Triangles(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

TRIANGLES MCQs

Total Questions : 83 | Page 8 of 9 pages

Question 71. Let ΔABC and ΔABD be on the same base AB and between the same parallels AB and CD. Then the relation between areas of triangles ABC and ABD will be

ΔABD = $$\frac{1}{3}$$ ΔABC
ΔABD = $$\frac{1}{2}$$ ΔABC
ΔABC = $$\frac{1}{2}$$ ΔABD
ΔABC = ΔABD

Discuss Question

Answer: Option D. -> ΔABC = ΔABD
The height of ΔABC and ΔABD are same and have same base.
∴ Area ΔABC = Area ΔABD

Question 72. G is the centroid of ΔABC. If AB = BC = AC, then measure of ∠BGC is:

45°
60°
90°
120°

Discuss Question

Answer: Option D. -> 120°
∴ AB = BC = CA
∠BAC = 60°
So, ∠BGC = 90 + $$\frac{{60}}{2}$$
∠BGC = 120°

Question 73. In a triangle ABC, if ∠A + ∠C = 140° and ∠A + 3∠B = 180°, then ∠A is equal to:

80°
40°
60°
20°

Discuss Question

Answer: Option C. -> 60°
∠A + ∠B + ∠C = 180°
(for a triangle)
∠A + ∠C = 140°
then, ∠B = 40°
∠A + 3∠B = 180°
∠A + 120° = 180°
⇒ ∠A = 60°

Question 74. 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

5 cm
6 cm
$$\frac{{31}}{3}$$ cm
$$\frac{{41}}{3}$$ cm

Discuss Question

Answer: Option C. -> $$\frac{{31}}{3}$$ cm
According to question,
Given :
PT = 5 cm
PS = 3 cm
TQ = 3 cm
SR = ?
ΔPQR ∼ ΔPST
$$\eqalign{
& \frac{{PR}}{{PT}} = \frac{{PQ}}{{PS}} \cr
& \frac{{PR}}{5} = \frac{8}{3} \cr
& PR = \frac{{40}}{3} \cr
& \therefore SR = PR - PS \cr
& SR = \frac{{40}}{3} - 3 \cr
& SR = \frac{{40 - 9}}{3} \cr
& SR = \frac{{31}}{3}\,{\text{cm}} \cr} $$

Question 75. 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 :

Similar only if D lies outside the line segment AB
Congruent only If D lies out side the line segment AB
Always similar
Always congruent

Discuss Question

Answer: Option C. -> Always similar
According to question,
Give :
'D' and 'E' are the points on AB and BC
AC || DE
∠D = ∠A
∠E = ∠C
∴ ΔBDE ∼ ΔBAC

Question 76. In ΔABC, ∠A + ∠B = 65°, ∠B + ∠C = 140°, then find ∠B.

40°
25°
35°
20°

Discuss Question

Answer: Option B. -> 25°
According to question,
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°

Question 77. In a ΔABC, AB = AC and BA is produced to D such that AC = AD. Then the ∠BCD is :

100°
60°
80°
90°

Discuss Question

Answer: Option D. -> 90°
According to question,
ABC is an isosceles triangle.
∴ ∠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°

Question 78. If ABC is an equilateral triangle and P, Q, R respectively denote the middle points of AB, BC, CA then

PQR must be an equilateral triangle
PQ + QR = PQR + AB
PQ + QR = PR + 2AB
PQR must be a right angled

Discuss Question

Answer: Option A. -> PQR must be an equilateral triangle
According to question,
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

Question 79. ABC is a triangle in which ∠A = 90°. Let P be any point on side AC. If BC = 10 cm, AC = 8 cm and BP = 9 cm, then AP = ?

$$2\sqrt 5 $$ cm
$$3\sqrt 5 $$ cm
$$2\sqrt 3 $$ cm
$$3\sqrt 3 $$ cm

Discuss Question

Answer: Option B. -> $$3\sqrt 5 $$ cm
According to question,
BC = 10 cm, AC = 8 cm, BP = 9 cm
In ΔCAB,
BC2 = AB2 + AC2
102 = AB2 + 82
AB2 = 100 - 64
AB2 = 36
AB = 6 cm
In ΔPAB
BP2 = AB2 + AP2
92 = 62 + AP2
AP2 = 81 - 36
AP2 = 45
AP = $$3\sqrt 5 $$ cm

Question 80. ∠A + $$\frac{1}{2}$$ ∠B + ∠C = 140°, then ∠B is :

50°
80°
40°
60°

Discuss Question

Answer: Option B. -> 80°
According to question,
∠A + $$\frac{1}{2}$$ ∠B + ∠C = 140° . . . . . . . . (i)
As we know that
∠A + ∠B + ∠C = 180° . . . . . . . . . (ii)
Compare equation (i) and (ii)
$$\frac{1}{2}$$ ∠B = 40°
∴ ∠B = 80°