Triangles - Aptitude MCQ Questions and Solutions with Explanations

According to question,

AB + BC = 12 cm
BC + CA = 14 cm
CA + AB = 18 cm
2(AB + BC + CA) = 44 cm
AB + BC + CA = $$\frac{{44}}{2}$$ cm
AB + BC + CA = 22 cm
Perimeter of triangle = 22 cm
Perimeter of triangle = Perimeter of circle
22 = 2πr
2 × $$\frac{{22}}{7}$$ × r = 22
r = $$\frac{7}{2}$$ cm

According to question,

AB = AC,         ∠B = ∠C
∠A + ∠B + ∠C = 180°
∠A + 2∠B = 180°
∠A = 180° - 70°
∠A = 110°
Note : In isosceles triangle median bisects the opposite side and make angle 90° on opposite side. It also bisects the vertex angle.
∠BAD = $$\frac{{\angle A}}{2}$$
∠BAD = $$\frac{{{{110}^ \circ }}}{2}$$
∠BAD = 55°

According to question,

Given :
∠A = 40°       AB = AC
∴ ∠B = ∠C
In ΔABC
∠A + ∠B + ∠C = 180°
40° + 2∠B = 180°
2∠B = 180° - 40°
2∠B = 140°
∠B = 70°
∴ External angle at B = 180° - 70° = 110°

According to question,

ΔABC ∼ ΔACD
$$\eqalign{ & \therefore \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{B{C^2}}}{{A{D^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{{8^2}}}{{{6^2}}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{64}}{{36}} \cr & \frac{{{\text{area}}\,{\text{of}}\,\Delta ABC}}{{{\text{area}}\,{\text{of}}\,\Delta ACD}} = \frac{{16}}{9} \cr} $$
area ΔABC : area ΔACD = 16 : 9

According to question,

Given : ∠BOC = 60°
As we know that
∴ ∠O = 90 - $$\frac{1}{2}$$ ∠A
⇒ $$\frac{1}{2}$$ ∠A = 90° - 60°
⇒ $$\frac{1}{2}$$ ∠A = 30°
⇒ ∠A = 60°

ar ADE = ar DEBC
So, ar ΔADE = 1 unit² and ar ABC = 2 unit²
$$\eqalign{ & \frac{{{\text{ar}}\,\Delta ADE}}{{{\text{ar}}\,\Delta ABC}} = \frac{{A{D^2}}}{{A{B^2}}} \cr & \frac{1}{2} = {\left( {\frac{{AD}}{{AB}}} \right)^2} \cr & \frac{1}{{\sqrt 2 }} = \frac{{AD}}{{AB}} \cr & \therefore \frac{{AD}}{{DB}} = \frac{1}{{\sqrt 2 - 1}} \cr & \left( {\therefore DB = AB - AD = \sqrt 2 - 1} \right) \cr & {\text{So,}}\,AD:BD = 1:\sqrt 2 - 1 \cr} $$

According to question,

Let AB = AC = 2x
∵ AQ = QC = x
∴ AB is a secant
∴ AP × AB = AQ²
AP × 2x = x²
$$\eqalign{ & AP = \frac{x}{2} \cr & \frac{{AP}}{{AB}} = \frac{x}{{2 \times 2x}} = \frac{1}{4} \cr & \frac{{AP}}{{AB}} = \frac{1}{4} \cr & AP:AB = 1:4 \cr} $$

The sum of two sides of a triangle should be greater than the third side. There are only two possible pairs (2, 5, 6) and (3, 5, 6)

According to question,
If the median drawn on the base of a triangle is half of its base of the triangle then the triangle will be right angled triangle.

According to question,

Given :
∠ABC = 60°
∠BCA = 80°
∠BIC = ?
∠BAC = 40°
∴ ∠BIC = 90° + $$\frac{1}{2}$$ × 40°
∠BIC = 110°