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Directions: Study the following questions carefully and choose right answer:
1
In ΔABC, ∠ABC = 70°, ∠BCA = 40°. O is the point of intersection of the perpendicular bisectors of the sides, then the angle ∠BOC is
» Explain it
D

 
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 × ∠BAC = 2 × 70° = 140°

Hence, option D is correct.
 
2
A, B, C are three points on the circumference of a circle and if AB = AC = 5 √2 cm and ∠BAC = 90°, find the radius. 
» Explain it
B
 
AB = AC = 5√2 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 × ∠BAC = 2 × 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.

3
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is
 
» Explain it
D

∠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.
 
4
Two chords of lengths a metre and b metre subtend angles 60° and 90° at the centre of the circle respectively. Which of the following is true ?
» Explain it
A
 
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.

5
Two circles touch externally at P, QR is a common tangent of the circles touching the circles at Q and R. Then measure of ∠QPR is
» Explain it
C

∠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.