MCQ Questions on Geometry for Competitive Exams

ABCD is cyclic quadrilateral
∴ 3p + 2p = 180°
5p = 180°
p = 36°
∠ADC = 3p = 3 × 36 = 108°
∠AOC = $$\frac{1}{2}$$∠ADC
= $$\frac{1}{2}$$ × 108°
= 54°
Reflex of ∠AOC = 360° - 54° = 206°

∠ADB = 180° - 62° - 62° = 56°

Two triangles are similar if their corresponding sides are proportional.

According to question
Given:

∠BDC = 120°, ∠ABP =?
∴ ∠CDP = 180° - ∠BDC
∠CDP = 180° - 120°
∠CDP = 60°
CD || AB
∴ ∠CDP = ∠ABP = 60°

$$\eqalign{ & R{T^2} = {25^2} - {20^2} \cr & RT = 15 \cr & {\text{Inradius }}\left( {\text{r}} \right) = \frac{{20 + 15 - 25}}{2} = \frac{{10}}{2} = 5 \cr & {\text{In }}\Delta {O_1}A{O_2} \cr & {O_1}O_2^2 = {O_1}{A^2} + AO_2^2 \cr & = {\left( 5 \right)^2} + {\left( {15} \right)^2} \cr & = 25 + 225 \cr & = 250 \cr & {O_1}{O_2} = 5\sqrt {10} {\text{ cm}} \cr} $$

$$\eqalign{ & \angle A = {66^ \circ } \cr & \angle BOC = ? \cr & \angle BOC = {90^ \circ } - \frac{{\angle A}}{2} \cr & = {90^ \circ } - \frac{{{{66}^ \circ }}}{2} \cr & = {90^ \circ } - {33^ \circ } \cr & = {57^ \circ } \cr} $$

$$\eqalign{ & OB = 7\,{\text{cm}} \cr & {\text{Sine Rule in }}\Delta BOC \cr & \frac{{OB}}{{\sin {{30}^ \circ }}} = \frac{{BC}}{{\sin {{30}^ \circ }}} \cr & BC = 7\,{\text{cm}} \cr} $$

So, ΔAOB is an equilateral triangle
So, ∠APB + ∠AOB = 180°
∠APB = 120°
So PA and PB are parallel to OC and OD
So ∠APB = ∠COD
∴ ∠COD = 120°

2α + 2β + 54° = 180°
2(α + β) = 126°
α + β = 63°
∠PCA = ∠CAB + ∠CBA
= α + β
= 63°