Triangles - Aptitude MCQ Questions and Solutions with Explanations

According to question,
Length of the three sides of a triangle are 6 cm, 8 cm and 10 cm, this is right angle triangle.

Note: In right angle triangle median divides the hypotenuse in two equal parts
∴ BD = $$\frac{{\text{H}}}{2}$$
BD = $$\frac{{10}}{2}$$
BD = 5 cm

According to question,
Given:
∠BAC = 85°
∠BCA = 75°
∠OAC = ?

∠ABC + ∠BCA + ∠CAB = 180°
∠ABC = 20°
∴ ∠COA = 2 × ∠ABC
    ∠COA = 2 × 20 = 40°
In ΔAOC
We know OC = OA
∴ ∠OAC = ∠OCA
∴ ∠OAC + ∠OCA + ∠COA = 180°
2∠OAC = 180° - 40°
2∠OAC = 140°
∠OAC = 70°

Inradius = $$\frac{a}{{2\sqrt 3 }}$$ (a = side of Δ)
3 = $$\frac{a}{{2\sqrt 3 }}$$
a = 6$$\sqrt 3 $$ cm

According to question,
Given: AC = 9 cm

area of ΔABC = 40 cm²
area of ΔADC = 10 cm²
ΔABC ∼ ΔADC


$$\frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ADC}} = \frac{{A{B^2}}}{{A{D^2}}} = \frac{{B{C^2}}}{{A{C^2}}}$$
(In similar Δratio of their area is square of ratio of corresponding sides)
$$\eqalign{ & \frac{{40}}{{10}} = \frac{{B{C^2}}}{{{{\left( 9 \right)}^2}}} \cr & \frac{{40}}{{10}} \times 81 = B{C^2} \cr & BC = 18\,{\text{cm}} \cr} $$

The orthocentre of a right angled triangle lies at the right angular vertex

According to question,
Given:

∴ ∠BOC = 90° + $$\frac{1}{2}$$ ∠A
∠BOC = 90° + $$\frac{1}{2}$$ × 30°
∠BOC = 90° + 15°
∠BOC = 105°

According to question,
Given: ∠C = 90°

BC = AC = 5 cm (Isosceles triangle)
By Pythagoras theorem
AB² = AC² + BC²
AB² = 5² + 5²
AB² = 25 + 25
AB² = 50
AB = 5$$\sqrt 2 $$ cm

According to question,

Given: BAC is a right angle triangle
AD ⊥ BC
AD = 6 cm
BD = 4 cm
BC = ?
In ΔBAD
$$\eqalign{ & AB = \sqrt {B{D^2} + A{D^2}} \cr & AB = \sqrt {{4^2} + {6^2}} \cr & AB = \sqrt {52} \,cm \cr} $$
ΔBAC ∼ ΔBDA


$$\eqalign{ & \therefore \frac{{BC}}{{AB}} = \frac{{AB}}{{BD}} \cr & \therefore \frac{{BC}}{{\sqrt {52} }} = \frac{{\sqrt {52} }}{4} \cr & BC = \frac{{52}}{4} \cr & BC = 13\,cm \cr} $$

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Alternate :
$$\eqalign{ & A{B^2} = BD.BC \cr & {\left( {\sqrt {B{D^2} + A{D^2}} } \right)^2} = BD.BC \cr & {\left( {\sqrt {{4^2} + {6^2}} } \right)^2} = 4.BC \cr & \frac{{52}}{4} = BC, \cr & \therefore BC = 13\,cm \cr} $$

According to question,

Given : ∠BOD = 15°
∴ ∠BDO + ∠DOB + ∠DBO = 180°
∠DBO = 75°
∠ABC = 2 × ∠DBO
∠ABC = 2 × 75°
∠ABC = 150°

Circumcentre of a triangle lies outside then triangle is obtuse angled triangle.