Math Olympiad Test: Circles- 2 - Class 9 MCQ

In ∆OAB, 
OA = OB
∴ ∠OAB = ∠OBA …(i)
∵ ∠OBC + ∠OBA = ∠ABC
∠OBC + ∠BAC = ∠ABC
∠OBC + ∠BAC = 90°
[∵ ∠ABC is the angle in semicircle]

∵ AB = BC
∴ ∠ABC = ∠ACB = 64° 
∴ ∠BAC = 180° – 64° × 2 = 52° 
∴ ABCD is a cyclic quadrilateral
∴ ∠A + ∠E = 180°
∴ ∠A + ∠E = 180°
= 128° 

Let ∠CAB = x, 
∴ ∠ACD = x (Alternate ∠S) 
In ∆ACD, 
∠D = 180° – (30° + x)
= 150° – x
∵ ABCD is a cyclic quadrilateral
∴ ∠D + ∠B = 180° 
⇒ 150° – x + 65° = 180° 
⇒ x = 35°

∵ ABCD is a cyclic quadrilateral
∴ ∠A + ∠C = ∠B + ∠D = 180° …(i)
Also, 
∠A + ∠D = ∠B + ∠C = 180° …(ii) 
∴     ∠B = ∠C [using (i) and (ii)]
∵ ABCD is a trapezium having ∠C = ∠D
∴ ABCD should be an isosceles trapezium
⇒ AD = BC

In ∆CBA, 
∠A + ∠C + ∠B = 180°
⇒ ∠B = 180° – 35° – 90°
[∵ ∠C = 90°, i.e., angle in a semicircle]
⇒ ∠B = 55°
∵ ABCD is a cyclic quadrilateral 
∴ ∠B + ∠D = 180°  
⇒ x = 180° – 55° 
= 125° 

∵ AB = CD
∴ OL = OM, as the distance of equal chords from the center of the circle, should be equal
⇒ ∠OLM = ∠OML

∵ Diagonals of the quadrilateral ABCD intersect at right angles and are also of equal length i.e., 2r

∴ ABCD is a square. 

(angle subtended at the center is double the angle at the circumference) 
⇒ ∠BAD = 2∠DPB
∵ ABCD is a parallelogram
∴ ∠BAD = ∠BCD
⇒ ∠BCD = 2∠DPB
∠DEB = ∠BPD (angles in the same segment)

According to the question,

OA = r = 26 cm
OB′ = 10 cm 

(Using Pythagoras theorem)

∴ AB = 2 × AB′ = 2 × 24 cm = 48 cm.

For an equilateral triangle ABC, O lies on the perpendicular from any vertex to the opposite side.

Also, 
AO : OD = 2 : 1 (for equilateral triangle)

(using componendo-dividendo) 

In ∆ABM and ∆ACM
AB ≅ AC (given)
AM = AM (common)
OM ⊥ BC (given)
∴ ∆ABM ≅ ∆ACM
∴  BM = CM

In ∆ODE,

OD = OE
∴ ∠ODE = ∠OED ……(i) 
Now, 
∠ODA = ∠ODE = 90° ….(ii)
Subtracting eq.(i) from (ii), we get
∠ODA – ∠ODE = ∠OEA – ∠OED
∠ADE = ∠AED 
∴ AD = AE ⇒ ∆ADE is an isosceles triangle. 

In ∆ABC, 
AB = AC,
⇒ ∠ABC = ∠ACB = 40°
Also, 
∠ABC + ∠BAC + ∠ACB = 180° 
⇒ ∠BAC = 180° – 40° × 2
= 100° 
∴ ∠BAC = ∠BDC = 100°
(angles in the same segment are equal). 

∵ ∠APC is an exterior angle for ∆ABP
∴ ∠ABP + ∠PAB = 90°
⇒ ∠ABP = 90° – 44° = 46° 
∠ADC = ∠ABP
(Angles in the same segment)
∴ ∠ABP = 46° 

∵ ABCP is a cyclic quadrilateral. 
∴ ∠B + ∠P = 180° 
⇒ ∠P = 180° – 120° = 60°
∠AOC = 2 ∠P = 2 × 60° = 120°