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Circle Quiz 5
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Circle Quiz 1
Directions: Study the following questions carefully and choose right answer:
important for :
CGL Tier 2
CGLTier 1
SSC 10+2
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
A.
100°
B.
120°
C.
130°
D.
140°
Ans
D
Explain it
Answer:
Option
D
Explanation:
OA = OB = OC = Circumradius
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.
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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.
A.
10 cm
B.
5 cm
C.
20 cm
D.
15 cm
Ans
B
Explain it
Answer:
Option
B
Explanation:
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
AB
^{2}
= OA
^{2}
+ OB
^{2}
(5√2)
^{2}
= r
^{2}
+ r
^{2}
50 = 2r
^{2}
r
^{2}
= 25
r = 5 cm
Hence, option B is correct.
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3.
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is
A.
30°
B.
40°
C.
20°
D.
70°
Ans
D
Explain it
Answer:
Option
D
Explanation:
∠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.
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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 ?
A.
b = √2a
B.
a = √2b
C.
a = 2b
D.
b = 2a
Ans
A
Explain it
Answer:
Option
A
Explanation:
OA = OB = OC = OD = r units (radii)
AB = a metre and CD = b metre
∠AOB = 60° and ∠COD = 90°
In ΔCOD, By pythagoras theorem
CD
^{2}
= OC
^{2}
+ OD
^{2}
b
^{2}
= r
^{2}
+ r
^{2}
= 2r
^{2}
...(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.
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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
A.
60°
B.
30°
C.
90°
D.
45°
Ans
C
Explain it
Answer:
Option
C
Explanation:
∠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.
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