MCQ Questions on Geometry for Competitive Exams

CD = AB, AC = BD
OC = OB, AO = OD
AE = DF
ΔABC ≅ ΔDCB
ΔAOE ≅ ΔDOF
ΔAEB ≅ ΔDFC
ΔADC ≅ ΔABD (×) → Should be ΔADB

∠COD = ∠ABF (by the property of cyclic Quadrilateral)
∠DCF = ∠EAB
∠ABF + ∠EAB = 85° + 94° = 179°

a = 7 cm
2S = a + b + c = 18
S = 9 cm
a + b + c = 18
7 + b + c = 18
b + c = 11
b = 11 - c
$$\eqalign{ & \Delta = \sqrt {{\text{S}}\left( {{\text{S}} - {\text{A}}} \right)\left( {{\text{S}} - {\text{B}}} \right)\left( {S - {\text{C}}} \right)} \cr & \Delta = \sqrt {9\left( {9 - 7} \right)\left( {9 - 11 - {\text{C}}} \right)\left( {9 - {\text{C}}} \right)} \cr & \Delta = \sqrt {9 \times 2\left( {{\text{C}} - 2} \right)\left( {9 - {\text{C}}} \right)} \cr & \sqrt {108} = \sqrt {18\left( {{\text{C}} - 2} \right)\left( {9 - {\text{C}}} \right)} \cr} $$
6 = 9C - C² - 18 + 2C
24 = C² + 11C
C² - 11C + 24 = 0
C² - 8C - 3C + 24 = 0
C(C - 8) - 3(C - 8) = 0
(C - 3)(C - 8) = 0
C = 3, 8
B = 8, 3
Side = 3, 8

AB = 5 unit
AO = 2.5
OP = 2.5
OX = OB - BX
= 2.5 - 2
= 0.5
OP = 2.5 unit
2.5 → 5
1 → 2
0.5 → 0.5 × 2 = 1
In ΔOPX

$$\eqalign{ & {\text{PX}} = \sqrt {25 - 1} = \sqrt {24} = 2\sqrt 6 \cr & {\text{PQ}} = 2{\text{PX}} = 2 \times 2\sqrt 6 = 4\sqrt 6 \cr} $$

∠EDC + ∠DCF = ∠CEB
∠EDC + 25° = 128°
∠EDC = 103°
∠EDC = ∠BAC (all angles are equal same sector)
∠BAC = 103°

Number of possible triangles when perimeter is odd
$$\eqalign{ & = \frac{{{{\left( {P + 3} \right)}^2}}}{{48}} \cr & = \frac{{{{\left( {13 + 3} \right)}^2}}}{{48}} \cr & = \frac{{16 \times 16}}{{48}} \cr & = \frac{{16}}{3} \cr & = 5\,... \cr & = 5 \cr} $$

∠ABC = 108° {Internal Angle of regular pentagon}
∠EGD = 108° - (∠E + ∠D)
= 108° - (72° + 72°)
= 36°
∠BAJ = 72°
$$\eqalign{ & \frac{{\angle {\text{ABC}} + 2\angle {\text{EGD}} + 3\angle {\text{BAJ}}}}{6} \cr & = \frac{{{{72}^ \circ } + 2 \times {{36}^ \circ } + 3 \times {{72}^ \circ }}}{6} \cr & = {18^ \circ } + {12^ \circ } + {36^ \circ } \cr & = {66^ \circ } \cr} $$

MO = 10² - 8² = 6
ON = 6
AB = 2AN
= 2 × 8
= 16

⇒ Given that, XO = $$\frac{1}{2}$$YZ and, XO is median, then YO = OZ
⇒ ∠OXY = 30° [Given]
and, OX = OY
So, ∠OXY = ∠XYZ
30° = ∠XYZ Answer

$$\eqalign{ & \angle {\text{QOR}} = {90^ \circ } \times \frac{{\angle {\text{P}}}}{2} \cr & = {90^ \circ } + \frac{{{{42}^ \circ }}}{2} \cr & = {111^ \circ } \cr} $$