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Question 1:

A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of the chord from the centre of the circle.

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.

Question 2:

Find the length of a chord which is at a distance of 3 cm from the centre of a circle of radius 5 cm.

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.

Question 3:

A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find out the radius of the circle.

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.

Question 4:

In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. Calculate the distance between the chords if they are
(i) on the same side of the centre
(ii) on the opposite sides of the centre.

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

Question 5:

Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of radius 17 cm. Find the distance between the chords.

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

Question 6:

In the given figure, the diameter CD of a circle with centre O is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, calculate the radius of the circle.

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.

Question 7:

In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. Find the radius of the circle.

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.

Question 8:

In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC || CD and AC = 2 × OD.

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.

Question 9:

In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠BPD. Prove that AB = CD.

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

Question 10:

Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

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.

Question 11:

Prove that two different circles cannot intersect each other at more than two points.

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.

Question 12:

Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

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 $\left(8+2\sqrt{7}\right)\mathrm{cm}$.

Question 13:

Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB.

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)

Question 14:

If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

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

Question 15:

In the adjoining figure, two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

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 $4\sqrt{6}\mathrm{cm}$.

Question 16:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ACD = y° and ∠AOD = x°, prove that x = 3y.

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

Question 17:

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre then prove that 4q2 = p2 + 3r2.

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.

Question 18:

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥  AB and OQAC, prove that PB = QC.

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)

Question 19:

In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

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

Question 20:

An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

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 $3\sqrt{3}\mathrm{cm}$.

Question 21:

In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of ∠BAC.

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.

Question 22:

In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square at X and Y. Prove that QX = QY.

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.

Question 23:

Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO'.

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'.

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.

(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°

Question 2:

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

(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°

Question 3:

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

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.

Question 4:

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

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°

Question 5:

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

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°

Question 6:

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

(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.

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°

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.

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°

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.

(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.

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°

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.

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°

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

(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.

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.

Question 14:

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

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.

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.

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.

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.

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.

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.

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

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).

Question 1:

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that DBC = 60° and ∠BAC = 40°. Find (i) ∠BCD, (ii) ∠CAD.

(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°

Question 2:

In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If PSR = 150°, find ∠RPQ.

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°

Question 3:

In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is extended to P, find ∠PBC.

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.

Question 4:

In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If ABC = 92° and ∠FAE = 20°, find ∠BCD.

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°

Question 5:

In the given figure, BD = DC and CBD = 30°, find m(∠BAC).

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°

Question 6:

In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine ∠ADC and ∠ABC.

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°

Question 7:

In the given figure, ABC is equilateral. Find (i) ∠BDC, (ii) ∠BEC.

(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°

Question 8:

In the adjoining figure, ABCD is a cyclic quadrilateral in which BCD = 100° and ∠ABD = 50°. Find ∠ADB.

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°

Question 9:

In the given figure, O is the centre of a circle and BOD = 150°. Find the values of x and y.

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°

Question 10:

In the given figure, O is the centre of the  circle and DAB = 50°. Calculate the values of x and y.

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°

Question 11:

In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If CBF = 130° and ∠CDE = x°, find the value of x.

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°

Question 12:

In the given figure, AB is a diameter of a circle with centre O and DO || CB.
If BCD = 120°, calculate
(ii) ∠ABD
(iii) ∠CBD
Also, show that ∆OAD is an equilateral triangle.

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.

Question 13:

Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 6 cm, BP = 2 cm and PD = 25 cm, find CD.

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

Question 14:

In the given figure, O is the centre of a circle. If AOD = 140° and ∠CAB = 50°, calculate
(i) ∠EDB,
(ii) ∠EBD.

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°

Question 15:

In the given figure ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.

ABC is an isosceles triangle.
Here, AB = AC
∴ ∠ACB = ∠ABC   ...(i)
= ∠ABC   [from(i)]
∴ ∠ADE = ∠ABC  (Corresponding angles)
Hence, DE || BC

Question 16:

In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that AEB is isosceles.

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.

Question 17:

In the given figure, BAD = 75°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.

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°

Question 18:

In the given figure, ABCD is a quadrilateral in which AD = BC and ADC = ∠BCD. Show that the points A, B, C, D lie on a circle.

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.

Question 19:

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

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.

Question 20:

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

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.

Question 21:

ABCD is a rectangle. Prove that the centre of the circle thought A, B, C, D is the point of intersection of its diagonals.

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.

Question 22:

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

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.

Question 23:

In a cyclic quadrilateral ABCD, if (B − ∠D) = 60°, show that the smaller of the two is 60°.

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°.

Question 24:

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

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

Question 25:

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.

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.

Question 26:

ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = $\frac{1}{2}$BAD.

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.

Question 1:

The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is
(a) 11.5 cm
(b) 12 cm
(c)
(d) 23 cm

(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.

Question 2:

A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
(a) 25 cm
(b) 12.5 cm
(c) 30 cm
(d) 9 cm

(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.

Question 3:

In the given figure, BOC is a diameter of a circle and AB = AC. Then, ABC = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

(b) 45°

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

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

Question 4:

In the given figure, O is the centre of a circle and ACB = 30°. Then, ∠AOB = ?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
Figure

(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°

Question 5:

In the given figure, O is the centre of a circle. If OAB = 40° and C is a point on the circle, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 80°
(d) 100°

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

Question 6:

In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is
(a) 8 cm
(b) 15 cm
(c) 18 cm
(d) 6 cm

(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

Question 7:

AB and CD are two equal chords of a circle with centre O such that AOB = 80°, then ∠COD = ?
(a) 100°
(b) 80°
(c) 120°
(d) 40°

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

Question 8:

In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circles is
(a) 6 cm
(b) 9 cm
(c) 7.5 cm
(d) 8 cm

(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.

Question 9:

In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
(a) 10 cm
(b) 12 cm
(c) 6 cm
(d) 8 cm

(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.

Question 10:

In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. If AB = 10 cm, then CD = ?
(a) 5 cm
(b) 12.5 cm
(c) 15 cm
(d) 10 cm

(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

Question 11:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ACD = 25°, then ∠AOD = ?
(a) 50°
(b) 75°
(c) 90°
(d) 100°

(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°

Question 12:

In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If OD AB such that OD = 6 cm, then AC = ?
(a) 9 cm
(b) 12 cm
(c) 15 cm
(d) 7.5 cm

(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

Question 13:

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is
(a) 3 cm
(b)
(c)
(d) 6 cm
Figure

(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 = $\frac{2}{3}\mathrm{AD}=\left(\frac{2}{3}×\frac{9\sqrt{3}}{2}\right)\mathrm{cm}=3\sqrt{3}\mathrm{cm}$

Question 14:

The angle in a semicircle measures
(a) 45°
(b) 60°
(c) 90°
(d) 36°
Figure

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

Question 15:

Angles in the same segment of a circle area are
(a) equal
(b) complementary
(c) supplementary
(d) none of these
Figure

(a) equal
The angles in the same segment of a circle are equal.

Question 16:

In the given figure, ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°.
(a) 5
(b) 60°
(c) 70°
(d) 80°

(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°

Question 17:

In the given figure, BOC is a diameter of a circle with centre O. If BCA = 30°, then ∠CDA = ?
(a) 30°
(b) 45°
(c) 60°
(d) 50°

(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)

Question 18:

In the given figure, O is the centre of a circle. If OAC = 50°, then ∠ODB = ?
(a) 40°

(b) 50°
(c) 60°
(d) 75°

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

Question 19:

In the given figure, O is the centre of a circle in which OBA = 20° and ∠OCA = 30°. Then, ∠BOC = ?
(a) 50°

(b) 90°
(c) 100°
(d) 130°

(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°

Question 20:

In the given figure, O is the centre of a circle. If AOB = 100° and ∠AOC = 90°, then ∠BAC = ?
(a) 85°

(b) 80°
(c) 95°
(d) 75°

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

Question 21:

In the given figure, O is the centre of a circle. Then, OAB = ?
(a) 50°

(b) 60°
(c) 55°
(d) 65°

(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°

Question 22:

In the given figure, O is the centre of a circle and AOC = 120°. Then, ∠BDC = ?
(a) 60°

(b) 45°
(c) 30°
(d) 15°

(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
$\angle \mathrm{BDC}=\frac{1}{2}\angle \mathrm{COB}=\left(\frac{1}{2}×60°\right)=30°$

Question 23:

In the given figure, O is the centre of a circle and OAB = 50°. Then , ∠CDA = ?
(a) 40°

(b) 50°
(c) 75°
(d) 25°

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

Question 24:

In the give figure, AB and CD are two intersecting chords of a circle. If CAB = 40° and BCD = 80°, then ∠CBD = ?
(a) 80°

(b) 60°
(c) 50°
(d) 70°
Figure

(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°

Question 25:

In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If AEB = 110° and ∠CBE = 30°, then ∠ADB = ?
(a) 70°

(b) 60°
(c) 80°
(d) 90°

(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°

Question 26:

In the given figure, O is the centre of a circle in which OAB = 20° and ∠OCB = 50°. Then, ∠AOC = ?
(a) 50°

(b) 70°
(c) 20°
(d) 60°

(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°

Question 27:

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ADC = 120°, then ∠BAC = ?
(a) 60°

(b) 30°
(c) 20°
(d) 45°

(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°

Question 28:

In the given figure ABCD is a cyclic quadrilateral in which AB || DC and BAD = 100°. Then, ∠ABC = ?
(a) 80°

(b) 100°
(c) 50°
(d) 40°

(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°

Question 29:

In the given figure, O is the centre of a circle and AOC = 130°. Then, ∠ABC = ?
(a) 50°

(b) 65°
(c) 115°
(d) 130°

(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°

Question 30:

In the given figure, AOB is  a diameter of a circle and CD || AB. If BAD = 30°, then ∠CAD = ?
(a) 30°

(b) 60°
(c) 45°
(d) 50°

(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°

Question 31:

In the given figure, O is the centre of a circle in which AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
(a) 50°

(b) 40°
(c) 25°
(d) 80°

(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°

Question 32:

In the given figure, O is the centre of a circle and OAB = 50°. Then, ∠BOD = ?
(a) 130°

(b) 50°
(c) 100°
(d) 80°

(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°

Question 33:

In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and CBD = 35°. Then, ∠BAD = ?
(a) 65°

(b) 70°
(c) 110°
(d) 90°

(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°

Question 34:

In the given figure, equilateral ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then, ∠BDC = ?
(a) 90°

(b) 60°
(c) 120°
(d) 150°

(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°

Question 35:

In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If CBE = 100°, then ∠CDE = ?
(a) 100°

(b) 80°
(c) 130°
(d) 90°

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

Question 36:

In the given figure, O is the centre of a circle and AOB = 140°. Then, ∠ACB = ?
(a) 70°

(b) 80°
(c) 110°
(d) 40°

(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°

Question 37:

In the given figure, O is the centre of a circle and AOB = 130°. Then, ∠ACB = ?
(a) 50°

(b) 65°
(c) 115°
(d) 155°

(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°

Question 38:

In the given figure, ABCD and ABEF are two cyclic   quadrilaterals. If BCD = 110°, then ∠BEF = ?
(a) 55°

(b) 70°
(c) 90°
(d) 110°

(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°

Question 39:

In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ADC = 95° and ∠ECF = 20°. Then, ∠BAD = ?
(a) 95°

(b) 85°
(c) 105°
(d) 75°

(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°

Question 40:

Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD = ?
(a) 10.5 cm
(b) 9.5 cm
(c) 8.5 cm
(d) 7.5 cm

(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

Question 41:

In the given figure, A and B are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points P and Q respectively. If AB = 4 cm, then the length of common chord PQ is
(a) 3 cm
(b) 6 cm
(c) 7.5 cm
(d) 9 cm

(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

Question 42:

In the given figure, AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°

(b) 45°
(c) 60°
(d) 90°

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