MCQ Questions on Geometry for Competitive Exams

ΔADC and ΔEBC
AD = EC
∠A = ∠C = 60°
AC = BC
ΔADC ∽ ΔEBC   [By SAS]
Let ∠ECF = ∠EBC = x
∠FCB = 60° - x
In ΔBFC
∠B + ∠C + ∠F = 180°
x + 60° - x + ∠BFC = 180°
∠BFC = 180° - 60° = 120°
∠CFB = 120°

Since $$\frac{{{\text{AB}}}}{{{\text{AC}}}} = \frac{{{\text{BD}}}}{{{\text{DC}}}}$$
Hence AD is an angle Bisector
Hence ∠BAD $$ = \frac{{{{180}^ \circ } - \left( {{{75}^ \circ } + {{45}^ \circ }} \right)}}{2} = \frac{{{{60}^ \circ }}}{2} = {30^ \circ }$$

PSQ = 90° (Angle in semicircle)
∠TPS = $$\frac{1}{2}$$∠ROS (Angle of same arc)
∠TPS = $$\frac{1}{2}$$ × 42° = 21°
In ΔPTS
21° + ∠PTS + 90° = 180°
∠PTG = 69° Answer

$$\eqalign{ & \angle FOH = \frac{{\angle FOB}}{2} = {90^ \circ } \cr & \left[ {\therefore FOB{\text{ is a straight line}}} \right] \cr & \angle HDF = \frac{{\angle FOH}}{2} = \frac{{{{90}^ \circ }}}{2} = {45^ \circ } \cr} $$

1 : 2 : 1 → 4x = 180
                    x = 45

Ratio will be = $$\boxed{1:\sqrt 2 :1}$$

$$\eqalign{ & {\text{Height of }}\Delta {\text{MND}} \cr & = \sqrt {{{\left( {16\sqrt 5 } \right)}^2} - {{\left( {\frac{{32}}{2}} \right)}^2}} \cr & = \sqrt {{{\left( {16\sqrt 5 } \right)}^2} - {{\left( {16} \right)}^2}} \cr & = 16\sqrt {5 - 1} \cr & = 16 \times 2 \cr & = 32 \cr & {\text{R}} = \frac{{{\text{abc}}}}{{4\Delta }} = \frac{{16\sqrt 5 \times 16\sqrt 5 \times 32}}{{4 \times \frac{1}{2} \times 32 \times 32}} = 20 \cr} $$

ΔAPB = 3
ΔABC = 7 + 7 = 14
(median divide into two equal part)
ΔAPB : ΔABC = 3 : 14

AC $$ = \sqrt {{3^2} + {4^2}} = \sqrt {9 + 16} $$
AC = 5
∵ D is circumcenter, it is equidistant from all the center
So AD = CD = BD = $$\frac{5}{2}$$ = 2.5 cm

According to question
Given:

∠AOD = 100°, ∠BOC = 70°
∴ ∠ACD = ∠ACP = $$\frac{{{{100}^ \circ }}}{2}$$ = 50°
∴ The angle subtended at the centre is twice to that of angle subtended at the circumference by the same arc
∠BOC = 70°
∴ ∠BDC = ∠BAC = $$\frac{{{{70}^ \circ }}}{2}$$ = 35°
In ΔAPC
∠PAC + ∠ACP + ∠APC = 180°
∠APC = 180° - 50° - 35°
∠APC = 95°