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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-10 section-1

Each section contains 100 MCQs on Triangles .
91
If the measures of the sides of triangle are (x2 - 1), (x2 + 1) and 2x cm, then the triangle would be :
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
Sides AB = x2 - 1
          BC = 2x
          AC = x2 + 1
By using Pythagoras theorem
AC2 = AB2 + BC2
(x2 + 1)2 = (x2 - 1)2 + (2x)2
x4 + 1 + 2x2 = x2 + 1 - 2x2 + 4x2
(x2 + 1)2 = (x2 + 1)2
∴ The triangle is right angle Δ
92
In ΔABC, ∠C is an obtuse angle. The bisectors of the exterior angles at A and B meet BC and AC produced at D and E respectively. If AB = AD = BE, then ∠ACB = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Triangles mcq solution image
Let ∠CAB = x and ∠CBA = y
$$\eqalign{ & \Rightarrow \angle CAD = \frac{{180 - x}}{2} = 90 - \frac{x}{2} \cr & {\text{and}} \cr & \Rightarrow \angle EBC = \frac{{180 - y}}{2} = 90 - \frac{y}{2} \cr & {\text{also}}\,\angle AEB = \angle EAB = x \cr} $$
(∵ AB = EB ⇒ ABE is an isosceles triangle)
and ∠ADB = ∠ABD = y
(∵ AB = AD ⇒ ADB is an isosceles triangle)
In ΔAEB,
∠AEB + ∠ABE + ∠BAE = 180°
x + x + y + 90 - $$\frac{y}{2}$$ = 180°
⇒ 4x + y = 180°
Similarly in ΔADB
4y + x = 180°
⇒ 4y + x + 4x + y = 180 + 180
⇒ 5x + 5y = 360°
⇒ x + y = 72°
In triangle ABC,
∠ACB + x + y = 180°
⇒ ∠ACB = 180 - 72
⇒ ∠ACB = 108°
93
Let ABC be an equilateral triangle and AX, BY, CZ be the altitude. Then the right statement out of the four give responses is :
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
In an equilateral triangle
AB = BC = AC
∠A = ∠B = ∠C = 60°
∴ AX = BY = CZ
(All altitudes are same in an equilateral triangles)
94
In ΔABC, DE || AC, D and E are two points on AB and CB respectively. If AB = 10 cm and AD = 4 cm, then BE : CE is
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
Triangles mcq solution image
Given :
AB = 10 cm
AD = 4 cm
DE || AC
ΔABC ∼ ΔDBE
$$\eqalign{ & \therefore \frac{{BD}}{{AD}} = \frac{{BE}}{{CE}} \cr & \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{6}{4} \cr & \,\,\,\,\,\,\frac{{BE}}{{CE}} = \frac{3}{2} \cr & \,\,\,\,\,\,BE:CE = 3:2 \cr} $$
95
If the three angles of a triangle are: $${\left(x + 15 \right)^ \circ },$$   $${\left({\frac{{6x}}{5} + 6} \right)^ \circ }$$  and $${\left({\frac{{2x}}{3} + 30} \right)^ \circ }$$   then the triangle is:
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
 $$ \Rightarrow \left( {x + {{15}^ \circ }} \right) + {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } + $$      $${\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = $$    $${180^ \circ }$$
 $$\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\angle A + \angle B + \angle C = {{180}^ \circ }} \right\}$$
 $$ \Rightarrow x + \frac{{6x}}{5} + \frac{{2x}}{3} = $$     $${180^ \circ } - \left(15 + 6 + 30\right)$$
$$\eqalign{ & \Rightarrow \frac{{15x + 18x + 10x}}{{15}} = 180 - 51 \cr & \Rightarrow 43x = 129 \times 15 \cr & \Rightarrow x = {45^ \circ } \cr & \Rightarrow {\text{each}}\,{\text{angle}} \cr & \Rightarrow {\left( {x + 15} \right)^ \circ } = 45 + 15 = {60^ \circ } \cr & \Rightarrow {\left( {\frac{{6x}}{5} + 6} \right)^ \circ } = {60^ \circ } \cr & \Rightarrow {\left( {\frac{{2x}}{3} + 30} \right)^ \circ } = {60^ \circ } \cr} $$
∵ All three angles are equal 60°
⇒ Triangle will be equilateral triangle
96
If in a triangle ABC, D and E are on the sides AB and AC, such that, DE is parallel to BC and $$\frac{{AD}}{{BD}}$$ = $$\frac{3}{5}$$. If AC = 4 cm, then AE is
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
Given :
AD = 3
BD = 5
AB = 8
AC = 4
AE = ?
By applying B. P. T
$$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr & \frac{3}{8} = \frac{{AE}}{4} \cr & AE = \frac{3}{2} = 1.5\,{\text{cm}} \cr} $$
97
In a ΔABC, ∠A + ∠B = 118°, ∠A + ∠C = 96°. Find the value of ∠A.
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
Triangles mcq solution image
   ∠A + ∠B = 118°
   ∠A + ∠C = 96°
   ∠A = ?
As we know that
   ∠A + ∠B + ∠C = 180°
   ∠C = 180° - (∠A + ∠B)
   ∠C = 180° - 118°
∴ ∠C = 62°
   ∠A = 96° - 62°
   ∠A = 34°
98
For a triangle ABC, D and E are two points on AB and AC such that AD = $$\frac{1}{4}$$ AB, AE = $$\frac{1}{4}$$ AC. If BC = 12 cm, then DE is :
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
By using B.P.T
$$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr & \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} \cr & \Rightarrow \,\frac{1}{4} = \frac{{DE}}{{12}} \cr & \Rightarrow DE = 3\,{\text{cm}} \cr} $$
99
In triangle ABC a straight line parallel to BC intersects AB and AC at D and E respectively. If AB = 2AD, then DE : BC is
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
Given :
AB = 2AD
$$\frac{{AB}}{{AD}} = \frac{2}{1}$$
By applying B. P. T
$$\eqalign{ & \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} = \frac{{AE}}{{AC}} \cr & \frac{{DE}}{{BC}} = \frac{1}{2} \cr & \therefore DE:BC = 1:2 \cr} $$
100
ABC is a triangle and the sides AB, BC and CA are produced to E, F and G respectively. If ∠CBE = ∠ACF = 130°, then the value of ∠GAB is :
Discuss
Answer & Solution
Answer: Option A
Solution:
We know that
⇒ Add of total exterior angle of a triangle (polygon) = 360°
Triangles mcq solution image
⇒ So, 130° + 130° + x° = 360°
x° = 100°