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Page No 400:

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.

Answer:

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 = 162 cm=8 cm
In the right  ΔOMB, we have:
OB2= OM2 + MB2   (Pythagoras theorem)
⇒ 102 = OM2 + 82
⇒ 100 = OM2 + 64
OM2 = (100 - 64) = 36
OM=36 cm=6 cm
Hence, the distance of the chord from the centre is 6 cm.

Page No 400:

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.

Answer:

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
MB=16 cm=4 cm
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 400:

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.

Answer:

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.
MB=AB2=302 cm=15 cm
From the right ΔOMB, we have:
OB2= OM2 + MB2
OB2 = 82 + 152
OB2 = 64 + 225
OB2 = 289
OB=289 cm=17 cm
Hence, the  required length of the radius is 17 cm.

Page No 400:

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.

Answer:

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=AB2=82=4cm
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=CD2=62=3cm
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=AB2=82=4cm
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=CD2=62=3cm
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 400:

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.

Answer:

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=302=15 cmAL=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=162=8 cmCM=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 400:

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.

Answer:

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:
AE=AB2=122 cm=6 cm
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=456 cm=7.5 cm
r = 7.5 cm
Hence, the required radius of the circle is 7.5 cm.



Page No 401:

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.

Answer:

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=808 cm=10 cm
r = 10 cm
Hence, the required radius of the circle is 10 cm.

Page No 401:

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.

Answer:


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.
i.e., AD = BD
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.
i.e., ODAC and OD=12AC
AC = 2 × OD

Hence, proved.

Page No 401:

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.

Answer:


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 401:

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.

Answer:


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 401:

Question 11:

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

Answer:

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 401:

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.

Answer:



Given: OA = 10 cm, O'A = 8 cm and AB = 12 cm
AD=AB2=122=6 cm
Now, in right angled ΔADO, we have:
OA2 = AD2 + OD2
OD2 = OA2 - AD2
             = 102 - 62
             = 100 - 36 = 64
OD = 8 cm

Similarly, in right angled ΔADO', we have:
O'A2 = AD2 + O'D2
O'D2= O'A2 - AD2
              = 82 - 62
              = 64 - 36
              = 28
O'D=28=27 cm
Thus, OO' = (OD + O'D)
                =  8+27 cm
Hence, the distance between their centres is 8+27cm.

Page No 401:

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.

Answer:

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 402:

Question 14:

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

Answer:

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 402:

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.

Answer:

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:
AL=AB2=22=1 cm
Now, in right angled ΔPLA, we have:
AP2 = AL2 + PL2
PL2 = AP2 - AL2
            =  52 - 12
            = 25 - 1 = 24
PL=24=26 cm
Thus PQ = 2 × PL
              =  2×26=46cm
Hence, the required length of PQ is 46cm.

Page No 402:

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.

Answer:

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 402:

Question 17:

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.

Answer:

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)
12AB=12AC
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)
 and OP = OQ (Radii)
⇒ 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 402:

Question 18:

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.

Answer:

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 402:

Question 19:

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

Answer:

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=BC2=92=4.5cm
In right angled ΔADB, we have:
AB2 = AD2 + BD2
AD2 = AB2 - BD2
AD=AB2-BD2
         =92-922cm
         =932cm
In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.
Now, radius = AG=23AD
AG=23×932=33cm
∴ The radius of the circle is 33cm.

Page No 402:

Question 20:

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.

Answer:

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 403:

Question 21:

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.

Answer:

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 420:

Question 1:

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

Answer:

(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
BAC=BOC2=160°2=80°
Hence, BAC = 80°



Page No 421:

Question 2:

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

Answer:


(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
OCA=AOB2=70°2=35°

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

Page No 421:

Question 3:

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

Answer:

From the given diagram, we have:


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

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

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

Page No 421:

Question 4:

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

Answer:


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

Page No 421:

Question 5:

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

Answer:


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 421:

Question 6:

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

Answer:


(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 421:

Question 7:

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

Answer:


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



Page No 422:

Question 8:

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

Answer:


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°
CED=12COD
CED=12×80°=40°
CED = 40°

Page No 422:

Question 9:

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

Answer:


(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 422:

Question 10:

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

Answer:


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]
DCB=12AOB
DCB=12×40°=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 422:

Question 11:

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

Answer:

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
ACB=12AOB=12×130°=65°
Here,ACB = ECB
∴ ECB = 65°   ...(i)

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

Page No 422:

Question 12:

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

Answer:


(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 422:

Question 13:

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

Answer:


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 423:

Question 14:

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

Answer:


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 438:

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.

Answer:

(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 439:

Question 2:

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

Answer:


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 439:

Question 3:

In the given figure, ABCD is a cyclic quadrilateral in which AB || DC.
If BAD = 100°, find
(i) BCD (ii) ∠ADC (iii) ∠ABC.

Answer:


In cyclic quadrilateral ABCD, AB is parallel to DC and BAD = 100°.
(i)
BCD + BAD = 180°   (ABCD is a cyclic quadrilateral )
BCD + 100° = 180°
BCD = 180° – 100° = 80°
BCD = 80°

(ii)
ADC +BAD = 180°   (Interior angles on the same side of transversal are supplementary)
ADC + 100° = 180°
ADC = 80°

(iii)
ABC + BCD =  180°    (Interior angles on the same side of transversal are supplementary)
ABC = 180°- 80° = 100°

Page No 439:

Question 4:

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.

Answer:

Take any point D on the major arc CA and then join AD and DC.

Since the angle subtended by an arc on the centre is twice the angle subtended by it on the circumference, we have:
AOC = 2ADC
⇒ 130° = 2ADC      [∵ AOC = 130°]
ADC = 65°             ...(i)
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
Hence, PBC = ADC = 65°

Page No 439:

Question 5:

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.

Answer:


Given: ABCD is a cyclic quadrilateral.

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

Page No 439:

Question 6:

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

Answer:



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 439:

Question 7:

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

Answer:


We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.
Thus, AOC = 2ADC
⇒ 100° = 2ADC
ADC = 50°
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, ADC +ABC = 180°
⇒ 50° + ABC = 180°
ABC = (180° – 50°) = 130°
ADC = 50° and ABC = 130°



Page No 440:

Question 8:

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

Answer:


(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 440:

Question 9:

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

Answer:


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°
Hence, ADB = 50°

Page No 440:

Question 10:

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

Answer:


O is the centre of the circle and BOD = 150°.
Thus, reflex angle BOD = (360° – 150°) = 210°
Now, x=12reflexBOD=12×210°=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 440:

Question 11:

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

Answer:


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.
ABCD is a cyclic quadrilateral.
Thus, DAB + BCD = 180°
BCD = (180° – 50°) = 130°
y = 130°
Hence, x = 100° and y = 130°

Page No 440:

Question 12:

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.

Answer:



ABCD is a cyclic quadrilateral.
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 441:

Question 13:

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

Answer:

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:
BCD + BAD = 180°
⇒ 120° + BAD = 180°
BAD = (180° – 120°) = 60°
BAD = 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)
ODA = OAD
          = BAD
= 60°
ODB = 90° - ODA = (90° - 60°) = 30°
Here, DO || BC (Given; alternate angles)
CBD = ODB = 30°
∠CBD = 30°
(iv)
ADC = ADB + CDB
            = 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 441:

Question 14:

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.

Answer:


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=9.752.5=3.9

Hence, CD = 3.9 cm

Page No 441:

Question 15:

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

Answer:


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 441:

Question 16:

In the given figure, ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet in E. Prove that ∆EBC ∼ ∆EDA.

Answer:


In ∆EBC and ∆EDA, we have:
Side AB of the cyclic quadrilateral ABCD is produced to E.
∠EBC = ∠CDA  (Opposite angles of a cyclic quadrilateral are equal)
∠EBC = ∠EDA                     ...(i)
Again, side DC of the cyclic quadrilateral ABCD is produced to E.
∠ECB = ∠BAD
∠ECB = ∠EAD                         ...(ii)
∠BEC = ∠DEA      (Each angle equal to ∠E)              ...(iii)
From (i), (ii) and (iii), we have:
∆EBC ∆EDA  (AAA criterion)

Page No 441:

Question 17:

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.

Answer:


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

Page No 441:

Question 18:

ABC is an isosceles triangle in which AB = AC. If D and E are midpoints of AB and AC respectively, prove that the points D, B, C, E are concyclic.

Answer:


ΔABC is an isosceles triangle in which AB = AC and D and E are the midpoints of AB and AC, respectively.
By mid point theorem, we have:
DE parallel to BC
⇒ ∠ADE = ∠ABC                   ...(i)
Also, AB = AC
⇒ ∠ABC = ∠ACB                    ...(ii)
From (i) and (ii), we have:
∠ADE = ∠ACB
Now, ∠ADE + ∠EDB = 180°    [∵ ADB is a straight line]
⇒ ∠ACB + ∠EDB = 180°          
Since,  Opposite angles of a quadrilateral are supplementary
Hence, DBCE  is a cyclic quadrilateral or the points D, B, C and E are concyclic.

Page No 441:

Question 19:

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

Answer:

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 442:

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.

Answer:

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 442:

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.

Answer:

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 442:

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.

Answer:

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 442:

Question 23:

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

Answer:

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 442:

Question 24:

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.

Answer:


ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD.
Draw DE ⊥ AB and CF ⊥ AB.
In ΔADE and ΔBCF, we have:
∠ADE = ADC - 90° = ∠BCD - 90° = ∠BCF   (Given: ∠ADC = ∠BCD)
AD = BC   (Given)
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 442:

Question 25:

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

Answer:


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 442:

Question 26:

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.

Answer:

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 442:

Question 27:

In the given figure, chords AB and CD of a circle are produced to meet at E. Prove that ∆EDB and ∆EAC are similar.

Answer:


It is given that chord AB of a circle is produced to point E.
If one side of a cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.
Exterior ∠BDE = ∠BAC = ∠EAC           ...(i)
Chord CD of a circle is produced to point E.
Exterior ∠DBE = ∠ACD = ∠ACE            ...(ii)
Consider the triangles ΔEDB and ΔEAC.
∠BDE = ∠CAE       [From (i)]
∠DBE = ∠ACE      [From (ii)]
∠ E = ∠E (Common)
∴ ΔEDB ~ ΔEAC  (AAA criterion)
Hence, proved.

Page No 442:

Question 28:

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.

Answer:


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 443:

Question 29:

In the given figure, AB is a diameter of a circle with centre O. If ADE and CBE are straight lines, meeting at E such that BAD = 35° and ∠BED = 25°, find
(i) ∠DBC
(ii) ∠DCB
(iii) ∠BDC.

Answer:


AB is a diameter of the circle with centre O.
ADE and CBE are straight lines that meet at E such that BAD = 35° and BED = 25°.
Join BD and AC.

(i)
Now, BDA = 90°           (Angle in the semicircle)
Also, EDB + BDA = 180°   (Linear pair)
Or EDB = 90°  

Now, EBD = {180° – (EDB + BED}  (Angle sum property)
                  = (180° – (90° + 25°)
                  = (180° – 115°) = 65°
DBC = (180°EBD) = (180° - 65°) = 115°      (Linear pair)
DBC = 115°

(ii)
Here, DCB = BAD        (Angles in the same segment)
BAD = 35°
 ∴ DCB = 35°
(iii)
BDC = 180° – (DBC + DCB)    (Angle sum property)
            = {180° – (DBC + BAD)}
            = 180° – (115° + 35°)
            =180° – 150°
            = 30°
BDC = 30°



Page No 447:

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) 69 cm
(d) 23 cm

Answer:

(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 = 102cm=5cm
From the right ΔOMB, we have:
OB2= OM2 + MB2
⇒ 132 = OM2 + 52
⇒ 169 = OM2 + 25
⇒ OM2 = (169 - 25) = 144
OM=144cm=12cm
Hence, the distance of the chord from the centre is 12 cm.

Page No 447:

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

Answer:

(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
MB=225cm=15cm
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 447:

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°

Answer:

(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 448:

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

Answer:

(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 448:

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°

Answer:

(b) 50°
OA = OB
∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
ACB=12AOB=12×100°=50°

Page No 448:

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

Answer:

(a) 8 cm
Join OC. Then OC = radius = 17 cm

CL=12CD=12×30cm=15cm
In right ΔOLC, we have:
OL2 = OC2 - CL2 = (17)2 - (15)2 = (289 - 225) = 64
OL=64=8cm
∴ Distance of CD from AB = 8 cm

Page No 448:

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°

Answer:

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

Page No 448:

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

Answer:

(c) 7.5 cm
Let OA = OC = r cm.
Then OE = (r - 3) cm and AE=12AB=6cm
Now, in right ΔOAE, we have:
OA2 = OE2 +AE2 
⇒ (r)2 = (r - 3)2 + 62
r2 = r2 + 9 - 6r + 36
⇒ 6r = 45
⇒ r=456=7.5 cm
Hence, the required radius of the circle is 7.5 cm.

Page No 448:

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

Answer:

(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 449:

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

Answer:

(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 449:

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°

Answer:

(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 449:

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

Answer:

(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.
OD=12AC  (By mid point theorem)
⇒ AC = 2OD = (2 × 6) cm = 12 cm

Page No 449:

Question 13:

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is
(a) 3 cm
(b) 32 cm
(c) 33 cm
(d) 6 cm
Figure

Answer:

(c) 33 cm

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
AD=AB2-BD2=92-922=81-814=324-814=2434=932cm
Let G be the centroid of ΔABC.
Then AG : GD = 2 : 1
∴ Radius =  AG = 23AD=23×932cm=33cm

Page No 449:

Question 14:

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

Answer:

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

Page No 449:

Question 15:

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

Answer:

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

Page No 449:

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°

Answer:

(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 449:

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°

Answer:

(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 450:

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°

Answer:

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

Page No 450:

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°

Answer:

(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 450:

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°

Answer:

(a) 85°
We have:
∠BOC + ∠BOA + ∠AOC = 360°
∠BOC + 100° + 90° = 360°
∠BOC = (360° - 190°) = 170°
BAC=12×BOC=12×170°=85°

Page No 450:

Question 21:

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

(b) 60°
(c) 55°
(d) 65°

Answer:

(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=1302°=65°
Hence, OAB = 65°

Page No 450:

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°

Answer:

(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
BDC=12COB=12×60°=30°

Page No 450:

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°

Answer:

(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 450:

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

Answer:

(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 450:

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°

Answer:

(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 451:

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°

Answer:

(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  
(Radii of a circle)
∠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 451:

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°

Answer:

(b) 30°
We have:
∠ABC + ∠ADC = 180°     (Opposite angles of a cyclic quadrilateral)
∠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 451:

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°

Answer:

(b) 100°
Since ABCD is a cyclic quadrilateral, we have:
∠BAD + ∠BCD = 180°     (Opposite angles of a cyclic quadrilateral)
⇒ 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 451:

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°

Answer:

(c) 115°
Take a point D on the remaining part of the circumference.
Join AD and CD.

Then ADC=12AOC=12×130°=65°
In cyclic quadrilateral ABCD, we have:
∠ABC + ∠ADC = 180°     (Opposite angles of a cyclic quadrilateral)
∠ABC + 65° = 180°
∠ABC  = (180° - 65°) = 115°

Page No 451:

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°

Answer:

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

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

Page No 451:

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°

Answer:

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.

Then AEC=12AOC=12×100°=50°
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°

Page No 451:

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°

Answer:

(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 452:

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°

Answer:

(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 + ∠BCD = 180°

∠BAD + 110° = 180°
∠BAD = (180° - 110°) = 70°

Page No 452:

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°

Answer:

(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 452:

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°

Answer:

(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 452:

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°

Answer:

(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.

∠AOB = 2∠ADB
ADB=12AOB=12×140°=70°
In the cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°

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

Page No 452:

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°

Answer:

(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.

∠AOB = 2∠ADB
ADB=12AOB=12×130°=65°
In cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°

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

Page No 452:

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°

Answer:

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

Page No 452:

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°

Answer:

(c) 105°
We have:
∠ABC + ∠ADC = 180°
∠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 + 75° = 180°
∠BAD = (180° - 75°) = 105°



Page No 453:

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

Answer:

(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=14×33.5=423.5=12
x = (12 - 3.5) cm = 8.5 cm
Hence, CD = 8.5 cm

Page No 453:

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

Answer:

(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 453:

Question 42:

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

(b) 45°
(c) 60°
(d) 90°

Answer:

(c) 60°
We have:
∠AOB = 2∠ACB
ACB=12AOB=12×90°=45°
∠COA = 2∠CBA = (2 × 30°) = 60°
∠COD = 180° - ∠COA = (180° - 60°) = 120°
CAO=12COD=12×120°=60°

Page No 453:

Question 43:

Three statements are given below:
I. If a diameter of a circle bisects each of the two chords of a circle, then the chords are parallel.
II. Two circles of radii 10 cm and 17 cm intersect each other and the length of the common chord is 16 cm. Then, the distance between their centres is 23 cm.
III. L is the line intersecting two concentric circles with centre O at points A, B, C and D as shown Then, AC = DB.
Which is true?
(a) I and II
(b) I and III
(c) II and III
(d) II only

Answer:

Option (b) is correct.
Parts (I) and (III) are clearly true.
Let us examine (II).
Let B and C be the centres of two circles of radii 10 cm and 17 cm, respectively. Let AD be the common chord cutting BC at E.
Then AE = ED = 8 cm

Thus, we have:
BE=102-82=100-64=36=6cm
CE=172-82=289-64=225=15cm
∴ BC = BE + CE = (6 + 15) cm = 21 cm.
But it is given that BC is 23 cm.
So, the given statement is wrong.



Page No 454:

Question 44:

Is ABCD a cyclic quadrilateral?
I. Points A, B, C and D lie on a circle.
II. B + ∠D = 180°.

(a) if the given question can be answered by any one of the statements but not the other;
(b) if the given question can be answered by using either statement alone;
(c) if the given question can be answered by using both the statements together but cannot be answered by using either statement;
(d) if the given question cannot be answered by using both the statements together.

Answer:

(b) if the given question can be answered by using either statement alone
We have,
I. If A, B, C and D lie on a circle, then ABCD is a cyclic quadrilateral.
II. ∠A + ∠B + ∠C + ∠D = 360° and ∠B + ∠D = 180°
∠A + ∠C = 180°
∴ A, B, C and D lie on a circle.
Thus, each one of I and II gives answer to the given question.
Hence, the correct option is (b).

Page No 454:

Question 45:

Is ∆ABC right-angled at B?
I. ABCD is a cyclic quadrilateral.
II. D = 90°.

(a) if the given question can be answered by any one of the statements but not the other;
(b) if the given question can be answered by using either statement alone;
(c) if the given question can be answered by using both the statements together but cannot be answered by using either statement;
(d) if the given question cannot be answered by using both the statements together.

Answer:

(c) The given question can be answered by using both the statements together but cannot be answered by using either of the statements.
I gives ∠B + ∠D = 180° and ∠A + ∠C = 180°.
II gives ∠D = 90°.
I and II together give ∠B = 90°.
Thus, the given question can be answered by using both statements.

Page No 454:

Question 46:

Assertion: The circle drawn taking any one of the equal sides of an isosceles right triangle as diameter bisects the base.
Reason: The angle in a semicircle is 1 right angle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both the assertion and the reason are true and the reason is a correct explanation of the assertion.

Assertion (A): Let ABC be a triangle in which AB  = AC and let O be the mid point of AB. With O as centre and OA as radius, draw a circle meeting BC at D.

In Δ ADB, ∠ADB =90°  (Angle in a semicircle)
Also, ∠ADB + ∠ADC = 180°
⇒ ∠ADC = 90°
In ΔADB and ΔADC, we have:
AB = AC (Given)
AD = AD  (Common)
and ∠ADB = ∠ADC = 90°
∴ ΔADB ≅ ΔADC  (RHS criterion)
∴ BD = DC
Hence, the given circle bisects the base.
Thus, assertion (A) is true.
Reason (R): Let ∠BAC be an angle in a semicircle with centre O and diameter BC.

Now, the angle subtended by arc BOC at the centre is ∠BOC = 180°
∴  ∠BOC = 2∠BAC
⇒ 2∠BAC = 180°
⇒ ∠BAC = 90°
Thus, reason (R) is true.
Clearly, reason (R) gives assertion (A).
Hence, the correct option is (a).

Page No 454:

Question 47:

Assertion: The radius of a circle is 10 cm and the length of one of its chords is 16 cm. Then, the distance of the chord from the centre is 6 cm.
Reason: The perpendicular from the centre of a circle to a chord (other than the diameter) bisects the chord.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both the assertion and the reason are true and the reason is a correct explanation of the assertion.
Assertion(A): Let O be the centre of a circle and AB be the chord. Draw OL ⊥ AB.
Then L is the mid point of AB.
Now, OA = 10 cm and AL = 12AB=8cm

Thus, OL=OA2-AL2=102-82=36=6cm
i.e., assertion(A) is true.
Clearly, reason (R) gives assertion (A).



Page No 455:

Question 48:

Assertion: In a circles of radius 13 cm, there is a chord of length 10 cm at a distance of 12 cm from the centre of the circle.
Reason: A unique circle can be drawn to pass through three give non-collinear points.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(b) Both the assertion and the reason are true, but the reason is not a correct explanation of the assertion.
Clearly, reason (R) is true.
Assertion (A):
OA = 13 cm and OL = 12 cm

AL2 = OA2 - OL2
⇒ AL2 = (13)2 - (12)2 = (169 - 144) = 25
⇒ AL = 5 cm
⇒ AB = 2 × AL = 10 cm
Thus, assertion (A) is true.
∴ Reason (R) and assertion (A) are both true, but reason (R) does not give assertion (A).

Page No 455:

Question 49:

Assertion: In the given figure, ABC = 70° and ∠ACB = 30°. Then, ∠BDC = 70°.

Reason: In the given figure, AOC = 130°, then ∠ABC = 115°.

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) The assertion is false and the reason is true.

Reason (R):
We know that ADC=12AOC=12×130°=65°
Also, ∠ABC + ∠ADC = 180°   (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC = (180° - 65°) = 115°, which is true.
∴ Reason (R) is true.

Assertion(A):
∠ABC + ∠BAC + ∠BCA = 180°
⇒ 70° + ∠BAC + 30° = 180°
⇒ ∠BAC  = (180° - 100°) = 80°
∴ ∠BDC = ∠BAC = 80°    (Angles in the same segment)
This is false.
Thus, assertion (A) is false and reason (R) is true.

Page No 455:

Question 50:

Assertion: A cyclic parallelogram is a square.
Reason: Diameter is the largest chord in a circle.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(d) The assertion is false and the reason is true.
Clearly, assertion (A) is false and reason (R) is true.
Hence, the correct answer is option (d).

Page No 455:

Question 51:

Assertion: If two circles intersect at two points, then the line joining their centres is perpendicular to the common chord.
Reason: The perpendicular bisectors of two chords of a circle intersect at its centre.
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Answer:

(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.

Clearly, Assertion (A) and Reason (R) are both correct and Reason (R) is the correct explanation of Assertion (A).

Page No 455:

Question 52:

Write T for true and F for false
(i) The degree measures of a semicircle is 180°.
(ii) The perimeter of a circle is called its circumference.
(iii) A circle divides the plane into three parts.
(iv) Let O be the centre of a circle with radius r. Then a point P such that OP < r is called an interior point of the circle.
(v) A circle can have only a finite number of equal chords.

Answer:

(i) True.

(ii) True.

(iii) True, a circle divides a plane into three parts: 
a) Points outside the circle.
b) Points inside the circle.
c) Points on the circle.

(iv) True.

(v) False. On a circle, there are infinite points. So, we can draw an infinite number of chords of a given length. Hence, a circle has infinite number of equal chords.



Page No 456:

Question 53:

Match the following columns:

Column I Column II
(a) Angle in a semicircle measures (p) 40°
(b) In the given figure, O is the centre of a circle. If AOB = 120°, then ∠ACB = ?
(q) 80°
(c) In the given figure, O is the centre of a circle. If POR = 90° and ∠POQ = 110°, then ∠QPR = ?
(r) 90°
(d) In cyclic quadrilateral ABCD, it is given that ADC = 130° and AOB is a diameter of the circle through A, B, C and D. Then, ∠BAC = ?
(s) 60°
(a) .....,
(b) .....,
(c) .....,
(d) .....,

Answer:

(a) Angle in a semicircle measures 90°.
(b) Now, chord AB subtends ∠AOB at the centre and ∠ACB at a point C of the remaining part of a circle.
      
AOB =2ACB
       ACB=12AOB =12×120°=60°
(c) ∠QOR = {360° - (110° + 90° )} = (360° - 200°) = 160°
    
∠QPR = 12QOR=12×160°=80°
(d) Since ∠ADC + ∠ABC = 180°  (Opposite angles in a cyclic quadrilateral)
     
⇒ 130° + ∠ABC = 180°
     
∠ABC = (180° - 130°) = 50°
      and ∠ACB = 90°    (Angle in a semicircle)
      In
ΔABC, we have:
      ∠ABC + ∠ACB + ∠BAC = 180°   (Angle sum property of a triangle)
     
⇒ 50° + 90° + ∠BAC = 180°
      
∠BAC = (180° - 140°) = 40°

Hence, (a) - (r), (b) - (s), (c) - (q) and (d) - (p)  

Page No 456:

Question 54:

Fill in the blanks
(i) Two circles having the same centre and different radii are called ...... circles.
(ii) Diameter is the ...... chord of a circle.
(iii) A continuous piece of a circles is called the ...... of the circle.
(iv) An arc of a circle is called a ...... if the ends of the arc are the ends of a diameter.
(v) A segment of a circle is the region between an arc and a ...... of the circle.
(vi) A line segment joining the centre to any point on the circle is called its ...... .

Answer:

(i) Two circles having the same centre and different radii are called concentric circles.
(ii) Diameter is the longest chord of a circle.
(iii) A continuous piece of a circles is called the arc of the circle.
(iv) An arc of a circle is called a semicircle if the ends of the arc are the ends of a diameter.
(v) A segment of a circle is the region between an arc and a chord of the circle.
(vi) A line segment joining the centre to any point on the circle is called its radius.



Page No 464:

Question 1:

In the given figure, ECB = 40° and ∠CEB = 105°. Then, ∠EAD = ?
(a) 50°
(b) 35°
(c) 20°
(d) 40°

Answer:

(b) 35°

∠AED = ∠BEC = 105°   (Vertically opposite angles)
∠ADE = ∠ACB = 40°     (Angles in the same segment)
In
ΔAED, we have:
∠AED + ∠ADE + ∠EAD = 180° 
105° + 40° + ∠EAD = 180° 
∠EAD = (180° - 145° ) = 35° 

Page No 464:

Question 2:

In the given figure, O is the centre of a circle, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(c) 60°

OA = OB
∠OAB = ∠OBA = 45°
ACB=12AOB=12×90°=45°
In
Δ ABC, we have:
∠ACB + ∠CAB + ∠ABC = 180°
⇒ 45° + ∠CAB + 30° = 180°
∠CAB = (180° - 75°) = 105°
Now, ∠CAO + ∠OAB = 105°

∠CAO + 45° = 105°
∠CAO = (105° - 45°) = 60°

Page No 464:

Question 3:

In the given figure, O is the centre of a circle. If ∠OAB = 40°, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer:

(b) 50°

OA = OB
⇒ ∠OBA = ∠OAB = 40°
In
ΔAOB, we have:
∠AOB + ∠OAB + ∠OBA = 180°
⇒ ∠AOB + 40° + 40° = 180°
∠AOB = (180° - 80°) = 100°
ACB=12AOB=12×100°=50°

Page No 464:

Question 4:

In the given figure, ∠DAB = 60° and ∠ABD = 50°, then ∠ACB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer:

(c) 70°

In ΔABD, we have:
∠DAB + ∠ABD + ∠BDA = 180°
⇒ 60° + 50° + ∠BDA = 180°
∠BDA = (180° - 110°) = 70°
ACB = ∠BDA = 70°    (Angles in the same segment)

Page No 464:

Question 5:

In the given figure, O is the centre of a circle, BC is a diameter and ∠BAO = 60°. Then, ∠ADC = ?
(a) 30°
(b) 45°
(c) 60°
(d) 120°

Answer:

(c) 60°

OA = OB
ABO = ∠BAO = 60°
∠ADC = ABO = 60°   (Angles in the same segment)

Page No 464:

Question 6:

Find the length of a chord which is at a distance of 9 cm from the centre of a circle of radius 15 cm.

Answer:

Let AB be the chord of the given circle with centre O and a radius of 15 cm.
From O, draw OM perpendicular to AB.
Then, OM = 9 cm and OB = 15 cm

From the right ΔOMB, we have:
⇒ OB2= OM2 + MB2
⇒ 152 = 92 + MB2
⇒ 225 = 81 + MB2
⇒ MB2 = (225 - 81) = 144
MB=144cm=12cmMB=16cm=4cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 12) cm = 24 cm
Hence, the required length of the chord is 24 cm.

Page No 464:

Question 7:

Prove that equal chords of a circle are equidistant from the centre.

Answer:

Given: A circle (O,r) in which chord AB = chord CD, OL ⊥ AB and OM ⊥ CD.
To Prove: OL = OM
Construction: Join OA and OC.

Proof: We know that the perpendicular from the centre of a circle to a chord bisects the chord.
AL=12AB and CM=12CD
Now, AB = CD
12AB=12CDAL=CM          ...(i)
Now, in the right angled ΔOLA and ΔOMC, we have:
AL = CM                     [From (i)]
OA = OC       [Each equal to r]
∴ ΔOLA ≅ ΔOMC      [By RHS congruency]
i.e., OL =  OM
Hence, AB and CD are equidistant from O.

Page No 464:

Question 8:

Prove that an angle in a semicircle is a right angle.

Answer:

Given: AB is a diameter of a circle C(O,r) and ∠ACB is an angle in a semicircle.
To Prove: ∠ACB = 90°

Proof: We know that the angle subtended by an arc at the centre of a circle is twice the angle formed by it at any point on the remaining part of the circle.
∴ ∠AOB = 2∠ACB                          [∵ angle subtended by arc AB = 2 × angle formed by it at C.]
⇒ 2∠ACB = ∠AOB  = 180°              [∵ ∠AOB is a straight angle]
∴ ∠ACB = 90°

Page No 464:

Question 9:

Prove that a diameter is the largest chord in a circle.

Answer:

Given: A circle C(O,r) in which AB is a diameter and CD is any other chord.
To Prove: AB > CD

Proof: Clearly, the diameter AB is nearer to the centre than any other chord CD.
We know that, between any two chords of a circle, the one which is nearer to the centre is longer.
i.e., AB > CD
Thus, AB is longer than every other chord.
Hence, a diameter is the longest chord in a circle.

Page No 464:

Question 10:

A circle with centre O is given in which OBA = 30° and ∠OCA = 40°. Find ∠BOC.

Answer:

We have:
OA = OB        (Radii of a circle)
⇒ ∠OAB = ∠OBA = 30°
Similarly, OA = OC    (Radii of a circle)
⇒ ∠OAC = ∠OCA = 40°
∴ ∠BAC = ∠OAB + ∠OAC
              = (30° + 40°) = 70°
Now, ∠BOC = 2∠BAC
                  = (2 × 70°) = 140°
∴ ∠BOC = 140°



Page No 465:

Question 11:

In the given figure, AOC is a diameter of a circle with centre O and arc AXB=12 arc BYC. Find BOC.

Answer:

We have:
arc AXB=12 arc BYC

AOB=12BOC

But ∠AOB + ∠BOC = 180°

12BOC+BOC=180°

32BOC=180°

BOC=180°×23=120°

∴ ∠BOC = 120°

Page No 465:

Question 12:

In given figure, O is the centre of a circle and ABC = 45°. Prove that OAOC.

Answer:

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.
Thus, 12AOC=ABC
                      = 45°
⇒ ∠AOC = 90°
∴ OA ⊥ OC

Page No 465:

Question 13:

In the given figure, O is the centre of a circle, ADC = 130° and chord BC = chord BE. Find ∠CBE.

Answer:

ABCD is a cyclic quadrilateral.
∴ ∠ABC + ∠ADC = 180°
⇒ ∠ABC + 130° = 180°
⇒ ∠ABC = (180° - 130°) = 50°

In Δ CFB, we have:
∠CFB + ∠CBF + ∠BCF = 180°   (Angle sum property of a triangle)
⇒ 90° + 50° + ∠BCF = 180°        (Since ∠CFB = 90° and  ∠CBF = ∠ABC = 50°)    
⇒ ∠BCF = (180° - 140°) = 40°      ...(i)
In Δ BCF, we have:
BC =  BE   (Given)
⇒ ∠BEC = ∠BCF = 40°      ...(ii)
In Δ  BCE, we have:
∠BCE + ∠BEC + ∠CBE = 180°    (Angle sum property of a triangle)
 ⇒ 40° + 40° + ∠CBE = 180    [From (i) and (ii)]
 ⇒ ∠CBE = (180° - 80°) = 100°
∴ ∠CBE = 100°

Page No 465:

Question 14:

In the given figure, O is the centre of a circle, ACB = 40°. Find ∠OAB.

Answer:

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 × 40°      [Given]
AOB = 80°         ...(i)
Let us consider the triangle ΔOAB.
OA = OB         (Radii of a circle)
OAB = OBA 
In Δ OAB, we have:
AOB + OAB + OBA = 180°  (Angle sum property of a triangle)
⇒ 80° + OAB + OAB = 180°
⇒ 80° + 2OAB = 180°
⇒ 2OAB = 180° – 80° = 100°
OAB = 50°
OAB = 50°

Page No 465:

Question 15:

In the given figure, O is the centre of a circle, OAB = 30° and OCB = 55°. Find ∠BOC and ∠AOC.

Answer:

We have:
OC = OB    (Radii of a circle)
⇒ ∠OBC = ∠OCB = 55°

In ΔOCB, we have:
∠BOC + ∠OCB + ∠OBC = 180°   (Angle sum property of a triangle)
⇒ ∠BOC + 55° + 55° = 180°
⇒ ∠BOC = (180° - 110°) = 70°
∴ ∠BOC = 70°

Again, we have:
OA = OB
∠OBA = ∠OAB = 30°

In ΔOAB, we have:
∠AOB + ∠OBA + ∠OAB = 180°   (Angle sum property of a triangle)
⇒ ∠AOB + 30° + 30° = 180°
⇒ ∠AOB = (180° - 60°) = 120°
Thus, ∠AOB  = 120°

Now, ∠AOC + ∠BOC = 120°    [Since ∠AOB = ∠AOC + ∠BOC]
⇒ ∠AOC + 70° = 120°
⇒ ∠AOC = (120° - 70°) = 50°
∴ ∠AOC = 50°

Page No 465:

Question 16:

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

Answer:

We have:
OD = OB       (Radius of a circle)
∴ BD =  OD = OB
⇒ ∠OBD = ∠ODB = ∠BOD = 60°
Arc BD subtends ∠BOD at the centre and ∠BCD at a point on the remaining part of the circle.
Then BCD=12BOD=12×60°=30°
Thus, ∠BCE = ∠BCD = 30°

In Δ BCE, we have:
∠BCE + ∠BEC + ∠CBE = 180°        (Angle sum property of a triangle)
⇒ 30° + 90° + ∠CBE = 180°
⇒ ∠CBE = (180° - 120°) = 60°
Thus, ∠CBA = ∠CBE = 60°
Also, ∠ ACB = 90°        (Angles in a semicircle)

In ΔABC, we have:
∠ACB + ∠CAB + ∠CBA = 180°    (Angle sum property of a triangle)
⇒  90° + ∠CAB + 60° = 180°
⇒ ∠CAB = (180° - 150°) = 30°
Hence, ∠CAB = 30°

Page No 465:

Question 17:

In the given figure, ABCD is a cyclic quadrilateral. A circle passing through A and B meets AD and BC in the points E and F respectively. Prove that EF || DC.

Answer:

Now, ABFE being a cyclic quadrilateral, we have:
∠ABF + ∠AEF = 180°               ...(i)
Also, ABCD being a cyclic quadrilateral, we have:
∠ABC + ∠ADC = 180°
or ∠ABF + ∠ADC = 180°            ...(ii)
From (i) and (ii), we have:
∠ABF + ∠AEF = ∠ABF + ∠ADC
⇒ ∠AEF = ∠ADC

∴ EF || DC      (∵ Corresponding angles are equal)



Page No 466:

Question 18:

In the given figure, AOB is a diameter of the circle and C, D, E are any three points on the semicircle. Find the value of ACD + ∠BED.

Answer:

Join AE.

AOB is the diameter of the circle.
C, D and E are the points on the semicircle.
ACDE is a cyclic quadrilateral.
Thus, ACD + DEA = 180°    ...(i)     (Sum of opposite angles of a cyclic quadrilateral = 180°)
Also, AEB = 90°                   ...(ii)      (Angle in a semicircle)
On adding (i) and (ii), we get:
(ACD + DEA) + AEB = 180° + 90°  = 270°
ACD + (DEA + AEB) = 270°
ACD + DEB = 270°         [∵ DEA + AEB = DEB]
Hence, ACD + DEB = 270°

Page No 466:

Question 19:

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

Answer:

In Δ EOC, we have:
∠EOC + ∠OEC + ∠ECO = 180°     (Angle sum property of a triangle)
⇒ ∠EOC + 90° + 30° = 180°       [∵ ∠OEC = 90° and ∠ECO = ∠BCO = 30°]
⇒ ∠EOC = (180° - 120°) = 60°
 Thus, ∠COD = (90° - 60°) = 30°
Since arc CD subtends ∠COD at the centre and ∠CBD on the remaining part of the circle.
i.e., CBD=12COD=12×30°=15°
Hence, y = ∠CBD = 15°

Similarly, arc AD subtends ∠AOD at the centre and ∠ABD on the remaining part of the circle.
i.e., ABD=12AOD=12×90°=45°       [∵ ∠AOD = 90°]
⇒ ∠ABE = (45° + 15°) = 60°
⇒ ∠AEB = 90°
Now, in Δ ABE, we have:
∠ABE + ∠AEB + ∠BAE = 180°
⇒ 60° + 90° + x = 180°
x = (180° - 150°) = 30°
Hence, x = 30° and y = 15°

Page No 466:

Question 20:

PQ and RQ are the chords of a circle equidistant from the centre. Prove that the diameter passing through Q bisects PQR and ∠PSR.

Answer:

We know that the chords of a circle that are equidistant from the centre are equal.
∴ PQ = RQ
Now, in ΔPQS and ΔRQS, we have:
PQ = RQ                        (Equal chords)
QPS = QRS = 90°    (Angles in semicircle)
QS = QS                        (Common hypotenuse)
Thus, ΔPQS ≅ ΔRQS    (RHS criterion)
i.e., PQS = RQS and PSQ = RSQ   (CPCT)
Hence, the diameter QOS bisects PQR and PSR.

Page No 466:

Question 21:

Prove that there is one and only one circle passing through three non-collinear points.

Answer:

Given: Three non-collinear points P, Q and R
To prove: There is one and only one circle passing through the points P, Q and R.
Construction: Join PQ and QR.
Draw the perpendicular bisectors AB of PQ and CD of QR.
Let the perpendicular bisectors intersect at the point O.
Now, join OP, OQ and OR. A circle is obtained that passes through the points P, Q and R.

Proof:
We know that each and every point on the perpendicular bisector of a line segment is equidistant from its end points.
Thus, OP = OQ           (∵ O lies on the perpendicular bisector of PQ)
Also, OQ = OR           (∵ O lies on the perpendicular bisector of QR)
∴ OP = OQ = OR
Let OP = OQ = OR = r      (Radius of a circle)
Now, draw a circle C(O, r) with O as the centre and r as the radius.
Then, circle C(O, r) passes through the points P, Q and R.
Now, we have to show that this is the only circle that passes through P, Q and R.

Let us suppose that there is an another circle C(O′, t) that passes through P, Q and R.
Then O′ will lie on the perpendicular bisectors AB and CD.
But O is the point of intersection of the perpendicular bisectors AB and CD.
∴ O′ must coincide with the point O.      (Since two lines can not intersect at more than one point)
As O′P = t and OP = r  and O′ coincides with O, we get t = r.
∴ C(O, r) and C(O, t) are congruent.
Hence, there is one and only one circle passing through three given non-collinear points.

Page No 466:

Question 22:

In the give figure, OPQR is square. A circle drawn with centre O cuts the square in X and Y. Prove that OX = QY.

Answer:

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)
∴ ΔOXP ≅ ΔOYR    (By RHS congruency rule)
⇒ PX = RY                 (by CPCT)
⇒  PQ − PX = QR − RY   (∵ PQ = QR, sides of the given square)
Hence, QX = QY

Page No 466:

Question 23:

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

Answer:

Given: AB and AC are two equal chords of a circle with centre O.
To prove: ∠OAB = ∠OAC

Proof:
In ΔOAB and ΔOAC, we have:
AB = AC         (Given)
OA = OA        (Common)
OB = OC         (Radii of a circle)
∴ ΔOAB ≅ ΔOAC  (By SSS congruency rule)
i.e.,∠OAB = ∠OAC    (CPCT)
Hence, point O lies on the bisector of ∠BAC.



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