nrop: RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES EXERCISE 12B

EXERCISE 12B

PAGE NO-456

Question 1:

(i) In Figure (1), O is the centre of the circle. If ∠OAB = 40° and ∠OCB = 30°, find ∠AOC.
(ii) In Figure (2), A, B and C are three points on the circle with centre O such that ∠AOB = 90° and ∠AOC = 110°. Find ∠BAC.

Answer 1:

(i) Join BO.

In ΔBOC, we have:
OC = OB (Radii of a circle)
⇒ ∠OBC = ∠OCB
∠OBC = 30°                 ...(i)
In ΔBOA, we have:
OB = OA (Radii of a circle)
⇒∠OBA = ∠OAB    [∵ ∠OAB = 40°]
⇒∠OBA = 40°       ...(ii)
Now, we have:
∠ABC = ∠OBC + ∠OBA
          = 30° + 40°    [From (i) and (ii)]
∴ ∠ABC = 70°
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., ∠AOC = 2∠ABC
           = (2 × 70°) = 140°
(ii)

Here, ∠BOC = {360° - (90° + 110°)}
            = (360° - 200°) = 160°
We know that ∠BOC = 2∠BAC
$\Rightarrow ∠ B A C = \frac{∠ B O C}{2} = (\frac{160 °}{2}) = 80 °$ $\Rightarrow ∠ B A C = \frac{∠ B O C}{2} = (\frac{160 °}{2}) = 80 °$
Hence, ∠BAC = 80°

Question 2:

In the given figure, O is the canter of the circle and ∠AOB = 70°.
Calculate the values of (i) ∠OCA, (ii) ∠OAC.

Answer 2:

(i)
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
Thus, ∠AOB = 2∠OCA
$\Rightarrow ∠ O C A = (\frac{∠ A O B}{2}) = (\frac{70 °}{2}) = 35 °$ $\Rightarrow ∠ O C A = (\frac{∠ A O B}{2}) = (\frac{70 °}{2}) = 35 °$

(ii)
OA = OC (Radii of a circle)
∠OAC = ∠OCA [Base angles of an isosceles triangle are equal]
= 35°

PAGE NO-457

Question 3:

In the given figure, O is the centre of the circle. if ∠PBC = 25° and ∠APB = 110°, find the value of ∠ADB.

Answer 3:

From the given diagram, we have:

∠ACB = ∠PCB
∠BPC = (180° - 110°) = 70° (Linear pair)

Considering ΔPCB, we have:
∠PCB + ∠BPC + ∠PBC = 180° (Angle sum property)
⇒ ∠PCB + 70° + 25° = 180°
⇒ ∠PCB = (180° – 95°) = 85°
⇒ ∠ACB = ∠PCB = 85°

We know that the angles in the same segment of a circle are equal.
∴ ∠ADB = ∠ACB = 85°

Question 4:

In the given figure, O is the centre of the circle. If ∠ABD = 35° and ∠BAC = 70°, find ∠ACB.

Answer 4:

It is clear that BD is the diameter of the circle.
Also, we know that the angle in a semicircle is a right angle.
i.e., ∠BAD = 90°
Now, considering the ΔBAD, we have:
∠ADB + ∠BAD + ∠ABD = 180° (Angle sum property of a triangle)
⇒ ∠ADB + 90° + 35° = 180°
⇒ ∠ADB = (180° - 125°) = 55°
Angles in the same segment of a circle are equal.
Hence, ∠ACB = ∠ADB = 55°

Question 5:

In the given figure, O is the centre of the circle. If ∠ACB = 50°, find ∠OAB.

Answer 5:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
∠AOB = 2∠ACB
= 2 × 50° [Given]
∠AOB = 100° ...(i)
Let us consider the triangle ΔOAB.
OA = OB (Radii of a circle)
Thus, ∠OAB = ∠OBA
In ΔOAB, we have:
∠AOB + ∠OAB + ∠OBA = 180°
⇒ 100° + ∠OAB + ∠OAB = 180°
⇒ 100° + 2∠OAB = 180°
⇒ 2∠OAB = 180° – 100° = 80°
⇒ ∠OAB = 40°
Hence, ∠OAB = 40°

Question 6:

In the given figure, ∠ABD = 54° and ∠BCD = 43°, calculate (i) ∠ACD (ii) ∠BAD (iii) ∠BDA.

Answer 6:

(i)
We know that the angles in the same segment of a circle are equal.
i.e., ∠ABD = ∠ACD = 54°

(ii)
We know that the angles in the same segment of a circle are equal.
i.e., ∠BAD = ∠BCD = 43°

(iii)
In ΔABD, we have:
∠BAD + ∠ADB + ∠DBA = 180° (Angle sum property of a triangle)
⇒ 43° + ∠ADB + 54° = 180°
⇒ ∠ADB = (180° – 97°) = 83°
⇒ ∠BDA = 83°

Question 7:

In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If ∠CBD = 60°, calculate ∠CDE.

Answer 7:

Angles in the same segment of a circle are equal.
i.e., ∠CAD = ∠CBD = 60°
We know that an angle in a semicircle is a right angle.
i.e., ∠ADC = 90°
In ΔADC, we have:
∠ACD + ∠ADC + ∠CAD = 180° (Angle sum property of a triangle)
⇒ ∠ACD + 90° + 60° = 180°
⇒∠ACD = 180° – (90° + 60°) = (180° – 150°) = 30°
⇒∠CDE = ∠ACD = 30° (Alternate angles as AC parallel to DE)
Hence, ∠CDE = 30°

Question 8:

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ∠ABC = 25°, calculate ∠CED.

Answer 8:

∠BCD = ∠ABC = 25° (Alternate angles)
Join CO and DO.
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.
Thus, ∠BOD = 2∠BCD
⇒∠BOD = 2 × 25° = 50°
Similarly, ∠AOC = 2∠ABC
⇒ ∠AOC = 2 × 25° = 50°
AB is a straight line passing through the centre.
i.e., ∠AOC + ∠COD + ∠BOD = 180°
⇒ 50° + ∠COD + 50° = 180°
⇒ ∠COD = (180° – 100°) = 80°
$\Rightarrow ∠ C E D = \frac{1}{2} ∠ C O D$ $\Rightarrow ∠ C E D = \frac{1}{2} ∠ C O D$
$\Rightarrow ∠ C E D = (\frac{1}{2} \times 80 °) = 40 °$ $\Rightarrow ∠ C E D = (\frac{1}{2} \times 80 °) = 40 °$
∴ ∠CED = 40°

PAGE NO-458

Question 9:

In the given figure, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find (i) ∠DCE, (ii) ∠ABC.

Answer 9:

(i)
∠CED = 90° (Angle in a semi circle)
In ΔCED, we have:
∠CED +∠EDC + ∠DCE = 180° (Angle sum property of a triangle)
⇒ 90° + 40° + ∠DCE = 180°
⇒ ∠DCE = (180° – 130°) = 50°               ...(i)
∴ ∠DCE = 50°

(ii)
As ∠AOC and ∠BOC are linear pair, we have:
∠BOC = (180° – 80°) = 100°                    ...(ii)
In Δ BOC, we have:
∠OBC + ∠OCB + ∠BOC = 180° (Angle sum property of a triangle)
⇒ ∠ABC + ∠DCE + ∠BOC = 180°     [∵ ∠OBC = ∠ABC and ∠OCB = ∠DCE]
⇒ ∠ABC = 180° – (∠BOC + ∠DCE)
⇒ ∠ABC = 180° – (100° + 50°)          [From (i) and (ii)]
⇒ ∠ABC = (180° - 150°) = 30°

Question 10:

In the given figure, O is the centre of a circle, ∠AOB 40° and ∠BDC = 100°, find ∠OBC.

Answer 10:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
∠AOB = 2∠ACB
= 2∠DCB [∵∠ACB = ∠DCB]
∴ $∠ D C B = \frac{1}{2} ∠ A O B$ $∠ D C B = \frac{1}{2} ∠ A O B$
$\Rightarrow ∠ D C B = (\frac{1}{2} \times 40 °) = 20 °$ $\Rightarrow ∠ D C B = (\frac{1}{2} \times 40 °) = 20 °$
Considering ΔDBC, we have:
∠BDC + ∠DCB + ∠DBC = 180°
⇒ 100° + 20° + ∠DBC = 180°
⇒ ∠DBC = (180° – 120°) = 60°
⇒ ∠OBC = ∠DBC = 60°
Hence, ∠OBC = 60°

Question 11:

In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If ∠OAB = 25°, calculate ∠EBC.

Answer 11:

OA = OB (Radii of a circle)
Thus, ∠OBA = ∠OAB = 25°
Join OB.

Now in ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180° (Angle sum property of a triangle)
$\Rightarrow$ $\Rightarrow$ 25° + 25° + ∠AOB = 180°
$\Rightarrow$ $\Rightarrow$ 50° + ∠AOB = 180°
$\Rightarrow$ $\Rightarrow$ ∠AOB = (180° – 50°) = 130°

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., ∠AOB = 2∠ACB
$\Rightarrow$ $\Rightarrow$ $∠ A C B = \frac{1}{2} ∠ A O B = (\frac{1}{2} \times 130 °) = 65 °$ $∠ A C B = \frac{1}{2} ∠ A O B = (\frac{1}{2} \times 130 °) = 65 °$
Here,∠ACB = ∠ECB
∴ ∠ECB = 65°   ...(i)

Considering the right angled ΔBEC, we have:
∠EBC + ∠BEC + ∠ECB = 180°     (Angle sum property of a triangle)
$\Rightarrow$ $\Rightarrow$ ∠EBC + 90° + 65° = 180°    [From(i)]
$\Rightarrow$ $\Rightarrow$ ∠EBC = (180° – 155°) = 25°
Hence, ∠EBC = 25°

Question 12:

In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 55°. Find (i) ∠BOC, (ii) ∠AOC

Answer 12:

(i)
OB = OC (Radii of a circle)
⇒ ∠OBC = ∠OCB = 55°
Considering ΔBOC, we have:
∠BOC + ∠OCB + ∠OBC = 180° (Angle sum property of a triangle)
⇒∠BOC + 55° + 55° = 180°
⇒∠BOC = (180° - 110°) = 70°

(ii)
OA = OB          (Radii of a circle)
⇒ ∠OBA = ∠OAB = 20°
Considering ΔAOB, we have:
∠AOB + ∠OAB + ∠OBA = 180°    (Angle sum property of a triangle)
⇒∠AOB + 20° + 20° = 180°
⇒∠AOB = (180° - 40°) = 140°
∴ ∠AOC = ∠AOB - ∠BOC
              = (140° - 70°)
           = 70°
Hence, ∠AOC = 70°

Question 13:

In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.

Answer 13:

In the given figure, OD is parallel to BC.

∴ ∠BCO = ∠COD    (Alternate interior angles)
⇒ $∠ C O D = 30 °$ $∠ C O D = 30 °$        ...(1)

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CD subtends ∠COD at the centre and ∠CBD at B on the circle.

∴ ∠COD = 2∠CBD
⇒ $∠ C B D = \frac{30 °}{2} = 15 °$ $∠ C B D = \frac{30 °}{2} = 15 °$            (from (1))

∴ $y = 15 °$ $y = 15 °$              ...(2)

Also, arc AD subtends ∠AOD at the centre and ∠ABD at B on the circle.

∴ ∠AOD = 2∠ABD
⇒ $∠ A B D = \frac{90 °}{2} = 45 °$ $∠ A B D = \frac{90 °}{2} = 45 °$          ...(3)

In ∆ABE,
x + y + ∠ABD + ∠AEB = 180^∘       (Sum of the angles of a triangle)
⇒ x + 15^∘ + 45^∘ + 90^∘ = 180^∘        (from (2) and (3))
⇒ x = 180^∘ − (90^∘+ 15^∘ + 45^∘)
⇒ x = 180^∘ − 150^∘
⇒ x = 30^∘

Hence, x = 30^∘ and y = 15^∘.

PAGE NO-459

Question 14:

In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Answer 14:

In the given figure, BD = OD and CD ⊥ AB.

Join AC and OC.

In ∆ODE and ∆DBE,
∠DOE = ∠DBE      (given)
∠DEO = ∠DEB = 90^∘
OD = DB     (given)
∴ By AAS conguence rule, ∆ODE ≌ ∆BDE,

Thus, OE = EB      ...(1)

Now, in ∆COE and ∆CBE,
CE = CE      (common)
∠CEO = ∠CEB = 90^∘
OE = EB     (from (1))
∴ By SAS conguence rule, ∆COE ≌ ∆CBE,

Thus, CO = CB      ...(2)

Also, CO = OB = OA (radius of the circle)         ...(3)

From (2) and (3),
CO = CB = OB
∴ ∆COB is equilateral triangle.
∴ ∠COB = 60^∘         ...(4)

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CB subtends ∠COB at the centre and ∠CAB at A on the circle.

∴ ∠COB = 2∠CAB
⇒ $∠ C A B = \frac{60 °}{2} = 30 °$ $∠ C A B = \frac{60 °}{2} = 30 °$            (from (4))

Hence, ∠CAB = 30^∘.

Question 15:

In the given figure, PQ is a diameter of a circle with centre O. If ∠PQR = 65°, ∠SPR = 40° and ∠PQM = 50°, find ∠QPR, ∠QPM and ∠PRS.

Answer 15:

Here, PQ is the diameter and the angle in a semicircle is a right angle.
i.e., ∠PRQ = 90°
In ΔPRQ, we have:
∠QPR + ∠PRQ + ∠PQR = 180° (Angle sum property of a triangle)
⇒ ∠QPR + 90° + 65° = 180°
⇒∠QPR = (180° – 155°) = 25°

In ΔPQM, PQ is the diameter.
∴∠PMQ = 90°
In ΔPQM, we have:
∠QPM + ∠PMQ + ∠PQM = 180° (Angle sum property of a triangle)
⇒∠QPM + 90° + 50° = 180°
⇒ ∠QPM = (180° – 140°) = 40°
Now, in quadrilateral PQRS, we have:
∠QPS + ∠SRQ = 180° (Opposite angles of a cyclic quadrilateral)
⇒∠QPR + ∠RPS + ∠PRQ + ∠PRS = 180°
⇒ 25° + 40° + 90° + ∠PRS = 180°
⇒ ∠PRS = 180° – 155° = 25°
∴ ∠PRS = 25°

Thus, ∠QPR = 25°; ∠QPM = 40°; ∠PRS = 25°

Question 16:

In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line.

If ∠APB = 150° and ∠BQD = x°, find the value of x.

Answer 16:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc AEB subtends ∠APB at the centre and ∠ACB at C on the circle.

∴ ∠APB = 2∠ACB
⇒ $∠ A C B = \frac{150 °}{2} = 75 °$ $∠ A C B = \frac{150 °}{2} = 75 °$            ...(1)

Since ACD is a straight line, ∠ACB + ∠BCD = 180^∘
⇒ ∠BCD = 180^∘ − 75^∘
⇒ ∠BCD = 105^∘            ...(2)

Also, arc BFD subtends reflex ∠BQD at the centre and ∠BCD at C on the circle.

∴ reflex ∠BQD = 2∠BCD
⇒ $reflex ∠ B Q D = 2 (105 °) = 210 °$ $reflex ∠ B Q D = 2 (105 °) = 210 °$            ...(3)

Now,
reflex ∠BQD + ∠BQD = 360^∘
⇒ 210^∘ + x = 360^∘
⇒ x = 360^∘ − 210^∘
⇒ x = 150^∘

Hence, x = 150^∘.

Question 17:

In the given figure, ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of ∆ABC whose centre is O.

Answer 17:

Join OB and OC.
∠BOC = 2∠BAC (As angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference)
            = 2 × 30°       [∵ ∠BAC = 30°]
            = 60°           ...(i)
Consider ΔBOC, we have:
OB = OC       [Radii of a circle]
⇒ ∠OBC = ∠OCB           ...(ii)
In ΔBOC, we have:
∠BOC + ∠OBC + ∠OCB = 180        (Angle sum property of a triangle)
⇒ 60° + ∠OCB + ∠OCB = 180°       [From (i) and (ii)]
⇒ 2∠OCB = (180° - 60°) = 120°
⇒ ∠OCB = 60°               ...(ii)
Thus we have:
∠OBC = ∠OCB = ∠BOC = 60°
Hence, ΔBOC is an equilateral triangle.
i.e., OB = OC = BC
∴ BC is the radius of the circumcircle.

Question 18:

In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E.
Prove that ∠AEC = $\frac{1}{2}$ $\frac{1}{2}$ (angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).

Answer 18:

Join AD

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc AXC subtends ∠AOC at the centre and ∠ADC at D on the circle.

∴ ∠AOC = 2∠ADC
⇒ $∠ A D C = \frac{1}{2} (∠ A O C)$ $∠ A D C = \frac{1}{2} (∠ A O C)$      ...(1)

Also, arc DYB subtends ∠DOB at the centre and ∠DAB at A on the circle.

∴ ∠DOB = 2∠DAB
⇒ $∠ D A B = \frac{1}{2} (∠ D O B)$ $∠ D A B = \frac{1}{2} (∠ D O B)$            ...(2)

Now, in ∆ADE,
∠AEC = ∠ADC + ∠DAB      (Exterior angle)
⇒ ∠AEC = $\frac{1}{2} (∠ A O C + ∠ D O B)$ $\frac{1}{2} (∠ A O C + ∠ D O B)$         (from (1) and (2))

Hence, ∠AEC = $\frac{1}{2}$ $\frac{1}{2}$ (angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).

RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES EXERCISE 12B