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#### Page No 436:

Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 cm From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
BM =
In the right  ΔOMB, we have:
OB2 = OM2 + MB2   (Pythagoras theorem)
⇒ 102 = OM2 + 82
⇒ 100 = OM2 + 64
OM2 = (100 - 64) = 36

Hence, the distance of the chord from the centre is 6 cm.

#### Page No 436:

Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm From the right ΔOMB, we have:
OB2 = OM2 + MB2        (Pythagoras theorem)
⇒ 52 = 32 + MB2
⇒ 25 = 9 + MB2
MB2 = (25 9) = 16

Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 4) cm = 8 cm
Hence, the required length of the chord is 8 cm.

#### Page No 436:

Let AB be the chord of the given circle with centre O. The perpendicular distance from the centre of the circle to the chord is 8 cm.
Join OB.
Then OM = 8 cm and AB = 30 cm We know that the perpendicular from the centre of a circle to a chord bisects the chord.

From the right ΔOMB, we have:
OB2 = OM2 + MB2
OB2 = 82 + 152
OB2 = 64 + 225
OB2 = 289

Hence, the  required length of the radius is 17 cm.

#### Page No 436:

We have:
(i)
Let AB and CD be two chords of a circle such that AB is parallel to CD on the same side of the circle.
Given: AB = 8 cm, CD = 6 cm and OB = OD = 5 cm
Join OL and OM. The perpendicular from the centre of a circle to a chord bisects the chord.
∴ $LB\mathit{=}\frac{\mathit{A}\mathit{B}}{\mathit{2}}=\left(\frac{8}{2}\right)=4\mathrm{cm}$
Now, in right angled ΔBLO, we have:
OB2 = LB2 + LO2
LO2 = OB2 LB2
⇒ LO2 = 52 − 42
⇒ LO2 = 25 − 16 = 9
LO = 3 cm

Similarly, $MD\mathit{=}\frac{\mathit{C}\mathit{D}}{\mathit{2}}=\left(\frac{6}{2}\right)=3\mathrm{cm}$
In right angled ΔDMO, we have:
OD2 = MD2 + MO2
MO2 = OD2 MD2
MO2   = 52 − 32
MO2 = 25 − 9 = 16
MO = 4 cm
∴ Distance between the chords = (MO LO) = (4 − 3) cm = 1 cm

(ii)
Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the centre.
Given: AB = 8 cm and CD = 6 cm
Draw OL AB and OM CD. Join OA and OC.
OA = OC = 5 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ $AL\mathit{=}\frac{\mathit{A}\mathit{B}}{\mathit{2}}=\left(\frac{8}{2}\right)=4\mathrm{cm}$
Now, in right angled ΔOLA, we have:
OA2 = AL2 + LO2
LO2 = OA2 − AL2
LO2 = 52 42
⇒ LO2 = 25 16 = 9
LO = 3 cm
Similarly, $CM\mathit{=}\frac{\mathit{C}\mathit{D}}{\mathit{2}}=\left(\frac{6}{2}\right)=3\mathrm{cm}$
In right angled ΔCMO, we have:
OC2 = CM2 + MO2
MO2 = OC2 − CM2
MO2 = 52 32
MO2 = 25 9 = 16
MO = 4 cm
Hence, distance between the chords = (MO + LO) = (4 + 3) cm = 7 cm

#### Page No 436:

Let AB and CD be two chords of a circle such that AB is parallel to CD and they are on the opposite sides of the centre.
Given: AB = 30 cm and CD = 16 cm
Draw OL AB and OM CD. Join OA and OC.
OA = OC = 17 cm (Radii of a circle)
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL=AB2=(82)=4cm
Now, in right angled ΔOLA, we have:
OA2 = AL2 + LO2
LO2 = OA2 AL2
LO2 = 172 − 152
LO2 = 289 − 225 = 64
LO = 8 cm

Similarly, CM=CD2=(62)=3cm
In right angled ΔCMO, we have:
OC2 = CM2 + MO2
MO2 = OC2 CM2
MO2 = 172 − 82
MO2  = 289 − 64 = 225
MO = 15 cm

Hence, distance between the chords = (LO + MO) = (8 + 15) cm = 23 cm

#### Page No 436:

CD is the diameter of the circle with centre O and is perpendicular to chord AB.
Join OA. Given: AB = 12 cm and CE = 3 cm
Let OA = OC = r cm   (Radii of a circle)
Then OE = (r - 3) cm
Since the perpendicular from the centre of the circle to a chord bisects the chord, we have:

Now, in right angled ΔOEA, we have:
⇒ OA2 = OE2 + AE2
⇒  r2 = (r − 3)2 + 62
⇒  r2 = r2 − 6r + 9 + 36
r2r2 + 6r = 45
⇒ 6r = 45

r = 7.5 cm
Hence, the required radius of the circle is 7.5 cm.

#### Page No 436:

AB is the diameter of the circle with centre O, which bisects the chord CD at point E.
Given: CE = ED = 8 cm and EB = 4 cm
Join OC. Let OC = OB = r cm   (Radii of a circle)
Then OE = (r − 4) cm
Now, in right angled ΔOEC, we have:
OC2 = OE2 + EC2      (Pythagoras theorem)
⇒  r2 = (r − 4)2 + 82
⇒  r2 = r2 − 8r + 16 + 64
r2r2 + 8r = 80
⇒ 8r = 80

r = 10 cm
Hence, the required radius of the circle is 10 cm.

#### Page No 437: Given:
BC is a diameter of a circle with centre O and OD AB.
To prove: AC parallel to OD and AC = 2 × OD
Construction: Join AC.
Proof:
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Here, OD AB
D is the mid point of AB.
Also, O is the mid point of BC.
i.e., OC = OB
Now, in ΔABC, we have:
D is the mid point of AB and O is the mid point of BC.
According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.

AC = 2 × OD

Hence, proved.

#### Page No 437: Given:
O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠BPD.
To prove: AB = CD
Construction: Draw OE AB and OF CD
Proof: In ΔOEP and ΔOFP, we have:
∠OEP = ∠OFP         (90° each)
OP = OP                   (Common)
OPE = ∠OPF         (∵ OP bisects ∠BPD )
Thus, ΔOEP ≅ ΔOFP      (AAS criterion)
OE = OF
Thus, chords AB and CD are equidistant from the centre O.
⇒  AB = CD         (∵ Chords equidistant from the centre are equal)
AB =  CD

#### Page No 437: Given:
AB and CD are two parallel chords of a circle with centre O.
POQ is a diameter which is perpendicular to AB.
To prove: PF CD and CF = FD
Proof:
AB || CD and POQ is a diameter.
PEB = 90°    (Given)
∠PFD = ∠PEB          (∵ AB || CD, Corresponding angles)
Thus, PF CD
OF CD
We know that the perpendicular from the centre to a chord bisects the chord.
i.e., CF = FD
Hence, POQ is perpendicular to CD and bisects it.

#### Page No 437:

Given: Two distinct circles
To prove: Two distinct circles cannot intersect each other in more than two points.
Proof: Suppose that two distinct circles intersect each other in more than two points.
∴ These points are non-collinear points.
Three non-collinear points determine one and only one circle.
∴ There should be only one circle.
This contradicts the given, which shows that our assumption is wrong.
Hence, two distinct circles cannot intersect each other in more than two points.

#### Page No 437: Given: OA = 10 cm, O'A = 8 cm and AB = 12 cm

Now, in right angled ΔADO, we have:
= 102 - 62
= 100 - 36 = 64
OD = 8 cm

Similarly, in right angled ΔADO', we have:
= 82 - 62
= 64 - 36
= 28
$O\mathit{\text{'}}D=\sqrt{28}=2\sqrt{7}$ cm
Thus, OO' = (OD + O'D)
=
Hence, the distance between their centres is $\left(8+2\sqrt{7}\right)\mathrm{cm}$.

#### Page No 437:

Given: Two equal circles intersect at point P and Q.
A straight line passes through P and meets the circle at points A and B.
To prove: QA = QB
Construction: Join PQ. Proof:
Two circles will be congruent if and only if they have equal radii.
Here, PQ is the common chord to both the circles.
Thus, their corresponding arcs are equal (if two chords of a circle are equal, then their corresponding arcs are congruent).
So, arc PCQ = arc PDQ
∴ ∠QAP = ∠QBP (Congruent arcs have the same degree in measure)
Hence,  QA = QB     (In isosceles triangle, base angles are equal)

#### Page No 437:

Given: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at points L and M.
To prove: AB || CD
Proof: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M. Then OL AB
Also, OM CD
∴ ∠ ALM = ∠ LMD = 90o
Since alternate angles are equal, we have:
AB|| CD

#### Page No 437:

Two circles with centres A and B of respective radii 5 cm and 3 cm touch each other internally.
The perpendicular bisector of AB meets the bigger circle at P and Q.
Join AP. Let PQ intersect AB at point L.
Here, AP = 5 cm
Then AB = (5 - 3) cm = 2 cm
Since PQ is the perpendicular bisector of AB, we have:

Now, in right angled ΔPLA, we have:
AP2 = AL2 + PL2
PL2 = AP2 - AL2
=  52 - 12
= 25 - 1 = 24

Thus PQ = 2 × PL
=  $\left(2×2\sqrt{6}\right)=4\sqrt{6}\mathrm{cm}$
Hence, the required length of PQ is $4\sqrt{6}\mathrm{cm}$.

#### Page No 438:

We have:
OB = OC, ∠BOC = ∠BCO = y
External ∠OBA = ∠BOC + ∠BCO = (2y)
Again, OA = OB, ∠OAB = ∠OBA = (2y)
External ∠AOD = ∠OAC + ∠ACO
Or x = ∠OAB + ∠BCO
Or x = (2y) + y = 3y
Hence, x = 3y

#### Page No 438:

Let AC = a.
Since, AB = 2AC, ∴ AB = 2a. From centre O, perpendicular is drawn to the chords AB and AC at points M and N, respectively.

It is given that OM = p and ON = q.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

AM = MB = a                     ...(1)
and AN = NC = $\frac{a}{2}$                   ...(2)

In ∆OAN,
(AN)2 + (NO)2 = (OA)2          (Pythagoras theorem)

In ∆OAM,
(AM)2 + (MO)2 = (OA)2          (Pythagoras theorem)

From eq. (3) and (4),
$4{r}^{2}-4{q}^{2}={r}^{2}-{p}^{2}\phantom{\rule{0ex}{0ex}}⇒4{r}^{2}-{r}^{2}+{p}^{2}=4{q}^{2}\phantom{\rule{0ex}{0ex}}⇒3{r}^{2}+{p}^{2}=4{q}^{2}$

Hence, 4q2 = p2 + 3r2.

#### Page No 438:

Given: AB and AC are chords of the circle with centre O. AB = AC, OP ⊥ AB and OQ ⊥ AC To prove: PB = QC
Proof:
AB = AC      (Given)
$\frac{1}{2}\mathrm{AB}=\frac{1}{2}\mathrm{AC}$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ MB = NC            ...(i)
Also, OM = ON    (Equal chords of a circle are equidistant from the centre)
⇒ OP - OM = OQ - ON
∴ PM = QN          ...(ii)
Now, in ΔMPB and ΔNQC, we have:
MB = NC                [From (i)]
∠PMB = ∠QNC     [90° each]
PM = QN                [From (ii)]
i.e., ΔMPB ≅ ΔNQC    (SAS criterion)
∴ PB = QC        (CPCT)

#### Page No 438:

Given: BC is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
TO prove: AB = CD
Construction: Draw OL AB and OM CD. Proof:
In ΔOLB and ΔOMC, we have:
∠OLB = ∠OMC        [90° each]
∠OBL = ∠OCD          [Alternate angles as AB || CD]
OB = OC                      [Radii of a circle]
∴ ΔOLB ≅ ΔOMC   (AAS criterion)
Thus, OL = OM   (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD

#### Page No 438:

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its median. Then, AD BC     (ΔABC is an equilateral triangle)
Also, $BD\mathit{=}\left(\frac{BC}{2}\right)=\left(\frac{9}{2}\right)=4.5\mathrm{cm}$
In right angled ΔADB, we have:
$⇒AD=\sqrt{A{B}^{2}-B{D}^{2}}$
$=\sqrt{{\left(9\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}\mathrm{cm}\phantom{\rule{0ex}{0ex}}$
$=\frac{9\sqrt{3}}{2}\mathrm{cm}$
In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.
Now, radius = $AG\mathit{=}\frac{\mathit{2}}{\mathit{3}}AD$
$⇒AG=\left(\frac{2}{3}×\frac{9\sqrt{3}}{2}\right)=3\sqrt{3}\mathrm{cm}$
∴ The radius of the circle is $3\sqrt{3}\mathrm{cm}$.

#### Page No 438:

Given: AB and AC are two equal chords of a circle with centre O. To prove: ∠OAB = ∠OAC
Construction: Join OA, OB and OC.
Proof:
In ΔOAB and ΔOAC, we have:
AB = AC         (Given)
OA = OA        (Common)
OB = OC         (Radii of a circle)
∴ Δ OAB Δ OAC  (By SSS congruency rule)
∠OAB = ∠OAC    (CPCT)
Hence, point O lies on the bisector of ∠BAC.

#### Page No 438:

Given: OPQR is a square. A circle with centre O cuts the square at X and Y.
To prove: QX = QY
Construction: Join OX and OY. Proof:
In ΔOXP and ΔOYR, we have:
∠OPX = ∠ORY      (90° each)
OX = OY                (Radii of a circle)
OP = OR                (Sides of a square)
∴ ΔOXPΔOYR    (BY RHS congruency rule)
PX = RY              (By CPCT)
PQ - PX = QR - RY   (PQ and QR are sides of a square)
QX = QY
Hence, proved.

#### Page No 439:

Given: Two circles with centres O and O' intersect at two points A and B. Draw a line PQ parallel to OO' through B, OX perpendicular to PQ, O'Y perpendicular to PQ, join all.

We know that perpendicular drawn from the centre to the chord, bisects the chord.

PX = XB and YQ = BY

PX + YQ = XB + BY

On adding XB + BY on both sides, we get

PX + YQ + XB + BY = 2(XB + BY)
PQ = 2(XY)
PQ = 2(OO')

Hence, PQ = 2OO'.

#### Page No 456:

(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 = 2ABC
= (2 × 70°) = 140°
(ii) Here, BOC = {360° - (90° + 110°)}
= (360° - 200°) = 160°
We know that BOC = 2BAC
$⇒\angle BAC\mathit{=}\frac{\mathit{\angle }\mathit{B}\mathit{O}\mathit{C}}{\mathit{2}}=\left(\frac{160°}{2}\right)=80°$
Hence, BAC = 80°

#### Page No 456: (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 = 2OCA
$⇒\angle OCA=\left(\frac{\angle AOB}{2}\right)=\left(\frac{70°}{2}\right)=35°$

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

#### Page No 457:

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.

#### Page No 457: It is clear that BD is the diameter of the circle.
Also, we know that the angle in a semicircle is a right angle.
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°

#### Page No 457: 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 = 2ACB
= 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° + 2OAB = 180°
⇒ 2OAB = 180° – 100° = 80°
OAB = 40°
Hence, OAB = 40°

#### Page No 457: (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°

#### Page No 457: 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.
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°

#### Page No 457: 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 = 2BCD
BOD = 2 × 25° = 50°
Similarly, AOC = 2ABC
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°
$⇒\angle CED=\frac{1}{2}\angle COD\phantom{\rule{0ex}{0ex}}$
$⇒\angle CED=\left(\frac{1}{2}×80°\right)=40°$
CED = 40°

#### Page No 458: (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°

#### Page No 458: 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 = 2ACB
= 2DCB       [∵ACB = DCB]
$\angle DCB=\frac{1}{2}\angle AOB\phantom{\rule{0ex}{0ex}}$
$⇒\angle DCB=\left(\frac{1}{2}×40°\right)=20°$
Considering ΔDBC, we have:
BDC + DCB + DBC = 180°
⇒ 100° + 20° + DBC = 180°
DBC = (180° – 120°) = 60°
OBC = DBC = 60°
Hence, OBC = 60°

#### Page No 458:

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)
$⇒$25° + 25° + AOB = 180°
$⇒$50° + AOB = 180°
$⇒$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 = 2ACB
$⇒$$\angle ACB\mathit{=}\frac{\mathit{1}}{\mathit{2}}\mathit{\angle }AOB=\left(\frac{1}{2}×130°\right)=65°$
Here,ACB = ECB
∴ ECB = 65°   ...(i)

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

#### Page No 458: (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°

#### Page No 458:

In the given figure, OD is parallel to BC.

∴ ∠BCO = ∠COD    (Alternate interior angles)
$\angle COD=30°$       ...(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
$\angle CBD=\frac{30°}{2}=15°$
(from (1))

$y=15°$             ...(2)

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

∴ ∠AOD = 2∠ABD
$\angle ABD=\frac{90°}{2}=45°$
...(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:

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
$\angle CAB=\frac{60°}{2}=30°$
(from (4))

Hence, ∠CAB = 30.

#### Page No 459: 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°

#### Page No 459:

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
$\angle ACB=\frac{150°}{2}=75°$
...(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

...(3)

Now,
reflex ∠BQD + ∠BQD = 360
210 + x = 360
x = 360 210
x = 150

Hence, x = 150.

#### Page No 459: Join OB and OC.
BOC = 2BAC (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)]
⇒ 2OCB = (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.

#### Page No 459: 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.

$\angle ADC=\frac{1}{2}\left(\angle AOC\right)$
...(1)

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

∴ ∠DOB = 2∠DAB
$\angle DAB=\frac{1}{2}\left(\angle DOB\right)$
...(2)

AEC = ∠ADC + DAB      (Exterior angle)
⇒ ∠AEC = $\frac{1}{2}\left(\angle AOC+\angle DOB\right)$        (from (1) and (2))

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

#### Page No 482:

(i) ∠BDC = ∠BAC = 40°  (Angles in the same segment)
In
ΔBCD, we have:
∠BCD + ∠DBC + ∠BDC = 180°  (Angle sum property of a triangle)
⇒ ∠BCD + 60° + 40° = 180°
⇒ ∠BCD = (180° - 100°) = 80°

(ii) ∠CAD = ∠CBD  (Angles in the same segment)
= 60°

#### Page No 482: In cyclic quadrilateral PQRS, we have:
∠PSR + ∠PQR = 180°
⇒ 150° + ∠PQR = 180°
⇒ ∠PQR = (180° – 150°) = 30°
∴ ∠PQR = 30°                ...(i)
Also, ∠PRQ = 90° (Angle in a semicircle)                 ...(ii)
Now, in ΔPRQ, we have:
∠PQR + ∠PRQ + ∠RPQ = 180°
⇒ 30° + 90° + ∠RPQ = 180°   [From(i) and (ii)]
⇒ ∠RPQ = 180° – 120° = 60°
∴ ∠RPQ = 60°

#### Page No 482:

Reflex ∠AOC + ∠AOC = 360
Reflex ∠AOC + 130 + x = 360
Reflex ∠AOC = 360 − 130
Reflex ∠AOC = 230

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 AC subtends reflex ∠AOC at the centre and ∠ABC at B on the circle.

∴ ∠AOC = 2∠ABC
$\angle ABC=\frac{230°}{2}=115°$
...(1)

Since ABP is a straight line, ∠ABC + PBC = 180
⇒ ∠PBC = 180 − 115
⇒ ∠PBC = 65            ...(2)

Hence, ∠PBC = 65.

#### Page No 482: Given: ABCD is a cyclic quadrilateral.

Then ABC + ADC = 180°
⇒ 92° + ADC = 180°
ADC = (180° – 92°) = 88°
Again, AE parallel to CD.
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
BCD = DAF
=  88° + 20° = 108°
Hence, BCD = 108°

#### Page No 482: BD = DC
BCD = CBD = 30°
In ΔBCD, we have:
BCD + CBD + CDB = 180°  (Angle sum property of a triangle)
⇒ 30° + 30° + CDB = 180°
CDB = (180° – 60°) = 120°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, CDB + BAC = 180°
⇒ 120° + BAC = 180°
BAC = (180° – 120°) = 60°
BAC = 60°

#### Page No 482: We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
⇒ 50° + ABC = 180°
ABC = (180° – 50°) = 130°
ADC = 50° and ABC = 130°

#### Page No 483: (i)
Given: ΔABC is an equilateral triangle.
i.e., each of its angle = 60°
BAC = ABC = ACB = 60°
Angles in the same segment of a circle are equal.
i.e., BDC = BAC = 60°
BDC = 60°
(ii)
The opposite angles of a cyclic quadrilateral are supplementary.
Then in cyclic quadrilateral ABEC, we have:
BAC + BEC = 180°
⇒ 60° + BEC = 180°
BEC = (180° – 60°) = 120°
BDC = 60° and BEC = 120°

#### Page No 483: Given: ABCD is a cyclic quadrilateral.
DAB + DCB = 180°   ( Opposite angles of  a cyclic quadrilateral are supplementary)
DAB + 100° = 180°
DAB = (180° – 100°) = 80°
Now, in ΔABD, we have:
DAB + ABD + ADB = 180°
⇒ 80° + 50° + ADB = 180°
ADB = (180° – 130°) = 50°

#### Page No 483: O is the centre of the circle and BOD = 150°.
Thus, reflex angle BOD = (360° – 150°) = 210°
Now, $x=\frac{1}{2}\left(\mathrm{reflex}\angle BOD\right)=\left(\frac{1}{2}×210°\right)=105°$
x = 105°
Again, x + y = 180° (Opposite angles of a cyclic quadrilateral)
⇒ 105° + y = 180°
y = (108° - 105°)= 75°
y = 75°
Hence,
x = 105° and y = 75°

#### Page No 483: O is the centre of the circle and DAB = 50°.
OA = OB (Radii of a circle)
OBA = OAB = 50°
In ΔOAB, we have:
OAB + OBA + AOB = 180°
⇒ 50° + 50° +AOB = 180°
AOB = (180° – 100°) = 80°
Since AOD is a straight line, we have:
x = 180°AOB
= (180° – 80°) = 100°
i.e., x = 100°
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, DAB + BCD = 180°
BCD = (180° – 50°) = 130°
y = 130°
Hence, x = 100° and y = 130°

#### Page No 483: We know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
CBF = CDA
CBF = (180°x)
⇒ 130° = 180°x   [∵ CBF = 130°]
x = (180° – 130°) = 50°
Hence, x = 50°

#### Page No 484:

We have,
AB is a diameter of the circle where O is the centre, DO || BC and BCD = 120°.
(i)
Since ABCD is a cyclic quadrilateral, we have:
⇒ 120° + BAD = 180°
BAD = (180° – 120°) = 60°
(ii)
BDA = 90° (Angle in a semicircle)
In Δ ABD, we have:
BDA + BAD + ABD = 180°
⇒ 90° + 60° + ABD = 180°
ABD = (180° – 150°) = 30°
ABD = 30°
(iii)
OD = OA (Radii of a circle)
= 60°
ODB = 90° - ODA = (90° - 60°) = 30°
Here, DO || BC (Given; alternate angles)
CBD = ODB = 30°
∠CBD = 30°
(iv)
= 90° + 30° = 120°
In ΔAOD, we have:
ODA + OAD +AOD = 180°
⇒ 60° + 60° + AOD = 180°
AOD = 180° – 120° = 60°

Since all the angles of ΔAOD are of 60° each, ΔAOD is an equilateral triangle.

#### Page No 484: AB and CD are two chords of a circle which intersect each other at P outside the circle.
AB = 6 cm, BP = 2 cm and PD = 2.5 cm
∴  AP × BP = CP × DP
⇒ 8 × 2 = (CD + 2.5) × 2.5  [∵ CP = CD + DP]
Let CD = x cm
Thus, 8 × 2 = (CD + 2.5) × 2.5
⇒ 16 = 2.5x + 6.25
⇒ 2.5x = (16 - 6.25) = 9.75

$x=\frac{9.75}{2.5}=3.9$

Hence, CD = 3.9 cm

#### Page No 484: O is the centre of the circle where AOD = 140° and CAB = 50°.
(i) BOD = 180°AOD
= (180° – 140°) = 40°
We have the following:
OB = OD (Radii of a circle)
OBD = ODB

In ΔOBD, we have:
BOD + OBD + ODB = 180°
BOD + OBD + OBD = 180°      [∵ OBD = ODB]
⇒ 40° +2OBD = 180°
⇒ 2OBD = (180° – 40°) = 140°
OBD = 70°
Since ABCD is a cyclic quadrilateral, we have:
CAB + BDC = 180°
CAB + ODB + ODC = 180°
⇒ 50° + 70° + ODC = 180°
ODC = (180° – 120°) = 60°
ODC = 60°
EDB = (180° – (ODC + ODB)
= 180° – (60° + 70°)
= 180° – 130° = 50°
∴ EDB = 50°

(ii) EBD = 180° - ∠OBD
= 180° - 70°
= 110°

#### Page No 484: ABC is an isosceles triangle.
Here, AB = AC
∴ ∠ACB = ∠ABC   ...(i)
= ∠ABC   [from(i)]
∴ ∠ADE = ∠ABC  (Corresponding angles)
Hence, DE || BC

#### Page No 484: AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E.
If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
∴ Exterior ∠EDC = ∠A   ...(i)
Exterior ∠DCE = ∠B    ...(ii)
Also, AB parallel to CD.
Then, ∠EDC = ∠B  (Corresponding angles)
and ∠DCE = ∠A  (Corresponding angles)
∴ ∠A = ∠B     [From(i) amd (ii)]
Hence, ΔAEB is isosceles.

#### Page No 484: We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
i.e., ∠BAD = ∠DCF = 75°
∠DCF = x  = 75°

Again, the sum of opposite angles in a cyclic quadrilateral is 180°.
Thus,
∠DCF + ∠DEF = 180°
75° + y = 180°
y = (180° - 75°) = 105°

Hence, x = 75° and y = 105°

#### Page No 485: Draw DE ⊥ AB and CF ⊥ AB.
In ΔADE and ΔBCF, we have:
and ∠AED = ∠BCF = 90°
∴ ΔADE ≅ ΔBCF  (By AAS congruency)
∠A = ∠B
Now,
∠A + ∠B + ∠C + ∠D = 360°
⇒ 2∠B + 2∠D = 360°
∠B + ∠D = 180°
Hence, ABCD is a cyclic quadrilateral.

#### Page No 485:

Let ABCD be the cyclic quadrilateral and PO, QO, RO and SO be the perpendicular bisectors of sides AB, BC, CD and AD. We know that the perpendicular bisector of a chord passes through the centre of the circle.
Since, AB, BC, CD and AD are the chords of a circle, PO, QO, RO and SO pass through the centre.
i.e., PO, QO, RO and SO are concurrent.
Hence, the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

#### Page No 485:

Let ABCD be the rhombus with AC and BD as diagonals intersecting at point O. The diagonals of a rhombus bisect each other at right angles.
i.e., ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
Now, circles with AB, BC, CD and DA as diameter passes through O (angle in a semi-circle is a right angle).
Hence, the circle with four sides of a rhombus as diameter, pass through O, i.e., the point of intersection of its diagonals.

#### Page No 485:

Given: ABCD is a cyclic rectangle whose diagonals intersect at O.
To prove: O is the centre of the circle.
Proof: Here, ∠BCD = 90°     [Since it is a rectangle]
So, BD is the diameter of the circle (if the angle made by the chord at the circle is right angle, then the chord is the diameter).
Also, diagonals of a rectangle bisect each other and are equal.
∴ OA =  OB = OC = OD
BD is the diameter.
∴ BO and OD are the radius.
Thus, O is the centre of the circle.
Also, the centre of the circle is circumscribing the cyclic rectangle.
Hence, O is the point of intersection of the diagonals of ABCD.

#### Page No 485:

Let A, B and C be the given points.
With B as the centre and a radius equal to AC, draw an arc.
With C as the centre and AB as radius, draw another arc intersecting the previous arc at D.
Then D is the desired point.
Proof: Join BD and CD. In ΔABC and ΔDCB, we have:
AB =  DC
AC = DB
BC =  CB
i.e., ΔABC ≅ ΔDCB
⇒ ∠BAC = ∠CDB
Thus, BC subtends equal angles ∠BAC and ∠CDB on the same side of it.
∴ Points A, B, C and D are cyclic.

#### Page No 485:

In cyclic quadrilateral ABCD, we have:
B + D = 180°            ...(i)     (Opposite angles of a cyclic quadrilateral )
B - D = 60°               ...(ii)     (Given)
From (i) and (ii), we get:
2B  = 240°
B = 120°
∠D = 60°
Hence, the smaller of the two angles is 60°.

#### Page No 485:

Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.
Let OL ⊥ AB such that LO produced meets CD at M. Then we have to prove that CM = MD
Clearly, ∠1 = ∠2   [Angles in the same segment]
∠2 + ∠3 = 90°   [∵ ∠OLB = 90°]
∠3 + ∠4= 90°    [∵ LOM is a straight line and ∠BOC = 90°]
∴ ∠2 + ∠3  = ∠3 + ∠4 ⇒∠2 = ∠4
Thus, ∠1 = ∠2 and ∠2 = ∠4 ⇒ ∠1 = ∠4
∴ OM = CM and, similarly, OM =  MD
Hence, CM =  MD

#### Page No 485:

Draw two right triangles ACB and ADB in a circle with centre O, where AB is the diameter of the circle. Join CO.

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           ...(1)

Also, arc CB subtends ∠COB at the centre and ∠CDB at D on the circle.

∴ ∠COB = 2∠CDB           ...(2)

Equating (1) and (2),
2∠CAB = 2∠CDB
CAB = ∠CDB

Hence, ∠BAC = ∠BDC.

#### Page No 485:

In the given figure, ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Join AC and BD.

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 ∠CAD at the centre and ∠CBD at B on the circle.

Also, arc CB subtends ∠CAB at the centre and ∠CDB at D on the circle.

∴ ∠CAB = 2∠CDB           ...(2)

Adding (1) and (2), we get
CAD + CAB = 2(∠CBD + CDB)
CBD + ∠CDB = $\frac{1}{2}$BAD

Hence, ∠CBD + ∠CDB = $\frac{1}{2}$BAD.

#### Page No 489:

(b) 12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ BM = $\left(\frac{10}{2}\right)\mathrm{cm}=5\mathrm{cm}$
From the right ΔOMB, we have:
OB2 = OM2 + MB2
⇒ 132 = OM2 + 52
⇒ 169 = OM2 + 25
⇒ OM2 = (169 - 25) = 144
$\mathrm{OM}=\sqrt{144}\mathrm{cm}=12\mathrm{cm}$
Hence, the distance of the chord from the centre is 12 cm.

#### Page No 489:

(c) 30 cm

Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm From the right ΔOMB, we have:
OB2 = OM2 + MB2
⇒ 172 = 82 + MB2
⇒ 289 = 64 + MB2
⇒ MB2 = (289 - 64) = 225
$\mathrm{MB}=\sqrt{225}\mathrm{cm}=15\mathrm{cm}$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AB = 2 × MB = (2 x 15) cm = 30 cm
Hence, the required length of the chord is 30 cm.

#### Page No 489:

(b) 45°

Since an angle in a semicircle is a right angle, ∠BAC = 90°
∴ ∠ABC + ∠ACB = 90°

Now, AB = AC       (Given)
⇒ ∠ABC = ∠ACB = 45°

#### Page No 489:

(c) 60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.
Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

#### Page No 489:

(b) 50°
OA = OB
∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
$\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}×100\right)°=50°$

#### Page No 489:

(a) 8 cm
Join OC. Then OC = radius = 17 cm $\mathrm{CL}=\frac{1}{2}\mathrm{CD}=\left(\frac{1}{2}×30\right)\mathrm{cm}=15\mathrm{cm}$
In right ΔOLC, we have:
OL2 = OC2 - CL2 = (17)2 - (15)2 = (289 - 225) = 64
$⇒\mathrm{OL}=\sqrt{64}=8\mathrm{cm}$
∴ Distance of CD from AB = 8 cm

#### Page No 489:

(b) 80°
Given: AB = CD
We know that equal chords of a circle subtend equal angles at the centre.
∠COD = ∠AOB = 80°

#### Page No 490:

(c) 7.5 cm
Let OA = OC = r cm.
Then OE = (r - 3) cm and $\mathrm{AE}=\frac{1}{2}\mathrm{AB}=6\mathrm{cm}$
Now, in right ΔOAE, we have:
OA2 = OE2 +AE2
⇒ (r)2 = (r - 3)2 + 62
r2 = r2 + 9 - 6r + 36
⇒ 6r = 45
⇒ $r=\frac{45}{6}=7.5$ cm
Hence, the required radius of the circle is 7.5 cm.

#### Page No 490:

(a) 10 cm
Let the radius of the circle be r cm.
Let OD = OB = r cm.
Then OE = (r - 4) cm and ED = 8 cm
Now, in right ΔOED, we have:
OD2 = OE2 +ED2
⇒ (r)2 = (r - 4)2 + 82
r2 = r2 + 16 - 8r + 64
⇒ 8r = 80
r = 10 cm
Hence, the required radius of the circle is 10 cm.

#### Page No 490:

(d) 10 cm Draw OE ⊥ AB and OF ⊥ CD.
In Δ OEB and ΔOFC, we have:
OB =  OC              (Radius of a circle)
∠BOE = ∠COF     (Vertically opposite angles)
∠OEB = ∠OFC     (90° each)
∴ ΔOEB ≅ ΔOFC (By AAS congruency rule)
∴ OE = OF
Chords equidistant from the centre are equal.
∴ CD = AB = 10 cm

#### Page No 490:

(b) 75°
OB = BC (Given)
⇒ ∠BOC = ∠BCO = 25°
Exterior ∠OBA = ∠BOC + ∠BCO = (25° + 25°) = 50°
OA = OB (Radius of a circle)
⇒ ∠OAB = ∠OBA  = 50°
In Δ AOC, side CO has been produced to D.
∴ Exterior ∠AOD = ∠OAC  + ∠ACO
= ∠OAB + ∠BCO
= (50° + 25°) = 75°

#### Page No 490:

(b) 12 cm
OD ⊥ AB
i.e., D is the mid point of AB.
Also, O is the mid point of  BC.
Now, in Δ BAC, D is the mid point of AB and O is the mid point of BC.
$\mathrm{OD}=\frac{1}{2}\mathrm{AC}$  (By mid point theorem)
⇒ AC = 2OD = (2 × 6) cm = 12 cm

#### Page No 490:

(c)

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its medians. Then AD ⊥ BC and BD = 4.5 cm
$\mathrm{AD}=\sqrt{{\mathrm{AB}}^{2}-{\mathrm{BD}}^{2}}=\sqrt{{\left(9\right)}^{2}-{\left(\frac{9}{2}\right)}^{2}}=\sqrt{81-\frac{81}{4}}=\sqrt{\frac{324-81}{4}}=\sqrt{\frac{243}{4}}=\frac{9\sqrt{3}}{2}\mathrm{cm}$
Let G be the centroid of ΔABC.
Then AG : GD = 2 : 1
∴ Radius =  AG = $\frac{2}{3}\mathrm{AD}=\left(\frac{2}{3}×\frac{9\sqrt{3}}{2}\right)\mathrm{cm}=3\sqrt{3}\mathrm{cm}$

#### Page No 490:

(c) 90°
The angle in a semicircle measures 90°.

#### Page No 490:

(a) equal
The angles in the same segment of a circle are equal. #### Page No 491:

(c) 70°
∠BDC = ∠BAC = 60°   (Angles in the same segment of a circle)
In Δ BDC, we have:
DBC + ∠BDC + ∠BCD = 180°    (Angle sum property of a triangle)
∴ 50° + 60° + ∠BCD  = 180°
∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°

#### Page No 491:

(c) 60°
Angles in a semi circle measure 90°.
∠BAC = 90°
In
Δ ABC, we have:
∠BAC + ∠ABC + ∠BCA = 180° (Angle sum property of a triangle)
∴ 90° + ∠ABC + 30° = 180°
∠ABC = (180° - 120°) = 60°
∠CDA = ∠ABC = 60° (Angles in the same segment of a circle)

#### Page No 491:

(b) 50°
∠ODB =∠OAC = 50° (Angles in the same segment of a circle)

#### Page No 491:

(c) 100°
In Δ OAB, we have:
OA = OB          (Radii of a circle)
⇒ ∠OAB = ∠OBA = 20°
In ΔOAC, we have:
OA = OC         (Radii of a circle)
⇒ ∠OAC = ∠OCA = 30°
Now, ∠BAC = (20° + 30°) = 50°
∠BOC = (2 × ∠BAC) = (2 × 50°) = 100°

#### Page No 491:

(a) 85°
We have:
∠BOC + ∠BOA + ∠AOC = 360°
∠BOC + 100° + 90° = 360°
∠BOC = (360° - 190°) = 170°
$\angle \mathrm{BAC}=\left(\frac{1}{2}×\angle \mathrm{BOC}\right)=\left(\frac{1}{2}×170°\right)=85°$

#### Page No 491:

(d) 65°
We have:
OA = OB (Radii of a circle)
Let OAB = ∠ OBA = x°
In Δ OAB, we have:
x° + x° + 50° = 180°   (Angle sum property of a triangle)
⇒ 2x° = (180° - 50°) = 130°
$x=\left(\frac{130}{2}\right)°=65°$
Hence, OAB = 65°

#### Page No 491:

(c) 30°

∠COB = 180° - 120° = 60°  (Linear pair)
Now, arc BC subtends ∠COB at the centre and ∠BDC at the point D of the remaining part of the circle.
∠COB = 2∠BDC
$\angle \mathrm{BDC}=\frac{1}{2}\angle \mathrm{COB}=\left(\frac{1}{2}×60°\right)=30°$

#### Page No 492:

(b) 50°
We have:
OA = OB (Radii of a circle)
∠OBA = ∠OAB = 50°
∠CDA = ∠OBA = 50°   (Angles in the same segment of  a circle)

#### Page No 492:

(b) 60°
We have:
∠CDB = ∠CAB = 40°  (Angles in the same segment of a circle)
In Δ CBD, we have:
∠CDB + ∠BCD +∠CBD = 180°   (Angle sum property of a triangle)
40° + 80° + ∠CBD = 180°
∠CBD = (180° - 120°) = 60°

#### Page No 492:

(c) 80°
We have:
∠AEB + ∠CEB = 180°     (Linear pair angles)
⇒ 110° + ∠CEB = 180°
∠CEB = (180° - 110°) = 70°
In
ΔCEB, we have:
∠CEB + ∠EBC + ∠ECB = 180
°   (Angle sum property of a triangle)
70° +  30° + ∠ECB = 180°
∠ECB = (180° - 100°) = 80°

The angles in the same segment are equal.
Thus, ADB  = ∠ECB = 80°

#### Page No 492:

(d) 60°
We have:
OA = OB (Radii of a circle)
∠OBA= ∠OAB = 20°
In
ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180°  (Angle sum property of a triangle)
⇒ 20° + 20° + ∠AOB = 180°
∠AOB = (180° - 40°) = 140°

Again, we have:
OB = OC
∠OBC = ∠OCB = 50°
In
ΔOCB, we have:
∠OCB + ∠OBC + ∠COB = 180°  (Angle sum property of a triangle)
⇒ 50° + 50° + ∠COB = 180°
∠COB = (180° - 100°) = 80°
Since ∠AOB = 140°, we have:
∠AOC + ∠COB  = 140°
∠AOC + 80°  = 140°
∠AOC = (180° - 80°) = 60°

#### Page No 492:

(b) 30°
We have:
∠ABC + 120° = 180°
∠ABC = (180° - 120°) = 60°
Also, ∠ACB = 90°     (Angle in a semicircle)
In
ΔABC, we have:
∠BAC + ∠ACB  + ∠ABC = 180°    (Angle sum property of a triangle)
∠BAC + 90° + 60° = 180°
∠BAC = (180° - 150°) = 30°

#### Page No 492:

(b) 100°
Since ABCD is a cyclic quadrilateral, we have:
⇒ 100° + ∠BCD = 180°
∠BCD = (180° - 100°) = 80°
Now,
AB || DC and CB is the transversal.
∠ABC + ∠BCD = 180°
∠ABC + 80° = 180°
∠ABC = (180° - 80°) = 100°

#### Page No 492:

(c) 115°
Take a point D on the remaining part of the circumference. Then $\angle \mathrm{ADC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}×130°\right)=65°$
In cyclic quadrilateral ABCD, we have:
∠ABC + 65° = 180°
∠ABC  = (180° - 65°) = 115°

#### Page No 493:

(a) 30°
∠ADB = 90°                    (Angle in semicircle)
∴ ∠CDB = (90° + 30°) = 120°
But ABCD being a cyclic quadrilateral, we have:

∠BAC + ∠CDB = 180°
⇒ 30° + ∠CAD  + 120° = 180°
⇒ ∠CAD  = (180° - 150°) = 30°

#### Page No 493:

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE. Then $\angle \mathrm{AEC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}×100°\right)=50°$
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°

#### Page No 493:

(c) 100°
OA = OB  (Radii of a circle)
∠OBA = ∠OAB = 50°
In
Δ OAB, we have:
∠ OAB + ∠OBA + ∠AOB = 180°    (Angle sum property of a triangle)
50° + 50° + ∠AOB = 180°
∠AOB = (180° - 100°) = 80°
Since ∠AOB + ∠BOD = 180°
(Linear pair)
∠BOD = (180° - 80°) = 100°

#### Page No 493:

(b) 70°
BC = CD (given)
BDC = ∠CBD = 35°
In
Δ BCD, we have:
∠BCD +  BDC + ∠CBD = 180°     (Angle sum property of a triangle)
∠BCD + 35° + 35° = 180°
∠BCD = (180° - 70°) = 110°
In cyclic quadrilateral ABCD, we have:

∠BAD = (180° - 110°) = 70°

#### Page No 493:

(c) 120°
Since ΔABC is an equilateral triangle, each of its angle is 60°.
BAC = 60°
In a cyclic quadrilateral ABCD, we have:
BAC + BDC = 180°
⇒ 60° + BDC = 180°
BDC = (180° - 60°) = 120°

#### Page No 493:

(b) 80°
In a cyclic quadrilateral ABCD, we have:
Interior opposite angle, ∠ADC = exterior ∠CBE = 100°
∠CDF = (180° - ∠ADC) = (180° - 100°) = 80°   (Linear pair)

#### Page No 493:

(c) 110°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle. $⇒\angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}×140°\right)=70°$
In the cyclic quadrilateral, we have:

⇒ 70° + ∠ACB = 180°
∠ACB = (180° - 70°) = 110°

#### Page No 494:

(c) 115°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle. $⇒\angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}×130°\right)=65°$

⇒ 65° + ∠ACB = 180°
∠ACB = (180° - 65°) = 115°

#### Page No 494:

(d) 110°
Since ABCD is a cyclic quadrilateral, we have:
∠BAD = (180° - 110°) = 70°
Similarly in
ABEF, we have:
⇒ 70° + ∠BEF = 180°
∠BEF  = (180° - 70°) = 110°

#### Page No 494:

(c) 105°
We have:
∠ABC + 95° = 180°
∠ABC = (180° - 95°) = 85°
Now,
CF || AB and CB is the transversal.
∠BCF = ∠ABC = 85°     (Alternate interior angles)
∠BCE = (85° + 20°) = 105°
∠DCB = (180° - 105°) = 75°
Now, ∠BAD + ∠BCD = 180°

∠BAD = (180° - 75°) = 105°

#### Page No 494:

(c) 8.5 cm
Join AC. Then AE : CE = DE : BE     (Intersecting secant theorem)
∴ AE × BE = DE × CE
Let CD = x cm
Then AE = (AB + BE) = (11 + 3) cm = 14 cm; BE = 3cm; CE = (x + 3.5) cm; DE = 3.5 cm
∴ 14 × 3 = (x + 3.5) × 3.5
$⇒x+3.5=\frac{14×3}{3.5}=\frac{42}{3.5}=12$
x = (12 - 3.5) cm = 8.5 cm
Hence, CD = 8.5 cm

#### Page No 494:

(b) 6 cm
We know that the line joining their centres is the perpendicular bisector of the common chord.
Join AP.
Then AP = 5 cm; AB = 4 cm
Also, AP2 = BP2 + AB2
Or BP2  = AP2 - AB2
Or BP2  = 52 - 42
Or BP = 3 cm
∴ ΔABP is a right angled and PQ = 2 × BP = (2 × 3) cm = 6 cm

#### Page No 494:

$⇒\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}×90°\right)=45°$
$⇒\angle \mathrm{CAO}=\frac{1}{2}\angle \mathrm{COD}=\left(\frac{1}{2}×120°\right)=60°$