MCQ Questions on Geometry for Competitive Exams

∠A = ∠B = ∠D = ∠C = 90°
ΔAMB is equilateral
Hence ∠A = 60°
∠DAM = 90° + 60° = 150°
∠ADM $$ = \frac{{{{180}^ \circ } - {{150}^ \circ }}}{2} = \frac{{{{30}^ \circ }}}{2} = {15^ \circ }$$
∠MDC = 90° - 15° = 75°

∠AOB = 60°
∠COD = 90°
⇒ length of chord AB = a
⇒ length of chord CD = b
⇒ AO = OB = AB = OD = OC = a
⇒ In ΔODC
⇒ OD² + OC² = CD²
⇒ a² + a² = b²
⇒ b = $$\sqrt 2 $$ a

∠BAC + ∠ACB = ∠CBD
80° = ∠CBD
∠CBD = 80°
Let, GH || BD
∠DEH = 35°
∠CGE = 80°
∠CGE + ∠GCE = ∠CEH
80° + 45° = ∠CEH
∠CEH = 125°
∠DEH + ∠CEH = ∠CED
125° + 35° = ∠CED
∠CED = 160°

From chord BP
∠BAP = ∠BQP = z
(Angle subtended by same chord on circumference)
Similarly from chord RB
∠RCB = ∠RQB = y
In ΔABC
2x + 2y + 2z = 180°
x + y + z = 90°
y + z = 90° - x
∠RQP = 90° - x
∠RQP = $${90^ \circ } - \frac{{\angle {\text{B}}}}{2}$$

DE = EB = 3 cm (tangent at same point)
AE + EB = AB = AC (tangent at same point)
AE + 3 = 15
AE = 12 cm

AD² = BD × CD
AD² = 4 × 5
AD = 2√5

a² + b² + c² = ab + bc + ca
This is true only when a = b = c
So, triangle will be equilateral.

$${\text{MN}} = \frac{{12 - 7.5}}{2} = \frac{{4.8}}{2} = 2.4$$

Let, Perpendicular = 7 cm
Base = 24 cm
Hypotenuse = 30 cm
Here, P² + B² < C²
(7)² + (24)² < (30)²
49 + 576 < 900
625 < 900
Therefore triangle is obtuse angled triangle.