Triangles - Aptitude MCQ Questions and Solutions with Explanations

1

If ABC and PQR are similar triangles in which ∠A = 47° and ∠Q = 83°, then ∠C is:
triangles mcq solution 1

A) 50° B) 70° C) 60° D) 80°

Discuss

Answer & Solution

Answer: Option A

Solution:

Since, ΔABC and ΔPQR are similar triangles.
then, ∠B = ∠Q = 83°
Thus, in ΔABC,
∠C = 180° - (∠A + ∠ B)
or, ∠C = 180° - (47° + 83°)
∠C = 50°

2

In the following figure which of the following statements is true?
triangles mcq question

A) AB = BD B) AC = CD C) BC + CD D) AD < Cd

Discuss

Answer & Solution

Answer: Option B

Solution:

In Triangle ABD,
∠BAD + ∠B + 90° = 180°
or, ∠BAD + ∠B = 90° - - - - - - - (1)

Now, in Triangle ABC,
∠ACB + ∠B + ∠A = 180°
∠ACB + ∠B = 180° - 90°
∠ACB = 90° - ∠B - - - - - (2)
From (1) and (2), ∠BAD = ∠ACB
So, AC = CD

3

In triangle PQR length of the side QR is less than twice the length of the side PQ by 2 cm. Length of the side PR exceeds the length of the side PQ by 10 cm. The perimeter is 40 cm. The length of the smallest side of the triangle PQR is :

A) 6 cm B) 8 cm C) 7 cm D) 10 cm

Discuss

Answer & Solution

Answer: Option B

Solution:

In δ PQR,
QR + 2 = 2PQ
QR = 2PQ - 2 - - - - - - - (1)
PR = PQ + 10 - - - - - - (2)
Perimeter = 40 m
PQ + QR + Rp = 40
Putting the value of PQ and QR from equation (1) and (2),
PQ + 2PQ - 2 + PQ + 10 = 40
4PQ = 32
PQ = 8 cm which is the smallest side of the triangle.

4

AB and CD bisect each other at O. If AD = 6 cm. Then BC is :
triangles mcq question 3

A) 5.9 cm B) 4.8 cm C) 6 cm D) 7 cm

Discuss

Answer & Solution

Answer: Option C

Solution:

Since, AB and CD bisects each other at O,
Hence, BC = AD = 6 cm.

5

In a triangle ABC,∠ A = 90°, AL is drawn perpendicular to BC, Then ∠BAL is equal to:
Triangles mcq question image

A) ∠ALC B) ∠ACB C) ∠BAC D) ∠B - ∠BAL

Discuss

Answer & Solution

Answer: Option B

Solution:

∠BAL + ∠B + 90° = 180°
or, ∠BAL + ∠B = 90°
or, ∠BAL = 90° - ∠B - - - - - - - - (1)
Now, in ΔABC,
∠ACB + ∠B + ∠A = 180°
∠ACB = 90°
-∠B - - - - - (2)
From, (1) and (2),
∠BAL = ∠ACB

6

Consider the triangle shown in the figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC.
What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC ?
triangles mcq Aptitude question4

A) 7 : 9 B) 8 : 9 C) 6 :9 D) 5 : 9 E) None of these

Discuss

Answer & Solution

Answer: Option A

Solution:

Here, ∠ACB = c + 180 - (2c - b) = 180 - (b + c)
So, We can say that ΔBCD and ΔABC will be similar.
According to property of similarity,
$$\frac{{{\text{AB}}}}{{12}} = \frac{{12}}{9}$$
Hence,
AB = 16
$$\frac{{{\text{AC}}}}{6} = \frac{{12}}{9}$$
AC = 8
Hence, AD = 7 and AC = 8
Now,
$$\eqalign{ & \frac{{{\text{Perimeter of Delta ADC}}}}{{{\text{Perimeter of Delta BDC}}}} \cr & = \frac{{6 + 7 + 8}}{{9 + 6 + 12}} \cr & = \frac{{21}}{{27}} \cr & = \frac{7}{9} \cr} $$

7

In a triangle ABC, the internal bisector of the angle A meets BC at D. If AB = 4, AC = 3 and ∠A = 60°, then length of AD is :

A) $$2\sqrt 3 $$ B) $$\frac{{12\sqrt 3 }}{7}$$ C) $$\frac{{15\sqrt 3 }}{8}$$ D) $$\frac{{6\sqrt 3 }}{7}$$ E) None of these

Discuss

Answer & Solution

Answer: Option B

Solution:

Let BC = x and Ad = y, then as per bisector theorem,
$$\frac{{{\text{BD}}}}{{{\text{DC}}}} = \frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{4}{3}$$
Hence, BD = $$\frac{{4{\text{x}}}}{7}$$ and DC = $$\frac{{3{\text{x}}}}{7}$$
Now, in ΔABD using cosine rule,
$$\cos {30^ \circ } = \frac{{{4^2} + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,4\sqrt {3{\text{y}}} = {16 + {{\text{y}}^2} - {\frac{{16{{\text{x}}^2}}}{{49}}} } $$ - - - - - - (i)
Similarly in ΔADC,
$$\cos {30^ \circ } = \frac{{{3^2} + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} }}{{2 \times 3 \times {\text{y}}}}$$
$${\text{or,}}\,3\sqrt {3{\text{y}}} = {9 + {{\text{y}}^2} - {\frac{{9{{\text{x}}^2}}}{{49}}} } $$ - - - - - - - - (ii)
From equation (i) and (ii), we get
$${\text{y}} = \frac{{12\sqrt 3 }}{7}$$

8

The point of intersection of the altitudes of a triangle is called its:

A) Incentre B) Excentre C) Orthocentre D) Centroid

Discuss

Answer & Solution

Answer: Option C

Solution:

Orthocentre.

9

In ΔPQR, PS is the bisector of ∠P and PT ⊥ OR, then ∠TPS is equal to:

A) ∠Q + ∠R B) 90° + $$\frac{1}{2}$$ ∠Q C) 90° - $$\frac{1}{2}$$ ∠R D) $$\frac{1}{2}$$ (∠Q - ∠R)

Discuss

Answer & Solution

Answer: Option D

Solution:

∠1 + ∠2 = ∠3 [PS is bisector.] - - - - - - (1)
∠Q = 90° - ∠1
∠R = 90° -∠2 - ∠3
So,
∠Q - ∠R = (90° - ∠1) - (90° - ∠2 - ∠3)
∠Q - ∠R = ∠2 + ∠3 - ∠1
∠Q - ∠R = ∠2 + (∠1 + ∠2) -∠1[using equation 1]
∠Q - ∠R = 2∠2
$$\frac{1}{2}$$ × (∠Q - ∠R) = ∠TPS

10

Two right angled triangles are congruent if :
I. The hypotenuse of one triangle is equal to the hypotenuse of the other.
II. A side for one triangle is equal to the corresponding side of the other.
III. Sides of the triangles are equal.
IV. An angle of the triangle are equal.
Of these statements, the correct ones are combination of:

A) I and II B) II and III C) I and III D) IV only

Discuss

Answer & Solution

Answer: Option A

Solution:

Two right angled triangle are congruent if the hypotenuse of one triangle is equal to the hypotenuse of the other and a side of one triangle is equal to the corresponding side of the other triangle.