MCQ Questions on Geometry for Competitive Exams

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

If $$\frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{{{\text{BD}}}}{{{\text{DC}}}}$$   then AD is bisector of ∠A.
∠A = 180° - (68° + 52°)
∠A = 60°
∠BAD = $$\frac{{{{60}^ \circ }}}{2}$$ = 30°

x + y = 40°
2x + 2y = 80°
∠AOC + ∠BOD = 80°

Suppose PMQ is isosceles triangle
Since, PQ + QS = SR
Hence MR = a
∠P + ∠Q + ∠R = 180°
120° + 2θ + θ = 180°
θ = 20°
∠Q = 2θ = 40°

Direct common tangent:
Transverse common tangent
$$\eqalign{ & \sqrt {{D^2} - {{\left( {{r_1} - {r_2}} \right)}^2}} :\sqrt {{D^2} - {{\left( {{r_1} + {r_2}} \right)}^2}} \cr & \sqrt {{{10}^2} - {{\left( {4 - 3} \right)}^2}} :\sqrt {{{10}^2} - {{\left( {4 + 3} \right)}^2}} \cr & \sqrt {99} :\sqrt {51} \cr & \sqrt {33} :\sqrt {17} \cr} $$

⇒ PT² = PD × PC . . . . . . (1)
PT² = PB × PA . . . . . . (2)
⇒ PD × PC = PB × PA
5 × 12.8 = 4 × (4 + AB)
5 × 3.2 = (4 + AB)
16 = 4 + AB
AB = 12 Answer

(External angle of ΔAFP)
∠ABC = 95°, ∠FED = 115°
In, $$\square $$ FABC,
∠AFC = 180° - 95° = 85°
In $$\square $$ FEDA,
∠FAD = 180° - 115° = 65°
So, ∠APC = 85° + 65° = 150°

Orthocenter is a point where the altitudes meet.

$$\eqalign{ & {\text{Given that,}} \cr & PQ = PR = 2a, \cr & QR = a, \cr & QX = \frac{a}{2} \cr & \Rightarrow {\left( {2a} \right)^2} = {\left( {PX} \right)^2} + {\left( {\frac{a}{2}} \right)^2} \cr & \Rightarrow 4{a^2} - \frac{{{a^2}}}{4} = {\left( {PX} \right)^2} \cr & \Rightarrow {\left( {PX} \right)^2} = \frac{{15{a^2}}}{4} \cr & \Rightarrow PX = \frac{{\sqrt {15} a}}{2}{\text{ Answer}} \cr} $$

Interior angle of a regular polygon
$$\eqalign{ & = \frac{{\left( {n - 2} \right) \times {{180}^ \circ }}}{n} \cr & = \frac{{\left( {5 - 2} \right) \times {{180}^ \circ }}}{5} \cr & = {108^ \circ } \cr} $$
∠DEA = ∠BAE = 108°
∠AEF = ∠EAF = 72°
In ΔEAF,
∠E + ∠A + ∠F = 180°
72 + 72 + ∠EFA = 180°
∠EFA = 36°