What Is the Distance From Point Y to Wx in the Figure Below

Mathematics Office Two Solutions Solutions for Grade 10 Math Chapter iii Circle are provided hither with uncomplicated step-by-footstep explanations. These solutions for Circle are extremely pop among Class 10 students for Math Circle Solutions come handy for speedily completing your homework and preparing for exams. All questions and answers from the Mathematics Part 2 Solutions Book of Form 10 Math Chapter 3 are provided here for you for free. You will also dear the ad-free experience on Meritnation'south Mathematics Office II Solutions Solutions. All Mathematics Part Ii Solutions Solutions for class Form 10 Math are prepared by experts and are 100% accurate.

Page No 55:

Question 1:

In the adjoining figure the radius of a circumvolve with eye C is 6 cm, line AB is a tangent at A. Reply the following questions.
(ane) What is the mensurate of ∠CAB ? Why ?
(2) What is the distance of point C from line AB? Why ?
(three) d(A,B) = 6 cm, find d(B,C).
(4) What is the measure of ∠ABC ? Why ?

Answer:

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

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

Thus, the measure of ∠CAB is 90º.

(two) Distance of signal C from AB = half dozen cm    (Radius of the circumvolve)

(three) ∆ABC is a correct triangle.

CA = half-dozen cm and AB = 6 cm

Using Pythagoras theorem, we have

$$\begin{gathered} {\mathrm{BC}}^2={\mathrm{AB}}^2+{\mathrm{CA}}^2 \\ \Rightarrow \mathrm{BC}=\sqrt{6^2+6^2} \\ \Rightarrow \mathrm{BC}=6\sqrt{\mathrm{two}}\mathrm{cm} \end{gathered}$$

Thus,d(B, C) =

$6\sqrt{\mathrm{two}}$

 cm

(4) In right ∆ABC, AB = CA = half dozen cm

∴ ∠ACB = ∠ABC      (Equal sides take equal angles contrary to them)

As well, ∠ACB + ∠ABC = 90º            (Using angle sum belongings of triangle)

∴ ii∠ABC = 90º

⇒ ∠ABC =

$\frac{90°}{2}$

 = 45º

Thus, the measure out of ∠ABC is 45º.

Page No 55:

Question two:

In the bordering effigy, O is the center of the circumvolve. 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 circumvolve = 5 cm, then
(1) What is the length of each tangent segment ?
(2) What is the measure of ∠MRO ?
(three) What is the measure of ∠ MRN ?

Answer:

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

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

OM = five cm and OR = 10 cm

In right ∆OMR,

$$\begin{gathered} {\mathrm{OR}}^2={\mathrm{OM}}^2+{\mathrm{MR}}^{\mathrm{ii}} \\ \Rightarrow \mathrm{MR}=\sqrt{{\mathrm{OR}}^2-{\mathrm{OM}}^2} \\ \Rightarrow \mathrm{MR}=\sqrt{{10}^{\mathrm{two}}-{\mathrm{v}}^{\mathrm{ii}}} \\ \Rightarrow \mathrm{MR}=\sqrt{100-25}=\sqrt{75}=\mathrm{five}\sqrt{\mathrm{iii}}\mathrm{cm} \end{gathered}$$

Tangent segments drawn from an external bespeak to a circle are coinciding.

∴ MR = NR =

$5\sqrt{\mathrm{iii}}$

 cm

(2) In right ∆OMR,

$$\begin{gathered} \tan \angle \mathrm{MRO}=\frac{\mathrm{OM}}{\mathrm{MR}} \\ \Rightarrow \tan \angle \mathrm{MRO}=\frac{5\mathrm{cm}}{5\sqrt{3}\mathrm{cm}}=\frac{\mathrm{one}}{\sqrt{3}} \\ \Rightarrow \tan \angle \mathrm{MRO}=\tan 30° \\ \Rightarrow \angle \mathrm{MRO}=30° \end{gathered}$$

Thus, the measure out of ∠MRO is 30º.

Similarly, ∠NRO = 30º

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

Thus, the measure of ∠MRN is 60º.

Folio No 55:

Question 3:

Seg RM and seg RN are tangent segments of a circle with centre O. Evidence that seg OR bisects ∠MRN likewise as ∠Mon.

Answer:

In ∆OMR and ∆ONR,

seg MR = seg NR        (Tangent segments drawn from an external indicate to a circumvolve are coinciding)

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

seg OR = seg OR        (Common)

∴ ∆OMR ≌ ∆ONR     (SSS congruence criterion)

⇒ ∠ORM = ∠ORN    (CPCT)

Likewise, ∠MOR = ∠NOR    (CPCT)

∴ seg OR bisects ∠MRN and ∠Monday.

Folio No 55:

Question 4:

What is the distance between two parallel tangents of a circle having radius iv.v cm ? Justify your reply.

Answer:

In the given figure, O is the center 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 bore of the circumvolve.

PQ = Distance between the parallel tangents PT and QR = 2 × Radius = 2 × iv.5 = nine cm

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

Page No 58:

Question 1:

Two circles having radii iii.5 cm and 4.8 cm touch each other internally. Observe the altitude between their centres.

Answer:

It is given that two circle having radii 3.5 cm and 4.viii cm touch each other internally.

Nosotros know, the altitude between the centres of the circles touching internally is equal to the departure of their radii.

∴ Distance between the centres of the two circles = 4.8 cm − iii.5 cm = 1.3 cm

Thus, the altitude betwixt their centres is ane.3 cm.

Page No 58:

Question 2:

Ii circles of radii v.5 cm and 4.2 cm affect each other externally. Notice the distance betwixt their centres.

Answer:

Information technology is given that 2 circle having radii 5.5 cm and 4.ii cm bear upon 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.five cm + 4.ii cm = 9.vii cm

Thus, the distance between their centres is 9.seven cm.

Page No 58:

Question 3:

If radii of 2 circles are 4 cm and 2.8 cm. Describe figure of these circles touching each other – (i) externally (two) internally.

Answer:

The radii of the two circles are four cm and 2.viii cm.

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

Distance between the centres = 4 cm + two.viii cm = six.viii cm

If 2 circles touch each other internally, and so the distance between their centres is equal to the divergence of the radii.

Distance between the centres = 4 cm − two.eight cm = 1.two cm

Folio No 58:

Question 4:

In the given effigy, the circles with centres P and Q bear upon 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°

Reply:

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

∴ ∠ARP = ∠RAP     .....(1)             (In a triangle, equal sides take 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), (two) and (3), we have

∠RAP = ∠RBQ

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

(ii)
In ∆April and ∆RQB,

∠RAP = ∠RBQ       (Proved)

∠ARP = ∠BRQ        (Vertically reverse angles)

∴ ∆April ~ ∆RQB     (AA similarity benchmark)

(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 Eastward. Line l is a mutual 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, half dozen cm.

Answer:

If two circles bear upon each other externally, then altitude between their centres is equal to the sum of their radii.

∴ AB = AE + EB = four cm + 6 cm = 10 cm

It is given that fifty is a mutual tangent which touches the circles at C and D.

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

Draw AF ⊥ BD.

Here, ACDF is a rectangle.

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

BF = BD − DF = BD − Air-conditioning = half dozen cm − 4 cm = 2 cm

In right ∆ABF,

$$\begin{gathered} {\mathrm{AB}}^{\mathrm{ii}}={\mathrm{AF}}^2+{\mathrm{BF}}^{\mathrm{ii}} \\ \Rightarrow \mathrm{AF}=\sqrt{{\mathrm{AB}}^{\mathrm{two}}-{\mathrm{BF}}^2} \\ \Rightarrow \mathrm{AF}=\sqrt{{10}^2-2^{\mathrm{ii}}} \\ \Rightarrow \mathrm{AF}=\sqrt{100-4}=\sqrt{96}=\mathrm{iv}\sqrt{6}\mathrm{cm} \end{gathered}$$

∴ CD = AF =

$\mathrm{iv}\sqrt{\mathrm{half\; dozen}}$

 cm

Thus, the length of seg CD is

$4\sqrt{6}$

 cm.

Page No 63:

Question 1:

In the given figure, points G, D, Eastward, F are concyclic points of a circle with centre C.
∠ ECF = lxx°, m(arc DGF) = 200° observe chiliad(arc DE) and thou(arc DEF).

Answer:

thousand(arc EF) = ∠ECF = 70º      (Measure of an arc is the measure of its primal angle)

Now,

m(arc DE) = 360º −g(arc EF) −grand(arc DGF)

⇒m(arc DE) = 360º − 70º − 200º = 90º

Also,

k(arc DEF) = yard(arc DE) +chiliad(arc EF)

⇒m(arc DEF) = 90º + 70º = 160º

Thus, thethousand(arc DE) andm(arc DEF) is 90º and 160º, respectively.

Page No 64:

Question ii:

In the given figure, ∆QRS is an equilateral triangle. Prove that,
(one) arc RS ≅ arc QS ≅ arc QR
(2) 1000(arc QRS) = 240°.

Respond:

(1)
It is given that ∆QRS is an equilateral triangle.

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

⇒yard(arc RS) =one thousand(arc QS) =m(arc QR)         .....(ane)          (Corresponding arcs of congruent chords of a circle are congruent)

(two)
1000(arc RS) +m(arc QS) +yard(arc QR) = 360º                      (Measure out of a complete circle is 360º)

⇒one thousand(arc RS) +k(arc RS) +g(arc RS) = 360º                  [Using (one)]

⇒ three ×m(arc RS) = 360º

⇒k(arc RS) = 120º

∴m(arc RS) =m(arc QR) = 120º

Now,

m(arc QRS) =chiliad(arc QR) +1000(arc RS)

⇒chiliad(arc QRS) = 120º + 120º = 240º

Page No 64:

Question 3:

In the given figure, chord AB ≅ chord CD, Prove that, arc Air conditioning ≅ 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) =yard(arc CBD)

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

⇒m(arc AC) =m(arc BD)

⇒ arc AC ≅ arc BD          (2 arcs are congruent if their measures are equal)

Page No 73:

Question ane:

In the given figure, in a circumvolve with centre O, length of chord AB is equal to the radius of the circle. Detect measure out of each of the post-obit.
(1) ∠ AOB (2)∠ ACB
(3) arc AB (4) arc ACB.

Reply:

(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 bending subtended by an arc at a bespeak on the circle is one-half of the measure of the angle subtended by the arc at the centre.

∠ACB =

$\frac{1}{2}$

∠AOB =

$\frac{1}{2}\times 60°$

 = 30º

Thus, the measure of ∠ACB is 30º.

(3)
grand(arc AB) = ∠AOB = 60º       (Measure out of an arc is the measure measure of its corresponding fundamental bending)

Thus, the mensurate of arc AB is 30º.

(iv)
grand(arc ACB) = 360º −one thousand(arc AB) = 360º − 60º = 300º

Thus, the measure of arc ACB is 300º.

Page No 73:

Question ii:

In the given effigy, ▢PQRS is cyclic. side PQ ≅ side RQ. ∠ PSR = 110°, Find–
(1) measure of ∠ PQR
(ii) m(arc PQR)
(iii) g(arc QR)
(four) mensurate of ∠ PRQ

Answer:

(one)
∠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.

(two)
∠PSR =

$\frac{\mathrm{one}}{2}$

m(arc PQR)      (Measure of an inscribed angle is half of the measure of the arc intercepted by it)

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

(iii)
side PQ ≅ side RQ

⇒ arc PQ ≅ arc RQ

⇒yard(arc PQ) =thousand(arc RQ)            .....(1)

Now,

yard(arc PQR) =m(arc PQ) +1000(arc QR)

⇒m(arc PQR) =m(arc QR) +1000(arc QR)          [From (i)]

⇒m(arc PQR) = 2 ×m(arc QR)

⇒chiliad(arc QR) =

$\frac{220°}{2}$

 = 110º                  [m(arc PQR) = 220º]

(4)
​∠PRQ =

$\frac{\mathrm{ane}}{\mathrm{two}}$

m(arc QR)      (Measure of an inscribed angle is half of the measure out of the arc intercepted by it)

⇒ ∠PRQ =

$\frac{1}{2}\times 110°$

 = 55º                    [m(arc QR) = 110º]

Page No 73:

Question 3:

▢MRPN is cyclic, ∠ R = (5x – 13)°, ∠ N = (4ten + 4)°. Observe measures of ∠ R and ∠ Due north.

Respond:

MRPN is a cyclic quadrilateral.

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

⇒ fivex − 13º + 4x + 4º = 180º

⇒ ninex − 9º = 180º

⇒ 9x = 180º + 9º = 189º

⇒x = 21º

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

∠North = 4x + 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 circumvolve. 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 bore of the circle with middle O.

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

In ∆PTS, ∠RPS is an outside 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 astute bending.

Page No 73:

Question v:

Show that, any rectangle is a cyclic quadrilateral.

Answer:

Suppose ABCD is a rectangle.

∴ ∠A = ∠B = ∠C = ∠D = 90º       (Each bending 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 circadian.

∴ Rectangle ABCD is a circadian quadrilateral.

Then, any rectangle is a circadian quadrilateral.

Page No 74:

Question 6:

In the given figure, altitudes YZ and XT of ∆WXY intersect at P. Testify that,
(1) ▢WZPT is cyclic.
(2) Points X, Z, T, Y are concyclic.

Reply:

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

∴ ∠WZY = 90º     .....(i)

∠WTX = 90º         .....(2)

Adding (1) and (2), nosotros 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, and then quadrilateral is circadian.

Therefore, quadrilateral WZPT is cyclic.

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

∴ ∠XZY = 90º and ∠XTY = 90º

 ⇒ ∠XZY = ∠XTY

And then, two points Ten and Y on the line XY subtends equal angles at ii distinct points Z and T which lie on the same side of the line XY.

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

Page No 74:

Question vii:

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 circumvolve, so the measure of the angle betwixt them is half the deviation in measures of the arcs intercepted by the angle.

∴ ∠NMS =

$\frac{1}{2}$

[thou(arc NS) −m(arc EF)]

⇒ ∠NMS =

$\frac{\mathrm{ane}}{2}\times \left(125°-37°\right)=\frac{1}{2}\times 88°$

 = 44º

Thus, the measure of ∠NMS is 44º.

Folio No 74:

Question 8:

In the given figure, chords Ac and DE intersect at B. If ∠ ABE = 108°, 1000(arc AE) = 95°, find m(arc DC).

Answer:

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

∴ ∠ABE =

$\frac{1}{\mathrm{ii}}$

[m(arc AE) +thou(arc DC)]

⇒m(arc AE) +m(arc DC) = ii∠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 effigy, ray PQ touches the circumvolve at point Q. PQ = 12, PR = 8, detect PS and RS.

Answer:

Using tangent secant segments theorem, we accept

PQ² = PR × PS

⇒ (12)²= 8 × PS

⇒ PS =

$\frac{144}{8}$

 = eighteen units

∴ RS = PS − PR = 18 − 8 = ten units

Page No 82:

Question 2:

In the given figure, chord MN and chord RS intersect at betoken D.
(ane) If RD = 15, DS = iv,
MD = viii find DN
(2) If RS = xviii, Dr. = nine,
DN = 8 find DS

Respond:

If two chords of a circumvolve intersect each other in the interior of the circumvolve, then the production of the lengths of the 2 segments of one chord is equal to the production of the lengths of the two segments of the other chord.

(i)
MD × DN = RD × DS

⇒ 8 × DN = xv × 4

⇒ DN =

$\frac{\mathrm{sixty}}{8}$

 = 7.five units

(2)
MD × DN = RD × DS

⇒ Dr. × DN = (RS − DS) × DS

⇒ 9 × viii = (xviii − DS) × DS

⇒ DSⁱⁱ − 18DS + 72 = 0

⇒ DS² − 12DS − 6DS + 72 = 0

⇒ DS(DS − 12) − 6(DS − 12) = 0

⇒ (DS − 12)(DS − vi) = 0

⇒ DS − 12 = 0 or DS − 6 = 0

⇒ DS = 12 units or DS = 6 units

Page No 82:

Question three:

In the given figure, O is the eye of the circle and B is a point of contact. seg OE ⊥ seg Advertizing, AB = 12, Air-conditioning = 8, find
(1) AD (2) DC (3) DE.

Respond:

(ane)
Using tangent secant segment theorem, we take

AB² = AC × AD

⇒ 12ⁱⁱ = 8 × AD

⇒ AD =

$\frac{144}{8}$

 = 18 units

(2)
DC = Advertisement − AC = 18 − 8 = 10 units

(3)
CD is the chord of the circle with heart O.

Also, OE ⊥ CD           (​seg OE ⊥ seg AD)

∴ DE = EC =

$\frac{\mathrm{DC}}{2}=\frac{10}{2}$

 = v units       (Perpendicular drawn from the eye of a circle on its chord bisects the chord)

Folio No 82:

Question 4:

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

Respond:

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

PT × PS = PR × PQ

⇒ (TS + PS) × PS = (QR + PQ) × PQ

⇒ (TS + 8) × 8 = (10 + six) × 6

⇒ 8TS + 64 = sixteen × half dozen = 96

⇒ 8TS = 96 − 64 = 32

⇒ TS =

$\frac{32}{8}$

 = iv units

Page No 82:

Question 5:

In the given figure, seg EF is a bore and seg DF is a tangent segment. The radius of the circle is r. Testify that, DE × GE = ivr ²

Answer:

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

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

In correct ∆DEF,

DE² = EF²+ DF² .....(1)

Using tangent secant segments theorem, nosotros have

DE × DG = DF^two         .....(2)

Subtracting (2) from (1), nosotros get

DE² − DE × DG = EFⁱⁱ+ DF² − DF²

⇒ DE × (DE − DG) = EF²

⇒ DE × GE = (2r)² =ivr ² (EF = 2r)

Hence, DE × GE = 4r ⁱⁱ

Page No 83:

Question ane:

Iv culling answers for each of the following questions are given. Choose the right alternative.
(ane) 2 circles of radii v.5 cm and 3.3 cm respectively touch each other. What is the distance between their centers ?

(A) four.4 cm (B) eight.8 cm (C) 2.2 cm (D) viii.8 or 2.2 cm

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

(A) 6 cm (B) 12 cm (C) 24 cm (D) tin can't say

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

(A) rectangle (B) rhombus (C) square (D) trapezium

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

(A) 25 cm (B) 24 cm (C) 7 cm (D) 14 cm

(5) If 2 circles are touching externally, how many common tangents of them can be drawn?

(A) One (B) Two (C) Iii (D) Four

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

(A) 65° (B) 130° (C) 295° (D) 230°

(seven) Chords AB and CD of a circle intersect inside the circle at bespeak Due east. If AE = 5.6, EB = 10, CE = 8, discover ED.

(A) seven (B) 8 (C) 11.two (D) ix

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

(A) 36 (B) 72 (C) 90 (D) 108

(nine) Points A, B, C are on a circumvolve, such that m(arc AB) = thousand(arc BC) = 120°. No point, except point B, is mutual to the arcs. Which is the type of ∆ABC?

(A) Equilateral triangle (B) Scalene triangle (C) Right angled triangle (D) Isosceles triangle

(ten) 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 tin can't exist a right angle. (three) ∠XYZ is an birdbrained angle. (iv) Can't make a definite statement for measure out of ∠XYZ.

(A) Only i (B) Only two (C) Only three (D) All

Reply:

(1)
The radii of the two circles are v.5 cm and 3.3 cm.

If 2 circles impact each other externally, distance betwixt their centres is equal to the sum of their radii.

∴ Distance between their centres = five.v cm + three.3 cm = 8.8 cm

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

∴ Distance between their centres = five.5 cm − 3.3 cm = 2.2 cm

Thus, the altitude betwixt their centres is 8.8 cm or 2.two cm

Hence, the correct answer is pick (D).

(2)
Let C_i and C₂ be the centres of the two circles.

Radius of circle with centre C_ane= Radius of circle with centre C₂ = Altitude between their centres = C₁C₂ = 12 cm

Thus, the radius of each circle is 12 cm.

Hence, the correct answer is option (B).

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

ABCD is a parallelogram.

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

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

Tangent segments fatigued from an external point to a circumvolve are congruent.

AE = AH           .....(3)

DG = DH          .....(4)

BE = BF            .....(5)

CG = CF           .....(6)

Adding (three), (iv), (v) and (half-dozen), we get

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

⇒ AB + CD = Advert + BC      .....(vii)

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

2AB = 2BC

⇒ AB = BC           .....(8)

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

AB = BC = CD = Advertizing

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

Hence, the correct answer is option (B).

(4)
Permit 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 betoken of a circumvolve is perpendicular to the radius through the point of contact.

∴ ∠ABO = 90º

In correct ∆ABO,

$$\begin{gathered} {\mathrm{OA}}^{\mathrm{ii}}={\mathrm{AB}}^{\mathrm{ii}}+{\mathrm{OB}}^2 \\ \Rightarrow \mathrm{OB}=\sqrt{{\mathrm{OA}}^2-{\mathrm{AB}}^2} \\ \Rightarrow \mathrm{OB}=\sqrt{{\left(12.5\right)}^2-{\left(12\right)}^{\mathrm{two}}} \\ \Rightarrow \mathrm{OB}=\sqrt{156.25-144} \\ \Rightarrow \mathrm{OB}=\sqrt{12.25}=3.\mathrm{v}\mathrm{cm} \end{gathered}$$

Radius of the circle = OB = three.v cm

∴ Diameter of the circle = 2 × Radius of the circumvolve = 2 × 3.5 = vii cm

Hence, the correct answer is pick (C).

(5)
If two circles bear upon each other externally, then three mutual tangents tin fatigued to the circles.

Hence, the right respond 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 selection (D).

(7)

If two chords of a circle intersect each other in the interior of the circle, then the production of the lengths of the 2 tangents of one chord is equal to the product of the lengths of the ii segments of the other chord.

∴ AE × EB = CE × ED

⇒ 5.half dozen × x = 8 × ED

⇒ ED =

$\frac{56}{8}$

 = 7 units

Hence, the correct respond is option (A).

(8)
ABCD is a cyclic quadrilateral.

2∠A = iii∠C        .....(1)        (Given)

Now,

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

⇒

$\frac{3}{2}$

∠C + ∠C = 180º    [From (1)]

​⇒

$\frac{5}{2}$

∠C = 180º

⇒ ∠C =

$\frac{2\times 180°}{5}$

 = 72º

Thus, the measure of ∠C is 72º.

Hence, the correct answer is option (B).

(9)

k(arc AB) =chiliad(arc BC) = 120º

Now,

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

⇒ 120º + 120º +m(arc CA) = 360º

⇒ 240º +yard(arc CA) = 360º

⇒m(arc CA) = 360º − 240º = 120º

∴thou(arc AB) =thou(arc BC) =one thousand(arc CA)

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

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

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

Hence, the correct respond is option (A).

(ten)
Let P be any point on the arc XZ.

XZ is the diameter of the circumvolve.

∴ ∠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 truthful.

Hence, the correct answer is option (C).

Folio No 84:

Question 2:

Line l touches a circumvolve with centre O at point P. If radius of the circle is 9 cm, answer the following.
(ane) What is d(O, P) = ? Why ?
(2) If d(O, Q) = 8 cm, where does the indicate Q prevarication ?
(3) If d(OQ) = fifteen cm, How many locations of point Q are line on line l? At what altitude will each of them be from signal P?

Answer:

Radius of the circle = 9 cm

(one)
It is given that linefifty is tangent to the circumvolve at P.

∴ OP = 9 cm     (Radius of the circle)

⇒d(O, P) = nine cm

(2)
d(O, Q) = 8 cm < Radius of the circle

∴ Point Q lies in the interior of the circle.

(3)
Ifd(OQ) = fifteen cm, then there are two locations of point Q on the linefifty. Ane on the left of point P and i on the right of point P.

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

∴ ∠OPQ = 90º

In correct ∆OPQ,

$$\begin{gathered} {\mathrm{OQ}}^2={\mathrm{OP}}^{\mathrm{two}}+{\mathrm{PQ}}^{\mathrm{two}} \\ \Rightarrow \mathrm{PQ}=\sqrt{{\mathrm{OQ}}^{\mathrm{ii}}-{\mathrm{OP}}^2} \\ \Rightarrow \mathrm{PQ}=\sqrt{{\left(15\right)}^{\mathrm{two}}-{\left(9\right)}^2} \\ \Rightarrow \mathrm{PQ}=\sqrt{225-81} \\ \Rightarrow \mathrm{PQ}=\sqrt{144}=12\mathrm{cm} \end{gathered}$$

Thus, the two locations of the point Q on linel, which are at a altitude of 12 cm from indicate P.

Folio No 84:

Question 3:

In the given figure, Thousand is the centre of the circle and seg KL is a tangent segment.
If MK = 12, KL =$\mathrm{six}\sqrt{\mathrm{iii}}$ and so find –
(1) Radius of the circle.
(2) Measures of ∠Thou and ∠M.

Answer:

(1)
The tangent at any indicate of a circle is perpendicular to the radius through the signal of contact.

∴ ∠MLK = 90º

In right ∆MLK,

$$\begin{gathered} {\mathrm{MK}}^{\mathrm{two}}={\mathrm{ML}}^2+{\mathrm{LK}}^2 \\ \Rightarrow \mathrm{ML}=\sqrt{{\mathrm{MK}}^{\mathrm{two}}-{\mathrm{LK}}^{\mathrm{two}}} \\ \Rightarrow \mathrm{ML}=\sqrt{{\left(12\right)}^2-{\left(6\sqrt{3}\right)}^2} \\ \Rightarrow \mathrm{ML}=\sqrt{144-108} \\ \Rightarrow \mathrm{ML}=\sqrt{36}=6\mathrm{cm} \end{gathered}$$

Thus, the radius of the circle is half dozen cm.

(two)
In right ∆MLK,

$$\begin{gathered} \tan \angle \mathrm{1000}=\frac{\mathrm{ML}}{\mathrm{KL}} \\ \Rightarrow \tan \angle \mathrm{K}=\frac{\mathrm{six}}{\mathrm{vi}\sqrt{3}}=\frac{\mathrm{one}}{\sqrt{3}} \\ \Rightarrow \tan \angle \mathrm{Grand}=\tan 30° \\ \Rightarrow \angle \mathrm{Grand}=\mathrm{thirty}° \end{gathered}$$

Using bending sum belongings, we accept

$$\begin{gathered} \angle \mathrm{K}+\angle \mathrm{L}+\angle \mathrm{M}=180° \\ \Rightarrow 30°+90°+\angle \mathrm{M}=180° \\ \Rightarrow 120°+\angle \mathrm{M}=180° \\ \Rightarrow \angle \mathrm{M}=180°-120°=60° \end{gathered}$$

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

Page No 84:

Question 4:

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

Reply:

O is the heart of the circle. Seg AB and seg Air-conditioning are tangent segments drawn from A to the circumvolve.

Join OB and OC.

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

∴ ∠OBA = ∠OCA = 90º

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

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

∴ AC = AB =r .....(ii)

From (one) and (two), we have

AB = OB = OC = AC

In quadrilateral ABOC,

AB = OB = OC = Air conditioning and ∠OBA = ∠OCA = 90º

∴ Quadrilateral ABOC is a square.          (All sides of foursquare 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, M, H are touching points. If AE = four.5, EB = five.5, find AD.

Reply:

ABCD is a parallelogram.

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

Ad = BC          .....(ii)       (Opposite sides of parallelogram are equal)

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

AE = AH           .....(three)

DG = DH          .....(4)

BE = BF            .....(v)

CG = CF           .....(half dozen)

Adding (three), (4), (5) and (6), nosotros get

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

⇒ AB + CD = AD + BC      .....(7)

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

2AB = 2AD

⇒ AB = AD

∴ Advertising = AB = AE + EB = 4.5 + 5.v = 10 units

Page No 85:

Question six:

In the given figure, circle with centre M touches the circle with centre N at point T. Radius RM touches the smaller circle at South. Radii of circles are nine cm and 2.5 cm. Observe the answers to the following questions hence find the ratio MS:SR.
(1) Detect the length of segment MT
(2) Discover the length of seg MN
(iii) Discover the measure of ∠NSM.

Answer:

Radius of circumvolve with centre M = 9 cm

Radius of circle with center N = 2.five cm

Bring together MT and NS.

If two circles bear upon each other, their point of contact lie on the line joining their centres. So, the points Yard, 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 nine cm.

(two)
Length of segment NT = ii.five cm         (Radius of circle with centre N)

∴ Length of segment MN = Length of segment MT − Length of segment NT = 9 − 2.five = half dozen.5 cm

Thus, the length of the segment MN is six.5 cm.

(3)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

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

∴ ∠NSM = 90º

In right ∆NSM,

$$\begin{gathered} {\mathrm{MN}}^{\mathrm{ii}}={\mathrm{NS}}^2+{\mathrm{SM}}^2 \\ \Rightarrow \mathrm{SM}=\sqrt{{\mathrm{MN}}^2-{\mathrm{NS}}^2} \\ \Rightarrow \mathrm{SM}=\sqrt{{\left(6.5\right)}^{\mathrm{two}}-{\left(2.5\right)}^2} \\ \Rightarrow \mathrm{SM}=\sqrt{42.25-6.25} \\ \Rightarrow \mathrm{SM}=\sqrt{36}=6\mathrm{cm} \end{gathered}$$

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

⇒ MS : SR = vi cm : 3 cm = 2 : ane

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

Page No 85:

Question vii:

In the bordering 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 consummate the proof.

Answer:

Construction: Draw segments XZ and

YZ

.

Proof: By theorem of touching circles, points Ten, Z, Y are

collinear

.

∴ ∠XZA ≅

∠BZY

(opposite angles)

Allow ∠XZA = ∠BZY =a                   .....(I)

Now, seg XA ≅ seg XZ                    ..... (

Radii of circle with centre 10

)

∴∠XAZ =

∠XZA

=a                     ..... (isosceles triangle theorem)    (Two)

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 viii:

In the given effigy, circles with centres X and Y bear on internally at betoken Z . Seg BZ is a chord of bigger circle and it itersects smaller circumvolve at signal A. Prove that, seg AX || seg BY.

Answer:

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

Join YZ.

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

At present, seg XA ≅ seg XZ            (Radii of circumvolve with heart 10)

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

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

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

From (ane) 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 examination)

Folio No 86:

Question 9:

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

Respond:

Information technology is given that linefifty touches the circle with centre O at signal P.

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

Join OR.

Now, chord RS ||l.

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

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

So, RQ = QS =

$\frac{12}{\mathrm{ii}}$

 = 6 units

Letr be the radius of the circle.

∴ OR = PQ =r (Radii of the circle)

Q is the mid-point of OP.

∴ OQ = PQ =

$\frac{r}{\mathrm{two}}$

In right ∆OQR,

$$\begin{gathered} {\mathrm{OR}}^2={\mathrm{OQ}}^2+{\mathrm{RQ}}^{\mathrm{ii}} \\ \Rightarrow r^2={\left(\frac{r}{2}\right)}^{\mathrm{two}}+{\left(6\right)}^2 \\ \Rightarrow r^{\mathrm{ii}}-\frac{r^{\mathrm{ii}}}{4}=36 \\ \Rightarrow \frac{3r^2}{4}=36 \\ \Rightarrow r=\sqrt{\frac{36\times 4}{3}}=\sqrt{48}=4\sqrt{\mathrm{iii}}\mathrm{units} \end{gathered}$$

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.

Reply:

Seg AB is a diameter of a circumvolve with middle C.

∴ Air conditioning = CB    (Radii of the circle)

Join CP, CT and CQ.

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

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

⇒ seg CT ⊥ line PQ

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

∴ seg AP || seg CT || seg BQ

Nosotros know, the ratio of the intercepts fabricated 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.

$$\begin{gathered} \therefore \frac{\mathrm{PT}}{\mathrm{TQ}}=\frac{\mathrm{AC}}{\mathrm{CB}} \\ \Rightarrow \frac{\mathrm{PT}}{\mathrm{TQ}}=1\left(\mathrm{AC}=\mathrm{CB}\right) \\ \Rightarrow \mathrm{PT}=\mathrm{TQ} \end{gathered}$$

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 coinciding triangles)

Hence proved.

Folio No 86:

Question 11:

Describe 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 = three cm

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

∴ AB = three cm + iii cm = 6 cm

BC =  3 cm + 3 cm = 6 cm

CA = 3 cm + 3 cm = 6 cm

Describe a line seg AB = 6 cm.

With A equally eye and radius = 6 cm, mark an arc.

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

Bring together Air conditioning and BC.

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

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

Page No 86:

Question 12:

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

Reply:

Let A, B and C exist whatever 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 exist parallel because 2 or more lines which are perpendicular to a given line are parallel to each other.

At present, 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 exist parallel, as parallel lines exercise not intersect at a point.

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

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

Page No 87:

Question 13:

In the given figure, line PR touches the circumvolve at point Q. Answer the post-obit questions with the aid of the effigy.
(1) What is the sum of ∠ TAQ and ∠ TSQ ?
(2) Discover the angles which are congruent to ∠ AQP.
(3) Which angles are congruent to ∠ QTS ?
(4) ∠TAS = 65°, discover 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            (Contrary angles of a cyclic quadrilateral are supplementary)

(2)
The bending 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 betwixt a tangent of a circumvolve and a chord drawn from the point of contact is congruent to the angle inscribed in the arc reverse to the arc intercepted by that angle.

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

∴ ∠QTS ≅ ∠SQR

⇒ ∠QTS ≅ ∠SAQ ≅ ∠SQR

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

The measure of an inscribed angle is one-half of the measure of the arc intercepted past it.

∴ ∠TAS =

$\frac{\mathrm{one}}{\mathrm{ii}}$

one thousand(arc TS)

⇒k(arc TS) = 2∠TAS = two × 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 signal of contact is congruent to the angle inscribed in the arc reverse 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 mensurate of ∠ATQ is 100º.

Folio 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) = lxxx°, discover –
(1) chiliad(arc PR)
(2) m(arc QS)
(three) g(arc QSR)

Reply:

O is the centre of the circle.

chord PQ ≅ chord RS          (Given)

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

⇒m(arc PQ) = m(arc RS)

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

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

(2)
thousand(arc PR) +thousand(arc PQ) +thou(arc QS) +m(arc RS) = 360º

⇒ 70º + 80º + k(arc QS) + 80º = 360º

⇒k(arc QS) = 360º − 230º = 130º

(3)
m(arc QSR) =m(arc QS) +m(arc RS) = 130º + 80º = 210º

Page No 87:

Question xv:

In the given figure, thousand(arc WY) = 44°, m(arc ZX) = 68°, then
(1) Find the measure out of ∠ ZTX.
(2) If WT = 4.8, TX = eight.0,
YT = six.iv, find TZ.
(iii) If WX = 25, YT = 8,
YZ = 26, observe WT.

Respond:

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

(1)
If two chords of a circumvolve intersect each other in the interior of a circumvolve then the measure of the angle between them is half the sum of measures of arcs intercepted by the angle and its contrary angle.

∴ ∠ZTX =

$\frac{\mathrm{one}}{2}$

[k(arc ZX) + chiliad(arc WY)] =

$\frac{\mathrm{i}}{2}\times \left(68°+44°\right)=\frac{1}{\mathrm{ii}}\times 112°$

 = 56º

Thus, the measure of ∠ZTX is 56º.

(2)
WT × TX = YT × TZ                  (Theorem of internal division of chords)

⇒ four.8 × viii = half dozen.4 × TZ

⇒ TZ =

$\frac{4.8\times \mathrm{viii}}{6.4}$

 = six

(3)
WT × TX = YT × TZ                  (Theorem of internal segmentation of chords)

⇒ WT × (WX − WT) = YT × (YZ − YT)

⇒ WT × (25 − WT) = 8 × (26 − 8)

⇒ 25WT  − WTⁱⁱ = 8 × eighteen = 144

⇒ WT²− 25WT + 144 = 0

⇒ WT^two− 16WT − 9WT + 144 = 0

⇒ WT(WT− sixteen) − 9(WT − 16) = 0

⇒ (WT− 16)(WT− 9) = 0

⇒ WT− 16 = 0 or WT− nine = 0

⇒ WT= xvi or WT= 9

Page No 88:

Question sixteen:

In the given effigy
(1) thousand(arc CE) = 54°,m(arc BD) = 23°, find measure of ∠CAE.
(2) If AB = 4.2, BC = 5.4, AE = 12.0, find Advertizing
(3) If AB = 3.6, Ac = 9.0, AD = 5.4, detect AE

Answer:

(1)
If two lines containing chords of a circle intersect each other outside the circumvolve, and so the measure of angle betwixt them is half the deviation in measures of the arcs intercepted past the angle.

∴ ∠CAE =

$\frac{1}{2}$

[yard(arc CE) −m(arc BD)] =

$\frac{\mathrm{one}}{\mathrm{two}}\times \left(54°-23°\right)=\frac{\mathrm{i}}{2}\times 31°$

 = 15.5º

(2)
Ac × AB = AE × Advertisement            (Theorem of external segmentation of chords)

⇒ (AB + BC) × AB = AE × AD

⇒ (4.two + v.4) × 4.2 = 12 × AD

⇒ AD =

$\frac{9.\mathrm{half\; dozen}\times 4.2}{12}$

 = iii.36

(3)
Air conditioning × AB = AE × Advertisement            (Theorem of external partition of chords)

⇒  9 × iii.vi = AE × five.4

⇒ AE =

$\frac{\mathrm{nine}\times 3.6}{5.\mathrm{iv}}$

 = half-dozen

Page No 88:

Question 17:

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

Answer:

Proof: Describe seg GF.

∠EFG = ​∠FGH                       .....

$\boxed{\mathrm{property}\mathrm{of}\mathrm{alternate}\mathrm{angles}\mathrm{of}\mathrm{parallel}\mathrm{lines}}$

      (I)

∠EFG = ​

$\boxed{\frac{1}{2}\mathrm{one\; thousand}\left(\mathrm{arc}\mathrm{EG}\right)}$

              .....inscribed angle theorem (Ii)

∠FGH = ​

$\boxed{\frac{1}{\mathrm{ii}}m\left(\mathrm{arc}\mathrm{FH}\right)}$

              .....inscribed angle theorem (III)

∴m(arc EG) =

$\boxed{m\left(\mathrm{arc}\mathrm{FH}\right)}$

        from (I), (2), (III)

chord EG ≅ chord FH              .....

$\boxed{\mathrm{Chords}\mathrm{corresponding}\mathrm{to}\mathrm{coinciding}\mathrm{arcs}\mathrm{of}\mathrm{a}\mathrm{circumvolve}\mathrm{are}\mathrm{coinciding}}$

Folio No 88:

Question 18:

In the given figure, P is the point of contact.
(1) If chiliad(arc PR) = 140°, ∠ POR = 36°, find grand(arc PQ)
(2) If OP = 7.2, OQ = 3.2, observe OR and QR
(three) If OP = 7.2, OR = 16.2, notice QR.

Answer:

Bring together PQ.

(1)
The measure of an inscribed bending is half of the mensurate of the arc intercepted past it.

∴ ∠PQR =

$\frac{1}{\mathrm{two}}$

m(arc PR) =

$\frac{1}{\mathrm{ii}}\times 140°$

 = 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 indicate of contact is coinciding to the angle inscribed in the arc opposite to the arc intercepted past that angle.

∴ ∠PRQ = ∠OPQ = 34º

Now,

∠PRQ =

$\frac{1}{2}$

grand(arc PQ)    (The mensurate of an inscribed angle is half of the measure of the arc intercepted by information technology)

⇒1000(arc PQ) = ii∠PRQ = 2 × 34º = 68º

(ii)
OP is the tangent and OQR is the secant.

∴ OQ × OR = OP² (Tangent secant segment theorem)

⇒ iii.2 × OR = (7.two)²

⇒ OR =

$\frac{7.2\times \mathrm{seven}.\mathrm{ii}}{3.\mathrm{ii}}$

 = sixteen.two

∴  QR = OR − OQ = 16.2 − 3.2 = 13

(iii)
OP is the tangent and OQR is the secant.

∴ OQ × OR = OP^two           (Tangent secant segment theorem)

⇒ OQ × 16.2 = (7.2)²

⇒ OQ =

$\frac{7.2\times 7.\mathrm{ii}}{16.2}$

 = iii.2

∴  QR = OR − OQ = 16.2 − iii.2 = 13

Folio No 88:

Question xix:

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

Answer:

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

Join ED.

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

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

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

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

∴ Point A is the mid-point of chord EB.            (Perpendicular drawn from the center 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 circumvolve at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

Answer:

Proof: Draw seg OD.

∠ACB =

$\boxed{90°}$

                 ..... bending inscribed in semicircle

∠ACB =

$\boxed{45°}$

                 ..... CD is the bisector of ∠C

m(arc DB) = 2∠ACB =

$\boxed{90°}$

.....inscribed bending theorem

∠DOB =

$\boxed{90°}$

.....definition of measure of an arc (I)

seg OA ≅ seg OB           .....

$\boxed{\mathrm{Radii}\mathrm{of}\mathrm{the}\mathrm{circle}}$

    (II)

∴ line OD is

$\boxed{\mathrm{perpendicular}\mathrm{bisector}}$

 of seg AB      .....From (I) and (Two)

∴ seg Advert ≅ 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 = nine and d(O,L) = v. Detect the radius of the circle.

Answer:

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

Draw OP ⊥ MN and join OM.

MP = PN =

$\frac{\mathrm{MN}}{\mathrm{ii}}=\frac{25}{\mathrm{two}}$

 units          (Perpendicular drawn from the centre of a circle on its chord bisects the chord)

∴ LP = MP − ML =

$\frac{25}{2}-9=\frac{\mathrm{vii}}{2}$

 units

In right ∆OPL,

$$\begin{gathered} {\mathrm{OL}}^{\mathrm{ii}}={\mathrm{LP}}^{\mathrm{ii}}+{\mathrm{OP}}^2 \\ \Rightarrow \mathrm{OP}=\sqrt{{\mathrm{OL}}^2-{\mathrm{LP}}^2} \\ \Rightarrow \mathrm{OP}=\sqrt{5^{\mathrm{two}}-{\left(\frac{7}{\mathrm{ii}}\right)}^2} \\ \Rightarrow \mathrm{OP}=\sqrt{25-\frac{49}{\mathrm{iv}}} \\ \Rightarrow \mathrm{OP}=\sqrt{\frac{51}{\mathrm{iv}}}=\frac{1}{2}\sqrt{51}\mathrm{units} \end{gathered}$$

In right ∆OPM,

$$\begin{gathered} {\mathrm{OM}}^2={\mathrm{MP}}^2+{\mathrm{OP}}^{\mathrm{two}} \\ \Rightarrow \mathrm{OM}=\sqrt{{\left(\frac{25}{2}\right)}^2+{\left(\frac{\sqrt{51}}{2}\right)}^2} \\ \Rightarrow \mathrm{OM}=\sqrt{\frac{625+51}{4}} \\ \Rightarrow \mathrm{OM}=\sqrt{\frac{676}{4}} \\ \Rightarrow \mathrm{OM}=\sqrt{169}=13\mathrm{units} \end{gathered}$$

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 South and R. Their common tangent PQ touches the circle at points P, Q.
Show that, ∠ PRQ + ∠ PSQ = 180°

Reply:

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 betoken of contact is coinciding to the angle inscribed in the arc reverse to the arc intercepted by that bending.

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

∴ ∠RPQ = ∠PSR      .....(i)

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

∴ ∠RQP = ∠QSR      .....(2)

Calculation (i) 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 effigy, two circles intersect at points Chiliad and N. Secants drawn through M and Northward intersect the circles at points R, S and P, Q respectively. Prove that : seg SQ || seg RP.

Answer:

Information technology is given that two circles intersect at points M and N. Secants fatigued through G and Northward intersect the circles at points R, S and P, Q.

Join MN.

Quadrilateral PRMN is a circadian quadrilateral.

∴ ∠PRM = ∠MNQ       .....(one)           (Outside bending of a cyclic quadrilateral is coinciding to the angle contrary to its adjacent interior bending)

Quadrilateral QSMN is a circadian quadrilateral.

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

Adding (ane) and (ii), nosotros go

∠PRM + ∠QSM = ∠MNQ + ∠MNP              .....(3)

Now,

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

From (three) and (four), nosotros 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 ii singled-out lines are supplementary, then the two lines are parallel)

Hence proved.

Page No xc:

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 2 circles intersect each other at points A and E.

Join AE, AB and AD.

The angle betwixt a tangent of a circle and a chord drawn from the point of contact is congruent to the bending inscribed in the arc contrary 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 (four), we become

∠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 xc:

Question 25:

In the given figure, seg AD ⊥ side BC, seg Exist ⊥ side AC, seg CF ⊥ side AB. Ponit O is the orthocentre. Prove that , bespeak 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 circadian.     (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)

⇒ ∠OAE = ∠OFE          .....(one)       (Angles inscribed in the same arc are coinciding)

∠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         .....(ii)       (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 bending sum belongings of triangle)

From (3) and (4), we go

∠DAC + ∠ACD = ∠BCE + ∠CBE

⇒ ∠DAC = ∠CBE       .....(v)

From (ane), (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 indicate of intersection of its angle bisectors)

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Source: https://www.meritnation.com/cbse-class-10/math/mathematics-part-ii-solutions-/circle/textbook-solutions/91_1_3385_24435