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#### Page No 55:

#### Question 1:

In the adjoining figure the radius of a circle with centre C is 6 cm, line AB is a tangent at A. Answer the following questions.

(1) What is the measure of ∠CAB ? Why ?

(2) What is the distance of point C from line AB? Why ?

(3) *d*(A,B) = 6 cm, find *d*(B,C).

(4) What is the measure of ∠ABC ? Why ?

#### Answer:

(1) It is given that line AB is tangent to the circle at A.

∴ ∠CAB = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Thus, the measure of ∠CAB is 90º.

(2) Distance of point C from AB = 6 cm (Radius of the circle)

(3) ∆ABC is a right triangle.

CA = 6 cm and AB = 6 cm

Using Pythagoras theorem, we have

${\mathrm{BC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{CA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=\sqrt{{6}^{2}+{6}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=6\sqrt{2}\mathrm{cm}$

Thus, *d*(B, C) = $6\sqrt{2}$ cm

(4) In right ∆ABC, AB = CA = 6 cm

∴ ∠ACB = ∠ABC (Equal sides have equal angles opposite to them)

Also, ∠ACB + ∠ABC = 90º (Using angle sum property of triangle)

∴ 2∠ABC = 90º

⇒ ∠ABC = $\frac{90\xb0}{2}$ = 45º

Thus, the measure of ∠ABC is 45º.

#### Page No 55:

#### Question 2:

In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then

(1) What is the length of each tangent segment ?

(2) What is the measure of ∠MRO ?

(3) What is the measure of ∠ MRN ?

#### Answer:

(1) It is given that seg RM and seg RN are tangent segments touching the circle at M and N, respectively.

∴ ∠OMR = ∠ONR = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

OM = 5 cm and OR = 10 cm

In right ∆OMR,

${\mathrm{OR}}^{2}={\mathrm{OM}}^{2}+{\mathrm{MR}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{MR}=\sqrt{{\mathrm{OR}}^{2}-{\mathrm{OM}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{MR}=\sqrt{{10}^{2}-{5}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{MR}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}\mathrm{cm}$

Tangent segments drawn from an external point to a circle are congruent.

∴ MR = NR = $5\sqrt{3}$ cm

(2) In right ∆OMR,

$\mathrm{tan}\angle \mathrm{MRO}=\frac{\mathrm{OM}}{\mathrm{MR}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\angle \mathrm{MRO}=\frac{5\mathrm{cm}}{5\sqrt{3}\mathrm{cm}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\angle \mathrm{MRO}=\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{MRO}=30\xb0$

Thus, the measure of ∠MRO is 30º.

Similarly, ∠NRO = 30º

(3) ∠MRN = ∠MRO + ∠NRO = 30º + 30º = 60º

Thus, the measure of ∠MRN is 60º.

#### Page No 55:

#### Question 3:

Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON.

#### Answer:

In ∆OMR and ∆ONR,

seg MR = seg NR (Tangent segments drawn from an external point to a circle are congruent)

seg OM = seg ON (Radii of the same circle)

seg OR = seg OR (Common)

∴ ∆OMR ≌ ∆ONR (SSS congruence criterion)

⇒ ∠ORM = ∠ORN (CPCT)

Also, ∠MOR = ∠NOR (CPCT)

∴ seg OR bisects ∠MRN and ∠MON.

#### Page No 55:

#### Question 4:

What is the distance between two parallel tangents of a circle having radius 4.5 cm ? Justify your answer.

#### Answer:

In the given figure, O is the centre of the circle. Line PT and line QR are two parallel tangents to the circle at P and Q, respectively.

∴ ∠OPT + ∠OQR = 180º (Sum of adjacent interior angles on the same side of the transversal is supplementary)

⇒ POQ is a straight line segment.

∴ PQ is the diameter of the circle.

PQ = Distance between the parallel tangents PT and QR = 2 × Radius = 2 × 4.5 = 9 cm

Thus, the distance between two parallel tangents of the circle is 9 cm.

#### Page No 58:

#### Question 1:

Two circles having radii 3.5 cm and 4.8 cm touch each other internally. Find the distance between their centres.

#### Answer:

It is given that two circle having radii 3.5 cm and 4.8 cm touch each other internally.

We know, the distance between the centres of the circles touching internally is equal to the difference of their radii.

∴ Distance between the centres of the two circles = 4.8 cm − 3.5 cm = 1.3 cm

Thus, the distance between their centres is 1.3 cm.

#### Page No 58:

#### Question 2:

Two circles of radii 5.5 cm and 4.2 cm touch each other externally. Find the distance between their centres.

#### Answer:

It is given that two circle having radii 5.5 cm and 4.2 cm touch each other externally.

We know, the distance between the centres of the circles touching externally is equal to the sum of their radii.

∴ Distance between the centres of the two circles = 5.5 cm + 4.2 cm = 9.7 cm

Thus, the distance between their centres is 9.7 cm.

#### Page No 58:

#### Question 3:

If radii of two circles are 4 cm and 2.8 cm. Draw figure of these circles touching each other – (i) externally (ii) internally.

#### Answer:

The radii of the two circles are 4 cm and 2.8 cm.

If two circles touch each other externally, then the distance between their centres is equal to the sum of the radii.

Distance between the centres = 4 cm + 2.8 cm = 6.8 cm

If two circles touch each other internally, then the distance between their centres is equal to the difference of the radii.

Distance between the centres = 4 cm − 2.8 cm = 1.2 cm

#### Page No 58:

#### Question 4:

In the given figure, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that – (1) seg AP || seg BQ,

(2) ∆APR ~ ∆RQB, and

(3) Find ∠ RQB if ∠ PAR = 35°

#### Answer:

(1)

In ∆APR, AP = RP (Radii of the same circle)

∴ ∠ARP = ∠RAP .....(1) (In a triangle, equal sides have equal angles opposite to them)

In ∆BQR, QR = QB (Radii of the same circle)

∴ ∠RBQ = ∠BRQ .....(2) (In a triangle, equal sides have equal angles opposite to them)

Also, ∠ARP = ∠BRQ .....(3) (Vertically opposite angles)

From (1), (2) and (3), we have

∠RAP = ∠RBQ

∴ seg AP || seg BQ (If a transversal intersect two lines such that a pair of alternate interior angles is equal, then the two lines are parallel)

(2)

In ∆APR and ∆RQB,

∠RAP = ∠RBQ (Proved)

∠ARP = ∠BRQ (Vertically opposite angles)

∴ ∆APR ~ ∆RQB (AA similarity criterion)

(3)

∠RBQ = ∠PAR = 35º

∴ ∠BRQ = ∠RBQ = 35º

In ∆RQB,

∠BRQ + ∠RBQ + ∠RQB = 180º (Angle sum property)

∴ 35º + 35º + ∠RQB = 180º

⇒ 70º + ∠RQB = 180º

⇒ ∠RQB = 180º − 70º = 110º

Thus, the measure of ∠RQB is 110º.

#### Page No 58:

#### Question 5:

In the given figure, the circles with centres A and B touch each other at E. Line *l *is a common tangent which touches the circles at C and D respectively. Find the length of seg CD if the radii of the circles are 4 cm, 6 cm.

#### Answer:

If two circles touch each other externally, then distance between their centres is equal to the sum of their radii.

∴ AB = AE + EB = 4 cm + 6 cm = 10 cm

It is given that *l* is a common tangent which touches the circles at C and D.

∴ ∠ACD = ∠BDC = 90º (Tangent at any point of a circle is perpendicular to the radius throught the point of contact)

Draw AF ⊥ BD.

Here, ACDF is a rectangle.

∴ CD = AF and DF = AC (Opposite sides of rectangle are equal)

BF = BD − DF = BD − AC = 6 cm − 4 cm = 2 cm

In right ∆ABF,

${\mathrm{AB}}^{2}={\mathrm{AF}}^{2}+{\mathrm{BF}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AF}=\sqrt{{\mathrm{AB}}^{2}-{\mathrm{BF}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AF}=\sqrt{{10}^{2}-{2}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AF}=\sqrt{100-4}=\sqrt{96}=4\sqrt{6}\mathrm{cm}$

∴ CD = AF = $4\sqrt{6}$ cm

Thus, the length of seg CD is $4\sqrt{6}$ cm.

#### Page No 63:

#### Question 1:

In the given figure, points G, D, E, F are concyclic points of a circle with centre C.

∠ ECF = 70°, *m*(arc DGF) = 200° find *m*(arc DE) and *m*(arc DEF).

#### Answer:

*m*(arc EF) = ∠ECF = 70º (Measure of an arc is the measure of its central angle)

Now,

*m*(arc DE) = 360º − *m*(arc EF) − *m*(arc DGF)

⇒ *m*(arc DE) = 360º − 70º − 200º = 90º

Also,

*m*(arc DEF) = *m*(arc DE) + *m*(arc EF)

⇒ *m*(arc DEF) = 90º + 70º = 160º

Thus, the *m*(arc DE) and *m*(arc DEF) is 90º and 160º, respectively.

#### Page No 64:

#### Question 2:

In the given figure, ∆QRS is an equilateral triangle. Prove that,

(1) arc RS ≅ arc QS ≅ arc QR

(2) m(arc QRS) = 240°.

#### Answer:

(1)

It is given that ∆QRS is an equilateral triangle.

∴ chord RS = chord QS = chord QR (Sides of an equilateral triangle are equal)

⇒ *m*(arc RS) = *m*(arc QS) = *m*(arc QR) .....(1) (Corresponding arcs of congruent chords of a circle are congruent)

(2)

*m*(arc RS) + *m*(arc QS) + *m*(arc QR) = 360º (Measure of a complete circle is 360º)

⇒ *m*(arc RS) + *m*(arc RS) + *m*(arc RS) = 360º [Using (1)]

⇒ 3 × *m*(arc RS) = 360º

⇒ *m*(arc RS) = 120º

∴ *m*(arc RS) = *m*(arc QR) = 120º

Now,

*m*(arc QRS) = *m*(arc QR) + *m*(arc RS)

⇒ *m*(arc QRS) = 120º + 120º = 240º

#### Page No 64:

#### Question 3:

In the given figure, chord AB ≅ chord CD, Prove that, arc AC ≅ arc BD

#### Answer:

chord AB ≅ chord CD (Given)

⇒ arc ACB ≅ arc CBD (Corresponding arcs of congruent chords of a circle are congruent)

⇒ *m*(arc ACB) = *m*(arc CBD)

⇒ *m*(arc AC) + *m*(arc CB) = *m*(arc CB) + *m*(arc BD)

⇒ *m*(arc AC) = *m*(arc BD)

⇒ arc AC ≅ arc BD (Two arcs are congruent if their measures are equal)

#### Page No 73:

#### Question 1:

In the given figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.

(1) ∠ AOB (2)∠ ACB

(3) arc AB (4) arc ACB.

#### Answer:

(1)

In the given figure, OA and OB are the radii of the circle.

OA = OB = AB (Given)

∴ ∆OAB is an equilateral triangle.

⇒ ∠AOB = ∠OAB = ∠OBA = 60º

Thus, the measure of ∠AOB is 60º.

(2)

The measure of an angle subtended by an arc at a point on the circle is half of the measure of the angle subtended by the arc at the centre.

∠ACB = $\frac{1}{2}$∠AOB = $\frac{1}{2}\times 60\xb0$ = 30º

Thus, the measure of ∠ACB is 30º.

(3)

*m*(arc AB) = ∠AOB = 60º (Measure of an arc is the measure measure of its corresponding central angle)

Thus, the measure of arc AB is 30º.

(4)

*m*(arc ACB) = 360º − *m*(arc AB) = 360º − 60º = 300º

Thus, the measure of arc ACB is 300º.

#### Page No 73:

#### Question 2:

In the given figure, ▢PQRS is cyclic. side PQ ≅ side RQ. ∠ PSR = 110°, Find–

(1) measure of ∠ PQR

(2) *m*(arc PQR)

(3) *m*(arc QR)

(4) measure of ∠ PRQ

#### Answer:

(1)

∠PSR + ∠PQR = 180º (Opposite angles of a cyclic quadrilateral are supplementary)

⇒ 110º + ∠PQR = 180º

⇒ ∠PQR = 180º − 110º = 70º

Thus, the measure of ∠PQR is 70º.

(2)

∠PSR = $\frac{1}{2}$ *m*(arc PQR) (Measure of an inscribed angle is half of the measure of the arc intercepted by it)

⇒ *m*(arc PQR) = 2 × ∠PSR = 2 × 110º = 220º

(3)

side PQ ≅ side RQ

⇒ arc PQ ≅ arc RQ

⇒ *m*(arc PQ) = *m*(arc RQ) .....(1)

Now,

*m*(arc PQR) = *m*(arc PQ) + *m*(arc QR)

⇒ *m*(arc PQR) = *m*(arc QR) + *m*(arc QR) [From (1)]

⇒ *m*(arc PQR) = 2 × *m*(arc QR)

⇒ *m*(arc QR) = $\frac{220\xb0}{2}$ = 110º [*m*(arc PQR) = 220º]

(4)

∠PRQ = $\frac{1}{2}$*m*(arc QR) (Measure of an inscribed angle is half of the measure of the arc intercepted by it)

⇒ ∠PRQ = $\frac{1}{2}\times 110\xb0$ = 55º [*m*(arc QR) = 110º]

#### Page No 73:

#### Question 3:

▢MRPN is cyclic, ∠ R = (5*x *– 13)°, ∠ N = (4*x *+ 4)°. Find measures of ∠ R and ∠ N.

#### Answer:

MRPN is a cyclic quadrilateral.

∴ ∠R + ∠N = 180º (Opposite angles of a cyclic quadrilateral are supplementary)

⇒ 5*x* − 13º + 4*x* + 4º = 180º

⇒ 9*x* − 9º = 180º

⇒ 9*x* = 180º + 9º = 189º

⇒ *x* = 21º

∴ ∠R = 5*x* − 13º = 5 × 21º* *− 13º = 105º* *− 13º = 92º

∠N = 4*x* + 4º = 4 × 21º* *+ 4º = 84º* *+ 4º = 88º

Thus, the measures of ∠R and ∠N are 92º and 88º, respectively.

#### Page No 73:

#### Question 4:

In the given figure, seg RS is a diameter of the circle with centre O. Point T lies in the exterior of the circle. Prove that ∠ RTS is an acute angle.

#### Answer:

Join RT and TS. Suppose RT intersect the circle at P.

It is given that seg RS is a diameter of the circle with centre O.

∴ ∠RPS = 90º (Angle in a semi-circle is 90º)

In ∆PTS, ∠RPS is an exterior angle and ∠PTS is its remote interior angle.

We know, an exterior angle of a triangle is greater than its remote interior angle.

∴ ∠RPS > ∠PTS

⇒ 90º > ∠PTS

Or ∠RTS < 90º (∠PTS = ∠RTS)

Thus, ∠RTS is an acute angle.

#### Page No 73:

#### Question 5:

Prove that, any rectangle is a cyclic quadrilateral.

#### Answer:

Suppose ABCD is a rectangle.

∴ ∠A = ∠B = ∠C = ∠D = 90º (Each angle of a rectangle is 90º)

⇒ ∠A + ∠C = 180º and ∠B + ∠D = 180º

We know, if a pair of opposite angles of a quadrilateral is supplementary, then quadrilateral is cyclic.

∴ Rectangle ABCD is a cyclic quadrilateral.

So, any rectangle is a cyclic quadrilateral.

#### Page No 74:

#### Question 6:

In the given figure, altitudes YZ and XT of ∆WXY intersect at P. Prove that,

(1) ▢WZPT is cyclic.

(2) Points X, Z, T, Y are concyclic.

#### Answer:

(1) It is given that, YZ ⊥ WX and XT ⊥ WY.

∴ ∠WZY = 90º .....(1)

∠WTX = 90º .....(2)

Adding (1) and (2), we get

∠WZY + ∠WTX = 90º + 90º = 180º

Or ∠WZP + ∠WTP = 90º + 90º = 180º

In quadrilateral WZPT,

∠WZP + ∠WTP = 180º

We know, if a pair of opposite angles of a quadrilateral is supplementary, then quadrilateral is cyclic.

Therefore, quadrilateral WZPT is cyclic.

(2) It is given that, YZ ⊥ WX and XT ⊥ WY.

∴ ∠XZY = 90º and ∠XTY = 90º

⇒ ∠XZY = ∠XTY

So, two points X and Y on the line XY subtends equal angles at two distinct points Z and T which lie on the same side of the line XY.

Therefore, the points X, Z, T and Y are concyclic.

#### Page No 74:

#### Question 7:

In the given figure, *m*(arc NS) = 125°, *m*(arc EF) = 37°, find the measure ∠ NMS.

#### Answer:

We know, if two lines containing chords of a circle intersect each other outside the circle, then the measure of the angle between them is half the difference in measures of the arcs intercepted by the angle.

∴ ∠NMS = $\frac{1}{2}$[*m*(arc NS) − *m*(arc EF)]

⇒ ∠NMS = $\frac{1}{2}\times \left(125\xb0-37\xb0\right)=\frac{1}{2}\times 88\xb0$ = 44º

Thus, the measure of ∠NMS is 44º.

#### Page No 74:

#### Question 8:

In the given figure, chords AC and DE intersect at B. If ∠ ABE = 108°, *m*(arc AE) = 95°, find *m*(arc DC).

#### Answer:

We know, if two chords of a circle intersect each other in the interior of a circle, then the measure of the angle between them is half the sum of measures of the arcs intercepted by the angle and its opposite angle.

∴ ∠ABE = $\frac{1}{2}$[*m*(arc AE) + *m*(arc DC)]

⇒ *m*(arc AE) + *m*(arc DC) = 2∠ABE

⇒ 95º + *m*(arc DC) = 2 × 108º

⇒ *m*(arc DC) = 216º − 95º = 121º

Thus, the measure of arc DC is 121º.

#### Page No 82:

#### Question 1:

In the given figure, ray PQ touches the circle at point Q. PQ = 12, PR = 8, find PS and RS.

#### Answer:

Using tangent secant segments theorem, we have

PQ^{2} = PR × PS

⇒ (12)^{2 }= 8 × PS

⇒ PS = $\frac{144}{8}$ = 18 units

∴ RS = PS − PR = 18 − 8 = 10 units

#### Page No 82:

#### Question 2:

In the given figure, chord MN and chord RS intersect at point D.

(1) If RD = 15, DS = 4,

MD = 8 find DN

(2) If RS = 18, MD = 9,

DN = 8 find DS

#### Answer:

If two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord.

(1)

MD × DN = RD × DS

⇒ 8 × DN = 15 × 4

⇒ DN = $\frac{60}{8}$ = 7.5 units

(2)

MD × DN = RD × DS

⇒ MD × DN = (RS − DS) × DS

⇒ 9 × 8 = (18 − DS) × DS

⇒ DS^{2} − 18DS + 72 = 0

⇒ DS^{2} − 12DS − 6DS + 72 = 0

⇒ DS(DS − 12) − 6(DS − 12) = 0

⇒ (DS − 12)(DS − 6) = 0

⇒ DS − 12 = 0 or DS − 6 = 0

⇒ DS = 12 units or DS = 6 units

#### Page No 82:

#### Question 3:

In the given figure, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find

(1) AD (2) DC (3) DE.

#### Answer:

(1)

Using tangent secant segment theorem, we have

AB^{2} = AC × AD

⇒ 12^{2} = 8 × AD

⇒ AD = $\frac{144}{8}$ = 18 units

(2)

DC = AD − AC = 18 − 8 = 10 units

(3)

CD is the chord of the circle with centre O.

Also, OE ⊥ CD (seg OE ⊥ seg AD)

∴ DE = EC = $\frac{\mathrm{DC}}{2}=\frac{10}{2}$ = 5 units (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

#### Page No 82:

#### Question 4:

In the given figure, if PQ = 6, QR = 10, PS = 8 find TS.

#### Answer:

Using the theorem of chords intersecting outside the circle, we have

PT × PS = PR × PQ

⇒ (TS + PS) × PS = (QR + PQ) × PQ

⇒ (TS + 8) × 8 = (10 + 6) × 6

⇒ 8TS + 64 = 16 × 6 = 96

⇒ 8TS = 96 − 64 = 32

⇒ TS = $\frac{32}{8}$ = 4 units

#### Page No 82:

#### Question 5:

In the given figure, seg EF is a diameter and seg DF is a tangent segment. The radius of the circle is *r. *Prove that, DE × GE = 4*r*^{2}

#### Answer:

In the given figure, seg EF is a diameter and seg DF is a tangent segment.

∴ ∠HFD = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

In right ∆DEF,

DE^{2} = EF^{2 }+ DF^{2} .....(1)

Using tangent secant segments theorem, we have

DE × DG = DF^{2} .....(2)

Subtracting (2) from (1), we get

DE^{2} − DE × DG = EF^{2 }+ DF^{2} − DF^{2}

⇒ DE × (DE − DG) = EF^{2}

⇒ DE × GE = (2*r*)^{2 }=^{ }4*r*^{2} (EF = 2*r*)

Hence, DE × GE = 4*r*^{2}

#### Page No 83:

#### Question 1:

Four alternative answers for each of the following questions are given. Choose the correct alternative.

(1) Two circles of radii 5.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers ?

(2) Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12, what is the radius of each circle ?

(3) A circle touches all sides of a parallelogram. So the parallelogram must be a, ................... .

(4) Length of a tangent segment drawn from a point which is at a distance 12.5 cm from the centre of a circle is 12 cm, find the diameter of the circle.

(5) If two circles are touching externally, how many common tangents of them can be drawn?

(6) ∠ACB is inscribed in arc ACB of a circle with centre O. If ∠ACB = 65°, find

*m*(arc ACB).

(7) Chords AB and CD of a circle intersect inside the circle at point E. If AE = 5.6, EB = 10, CE = 8, find ED.

(8) In a cyclic ▢ABCD, twice the measure of ∠A is thrice the measure of ∠C. Find the measure of ∠C?

(9) Points A, B, C are on a circle, such that

*m*(arc AB) =

*m*(arc BC) = 120°. No point, except point B, is common to the arcs. Which is the type of ∆ABC?

(10) Seg XZ is a diameter of a circle. Point Y lies in its interior. How many of the following statements are true ? (i) It is not possible that ∠XYZ is an acute angle. (ii) ∠XYZ can’t be a right angle. (iii) ∠XYZ is an obtuse angle. (iv) Can’t make a definite statement for measure of ∠XYZ.

#### Answer:

(1)

The radii of the two circles are 5.5 cm and 3.3 cm.

If two circles touch each other externally, distance between their centres is equal to the sum of their radii.

∴ Distance between their centres = 5.5 cm + 3.3 cm = 8.8 cm

If two circles touch each other internally, distance between their centres is equal to the difference of their radii.

∴ Distance between their centres = 5.5 cm − 3.3 cm = 2.2 cm

Thus, the distance between their centres is 8.8 cm or 2.2 cm

Hence, the correct answer is option (D).

(2)

Let C_{1} and C_{2} be the centres of the two circles.

Radius of circle with centre C_{1 }= Radius of circle with centre C_{2} = Distance between their centres = C_{1}C_{2} = 12 cm

Thus, the radius of each circle is 12 cm.

Hence, the correct answer is option (B).

(3)

ABCD is a parallelogram. A circle with centre O touches the parallelogram at E, F, G and H.

ABCD is a parallelogram.

∴ AB = CD .....(1) (Opposite sides of parallelogram are equal)

AD = BC .....(2) (Opposite sides of parallelogram are equal)

Tangent segments drawn from an external point to a circle are congruent.

AE = AH .....(3)

DG = DH .....(4)

BE = BF .....(5)

CG = CF .....(6)

Adding (3), (4), (5) and (6), we get

AE + BE + CG + DG = AH + DH + BF + CF

⇒ AB + CD = AD + BC .....(7)

From (1), (2) and (7), we ahve

2AB = 2BC

⇒ AB = BC .....(8)

From (1), (2) and (8), we have

AB = BC = CD = AD

∴ Parallelogram ABCD is a rhombus. (A rhombus is a parallelogram with all sides equal)

Hence, the correct answer is option (B).

(4)

Let O be the centre of the circle and AB be the tangent segment drawn from an external point A touching the circle at B.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠ABO = 90º

In right ∆ABO,

${\mathrm{OA}}^{2}={\mathrm{AB}}^{2}+{\mathrm{OB}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OB}=\sqrt{{\mathrm{OA}}^{2}-{\mathrm{AB}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OB}=\sqrt{{\left(12.5\right)}^{2}-{\left(12\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OB}=\sqrt{156.25-144}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OB}=\sqrt{12.25}=3.5\mathrm{cm}$

Radius of the circle = OB = 3.5 cm

∴ Diameter of the circle = 2 × Radius of the circle = 2 × 3.5 = 7 cm

Hence, the correct answer is option (C).

(5)

If two circles touch each other externally, then three common tangents can drawn to the circles.

Hence, the correct answer is option (C).

(6)

The measure of an inscribed angle is half of the measure of the arc intercepted by it.

∴∠ACB = $\frac{1}{2}$ *m*(arc AB)

⇒ *m*(arc AB) = 2∠ACB = 2 × 65º = 130º

∴ *m*(arc ACB) = 360º − *m*(arc AB) = 360º − 130º = 230º

Hence, the correct answer is option (D).

(7)

If two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two tangents of one chord is equal to the product of the lengths of the two segments of the other chord.

∴ AE × EB = CE × ED

⇒ 5.6 × 10 = 8 × ED

⇒ ED = $\frac{56}{8}$ = 7 units

Hence, the correct answer is option (A).

(8)

ABCD is a cyclic quadrilateral.

2∠A = 3∠C .....(1) (Given)

Now,

∠A + ∠C = 180º (Opposite angles of a cyclic quadrilateral are supplementary)

⇒ $\frac{3}{2}$∠C + ∠C = 180º [From (1)]

⇒ $\frac{5}{2}$∠C = 180º

⇒ ∠C = $\frac{2\times 180\xb0}{5}$ = 72º

Thus, the measure of ∠C is 72º.

Hence, the correct answer is option (B).

(9)

*m*(arc AB) = *m*(arc BC) = 120º

Now,

*m*(arc AB) + *m*(arc BC) + *m*(arc CA) = 360º

⇒ 120º + 120º + *m*(arc CA) = 360º

⇒ 240º + *m*(arc CA) = 360º

⇒ *m*(arc CA) = 360º − 240º = 120º

∴ *m*(arc AB) = *m*(arc BC) = *m*(arc CA)

⇒ arc AB ≅ arc BC ≅ arc CA (Two arcs are congruent if their measures are equal)

⇒ chord AB ≅ chord BC ≅ chord CA (Chords corresponding to congruent arcs of a circle are congruent)

∴ ∆ABC is an equilateral triangle. (All sides of equilateral triangle are equal)

Hence, the correct answer is option (A).

(10)

Let P be any point on the arc XZ.

XZ is the diameter of the circle.

∴ ∠XPZ = 90º (Angle in a semi-circle is 90º)

So, ∠XYZ cannot be a right angle.

In ∆YPZ,

∠XYZ > ∠YPZ (An exterior angle of a triangle is greater than its remote interior angle)

⇒ ∠XYZ > 90º (∠YPZ = ∠XPZ)

So, ∠XYZ is an obtuse angle. Therefore, it is not possible that ∠XYZ is an acute angle.

Thus, three of the following statements are true.

Hence, the correct answer is option (C).

#### Page No 84:

#### Question 2:

Line *l *touches a circle with centre O at point P. If radius of the circle is 9 cm, answer the following.

(1) What is *d*(O, P) = ? Why ?

(2) If *d*(O, Q) = 8 cm, where does the point Q lie ?

(3) If d(OQ) = 15 cm, How many locations of point Q are line on line l? At what distance will each of them be from point P?

#### Answer:

Radius of the circle = 9 cm

(1)

It is given that line *l* is tangent to the circle at P.

∴ OP = 9 cm (Radius of the circle)

⇒ *d*(O, P) = 9 cm

(2)

*d*(O, Q) = 8 cm < Radius of the circle

∴ Point Q lies in the interior of the circle.

(3)

If *d*(OQ) = 15 cm, then there are two locations of point Q on the line *l*. One on the left of point P and one on the right of point P.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠OPQ = 90º

In right ∆OPQ,

${\mathrm{OQ}}^{2}={\mathrm{OP}}^{2}+{\mathrm{PQ}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PQ}=\sqrt{{\mathrm{OQ}}^{2}-{\mathrm{OP}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PQ}=\sqrt{{\left(15\right)}^{2}-{\left(9\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PQ}=\sqrt{225-81}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PQ}=\sqrt{144}=12\mathrm{cm}$

Thus, the two locations of the point Q on line *l*, which are at a distance of 12 cm from point P.

#### Page No 84:

#### Question 3:

In the given figure, M is the centre of the circle and seg KL is a tangent segment.

If MK = 12, KL = $6\sqrt{3}$ then find –

(1) Radius of the circle.

(2) Measures of ∠K and ∠M.

#### Answer:

(1)

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠MLK = 90º

In right ∆MLK,

${\mathrm{MK}}^{2}={\mathrm{ML}}^{2}+{\mathrm{LK}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ML}=\sqrt{{\mathrm{MK}}^{2}-{\mathrm{LK}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ML}=\sqrt{{\left(12\right)}^{2}-{\left(6\sqrt{3}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ML}=\sqrt{144-108}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{ML}=\sqrt{36}=6\mathrm{cm}$

Thus, the radius of the circle is 6 cm.

(2)

In right ∆MLK,

$\mathrm{tan}\angle \mathrm{K}=\frac{\mathrm{ML}}{\mathrm{KL}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\angle \mathrm{K}=\frac{6}{6\sqrt{3}}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{tan}\angle \mathrm{K}=\mathrm{tan}30\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{K}=30\xb0$

Using angle sum property, we have

$\angle \mathrm{K}+\angle \mathrm{L}+\angle \mathrm{M}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+90\xb0+\angle \mathrm{M}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 120\xb0+\angle \mathrm{M}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{M}=180\xb0-120\xb0=60\xb0$

Thus, the measures of ∠K and ∠M are 30º and 60º, respectively.

#### Page No 84:

#### Question 4:

In the given figure, O is the centre of the circle. Seg AB, seg AC are tangent segments. Radius of the circle is *r *and *l*(AB) = *r *, Prove that, ▢ABOC is a square.

#### Answer:

O is the centre of the circle. Seg AB and seg AC are tangent segments drawn from A to the circle.

Join OB and OC.

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠OBA = ∠OCA = 90º

Now, OB = OC = *r* .....(1) (Radii of the circle)

Tangent segments drawn from an external point to a circle are congruent.

∴ AC = AB = *r* .....(2)

From (1) and (2), we have

AB = OB = OC = AC

In quadrilateral ABOC,

AB = OB = OC = AC and ∠OBA = ∠OCA = 90º

∴ Quadrilateral ABOC is a square. (All sides of square are equal and measure of each angle is 90º)

#### Page No 85:

#### Question 5:

In the given figure, ▢ABCD is a parallelogram. It circumscribes the circle with cnetre T. Point E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.

#### Answer:

ABCD is a parallelogram.

∴ AB = CD .....(1) (Opposite sides of parallelogram are equal)

AD = BC .....(2) (Opposite sides of parallelogram are equal)

Tangent segments drawn from an external point to a circle are congruent.

AE = AH .....(3)

DG = DH .....(4)

BE = BF .....(5)

CG = CF .....(6)

Adding (3), (4), (5) and (6), we get

AE + BE + CG + DG = AH + DH + BF + CF

⇒ AB + CD = AD + BC .....(7)

From (1), (2) and (7), we ahve

2AB = 2AD

⇒ AB = AD

∴ AD = AB = AE + EB = 4.5 + 5.5 = 10 units

#### Page No 85:

#### Question 6:

In the given figure, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at S. Radii of circles are 9 cm and 2.5 cm. Find the answers to the following questions hence find the ratio MS:SR.

(1) Find the length of segment MT

(2) Find the length of seg MN

(3) Find the measure of ∠NSM.

#### Answer:

Radius of circle with centre M = 9 cm

Radius of circle with centre N = 2.5 cm

Join MT and NS.

If two circles touch each other, their point of contact lie on the line joining their centres. So, the points M, N and T are collinear.

(1)

Length of segment MT = 9 cm (Radius of circle with centre M)

Thus, the length of the segment MT is 9 cm.

(2)

Length of segment NT = 2.5 cm (Radius of circle with centre N)

∴ Length of segment MN = Length of segment MT − Length of segment NT = 9 − 2.5 = 6.5 cm

Thus, the length of the segment MN is 6.5 cm.

(3)

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

In the given figure, seg RM is tangent to the circle with centre N at point S.

∴ ∠NSM = 90º

In right ∆NSM,

${\mathrm{MN}}^{2}={\mathrm{NS}}^{2}+{\mathrm{SM}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{SM}=\sqrt{{\mathrm{MN}}^{2}-{\mathrm{NS}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{SM}=\sqrt{{\left(6.5\right)}^{2}-{\left(2.5\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{SM}=\sqrt{42.25-6.25}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{SM}=\sqrt{36}=6\mathrm{cm}$

∴ SR = MR − SM = 9 − 6 = 3 cm (MR = Radius of the circle with centre M)

⇒ MS : SR = 6 cm : 3 cm = 2 : 1

Thus, the ratio MS : SR is 2 : 1.

#### Page No 85:

#### Question 7:

In the adjoining figure circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA || radius YB.

Fill in the blanks and complete the proof.

#### Answer:

**Construction**: Draw segments XZ and __YZ__.

**Proof**: By theorem of touching circles, points X, Z, Y are __collinear__.

∴ ∠XZA ≅ __∠BZY__ (opposite angles)

Let ∠XZA = ∠BZY = *a* .....(I)

Now, seg XA ≅ seg XZ ..... (__Radii of circle with centre X__)

∴∠XAZ = __∠XZA__ = *a* ..... (isosceles triangle theorem) (II)

Similarly, seg YB ≅ __seg YZ__ ..... (__Radii of circle with centre Y__)

∴∠BZY = __∠ZBY__ = *a* ..... (__isosceles triangle theorem__) (III)

∴ from (I), (II), (III),

∠XAZ = ∠ZBY

∴ radius XA || radius YZ ..... (__Alternate angle test__)

#### Page No 86:

#### Question 8:

In the given figure, circles with centres X and Y touch internally at point Z . Seg BZ is a chord of bigger circle and it itersects smaller circle at point A. Prove that, seg AX || seg BY.

#### Answer:

Circles with centres X and Y touch internally at point Z.

Join YZ.

By theorem of touching circles, points Y, X, Z are collinear.

Now, seg XA ≅ seg XZ (Radii of circle with centre X)

∴∠XAZ = ∠XZA (Isosceles triangle theorem) .....(1)

Similarly, seg YB ≅ seg YZ (Radii of circle with centre Y)

∴∠BZY = ∠ZBY (Isosceles triangle theorem) .....(2)

From (1) and (2), we have

∠XAZ = ∠ZBY

If a pair of corresponding angles formed by a transversal on two lines is congruent, then the two lines are parallel.

∴ seg AX || seg BY (Corresponding angle test)

#### Page No 86:

#### Question 9:

In the given figure, line *l *touches the circle with centre O at point P. Q is the mid point of radius OP. RS is a chord through Q such that chords RS || line *l*. If RS = 12 find the radius of the circle.

#### Answer:

It is given that line *l* touches the circle with centre O at point P.

∴ ∠OPX = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

Join OR.

Now, chord RS || *l*.

∴ ∠OQR = 90º (Pair of corresponding angles)

⇒ Q is the mid-point of RS. (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

So, RQ = QS = $\frac{12}{2}$ = 6 units

Let *r* be the radius of the circle.

∴ OR = PQ = *r* (Radii of the circle)

Q is the mid-point of OP.

∴ OQ = PQ = $\frac{r}{2}$

In right ∆OQR,

${\mathrm{OR}}^{2}={\mathrm{OQ}}^{2}+{\mathrm{RQ}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}={\left(\frac{r}{2}\right)}^{2}+{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}-\frac{{r}^{2}}{4}=36\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{3{r}^{2}}{4}=36\phantom{\rule{0ex}{0ex}}\Rightarrow r=\sqrt{\frac{36\times 4}{3}}=\sqrt{48}=4\sqrt{3}\mathrm{units}$

Thus, the radius of the circle is $4\sqrt{3}$ units.

#### Page No 86:

#### Question 10:

In the given figure, seg AB is a diameter of a circle with centre C. Line PQ is a tangent, which touches the circle at point T. seg AP ⊥ line PQ and seg BQ ⊥ line PQ. Prove that, seg CP ≅ seg CQ.

#### Answer:

Seg AB is a diameter of a circle with centre C.

∴ AC = CB (Radii of the circle)

Join CP, CT and CQ.

It is given that line PQ is a tangent, which touches the circle at point T.

∴ ∠CTP = ∠CTQ = 90º (Tangent at any point of a circle is perpendicular to the radius through the point of contact)

⇒ seg CT ⊥ line PQ

Also, seg AP ⊥ line PQ and seg BQ ⊥ line PQ.

∴ seg AP || seg CT || seg BQ

We know, the ratio of the intercepts made on a tranversal by three parallel lines is equal to the ratio of the corresponding intercepts made on any other transversal by the same parallel lines.

$\therefore \frac{\mathrm{PT}}{\mathrm{TQ}}=\frac{\mathrm{AC}}{\mathrm{CB}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{PT}}{\mathrm{TQ}}=1\left(\mathrm{AC}=\mathrm{CB}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{PT}=\mathrm{TQ}$

In ∆CPT and ∆CQT,

seg PT ≅ seg TQ (Proved)

∠CTP = ∠CTQ (90º each)

seg CT ≅ seg CT (Common)

∴ ∆CPT ≅ ∆CQT (RHS congruence criterion)

⇒ seg CP ≅ seg CQ (Corresponding parts of congruent triangles)

Hence proved.

#### Page No 86:

#### Question 11:

Draw circles with centres A, B and C each of radius 3 cm, such that each circle touches the other two circles.

#### Answer:

Radius of each circle = 3 cm

If two circles touch each other externally, then the distance between their centres is equal to the sum of their radii.

∴ AB = 3 cm + 3 cm = 6 cm

BC = 3 cm + 3 cm = 6 cm

CA = 3 cm + 3 cm = 6 cm

Draw a line seg AB = 6 cm.

With A as centre and radius = 6 cm, mark an arc.

With B as centre and radius = 6 cm, mark an arc intersecting the previous drawn arc at C.

Join AC and BC.

Now, with A, B and C as centres and radius = 3 cm, draw three circles.

It can be seen that, each circle touches the other two circles.

#### Page No 86:

#### Question 12:

Prove that any three points on a circle cannot be collinear.

#### Answer:

Let A, B and C be any three points on a circle. Suppose these three points A, B and C on the circle are collinear.

Therefore, the perpendicular bisectors of the chords AB and BC must be parallel because two or more lines which are perpendicular to a given line are parallel to each other.

Now, AB and BC are the chords of the circle. We know that the perpendicular bisector of the chord of a circle passing through its centre.

So, the perpendicular bisectors of the chords AB and BC must intersect at the centre of the circle.

This is a contradiction to our statement that the perpendicular bisectors of AB and BC must be parallel, as parallel lines do not intersect at a point.

Hence, our assumption that three points A, B and C on the circle are collinear is **not **correct.

Thus, any three points on a circle cannot be collinear.

#### Page No 87:

#### Question 13:

In the given figure, line PR touches the circle at point Q. Answer the following questions with the help of the figure.

(1) What is the sum of ∠ TAQ and ∠ TSQ ?

(2) Find the angles which are congruent to ∠ AQP.

(3) Which angles are congruent to ∠ QTS ?

(4) ∠TAS = 65°, find the measure of ∠TQS and arc TS.

(5) If ∠AQP = 42°and ∠SQR = 58° find measure of ∠ATS.

#### Answer:

(1)

Quadrilateral ATSQ is a cyclic quadrilateral.

∴ ∠TAQ + ∠TSQ = 180º (Opposite angles of a cyclic quadrilateral are supplementary)

(2)

The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

Here, PR is the tangent and AQ is the chord.

∴ ∠AQP ≅ ∠ATQ and ∠AQP ≅ ∠ASQ

⇒ ∠AQP ≅ ∠ATQ ≅ ∠ASQ

(3)

Angles inscribed in the same arc are congruent.

∴ ∠QTS ≅ ∠SAQ

Also, the angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

Here, PR is the tangent and QS is the chord.

∴ ∠QTS ≅ ∠SQR

⇒ ∠QTS ≅ ∠SAQ ≅ ∠SQR

(4)

∠TQS = ∠TAS = 65º (Angles inscribed in the same arc are congruent)

The measure of an inscribed angle is half of the measure of the arc intercepted by it.

∴ ∠TAS = $\frac{1}{2}$ *m*(arc TS)

⇒ *m*(arc TS) = 2∠TAS = 2 × 65º = 130º

Thus, the measure of ∠TQS is 65º and arc TS is 130º.

(5)

The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

Here, PR is the tangent and AQ is the chord.

∴ ∠ATQ = ∠AQP = 42º .

Also, PR is the tangent and QS is the chord.

∴ ∠QTS ≅ ∠SQR = 58º

∠ATQ = ∠ATQ + ∠QTS = 42º + 58º = 100º

Thus, the measure of ∠ATQ is 100º.

#### Page No 87:

#### Question 14:

In the given figure, O is the centre of a circle, chord PQ ≅ chord RS If ∠ POR = 70° and (arc RS) = 80°, find –

(1) *m*(arc PR)

(2) *m*(arc QS)

(3) *m*(arc QSR)

#### Answer:

O is the centre of the circle.

chord PQ ≅ chord RS (Given)

⇒ arc PQ ≅ arc RS (Correspondidng arcs of congruent chords of a circle are congruent)

⇒ *m*(arc PQ) = *m*(arc RS)

⇒ *m*(arc PQ) = 80º [*m*(arc RS) = 80º]

(1)

*m*(arc PR) = ∠POR = 70º (Measure of a minor arc is the measure of its central angle)

(2)

*m*(arc PR) + *m*(arc PQ) + *m*(arc QS) + *m*(arc RS) = 360º

⇒ 70º + 80º + *m*(arc QS) + 80º = 360º

⇒ *m*(arc QS) = 360º − 230º = 130º

(3)

*m*(arc QSR) = *m*(arc QS) + *m*(arc RS) = 130º + 80º = 210º

#### Page No 87:

#### Question 15:

In the given figure, *m*(arc WY) = 44°, *m*(arc ZX) = 68°, then

(1) Find the measure of ∠ ZTX.

(2) If WT = 4.8, TX = 8.0,

YT = 6.4, find TZ.

(3) If WX = 25, YT = 8,

YZ = 26, find WT.

#### Answer:

XW and YZ are two chords of a circle intersecting each other in the interior of the circle at T.

(1)

If two chords of a circle intersect each other in the interior of a circle then the measure of the angle between them is half the sum of measures of arcs intercepted by the angle and its opposite angle.

∴ ∠ZTX = $\frac{1}{2}$[*m*(arc ZX) + *m*(arc WY)] = $\frac{1}{2}\times \left(68\xb0+44\xb0\right)=\frac{1}{2}\times 112\xb0$ = 56º

Thus, the measure of ∠ZTX is 56º.

(2)

WT × TX = YT × TZ (Theorem of internal division of chords)

⇒ 4.8 × 8 = 6.4 × TZ

⇒ TZ = $\frac{4.8\times 8}{6.4}$ = 6

(3)

WT × TX = YT × TZ (Theorem of internal division of chords)

⇒ WT × (WX − WT) = YT × (YZ − YT)

⇒ WT × (25 − WT) = 8 × (26 − 8)

⇒ 25WT − WT^{2} = 8 × 18 = 144

⇒ WT^{2 }− 25WT + 144 = 0

⇒ WT^{2 }− 16WT − 9WT + 144 = 0

⇒ WT(WT^{ }− 16) − 9(WT − 16) = 0

⇒ (WT^{ }− 16)(WT^{ }− 9) = 0

⇒ WT^{ }− 16 = 0 or WT^{ }− 9 = 0

⇒ WT^{ }= 16 or WT^{ }= 9

#### Page No 88:

#### Question 16:

In the given figure

(1) *m*(arc CE) = 54°, *m*(arc BD) = 23°, find measure of ∠CAE.

(2) If AB = 4.2, BC = 5.4, AE = 12.0, find AD

(3) If AB = 3.6, AC = 9.0, AD = 5.4, find AE

#### Answer:

(1)

If two lines containing chords of a circle intersect each other outside the circle, then the measure of angle between them is half the difference in measures of the arcs intercepted by the angle.

∴ ∠CAE = $\frac{1}{2}$[*m*(arc CE) − *m*(arc BD)] = $\frac{1}{2}\times \left(54\xb0-23\xb0\right)=\frac{1}{2}\times 31\xb0$ = 15.5º

(2)

AC × AB = AE × AD (Theorem of external division of chords)

⇒ (AB + BC) × AB = AE × AD

⇒ (4.2 + 5.4) × 4.2 = 12 × AD

⇒ AD = $\frac{9.6\times 4.2}{12}$ = 3.36

(3)

AC × AB = AE × AD (Theorem of external division of chords)

⇒ 9 × 3.6 = AE × 5.4

⇒ AE = $\frac{9\times 3.6}{5.4}$ = 6

#### Page No 88:

#### Question 17:

In the given figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.

#### Answer:

**Proof:** Draw seg GF.

∠EFG = ∠FGH .....$\overline{)\mathrm{property}\mathrm{of}\mathrm{alternate}\mathrm{angles}\mathrm{of}\mathrm{parallel}\mathrm{lines}}$ (I)

∠EFG = $\overline{)\frac{1}{2}m\left(\mathrm{arc}\mathrm{EG}\right)}$ .....inscribed angle theorem (II)

∠FGH = $\overline{)\frac{1}{2}m\left(\mathrm{arc}\mathrm{FH}\right)}$ .....inscribed angle theorem (III)

∴ *m*(arc EG) = $\overline{)m\left(\mathrm{arc}\mathrm{FH}\right)}$ from (I), (II), (III)

chord EG ≅ chord FH .....$\overline{)\mathrm{Chords}\mathrm{corresponding}\mathrm{to}\mathrm{congruent}\mathrm{arcs}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{are}\mathrm{congruent}}$

#### Page No 88:

#### Question 18:

In the given figure, P is the point of contact.

(1) If *m*(arc PR) = 140°, ∠ POR = 36°, find *m*(arc PQ)

(2) If OP = 7.2, OQ = 3.2, find OR and QR

(3) If OP = 7.2, OR = 16.2, find QR.

#### Answer:

Join PQ.

(1)

The measure of an inscribed angle is half of the measure of the arc intercepted by it.

∴ ∠PQR = $\frac{1}{2}$ *m*(arc PR) = $\frac{1}{2}\times 140\xb0$ = 70º

In ∆POQ,

∠PQR = ∠POQ + ∠OPQ (Measure of an exterior angle of a triangle is equal to the sum of its remote interior angles)

⇒ 70º = 36º + ∠OPQ

⇒ ∠OPQ = 70º − 36º = 34º

The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

∴ ∠PRQ = ∠OPQ = 34º

Now,

∠PRQ = $\frac{1}{2}$ *m*(arc PQ) (The measure of an inscribed angle is half of the measure of the arc intercepted by it)

⇒ *m*(arc PQ) = 2∠PRQ = 2 × 34º = 68º

(2)

OP is the tangent and OQR is the secant.

∴ OQ × OR = OP^{2} (Tangent secant segment theorem)

⇒ 3.2 × OR = (7.2)^{2}

⇒ OR = $\frac{7.2\times 7.2}{3.2}$ = 16.2

∴ QR = OR − OQ = 16.2 − 3.2 = 13

(3)

OP is the tangent and OQR is the secant.

∴ OQ × OR = OP^{2} (Tangent secant segment theorem)

⇒ OQ × 16.2 = (7.2)^{2}

⇒ OQ = $\frac{7.2\times 7.2}{16.2}$ = 3.2

∴ QR = OR − OQ = 16.2 − 3.2 = 13

#### Page No 88:

#### Question 19:

In the given figure, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅ seg AB.

#### Answer:

Circles with centres C and D touch internally at point E.

Join ED.

By theorem of touching circles, points E, C and D are collinear.

Since D lies on the inner circle with centre C, therefore, ED is the diameter of the inner circle.

∴ ∠EAD = 90º (Angle inscribed in a semi-circle is a right angle)

EB is the chord of the outer circle with centre D.

∴ Point A is the mid-point of chord EB. (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

⇒ seg EA ≅ seg AB

#### Page No 89:

#### Question 20:

In the given figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

#### Answer:

**Proof: **Draw seg OD.

∠ACB = $\overline{)90\xb0}$ ..... angle inscribed in semicircle

∠ACB = $\overline{)45\xb0}$ ..... CD is the bisector of ∠C

*m*(arc DB) = 2∠ACB = $\overline{)90\xb0}$ .....inscribed angle theorem

∠DOB = $\overline{)90\xb0}$ .....definition of measure of an arc (I)

seg OA ≅ seg OB .....$\overline{)\mathrm{Radii}\mathrm{of}\mathrm{the}\mathrm{circle}}$ (II)

∴ line OD is $\overline{)\mathrm{perpendicular}\mathrm{bisector}}$ of seg AB .....From (I) and (II)

∴ seg AD ≅ seg BD

#### Page No 89:

#### Question 21:

In the given figure, seg MN is a chord of a circle with centre O. MN = 25, L is a point on chord MN such that ML = 9 and *d*(O,L) = 5. Find the radius of the circle.

#### Answer:

seg MN is a chord of a circle with centre O.

Draw OP ⊥ MN and join OM.

MP = PN = $\frac{\mathrm{MN}}{2}=\frac{25}{2}$ units (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

∴ LP = MP − ML = $\frac{25}{2}-9=\frac{7}{2}$ units

In right ∆OPL,

${\mathrm{OL}}^{2}={\mathrm{LP}}^{2}+{\mathrm{OP}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=\sqrt{{\mathrm{OL}}^{2}-{\mathrm{LP}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=\sqrt{{5}^{2}-{\left(\frac{7}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=\sqrt{25-\frac{49}{4}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OP}=\sqrt{\frac{51}{4}}=\frac{1}{2}\sqrt{51}\mathrm{units}$

In right ∆OPM,

${\mathrm{OM}}^{2}={\mathrm{MP}}^{2}+{\mathrm{OP}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OM}=\sqrt{{\left(\frac{25}{2}\right)}^{2}+{\left(\frac{\sqrt{51}}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OM}=\sqrt{\frac{625+51}{4}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OM}=\sqrt{\frac{676}{4}}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OM}=\sqrt{169}=13\mathrm{units}$

Thus, the radius of the circle is 13 units.

#### Page No 89:

#### Question 22:

In the given figure, two circles intersect each other at points S and R. Their common tangent PQ touches the circle at points P, Q.

Prove that, ∠ PRQ + ∠ PSQ = 180°

#### Answer:

It is given that two circles intersect each other at points S and R.

Join RS.

The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

PQ is the tangent to the smaller circle and PR is the chord.

∴ ∠RPQ = ∠PSR .....(1)

Also, PQ is the tangent to the bigger circle and RQ is the chord.

∴ ∠RQP = ∠QSR .....(2)

Adding (1) and (2), we get

∠RPQ + ∠RQP = ∠PSR + ∠QSR

⇒ ∠RPQ + ∠RQP = ∠PSQ .....(3)

In ∆PRQ,

∠RPQ + ∠RQP + ∠PRQ = 180º .....(4) (Angle sum property)

From (3) and (4), we get

∠PSQ + ∠PRQ = 180º

Hence proved.

#### Page No 90:

#### Question 23:

In the given figure, two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q respectively. Prove that : seg SQ || seg RP.

#### Answer:

It is given that two circles intersect at points M and N. Secants drawn through M and N intersect the circles at points R, S and P, Q.

Join MN.

Quadrilateral PRMN is a cyclic quadrilateral.

∴ ∠PRM = ∠MNQ .....(1) (Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle)

Quadrilateral QSMN is a cyclic quadrilateral.

∴ ∠QSM = ∠MNP .....(2) (Exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle)

Adding (1) and (2), we get

∠PRM + ∠QSM = ∠MNQ + ∠MNP .....(3)

Now,

∠MNQ + ∠MNP = 180º .....(4) (Angles in linear pair)

From (3) and (4), we get

∠PRM + ∠QSM = 180º

Now, line RS is transversal to the lines PR and QS such that

∠PRS + ∠QSR = 180º

∴ seg SQ || seg RP (If the interior angles formed by a transversal of two distinct lines are supplementary, then the two lines are parallel)

Hence proved.

#### Page No 90:

#### Question 24:

In the given figure, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points B and D intersect each other at point C. Prove that ▢ABCD is cyclic.

#### Answer:

It is given that two circles intersect each other at points A and E.

Join AE, AB and AD.

The angle between a tangent of a circle and a chord drawn from the point of contact is congruent to the angle inscribed in the arc opposite to the arc intercepted by that angle.

BC is the tangent to the smaller circle and BE is the chord.

∴ ∠EBC = ∠BAE .....(1)

Also, CD is the tangent to the bigger circle and ED is the chord.

∴ ∠EDC = ∠DAE .....(2)

Adding (1) and (2), we get

∠EBC + ∠EDC = ∠BAE + ∠DAE

⇒ ∠EBC + ∠EDC = ∠BAD .....(3)

In ∆BCD,

∠DBC + ∠BDC + ∠BCD = 180º .....(4) (Angle sum property)

From (3) and (4), we get

∠BAD + ∠BCD = 180º

In quadrilateral ABCD,

∠BAD + ∠BCD = 180º

Therefore, quadrilateral ABCD is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)

Hence proved.

#### Page No 90:

#### Question 25:

In the given figure, seg AD ⊥ side BC, seg BE ⊥ side AC, seg CF ⊥ side AB. Ponit O is the orthocentre. Prove that , point O is the incentre of ∆DEF.

#### Answer:

It is given that seg AD ⊥ side BC, seg BE ⊥ side AC and seg CF ⊥ side BC. O is the orthocentre of ∆ABC.

Join DE, EF and DF.

∠AFO + ∠AEO = 90º + 90º = 180º

∴ Quadrilateral AEOF is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)

⇒ ∠OAE = ∠OFE .....(1) (Angles inscribed in the same arc are congruent)

∠BFO + ∠BDO = 90º + 90º = 180º

∴ Quadrilateral BFOD is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)

⇒ ∠OBD = ∠OFD .....(2) (Angles inscribed in the same arc are congruent)

In ∆ACD,

∠DAC + ∠ACD = 90º .....(3) (Using angle sum property of triangle)

In ∆BCE,

∠BCE + ∠CBE = 90º .....(4) (Using angle sum property of triangle)

From (3) and (4), we get

∠DAC + ∠ACD = ∠BCE + ∠CBE

⇒ ∠DAC = ∠CBE .....(5)

From (1), (2) and (5), we get

∠OFE = ∠OFD

⇒ OF is the bisector of ∠EFD.

Similarly, OE and OD are the bisectors of ∠DEF and ∠EDF, respectively.

Hence, O is the incentre of ∆DEF. (Incentre of a triangle is the point of intersection of its angle bisectors)

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