∠A = 21°,       ∠C = 38°
As we know that
∠A + ∠B + ∠C
∠B = 180° - 21° - 38°
∠B = 121°
∴ The triangle is obtuse-angled triangle.
86
In ΔPQR, S and T are point on sides PR and PQ respectively such that ∠PQR = ∠PST, If PT = 5 cm, PS = 3 cm and TQ = 3 cm, then length of SR is
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
Given :
PT = 5 cm
PS = 3 cm
TQ = 3 cm
SR = ?
ΔPQR ∼ ΔPST
Triangles mcq solution image
$$\eqalign{ & \frac{{PR}}{{PT}} = \frac{{PQ}}{{PS}} \cr & \frac{{PR}}{5} = \frac{8}{3} \cr & PR = \frac{{40}}{3} \cr & \therefore SR = PR - PS \cr & SR = \frac{{40}}{3} - 3 \cr & SR = \frac{{40 - 9}}{3} \cr & SR = \frac{{31}}{3}\,{\text{cm}} \cr} $$
87
In a ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is :
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
ABC is an isosceles triangle.
Triangles mcq solution image
∴ ∠C = ∠B = θ
∠CAD = ∠C + ∠B
∠CAD = θ + θ
∠CAD = 2θ
ADC is a isosceles triangle
∠C + ∠D + ∠A = 180°
2∠C = 180° - 2θ°
(∠C = ∠D)
∠C = 90° - θ
∴ ∠BCD = θ + 90 - θ
∠BCD = 90°
88
In ΔABC, two points D and E are taken on the lines AB and BC respectively in such a way that AC is parallel to DE. Then ΔABC and ΔDBE are :
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Give :
'D' and 'E' are the points on AB and BC
Triangles mcq solution image
AC || DE
∠D = ∠A
∠E = ∠C
∴ ΔBDE ∼ ΔBAC
89
If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
Given : P, Q and R are the mid points of AB, BC and AC
PQ || AC and PQ = $$\frac{1}{2}$$ AC
PR || BC and PR = $$\frac{1}{2}$$ BC
RQ || AB and RQ = $$\frac{1}{2}$$ AB
              (mid point theorem)
∴ ΔPQR is an equilateral triangle
90
In ΔABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Triangles mcq solution image
Given :
∠A + ∠B = 65°
∠B + ∠C = 140°
We know that
∠A + ∠B + ∠C = 180°
∠C = 180° - (∠A + ∠B)
∠C = 180° - 65°
∠C = 115°
∠B = 140° - 115°
∠B = 25°