MCQ Questions on Geometry for Competitive Exams

∠ABD = ∠MBD = θ (angle bisector)
∴ SD ⊥ AM
∠BDA = ∠BDM = 90°
It happen only in equilateral and isosceles triangle
∴ AD = DM
i.e. AD = $$\frac{{{\text{AM}}}}{2}$$
Given DE || BC
From Thales theorem
E will be mid point of AC
∵ AC = 12 cm
So, AE = 6 cm

Let P is the mid point of AD
$$\eqalign{ & BP = \sqrt {A{P^2} + A{B^2}} \cr & = \sqrt {A{P^2} + 16} \cr & = \sqrt {{2^2} + 16} \cr & = \sqrt {20} \cr & {\text{Similarly }}CP = \sqrt {20} \cr & BP + CP = \sqrt {20} + \sqrt {20} = 4\sqrt 5 \cr} $$

AB is the Diameter of the circle, and Tangents PQ and RS are parallel to each other.

In ΔPTQ
∠QPT + ∠PTQ + ∠PQT = 180°
42° + 90° + ∠PQT = 180°
∠PQT = 180° - 132°
∠PQT = 48°

12θ → 180°
θ → 15°
6θ → 90°
Since sum of two angle is equal to the third angle then, it is a right angle triangle.
Hence BC = 24 cm

$$\eqalign{ & \frac{{{\text{AP}}}}{{{\text{PC}}}} = \frac{{{\text{BP}}}}{{{\text{PD}}}}\,\,\,\,\,\,\left( {{\text{Similarity}}} \right) \cr & \frac{{3{\text{x}} - 1}}{{5{\text{x}} - 3}} = \frac{{2{\text{x}} + 1}}{{6{\text{x}} - 5}} \cr} $$
18x² - 6x - 15x + 5 = 10x² + 5x - 6x - 3
8x² - 20x + 8 = 0
x = 2, $$\frac{1}{2}$$
at x = 2,
DB = 6x - 5 + 2x + 1
= 8x - 4
= 8 × 2 - 4
= 12

$$\eqalign{ & \Delta ABC \sim \Delta DAC \cr & \frac{{AC}}{{CD}} = \frac{{BC}}{{AC}} \cr & \frac{{12}}{{CD}} = \frac{8}{{12}} \cr & CD = 18 \cr} $$

In $$\square $$ POQB,
PBOQ is a square because
PB = BQ
PB = OQ
So,
BQ = OQ = OP = r
Circumference = 2πr = 2π × 9 = 18π

∠AOD = 2∠ABD
∠COB = 2∠CDB
∠AOD + ∠COB = 2(∠ABD + ∠CDB)
32° + 26° = 2∠APD
∠APD = 29°