Triangles - Aptitude MCQ Questions and Solutions with Explanations

Given in question AD = BE = CF
[DB = AF = EC] Because
AB = BC = CA
So, Triangle is equilateral

Assume, ∠A = x
∴ ∠B = $$\frac{3}{4}$$ x
∠A + ∠B = 112° (∵ sum of two interior angle is equal to the exterior angle of the third angle)
$$\eqalign{ & {x^ \circ } + \frac{3}{4}{x^ \circ } = {112^ \circ } \cr & \frac{{7{x^ \circ }}}{4} = {112^ \circ } \cr & {x^ \circ } = {64^ \circ } \cr & {\text{Hence,}} \cr & \angle B = \frac{3}{4} \times {64^ \circ } \cr & \angle B = {48^ \circ } \cr} $$

Perimeter of ΔABC
= 3 + 4 + 5
= 12
$$\eqalign{ & \frac{{{\text{Perimeter}}\,{\text{of}}\,\vartriangle ABC}}{{{\text{Perimeter}}\,{\text{of}}\,\vartriangle DEF}} = \frac{{AB}}{{DE}} \cr & \frac{{12}}{{{\text{Perimeter}}\,{\text{of}}\,\vartriangle DEF}} = \frac{3}{{12}} \cr} $$

Perimeter of ΔDEF = 48 cms

Circumcenter at the mid point of QR hence angle made by QR
= 2 × 90°
= 180°
Angle made by QR at In center
= 90° + $$\frac{1}{2}$$ × ∠P = 135°
Ortho center is at point 'P'
Hence angle made by QR = 90
Then ration C : I : O
= 180 : 135 : 90
= 4 : 3 : 2

∵ ΔPQR ∼ ΔPXY

$$\eqalign{ & \frac{{PX}}{{PQ}} = \frac{{XY}}{{QR}} \cr & \frac{5}{{\left( {5 + 6} \right)}} = \frac{{XY}}{{QR}} \cr & XY:QR = 5:11 \cr} $$

∵ AB = AC
   Point D is the incenter
∴ ∠BDC = 90° + $$\frac{1}{2}$$ ∠A
              = 90° + $$\frac{1}{2}$$ × 80°
              = 90° + 40°
              = 130°

The height of ΔABC and ΔABD are same and have same base.
∴ Area ΔABC = Area ΔABD

∴ AB = BC = CA
∠BAC = 60°
So, ∠BGC = 90 + $$\frac{{60}}{2}$$
∠BGC = 120°

∠C = 180° - 120° = 60°
∠A + ∠B + ∠C = 180°
∠A + $$\frac{1}{2}$$ ∠A + 60° = 180°
$$\frac{3}{2}$$ ∠A = 120°
∠A = 80°

∠A + ∠B + ∠C = 180°
            (for a triangle)
∠A + ∠C = 140°
then, ∠B = 40°
∠A + 3∠B = 180°
∠A + 120° = 180°
⇒ ∠A = 60°