Triangles(Quantitative Aptitude ) Questions and answers for exam Preparation

Quantitative Aptitude

TRIANGLES MCQs

Total Questions : 83 | Page 3 of 9 pages

Question 21. 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

1.5 cm
2.0 cm
1.8 cm
2.4 cm

Discuss Question

Answer: Option A. -> 1.5 cm
According to question,
Given :
AD = 3
BD = 5
AB = 8
AC = 4
AE = ?
By applying B. P. T
$$\eqalign{
& \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr
& \frac{3}{8} = \frac{{AE}}{4} \cr
& AE = \frac{3}{2} = 1.5\,{\text{cm}} \cr} $$

Question 22. 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 :

100°
80°
130°
90°

Discuss Question

Answer: Option A. -> 100°
We know that
⇒ Add of total exterior angle of a triangle (polygon) = 360°
⇒ So, 130° + 130° + x° = 360°
x° = 100°

Question 23. 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

2 : 3
2 : 1
1 : 2
1 : 3

Discuss Question

Answer: Option C. -> 1 : 2
According to question,
Given :
AB = 2AD
$$\frac{{AB}}{{AD}} = \frac{2}{1}$$
By applying B. P. T
$$\eqalign{
& \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} = \frac{{AE}}{{AC}} \cr
& \frac{{DE}}{{BC}} = \frac{1}{2} \cr
& \therefore DE:BC = 1:2 \cr} $$

Question 24. 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 :

5 cm
4 cm
3 cm
6 cm

Discuss Question

Answer: Option C. -> 3 cm
According to question,
By using B.P.T
$$\eqalign{
& \frac{{AD}}{{AB}} = \frac{{AE}}{{AC}} = \frac{{DE}}{{BC}} \cr
& \frac{{AD}}{{AB}} = \frac{{DE}}{{BC}} \cr
& \Rightarrow \,\frac{1}{4} = \frac{{DE}}{{12}} \cr
& \Rightarrow DE = 3\,{\text{cm}} \cr} $$

Question 25. In a ΔABC, ∠A + ∠B = 118°, ∠A + ∠C = 96°. Find the value of ∠A.

36°
40°
30°
34°

Discuss Question

Answer: Option D. -> 34°
According to question,
   ∠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°

Question 26. In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a :

Right isosceles triangle
Isosceles triangle
Right-angled triangle
Equilateral triangle

Discuss Question

Answer: Option A. -> Right isosceles triangle
∵ AB = BC = K
⇒ AC = $$\sqrt 2 $$ K
⇒ (AC)2 = (AB)2 + (BC)2
⇒ ($$\sqrt 2 $$ K)2 = K2 + K2
⇒ 2K2 = 2K2
⇒ Therefore ΔABC will be a right isosceles triangle.

Question 27. In ΔABC, ∠BAC = 90° and AD ⊥ BC. If BD = 3 cm and CD = 4 cm, then length of AD is :

$$2\sqrt 3 $$ cm
3.5 cm
6 cm
5 cm

Discuss Question

Answer: Option A. -> $$2\sqrt 3 $$ cm
We know that,
AD2 = CD × BD
AD2 = 4 × 3
AD2 = 12
AD = $$2\sqrt 3 $$ cm

Question 28. In ΔABC, AD ⊥ BC and AD2 = BD × DC. The measure of ∠BAC is :

75°
90°
45°
60°

Discuss Question

Answer: Option B. -> 90°
AD2 = BD.DC
ΔADC ∼ ΔCAB (Property of a right angle Δ)
∠BAC = ∠ADC = 90°

Question 29. Let ABC be an equilateral triangle and AD perpendicular to BC, then AB2 + BC2 + CA2 = ?

3AD2
5AD2
2AD2
4AD2

Discuss Question

Answer: Option D. -> 4AD2
AB2 = AD2 + BD2 . . . . . . . (i)
AC2 = AD2 + CD2 . . . . . . . (ii)
AB2 + AC2 = 2AD2 + BD2 + CD2
AB2 + AC2 + BC2 = 2AD2 + a2 + $$\frac{{{{\text{a}}^2}}}{4}$$ + $$\frac{{{{\text{a}}^2}}}{4}$$
AB2 + AC2 + BC2 = 4AD2
$$\left( {{{\text{a}}^2} - \frac{{{{\text{a}}^2}}}{4} = {\text{A}}{{\text{D}}^2}} \right)$$

Question 30. In a ΔABC, If 2∠A = 3∠B = 6∠C, then the value of ∠B is:

60°
30°
45°
90°

Discuss Question

Answer: Option A. -> 60°
According to question,
Given :
$$\eqalign{
& 2\angle A = 3\angle B \cr
& \frac{{\angle A}}{{\angle B}} = \frac{3}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,3\angle B = 6\angle C \cr
& \frac{{\angle B}}{{\angle C}} = \frac{6}{3} = \frac{2}{1} \cr} $$
To make angle ∠B same
∴ ∠A : ∠B : ∠C
3 : 2 : 1
As we know that
∠A + ∠B + ∠C = 180°
3x + 2x + x = 180°
x = 30°
∠B = 2x
∠B = 60°