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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-6 section-1

Each section contains 100 MCQs on Triangles .
51
In a triangle, if three altitudes are equal, then the triangle is
Discuss
Answer & Solution
Answer: Option B
Solution:
If three altitudes are equal then the triangle is equilateral
52
In an isosceles triangle, if the unequal angle is twice the sum of the equal angles, then each equal angle is
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Given :
Triangles mcq solution image
AB = AC
y = 2(x + x)
y = 4x
∴ x + x + y = 180°
2x + 4x = 180°
6x = 180°
x = $$\frac{{{{180}^ \circ }}}{6}$$
x = 30°
53
If the length of the sides of a triangle are in the ratio 4 : 5 : 6 and the inradius of the triangle is 3 cm, then the altitude of the triangle corresponding to the largest side as base is :
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
Area of Δ(OBA + OAC + OBC) = area of ΔABC
$$\frac{1}{2}$$ × 4x × 3 + $$\frac{1}{2}$$ × 5x × 3 + $$\frac{1}{2}$$ × 6x × 3 = $$\frac{1}{2}$$ × (6x × AD)
$$\frac{1}{2}$$ × 3(4x + 5x + 6x) = $$\frac{1}{2}$$ × (6x × AD)
45x = 6x × AD
AD = $$\frac{{15}}{2}$$
AD = 7.5 cm
54
In a right-angle ΔABC, ∠ABC = 90°, AB = 5 cm and BC = 12 cm. The radius of the circumcircle of the triangle ABC is
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
ABC is a right angled triangle
Triangles mcq solution image
∴ By using Pythagoras theorem
AC2 = AB2 + BC2
AC2 = (5)2 + (12)2
AC2 = 25 + 144
AC2 = 169
AC = $$\sqrt {169} $$
AC = 13 cm
BD = IR = Circumradius = $$\frac{{AC}}{2}$$
∴ IR = $$\frac{{13}}{2}$$
IR = 6.5 cm
55
In triangle PQR, points A, B and C are taken on PQ, PR and QR respectively such that QC = AC and CR = CB. If ∠QPR = 40°, then ∠ACB is equal to:
Discuss
Answer & Solution
Answer: Option D
Solution:
According to question,
Triangles mcq solution image
In ΔPQR
x + y + 40° = 180°
x + y = 140° . . . . . . . (i)
In ΔAQC
x + x + ∠C = 180°
∠C = 180° - 2x . . . . . . . . (ii)
In ΔBCR
y + y + ∠C = 180°
∠C = 180° - 2y . . . . . . . . (iii)
But ∠ACB = 180° - 180° + 2x - 180° +2y
∠ACB = 2x + 2y - 180°
∠ACB = 2(x + y) - 180° . . . . . . . . . (iv)
Put the value of equation (i) in equation (iv)
∠ACB = 2 × 140° - 180°
∠ACB = 280° - 180°
∠ACB = 100°
56
If ABC is an equilateral triangle and D is a point of BC such that AD ⊥ BC, then
Discuss
Answer & Solution
Answer: Option C
Solution:
According to question,
Triangles mcq solution image
In equilateral triangle 'AD' bisects the BC in two equal parts.
Let side of equilateral triangle is 2 cm
$$\eqalign{ & \therefore \frac{{AB}}{{BD}} = \frac{2}{1} \cr & AB:BD = 2:1 \cr} $$
57
ΔABC is an isosceles triangle and $$\overline {AB} $$  = $$\overline {AC} $$  = 2a unit, $$\overline {BC} $$  = a unit. Draw $$\overline {AD} $$ ⊥ $$\overline {BC} $$ , and find the length of $$\overline {AD} $$
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Given :
Triangles mcq solution image
AB = AC = 2a
BC = a
AD ⊥ BC
In isosceles triangle perpendicular sides bisects the opposite side of the length
∴ BD = $$\frac{{BC}}{2}$$
   BD = $$\frac{a}{2}$$
In ΔADB using Pythagoras theorem
AB2 = BD2 + AD2
(2a)2 = $${\left( {\frac{a}{2}} \right)^2}$$ + AD2
4a2 = $$\frac{{{a^2}}}{4}$$ + AD2
AD2 = 4a2 - $$\frac{{{a^2}}}{4}$$
AD = $$\frac{{\sqrt {15} }}{2}$$ a units
58
ABC is a triangle. The bisectors of the internal angle ∠B and external angle ∠C intersect at D. If ∠BDC = 50°, then ∠A is
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Triangles mcq solution image
Given :
∠D = 50°
∠BAC = 2∠BDC (property)
∴ ∠BAC = 2 × 50°
∠BAC = 100°
59
AD is the median of a triangle ABC and O is the centroid such that AO = 10 cm. The length of OD (in cm) is
Discuss
Answer & Solution
Answer: Option B
Solution:
According to question,
Triangles mcq solution image
AD is the median and 'C' is the centroid
∴ AO = 10 cm
2 units = 10
1 units = 5
∴ OD = 5 cm
60
The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of ∠PSR is
Discuss
Answer & Solution
Answer: Option A
Solution:
According to question,
Given :
Triangles mcq solution image
PQR is an equilateral triangle
QR = RS
PR = RS
∠SRP = 180° - 60° (Exterior ∠)
∠SRP = 120°
∴ ∠RPS = ∠RSP
∴ ∠RPS + ∠PRS + ∠RSP = 180°
2∠PSR = 180° - 120°
∠PSR = $$\frac{{{{60}^ \circ }}}{2}$$
∠PSR = 30°