Triangles - Aptitude MCQ Questions and Solutions with Explanations

11

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

A) $$2\sqrt 3 $$ cm B) 3.5 cm C) 6 cm D) 5 cm

Discuss

Answer & Solution

Answer: Option A

Solution:

We know that,
AD² = CD × BD
AD² = 4 × 3
AD² = 12
AD = $$2\sqrt 3 $$ cm

12

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

A) 3AD² B) 5AD² C) 2AD² D) 4AD²

Discuss

Answer & Solution

Answer: Option D

Solution:

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
$$\left( {{{\text{a}}^2} - \frac{{{{\text{a}}^2}}}{4} = {\text{A}}{{\text{D}}^2}} \right)$$

13

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

A) 60° B) 30° C) 45° D) 90°

Discuss

Answer & Solution

Answer: Option A

Solution:

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°

14

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

A) 75° B) 90° C) 45° D) 60°

Discuss

Answer & Solution

Answer: Option B

Solution:

AD² = BD.DC
ΔADC ∼ ΔCAB (Property of a right angle Δ)
∠BAC = ∠ADC = 90°

15

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

A) Right isosceles triangle B) Isosceles triangle C) Right-angled triangle D) Equilateral triangle

Discuss

Answer & Solution

Answer: Option A

Solution:

∵ 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.

16

In ΔABC and ΔPQR, ∠B = ∠Q, ∠C = ∠R. M is the midpoint on QR, If AB : PQ = 7 : 4, then $$\frac{{{\text{area}}\,\left( {\vartriangle ABC} \right)}}{{{\text{area}}\,\left( {\vartriangle PMR} \right)}}$$ is :
Triangles mcq question image

A) $$\frac{{35}}{8}$$ B) $$\frac{{35}}{{16}}$$ C) $$\frac{{49}}{{16}}$$ D) $$\frac{{49}}{8}$$

Discuss

Answer & Solution

Answer: Option D

Solution:

$$\eqalign{ & \frac{{{\text{area}}{\mkern 1mu} \left( {\vartriangle ABC} \right)}}{{{\text{area}}{\mkern 1mu} \left( {\vartriangle PMR} \right)}} \cr & = \frac{{{{\left( 7 \right)}^2}}}{{\frac{1}{2} \times {{\left( 4 \right)}^2}}} \cr & = \frac{{49}}{8} \cr} $$

17

In ΔABC, ∠B = 70° and ∠C = 30°, AD and AE are respectively the perpendicular on side BC and bisector of ∠A. The measure of ∠DAE is:

A) 24° B) 10° C) 15° D) 20°

Discuss

Answer & Solution

Answer: Option D

Solution:

∠A = 180° - (∠B + ∠C)
∠A = 180° - 100°
∠A = 80°
∠BAE = ∠EAC = $$\frac{1}{2}$$ ∠A = 40°
In ΔBAD
∠BAD = 90° - ∠B
∠BAD = 90° - 70°
∠BAD = 20°
∠DAE = ∠BAE - ∠BAD
∠DAE = 40° - 20°
∠DAE = 20°

18

In ΔABC and ΔDEF, if ∠A = 50°, ∠B = 70°, ∠C = 60°, ∠D = 60°, ∠E = 70° and ∠F = 50°, then

A) ΔABC ∼ ΔFED B) ΔABC ∼ ΔDFE C) ΔABC ∼ ΔEDF D) ΔABC ∼ ΔDEF

Discuss

Answer & Solution

Answer: Option A

Solution:

From figure it is clear
= ΔABC ∼ ΔFED

19

In ΔABC, ∠B = 60° and ∠C = 40°; AD and AE are respectively the bisector of ∠A and perpendicular on BC. The measure of ∠EAD is:

A) 9° B) 11° C) 10° D) 12°

Discuss

Answer & Solution

Answer: Option C

Solution:

As we know
Triangles mcq solution image

$$\eqalign{ & \angle EAD = \frac{{\angle B - \angle C}}{2} \cr & \angle EAD = \frac{{60 - 40}}{2} \cr & \angle EAD = {10^ \circ } \cr} $$

20

In ΔABC, the line parallel to BC intersect AB & AC at P & Q respectively. If AB : AP = 5 : 3, then AQ : QC is:

A) 3 : 2 B) 1 : 2 C) 3 : 5 D) 2 : 3

Discuss

Answer & Solution

Answer: Option A

Solution:

$$\eqalign{ & \frac{{AP}}{{PB}} = \frac{{AQ}}{{QC}} \cr & \frac{{AQ}}{{QC}} = \frac{3}{2} \cr & AQ:QC = 3:2 \cr} $$