Triangles - Aptitude MCQ Questions and Solutions with Explanations

According to question,
Given : ∠ABC = 90°
$$\frac{{AN}}{{NC}} = ?$$

ΔABC ∼ ΔBNC
ΔABC ∼ ΔANB
∴ ΔABC ∼ ΔBNC ∼ ΔANB
AB = 3, BC = 4, AC = 5

$$\eqalign{ & \frac{{AB}}{{BN}} = \frac{{AC}}{{BC}} \cr & BN = \frac{{AB \times BC}}{{AC}} = \frac{{3 \times 4}}{5} = 2.4 \cr & \frac{{BC}}{{NC}} = \frac{{AB}}{{NB}} \cr & \frac{4}{{NC}} = \frac{3}{{2.4}} \cr & NC = 3.2 \cr & \frac{{AB}}{{AN}} = \frac{{BC}}{{NB}}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{3}{{AN}} = \frac{4}{{2.4}} \cr & AN = 1.8 \cr & \frac{{AN}}{{NC}} = \frac{{1.8}}{{3.2}} \cr & \frac{{AN}}{{NC}} = \frac{9}{{16}} \cr & \therefore AN:NC = 9:16 \cr} $$

According to question,
Given : ∠BOC = 110°
∠BOC = 90° + $$\frac{1}{2}$$ ∠A

110° = 90° + $$\frac{{\angle A}}{2}$$
$$\frac{{\angle A}}{2}$$ = 20
∠A = 40°

According to question,
Given :
∠ACB = 55°
∠ABC = 65°
∠BIC = ?

∴ ∠ACB + ∠ABC + ∠BAC = 180°
∠BAC = 180° - 55° - 65°
∠BAC = 60°
We know that
∠BIC = 90 + $$\frac{1}{2}$$ ∠A
∠BIC = 90 + $$\frac{1}{2}$$ × 60
∠BIC = 90 + 30
∠BIC = 120°

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Alternate:
In ΔBIC,
$$\frac{1}{2}$$ ∠B + $$\frac{1}{2}$$ ∠C + ∠BIC = 180°
$$\frac{1}{2}$$ (65° + 55°) + ∠BIC = 180°
∠BIC = 180° - 60°
∠BIC = 120°

According to question,
Given : ABC is a right angle triangle

BC = 6$$\sqrt 3 $$
∴ sin 30° = $$\frac{P}{H}$$ = $$\frac{{AB}}{{6\sqrt 3 }}$$
AB = 3$${\sqrt 3 }$$

sin 60° = $$\frac{P}{H}$$ = $$\frac{{AN}}{{AB}}$$
$$\frac{{\sqrt 3 }}{2}$$ = $$\frac{{AN}}{{3\sqrt 3 }}$$
AN = $$\frac{9}{2}$$
AN = 4.5 cm

According to question,

PX || BD [mid point theorem]
∴ PX = $$\frac{1}{2}$$BD
Similarly, QY || BD
∴ QY = $$\frac{1}{2}$$BD
∴ PX : QY,           $$\frac{1}{2}$$BD : $$\frac{1}{2}$$BD
PX : QY = 1 : 1

According to question,
Given :
BAC is right angle triangle
AB = $$\frac{1}{2}$$BC

$$\eqalign{ & \frac{{AB}}{{BC}} = \frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{P}{H} = \frac{1}{2} \cr & \sin \theta = \frac{P}{H} = \frac{1}{2} \cr & \sin \,{30^ \circ } = \frac{1}{2} \cr & \therefore \theta = \angle ACB = {30^ \circ } \cr} $$

According to question,
ACB is a right angle triangle
∴ area of ΔACB

$$\frac{1}{2}$$ × AC × BC = $$\frac{1}{2}$$ × AB × PC
$$\frac{1}{2}$$ × b × a = $$\frac{1}{2}$$ × c × p
c = $$\frac{{ab}}{p}$$ . . . . . . . (i)
By using pythagoras theorem
AB² = AC² + BC²
c² = b² + a² . . . . . . . . . . (ii)
Put the value of C in equation (ii)
$$\eqalign{ & {\left( {\frac{{ab}}{p}} \right)^2} = {a^2} + {b^2} \cr & \Rightarrow \frac{{{a^2}{b^2}}}{{{p^2}}} = {a^2} + {b^2} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{{{a^2}}}{{{a^2}{b^2}}} + \frac{{{b^2}}}{{{a^2}{b^2}}} \cr & \Rightarrow \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$

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Alternate :
From figure,
$$\eqalign{ & P = \frac{{ab}}{c} \cr & P = \frac{{ab}}{{\sqrt {{a^2} + {b^2}} }}\,\left( {\because {a^2} + {b^2} = {c^2}} \right) \cr & {P^2} = \frac{{{a^2}{b^2}}}{{{a^2} + {b^2}}} \cr & \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} \cr} $$

In a right angled triangle orthocentre lies vertex

According to question,
ABC is a right angle triangle

∴ Apply Pythagoras theorem
AC² = AB² +BC²
(x + 2)² = (x - 2)² + x²
x² + 4 + 4x = x² + 4 - 4x + x²
x² = 8x
x = 8

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Alternate : From option approach
$$\eqalign{ & \left( {x - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x + 2} \right) \cr & \,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{Triplet}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{x}} = {\text{8}} \cr} $$

According to question,
ABC is a right angle triangle,

∴ AB = 8 cm
   BC = 6 cm
∴ AC² = AB² + BC²
   AC = 64 + 36
   AC = $$\sqrt {100} $$
   AC = 10 cm
In right triangle
Circum Radius I_R = $$\frac{{AC}}{2}$$ = $$\frac{{10}}{2}$$ = 5 cm