Mathematics Part II Solutions Solutions for Class 9 Maths Chapter 6 Circle are provided here with simple step-by-step explanations. These solutions for Circle are extremely popular among Class 9 students for Maths Circle Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class 9 Maths Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part II Solutions Solutions. All Mathematics Part II Solutions Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 79:

#### Question 1:

Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of the circle.

#### Answer:

Let the centre of the circle be O.

AB = 12 cm

OC = 12 cm

OC ⊥ AB

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

So, AC = 6 cm

In ⃤ OCA,

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {8}^{2}+{6}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AO}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AO}=10$

Thus, AO = 10 cm

Diameter = 2 × 10 cm = 20 cm.

#### Page No 79:

#### Question 2:

Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.

#### Answer:

Diameter = 26 cm

chord = 24 cm

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

OD ⊥ AB

So, AD = DB

AO = OC (Radii)

In ∆ODA,

${\mathrm{OD}}^{2}+{\mathrm{AD}}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OD}}^{2}+{12}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OD}}^{2}+144=169\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OD}}^{2}=169-144=25\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OD}=5\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Thus, the distance of the chord from the centre is 5 cm.

#### Page No 79:

#### Question 3:

Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord.

#### Answer:

Let the chord be AB.

O be the centre of the circle.

OC be the perpendicular drawn from the centre to the chord AB.

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

So, OC $\perp $ AB.

In $\u25b3$OCA,

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {30}^{2}+{\mathrm{AC}}^{2}={34}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 900+{\mathrm{AC}}^{2}=1156\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=256\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=16\mathrm{cm}$

Thus, AB = $2\times \mathrm{AC}=2\times 16=32$.

#### Page No 79:

#### Question 4:

Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle.

#### Answer:

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

PR = RQ = 40 unit

In $\u25b3$OPR,

${\mathrm{OR}}^{2}+{\mathrm{PR}}^{2}={\mathrm{OP}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OR}}^{2}+{40}^{2}={41}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OR}}^{2}+=1681-1600\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OR}}^{2}=81\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OR}=9\mathrm{unit}$

Thus, the distance of the chord from the centre of the circle is 9 units.

#### Page No 79:

#### Question 5:

In the given figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ

#### Answer:

Draw perpendicular from O to line AB.

So, OC $\perp $ PQ and OC $\perp $AB

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

So, PC = CQ .....(1)

and AC = CB .....(2)

(2) $-$ (1)

AC $-$ PC = CB $-$ CQ

$\Rightarrow $AP = BQ

Hence proved

#### Page No 79:

#### Question 6:

Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.

#### Answer:

Let AB and CD be the two chords of the circle. The centre of the circle be O through which the diameter PQ passes.

Let the diameter PQ bisect the two chords and point R and S.

We know that the when the line passing through the centre of the circle bisects the chord then the line is perpendicular to the chord.

Thus, $\angle \mathrm{ARS}=\angle \mathrm{DSR}=90\xb0$

AB will be thus parallel to CD and cut by the transversal PS as interior angle on the sam side of the transversal are equal.

#### Page No 82:

#### Question 1:

Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the distance of these chords from the centre of the circle ?

#### Answer:

Radius = AO = 10 cm

Let OE be the perpendicular drawn from the centre of the circle to the chord AB.

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, AE = EB = 8 cm

In $\u25b3$AEO,

${\mathrm{EO}}^{2}+{\mathrm{AE}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{EO}}^{2}+{8}^{2}={10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{EO}}^{2}+64=100\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{EO}}^{2}=36\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{EO}=6\mathrm{cm}$

we know that congruent chords of a circle are equidistant from the centre.

So, EO = OF = 6 cm

Distance of these chords from the centre is 6 + 6 = 12 cm.

#### Page No 82:

#### Question 2:

In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the centre. Find the lengths of the chords.

#### Answer:

Let OE be the perpendicular drawn from the centre of the circle to the chord AB.

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, AE = EB

In $\u25b3$AEO,

${\mathrm{EO}}^{2}+{\mathrm{AE}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {5}^{2}+{\mathrm{AE}}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AE}}^{2}+25=169\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AE}}^{2}=144\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AE}=12\mathrm{cm}$

And AE = EB = 12 + 12 = 24 cm

Thus, AB = CD = 24 cm as chords of a circle equidistant from the centre are congruent.

#### Page No 82:

#### Question 3:

Seg PM and seg PN are congruent chords of a circle with centre C. Show that the ray PC is the bisector of $\angle $NPM.

#### Answer:

Join CN and CM.

In $\u25b3$CPN and $\u25b3$CPM,

PM $\cong $ PN (Given)

PC = PC (Common)

CN = CM (Radii)

Thus, $\u25b3$CPN $\cong $ $\u25b3$CPM (By SSS congurency)

So, $\angle $CPN = $\angle $CPM (CPCT)

Thus, ray PC is the bisector of $\angle $NPM.

#### Page No 86:

#### Question 1:

Construct $\u2206$ ABC such that $\angle $B = ${100}^{\xb0}$ , BC = 6.4 cm, $\angle $C = ${50}^{\xb0}$ and construct its incircle.

#### Answer:

**Steps of construction:**

1. Draw BC = 6.4 cm

2. Draw$\angle $B = ${100}^{\xb0}$ and $\angle $C = ${50}^{\xb0}$ using the protractor. Where the rays YB and XC meet, name the point as A.

$\u2206$ABC is thus obtained.

3. Construct the angle bisectors of $\angle $B and $\angle $C and let them meet at point O.

4. Through point O, draw a perpendicular to line BC. Let the perpendicular meet the line BC at point P.

5. With O as centre and OP as radius, draw a circle inside the $\u2206$ABC.

This is the required incircle.

#### Page No 86:

#### Question 2:

Construct $\u2206$ PQR such that $\angle $P = ${70}^{\xb0}$ , $\angle $R = ${50}^{\xb0}$ , QR = 7.3 cm and constructs its circumcircle.

#### Answer:

By angle sum property of triangles,

In $\u2206$PQR,

$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0+\angle \mathrm{Q}+50\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{Q}=60\xb0$

Steps of construction:

1. Draw a line QR = 7.3 cm.

2. With Q as centre, draw an angle of 60$\xb0$ using the protractor. Similarly, draw an angle of 50$\xb0$ with R as centre.

Where the rays RY and XQ meet, name the point as P.

Thus, $\u2206$PQR is obtained.

3. Draw the perpendicular bisectors of the lines PR and QR. Let these perpendicular bisectors meet at point O.

4. With O as centre and OP as radius, construct a circle touching all the vertices of the $\u2206$PQR.

This circle is thus the required circumcircle.

#### Page No 86:

#### Question 3:

Construct $\u2206$XYZ such that XY = 6.7 cm ,YZ = 5.8 cm, XZ = 6.9 cm and constructs its incircle.

#### Answer:

**Steps of construction:**

1. Draw XZ = 6.9 cm

2. With X as centre and 6.7 cm as radius, draw an arc above the line XZ. Also, with Z as centre and 5.8 cm as radius, draw an arc cutting the previous drawn arc at point Y.

$\u2206$XYZ is thus obtained.

3. Construct the angle bisectors of $\angle $X and $\angle $Z and let them meet at point O.

4. Through point O, draw a perpendicular to line XZ. Let the perpendicular meet the line XZ at point P.

5. With O as centre and OP as radius, draw a circle inside the $\u2206$XYZ.

This is the required incircle.

** **

#### Page No 86:

#### Question 4:

In $\u2206$ LMN, LM = 7.2 cm, $\angle $M = ${105}^{\xb0}$ , MN = 6.4 cm , then draw $\u2206$ LMN and construct its circumcircle.

#### Answer:

**Steps of construction:**

1. Draw a line LM = 7.2 cm.

2. With M as centre, draw an angle of 105$\xb0$ using the protractor.

3. With M as centre and 6.4 cm as radius, cut an arc on ray MX and name it as N.

4. Join LN. $\u2206$LMN is thus obtained.

5. Draw the perpendicular bisectors of the lines NM and LM. Let these perpendicular bisectors meet at point O.

4. With O as centre and OM as radius, construct a circle touching all the vertices of the $\u2206$LMN.

This circle is thus the required circumcircle.

#### Page No 86:

#### Question 5:

Construct $\u2206$ DEF such that DE = EF = 6 cm, $\angle $F = ${45}^{\xb0}$ and constructs its circumcircle.

#### Answer:

**Steps of construction:**

1. Draw a line EF = 6 cm.

2. With F as centre, draw an angle of 45$\xb0$ using the protractor.

3. With E as centre and 6 cm as radius, cut an arc on ray FX and name it as D.

4. Join DE. $\u2206$DEF is thus obtained.

5. Draw the perpendicular bisectors of the lines DF and EF. Let these perpendicular bisectors meet at point O.

4. With O as centre and OE as radius, construct a circle touching all the vertices of the $\u2206$DEF.

This circle is thus the required circumcircle.

#### Page No 86:

#### Question 1:

(i) Radius of a circle is 10 cm and distance of a chord from the centre is 6 cm. Hence the length of the chord is .........

(ii) The point of concurrence of all angle bisectors of a triangle is called the ......

(iii) The circle which passes through all the vertices of a triangle is called .....

(iv) Length of a chord of a circle is 24 cm. If distance of the chord from the centre is 5 cm, then the radius of that circle is ....

(v) The length of the longest chord of the circle with radius 2.9 cm is .....

(vi) Radius of a circle with centre O is 4 cm. If

*l*(OP) = 4.2 cm, say where point P will lie.

(vii) The lengths of parallel chords which are on opposite sides of the centre of a circle are 6 cm and 8 cm. If radius of the circle is 5 cm, then the distance between these chords is .....

#### Answer:

(i)

Let the chord be AB.

O be the centre and OC be the perpendicular drawn from the centre of the circle to the chord AB.

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

AC = CB

OA is the radius = 10 cm

In $\u25b3$OAC,

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{OA}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {6}^{2}+{\mathrm{AC}}^{2}={10}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 36+{\mathrm{AC}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=64\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=8\mathrm{cm}$

AC = CB = 8 cm

AB = AC + CB = 8 + 8 = 16 cm

Hence, the correct answer is option A.

(ii) The point of concurrence of all angle bisectors of a triangle is called the incentre.

Hence, the correct answer is option C.

(iii) The circle which passes through all the vertices of a triangle is called circumcircle.

Hence, the correct answer is option A.

(iv)

Let the chord be AB = 24 cm

Distance of the chord from the centre O is 5 cm.

AO is the radius of the circle.

Perpendicular from the centre of the circle to the chord bisects the chord.

So, AC = CB

In $\u25b3$AOC,

${\mathrm{OC}}^{2}+{\mathrm{AC}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {5}^{2}+{12}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AO}}^{2}=25+144=169\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AO}=13\mathrm{cm}$

Thus, the radius of the circle is 13 cm.

Hence, the correct answer is option B.

(v) The longest chord of the circle is the diameter.

Radius = 2.9 cm.

So, diameter = 2.9 + 2.9 = 5.8 cm

Hence, the correct answer is option D.

(vi) Radius = 4 cm

OP = 4.2 cm

OP will be thus outside the circle as it is greater than the radius.

Hence, the correct answer is option C.

(vii)

Let the chords be AB = 6 cm and CD = 8 cm

O be the centre with OA = OC = 5 cm as radius.

OE $\perp $ AB and OF $\perp $ CD.

In $\u25b3$AEO,

${\mathrm{AE}}^{2}+{\mathrm{EO}}^{2}={\mathrm{AO}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {3}^{2}+{\mathrm{EO}}^{2}={5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 9+{\mathrm{EO}}^{2}=25\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{EO}}^{2}=16\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{EO}=4\mathrm{cm}$

Similarly, in In $\u25b3$OFC,

${\mathrm{OF}}^{2}+{\mathrm{CF}}^{2}={\mathrm{OC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OF}}^{2}+{4}^{2}={5}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OF}}^{2}+16=25\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{OF}}^{2}=9\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{OF}=3\mathrm{cm}$

Thus, the distance between the two chords is EO + OF = 4 + 3 cm = 7 cm.

Hence, the correct answer is option D.

#### Page No 87:

#### Question 2:

#### Answer:

**Steps of construction:**

1. Draw a line SP = 7.5 cm.

2. With S as centre and 7.5 cm as radius, draw an arc above the line SP.

3. With P as centre and 7.5 cm as radius, cut an arc on on the previous drawn arc and name the point of intersection as D.

4. Join DS and DP. $\u2206$DSP is thus obtained.

5. Draw the perpendicular bisectors of the lines DP and SP. Let these perpendicular bisectors meet at point O.

4. With O as centre and OP as radius, construct a circle touching all the vertices of the $\u2206$DSP.

This circle is thus the required circumcircle.

Radius = 4 cm

Steps of construction

1. Draw a line SP = 7.5 cm.

2. With S as centre and 7.5 cm as radius, draw an arc above the line SP.

3. With P as centre and 7.5 cm as radius, cut an arc on on the previous drawn arc and name the point of intersection as D.

4. Join DS and DP. $\u2206$DSP is thus obtained.

5. Draw the angle bisectors of angle S and P and let them meet at point O.

6. Draw the perpendicular from point O to the line SP. Join OA.

7. With O as centre and OA as radius, draw a circle touching all the sides of the triangle.

This is the required incircle.

Radius = 2 cm

ratio of radius of circumcircle to the radius of incircle = 4 : 2 = 2 : 1

#### Page No 87:

#### Question 3:

Construct $\u2206$NTS where NT = 5.7 cm, TS = 7.5 cm and $\angle $NTS = 110 ° and draw incircle and circumcircle of it.

#### Answer:

**Steps of construction:**

1. Draw a line ST = 7.5 cm.

2. With T as centre draw an angle $\angle $STX = 110°.

3. With T as centre and 5.7 cm as radius, cut an arc on XT and name the point of intersection as N.

4. Join SN. $\u2206$NTS is thus obtained.

5. Draw the perpendicular bisectors of the lines ST and NT. Let these perpendicular bisectors meet at point O.

4. With O as centre and OT as radius, construct a circle touching all the vertices of the $\u2206$NTS.

This circle is thus the required circumcircle.

**Steps of construction:**

1. Draw a line ST = 7.5 cm.

2. With T as centre draw an angle $\angle $STX = 110°.

3. With T as centre and 5.7 cm as radius, cut an arc on XT and name the point of intersection as N.

4. Join SN. $\u2206$NTS is thus obtained.

5. Draw the angle bisectors of angle S and T and let them meet at point O.

6. Draw the perpendicular from point O to the line ST. Join OP.

7. With O as centre and OP as radius, draw a circle touching all the sides of the triangle.

This is the required incircle.

#### Page No 87:

#### Question 4:

In the given figure, C is the centre of the circle. seg QT is a diameter CT = 13, CP = 5, find the length of chord RS.

#### Answer:

Join RC and CS.

Perpendicular drawn from the centre of the circle to the chord bisects the chord.

So, RP = PS

In $\u25b3$RPC,

${\mathrm{RP}}^{2}+{\mathrm{PC}}^{2}={\mathrm{RC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{RP}}^{2}+{5}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{RP}}^{2}=169-25=144\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{RP}=12\mathrm{cm}$

Thus, RP = PS = 12 cm

So, RS = RP + PS = 12 + 12 = 24 cm.

#### Page No 87:

#### Question 5:

#### Answer:

Construction: Draw perpendicular from point P to the chords CD and AB.

To prove: AB = CD

Proof: In $\u25b3$MEP and $\u25b3$NEP,

$\angle $EMP = $\angle $ENP = 90$\xb0$

$\angle $AEP = $\angle $DEP (Given)

Thus, by AAS congruence rule, $\u25b3$MEP$\cong $$\u25b3$NEP.

So, MP = NP (CPCT)

We know that chords equidistant from the centre are equal to each other.

Hence, AB = CD.

#### Page No 87:

#### Question 6:

In the given figure, CD is a diameter of the circle with centre O. Diameter CD is perpendicular to chord AB at point E. Show that$\u2206$ABC is an isosceles triangle.

#### Answer:

In $\u25b3$AEC and $\u25b3$BEC,

AE = BE (Perpendicular drawn from the centre of the circle to the chord bisects the chord)

CE = CE (Common)

$\angle $CEA = $\angle $CEB = 90$\xb0$

Thus, $\u25b3$AEC $\cong $ $\u25b3$BEC by SAS congurency

So, by CPCT, AC = CB

Thus, $\u2206$ABC is an isosceles triangle.

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