Triangles - Aptitude MCQ Questions and Solutions with Explanations

61

An isosceles triangle ABC is right-angled at B. D is a point inside the triangle ABC. P and Q are the feet of the perpendiculars drawn from D on the side AB and AC respectively of ΔABC. If AP = a cm, AQ = b cm and ∠BAD = 15°, sin 75° = ?

A) $$\frac{{2b}}{{\sqrt 3 a}}$$ B) $$\frac{a}{{2b}}$$ C) $$\frac{{\sqrt 3 a}}{{2b}}$$ D) $$\frac{{2a}}{{\sqrt 3 b}}$$

Discuss

Answer & Solution

Answer: Option C

Solution:

According to question from ΔAQD,
Triangles mcq solution image

∠A = $$\frac{{180 - 90}}{2}$$
∠A = 45°
∠DAQ = 30°
sin 60° = $$\frac{{AQ}}{{AD}}$$
$$\frac{{\sqrt 3 }}{2}$$ = $$\frac{b}{{AD}}$$
AD = $$\frac{{2b}}{{\sqrt 3 }}$$
From ΔAPD
$$\eqalign{ & \sin {75^ \circ } = \frac{{AP}}{{AD}} \cr & \sin {75^ \circ } = \frac{a}{{2b}} \times \sqrt 3 \cr & \sin {75^ \circ } = \frac{{\sqrt 3 a}}{{2b}} \cr} $$

62

In a triangle ABC, the side BC is extended up to D such that CD = AC. If ∠BAD = 109° and ∠ACB = 72° then the value of ∠ABC is

A) 35° B) 60° C) 40° D) 45°

Discuss

Answer & Solution

Answer: Option A

Solution:

According to question,
Given :
Triangles mcq solution image

   ∠BAD = 109°
   ∠ACB = 72°
∴ ∠ACD = 180° - 72°
   ∠ACD = 108°
∴      AC = CD
   ∠CAD = ∠CDA
In ΔCDA
∠CAD + ∠CDA + ∠DCA = 180°
2∠CAD + 108° = 180°
2∠CAD = 180° -108°
2∠CAD = 72°
∠CAD = $$\frac{{{{72}^ \circ }}}{2}$$
∠CAD = 36°
∴ ∠CAB = 109° - 36°
∠CAB = 73°
In ΔABC
∠ABC + ∠ACB + ∠CAB = 180°
∠ABC + 72° + 73° = 180°
∠ABC + 145° = 180°
∠ABC = 180° - 145°
∠ABC = 35°

63

The equidistant point from the vertices of a triangle is called its:

A) Centroid B) Incenter C) Circumcenter D) Orthocenter

Discuss

Answer & Solution

Answer: Option C

Solution:

The equidistant point from the vertices of a triangle is called circumcenter

64

Let ABC be an equilateral triangle and AX, BY, CZ be the altitudes. Then the right statement out of the four given responses is

A) AX = BY = CZ B) AX ≠ BY = CZ C) AX = BY ≠ CZ D) AX ≠ BY ≠ CZ

Discuss

Answer & Solution

Answer: Option A

Solution:

ABC is an equilateral triangle and AX, BY and CZ be the altitude so
AX = BY = CZ

65

ABC is an isosceles triangle with AB = AC. The side BA is produced to D such that AB = AD. If ∠ABC = 30°, then ∠BCD is equal to

A) 45° B) 90° C) 30° D) 60°

Discuss

Answer & Solution

Answer: Option B

Solution:

According to question,
Triangles mcq solution image

In ΔABC
Exterior angle CAD = ∠ABC + ∠ACB
CAD = 2∠ABC (∵ ∠ABC = ∠ACB)
CAD = 2 × 30°
CAD = 60°
In ΔCAD,
∠ACD = ∠ADC = $$\frac{{180 - \angle {\text{CAD}}}}{2}$$ = 60°
⇒ ∠BCD = ∠ACD + ∠BCA
⇒ ∠BCD = 60° + 30°
⇒ ∠BCD = 90°

66

The sum of three altitudes of a triangle is

A) Equal to the sum of three sides B) Less than the sum of sides C) Greater than the sum of sides D) Twice the sum of sides

Discuss

Answer & Solution

Answer: Option B

Solution:

According to question,
To see in the figure
Triangles mcq solution image

AB > AP
BC > BQ
AC > CR
∴ AP + BQ + CR < AB + BC + AC

67

In ΔABC, ∠B = 60° and ∠C = 40°. If AD and AE be respectively the internal bisector of ∠A and perpendicular on BC, then the measure of ∠DAE is

A) 5° B) 10° C) 40° D) 60°

Discuss

Answer & Solution

Answer: Option B

Solution:

According to question,
Triangles mcq solution image

Given :
∠B = 60°
∠C = 40°
As we know that
∠A + ∠B + ∠C = 180°
∠A = 180° - 60° - 40°
∠A = 80°
∴ ∠BAD = $$\frac{{{{80}^ \circ }}}{2}$$ = 40°
In ΔAEB
∠A + ∠B + ∠E = 180°
∠A = 180° - 60° - 90°
∠A = 30°
Then,
∠DAE = ∠DAB - ∠EAB
∠DAE = 40° - 30°
∠DAE = 10°

By Trick
$$\eqalign{ & \angle DAE = \frac{{\angle B - \angle C}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{{{60}^ \circ } - {{40}^ \circ }}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {10^ \circ } \cr} $$

68

In a right-angled triangle, the product of two sides is equal to half of the square of the third side i.e., hypotenuse. One of the acute angle must be

A) 60° B) 30° C) 45° D) 15°

Discuss

Answer & Solution

Answer: Option C

Solution:

According to question,
Given :
Triangles mcq solution image

ab = $$\frac{{{C^2}}}{2}$$ . . . . . . . . . . (i)
∴ In ΔABC
Using Pythagoras theorem
AC² = AB² + BC²
c² = a² + b² . . . . . . . . . . . (ii)
Put the value of C² in equation (i)
2ab = a² + b²
a² + b² - 2ab = 0
(a - b)² = 0
∴ a - b = 0
a = b
If a = b means ABC is isosceles right angle triangle it means
∠A = 45° ∠B = 45°

69

In a ΔABC ∠A : ∠B : ∠C = 2 : 3 : 4. A line CD drawn || to AB, then the ∠ACD is :

A) 40° B) 60° C) 80° D) 20°

Discuss

Answer & Solution

Answer: Option A

Solution:

According to question,
Triangles mcq solution image

Given :
∠A + ∠B + ∠C = 180°
2 + 3 + 4 = 9 units
∴ 9 units = 180°
   1 unit = 20°
∴ ∠A = 2 × 20° = 40°
   ∠B = 3 × 20° =60°
   ∠C = 4 × 20° = 80°
and AB || CD
   ∠B = ∠C
∴ ∠ACD = 180° - 60° - 80°
   ∠ACD = 40°

70

BL and CM are medians of ΔABC right-angled at A and BC = 5 cm. If BL = $$\frac{{3\sqrt 5 }}{2}$$ cm, then the length of CM is

A) $$2\sqrt 5 $$ cm B) $$5\sqrt 2 $$ cm C) $$10\sqrt 2 $$ cm D) $$4\sqrt 5 $$ cm

Discuss

Answer & Solution

Answer: Option A

Solution:

According to question,
Triangles mcq solution image

According to figure, when two medians intersect each other in a right angled triangle then we use this equation.
⇒ 4 (BL² + CM²) = 5BC²
⇒ 4 × $${\left( {\frac{{3\sqrt 5 }}{2}} \right)^2}$$ + 4CM² = 5BC²
⇒ 45 + 4CM² = 125
⇒ CM² = $$\frac{{125 - 45}}{4}$$
⇒ CM² = 20
⇒ CM = $$2\sqrt 5 $$ cm