Class 9 R S AGGARWAL AND V AGGARWAL Solutions Maths Chapter 12: Circles
Circles Exercise MCQ
Solution 1
Correct option: (b)
Solution 2
Correct option: (c)
Solution 3
Solution 4
Solution 5
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Solution 7
Solution 8
Solution 9
Solution 10
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Solution 12
Solution 13
Solution 14
Solution 15
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Solution 17
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Solution 42
Circles Exercise Ex. 12A
Solution 1
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Solution 17
Let O be the centre of a circle with radius r.
⇒ OB = OC = r
Let AC = x
Then, AB = 2x
Let OM ⊥ AB
⇒ OM = p
Let ON ⊥ AC
⇒ ON = q
In ΔOMB, by Pythagoras theorem,
OB2 = OM2 + BM2
In ΔONC, by Pythagoras theorem,
OC2 = ON2 + CN2
Solution 18
Solution 19
Solution 20
Solution 21
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Solution 23
Draw OM ⊥ PQ and O'N ⊥ PQ
⇒ OM ⊥ AP
⇒ AM = PM (perpendicular from the centre of a circle bisects the chord)
⇒ AP = 2AM ….(i)
And, O'N ⊥ PQ
⇒ O'N ⊥ AQ
⇒ AN = QN (perpendicular from the centre of a circle bisects the chord)
⇒ AQ = 2AN ….(ii)
Now,
PQ = AP + PQ
⇒ PQ = 2AM + 2AN
⇒ PQ = 2(AM + AN)
⇒ PQ = 2MN
⇒ PQ = 2OO' (since MNO'O is a rectangle)
Circles Exercise Ex. 12B
Solution 1
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Solution 13
Given, ∠AOD = 90° and ∠OEC = 90°
⇒ ∠AOD = ∠OEC
But ∠AOD and ∠OEC are corresponding angles.
⇒ OD || BC and OC is the transversal.
∴ ∠DOC = ∠OCE (alternate angles)
⇒ ∠DOC = 30° (since ∠OCE = 30°)
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠DOC = 2∠DBC
Now, ∠ABE = ∠ABC = ∠ABD + ∠DBC = 45° + 15° = 60°
In ΔABE,
∠BAE + ∠AEB + ∠ABE = 180°
⇒ x + 90° + 60° = 180°
⇒ x + 150° = 180°
⇒ x = 30°
Solution 14
Construction: Join AC
Given, BD = OD
Now, OD = OB (radii of same circle)
⇒ BD = OD = OB
⇒ ΔODB is an equilateral triangle.
⇒ ∠ODB = 60°
We know that the altitude of an equilateral triangle bisects the vertical angle.
Now, ∠CAB = ∠BDC (angles in the same segment)
⇒ ∠CAB = ∠BDE = 30°
Solution 15
Solution 16
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠APB = 2∠ACB
Now, ACD is a straight line.
⇒ ∠ACB + ∠DCB = 180°
⇒ 75° + ∠DCB = 180°
⇒ ∠DCB = 105°
Again,
Solution 17
Solution 18
Join AC and BC
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠AOC = 2∠ABC ….(i)
Similarly, ∠BOD = 2∠BCD ….(ii)
Adding (i) and (ii),
∠AOC + ∠BOD = 2∠ABC + 2∠BCD
⇒ ∠AOC + ∠BOD = 2(∠ABC + ∠BCD)
⇒ ∠AOC + ∠BOD = 2(∠EBC + ∠BCE)
⇒ ∠AOC + ∠BOD = 2(180° - ∠CEB)
⇒ ∠AOC + ∠BOD = 2(180° - [180° - ∠AEC])
⇒ ∠AOC + ∠BOD = 2∠AEC
Circles Exercise Ex. 12C
Solution 1
Solution 2
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Solution 13
