MCQ: Circles - 3 - SSC CGL MCQ

We know that, Exterior angle is equal to the sum of two interior opposite angles.
∴ ∠BEC = ∠EDC + ∠ECD
130° = ∠EDC + 20°
∠EDC = 110°
∴ ∠BAC = ∠EDC = 110°
[∵ Angles on the same arc]
Hence, option D is correct.

∠POQ = ∠POR = 90°
OQ = OP = OR
[∵ Tangent drawn from the same external point]
∴ ∠OQP = ∠OPQ = ∠ORP = ∠OPR
In ΔPOQ, we know that
∠POQ + ∠OQP + ∠OPQ = 180°
90° + ∠OPQ + ∠OPQ = 180°
2∠OPQ = 180° – 90° = 90°
∠OPQ = 45°
Similarly in ΔPOR, we get
∠ORP = 45°
∴ ∠QPR = ∠OPQ + ∠ORP = 45° + 45° = 90°
Hence, option C is correct.

∠ONY = 50° and ∠OMY = 15°
In ΔONY,
ON = OY (radii)
∠OYN = ∠ONY = 50°
∴ ∠NOY = 180° – ∠ONY – ∠OYN = 180° – 50° – 50° = 80°
In ΔOMY,
OM = OY (radii)
∠OYM = ∠OMY = 15°
∴ ∠MOY = 180° – ∠OMY – ∠OYM = 180° – 15° – 15° = 150°
∴ ∠MON = ∠MOY – ∠NOY = 150° – 80° = 70°
Hence, option D is correct.

Radius = 7 cm
Diameter, AB = 14 cm
PB = 12 cm

∠APB = 90° [∵ angle in the semi circle]
In ΔAPB, By pythagoras theorem

Let, AN = x cm ⇒ NB = (14 – x) cm
In ΔAPN, By pythagoras theorem
PN2 = AP2 – AN2 = 52 – x2 ...(i)
Again, In ΔPNB, By pythagoras theorem
PN2 = PB2 – NB2 = 144 – (14 – x)2 ...(ii)
From Equation (i) and (ii),
52 – x2 = 144 – 196 + 28x – x2
28x = 104

Hence, option D is correct.

Here, ΔAOP,
AO = OP
⇒ ∠PAO = ∠APO = 35°
⇒ ∠AOP = 180° – (2 × 35°) = 110°
⇒ ∠POB = 180° – 110° = 70°
Also, In ΔPOB, BO = OP

Hence, option B is correct.

OO' = 7 cm
r₁ + r₂ = 7
4 + r₂ = 7
r₂ = 7 – 4 = 3 cm
Hence, option B is correct.

Area of the shaded portion (A + B) = Area of the bigger semicircle (B + C) as the two smaller semicircle 'A' and
'C' will have equal area
Radius of the smaller semicircle = 1cm
Now, Radius of the bigger semicircle = Diameter of the smaller semicircle
∴ Radius of the bigger semicircle = 2 cm

OC = 3 cm and OA = 5 cm
In ΔAOC, By pythagoras theorem,

Hence, option D is correct.

AB = AC = 5√ cm, ∠BAC = 90°
Note : The angle subtended by an arc of a circle at the centre is double the
angle subtended by it at any point on the remaining part of the circle.
∴ Exterior ∠BOC = 2 x ∠BAC = 2 x 90° = 180°
∴ ∠BOC = 360° – Exterior ∠BOC = 360° – 180° = 180°
OA = OB = OC = r cm (radii)
AB = AC
∴ ∠AOB = ∠AOC = 90°
In ΔAOB, By pythagoras theorem
AB² = OA² + OB²
(5√2)² = r² + r²
50 = 2r²
r² = 25
r = 5 cm
Hence, option B is correct.

OA = OB = OC = Circum-radius
In ΔABC, we know that
∠ABC + ∠BCA + ∠BAC = 180°
∠BAC = 180° – 70° – 40° = 70°

Note : The angle subtended by an arc of a circle at the centre is double the angle subtended by it
at any point on the remaining part of the circle.
∴ ∠BOC = 2 x ∠BAC = 2 x 70° = 140°
Hence, option D is correct.

∵ x° is an angle in the alternate segment for ∠BAT.
∴ ∠BAT = x = 24°
∵ y° is the angle at the centre and x° is angle on the arc
∴ y° = 2x = 2 × 24 = 48°
∵ In ΔOAB, ∠OBA = z° = ∠OAB
∴ z° + 48° + z° = 180°
or, 2z° = (180° – 48°)

Approach II: OA is perpendicular on PT at A.
⇒ ∠z° = 90 – 24 = 66°
⇒ ∠y° = 180 – (66° + 66°) = 48°

OA = OB = OC = OD = r units (radii)
AB = a metre and CD = b metre
∠AOB = 60° and ∠COD = 90°
In ΔCOD, By pythagoras theorem
CD² = OC² + OD²
b² = r² + r² = 2r² ...(i)
In ΔAOB,
OA = OB
∴ ∠ABO = ∠OAB
∠AOB + ∠ABO + ∠OAB = 180°
60° + ∠OAB + ∠OAB = 180°
2∠OAB = 180° – 60° = 120°
∠OAB = 60° = ∠ABO
∴ ΔAOB is an equilateral triangle.
OA = OB = AB ⇒ a = r
From equation (i),
b = √2r
b = √2a
Hence, option A is correct.

∠ODC = ∠BAC = 52° (∠ s in the same segment).
But OC = OD ⇒ ∠OCD = ∠ODC = 52°.
Hence, option A is correct.

We know that,
∠BOA + ∠AOC + ∠BOC = 360°
90° + 110° + ∠BOC = 360°
∠BOC = 360° – 200° = 160°
Note : The angle subtended by an arc of a circle at the centre is double the angle subtended by it
at any point on the remaining part of the circle.

Hence, option B is correct.