##### Circles Class 9th Mathematics Part Ii MHB Solution

**Practice Set 6.1**

- Distance of chord AB from the center of a circle is 8 cm. Length of the chord AB is 12…
- Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the…
- Radius of a circle is 34 cm and the distance of the chord from the center is 30 cm,…
- Radius of a circle with center O is 41 units. Length of a chord PQ is 80 units, find…
- In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the…
- Prove that, if a diameter of a circle bisects two chords of the circle then those two…

**Practice Set 6.2**

- Radius of circle is 10 cm. There are two chords of length 16 cm each. What will be the…
- In a circle with radius 13 cm, two equal chords are at a distance of 5 cm from the…
- Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC…

**Practice Set 6.3**

- Construct Î”ABC such that ∠B = 100^0 , BC = 6.4, ∠C = 50^0 and construct its incircle.…
- Construct Î”PQR such that ∠70^0 , ∠R = 50^0 , QR = 7.3cm, and construct its…
- Construct Î”XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.…
- In Î”LMN, LM = 7.2cm, ∠M = 105^0 , MN = 6.4cm, then draw Î”LMN and construct its…
- Construct Î”DEF such that DE = EF = 6 cm, ∠F = 45^0 and construct its circumcircle.…

**Problem Set 6**

- Radius of a circle is 10 cm and distance of a chord from the center is 6 cm. Hence the…
- The point of concurrence of all angle bisectors of a triangle is called the ......…
- The circle which passes through all the vertices of a triangle is called ..... Choose…
- Length of a chord of a circle is 24 cm. If distance of the chord from the center is 5…
- The length of the longest chord of the circle with radius 2.9 cm is ..... Choose…
- Radius of a circle with center O is 4 cm. If l(OP) = 4.2 cm, say where point P will…
- The lengths of parallel chords which are on opposite sides of the center of a circle…
- Construct incircle and circumcircle of an equilateral Î”DSP with side 7.5 cm. Measure…
- Construct Î”NTS where NT = 5.7 cm, TS = 7.5 cm and ∠NTS = 110^0 and draw incircle and…
- In the figure 6.19, C is the center of the circle. seg QT is a diameter CT = 13, CP =…
- In the figure 6.20, P is the center of the circle. chord AB and chord CD intersect on…
- In the figure 6.21, CD is a diameter of the circle with center O. Diameter CD is…

###### Practice Set 6.1

**Question 1.**

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

**Answer:**

Given that OP = 8 cm

And AB = 12 cm

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

∴ AP = PB = 6 cm

In the right angled Î”OAP using Pythagoras theorem,

⇒ OA^{2} = OP^{2} + AP^{2}

⇒ OA^{2} = 8^{2} + 6^{2}

⇒ OA^{2} = 64 + 36

⇒ OA^{2} = 100

⇒ OA = 10cm

So, the diameter of the circle is (2×10) = 20cm (Diameter = 2×Radius).

**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 center.

**Answer:**

Given that diameter = 26cm

Radius = Diameter / 2 = 26 /2 = 13cm

So, OA = 13cm

And AB = 24 cm

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

∴ AP = PB = 12 cm

In the right angled Î”OAP using Pythagoras theorem,

⇒ OA^{2} = OP^{2} + AP^{2}

⇒ 13^{2} = OP^{2} + 12^{2}

⇒ 169 = OP^{2} + 144

⇒ OP^{2} = 25

⇒ OP = 5cm

So, the distance of chord from the center is 5cm.

**Question 3.**

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

**Answer:**

Given that

Radius = 34cm

So, OA = 34cm

And OP = 30 cm

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

∴ AP = PB,

AB = 2PB

In the right angled Î”OAP using Pythagoras theorem,

⇒ OA^{2} = OP^{2} + AP^{2}

⇒ 34^{2} = 30^{2} + AP^{2}

⇒ 1156 = 900 + AP^{2}

⇒ AP^{2} = 256

⇒ AP = 16cm

So, the length of chord is 16× 2 = 32cm (AB = 2AP)

**Question 4.**

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

**Answer:**

Given that

Radius = 41 units

So, OP = 41 units

And PQ = 80 units

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

∴ PM = MQ = 40 cm

In the right angled Î”OAP using Pythagoras theorem,

⇒ OP^{2} = OM^{2} + PM^{2}

⇒ 41^{2} = OM^{2} + 40^{2}

⇒ 1681 = OM^{2} + 1600

⇒ OM^{2} = 81

⇒ OM = 9 units

So, the distance of chord from the center is 9 units.

**Question 5.**

In figure 6.9, center of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ

**Answer:**

We draw a perpendicular on chord AB from O.

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

Therefore,

AM = MB …….(1)

OM is also perpendicular to chord PQ of smaller circle.

Therefore,

PM = MQ ………….(2)

Subtracting (2) from (1)

AM-PM = MB-MQ

⇒ AP = BQ

Hence Proved.

**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:**

We draw a circle with center O and AB, CD are the chords of this circle. Diameter PQ bisects AB and CD at M and N respectively.

We know that the line from the center bisecting the chord is perpendicular to the chord.

Therefore,

∠ OMA = ∠ OMB = 90°

Also, ∠ ONC = ∠ OND = 90°

∠ OMA + ∠ ONC = 90° + 90° = 180°

Hence the two chords, AB and CD are parallel to each other.

###### Practice Set 6.2

**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 center of the circle?

**Answer:**

Given radius of circle is 10cm

OA = OD = 10cm

AB = CD = 16cm

We know that a perpendicular drawn from the center of a circle on its chord bisects

the chord.

CQ = QD = 8cm

In right angled Î”OQD using the Pythagoras theorem

OD^{2} = OQ^{2} + QD^{2}

10^{2} = OQ^{2} + 8^{2}

100 = OQ^{2} + 64

OQ^{2} = 36

OQ = 6cm

Therefore the chord CD is at 6cm from the center.

We know that Congruent chords of a circle are equidistant from the center of the circle.

As AB and CD are equal in length, they are equidistant.

∴ OP = OQ = 6cm

**Question 2.**

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

**Answer:**

Given radius of circle is 13cm

OA = OD = 13cm

OQ = OP = 16cm

We know that a perpendicular drawn from the centre of a circle on its chord bisects

the chord.

CQ = QD

CD = 2×QD

In right angled Î”OQD using the Pythagoras theorem

OD^{2} = OQ^{2} + QD^{2}

13^{2} = 5^{2} + QD^{2}

169 = 25 + QD^{2}

QD^{2} = 144

QD = 12cm

Therefore the length of chord CD = 2×12 = 24cm

We know that The chords of a circle equidistant from the center of a circle are congruent

As AB and CD are equidistant, they are equal in length.

∴ AB = CD = 24cm

**Question 3.**

Seg PM and seg PN are congruent chords of a circle with center C. Show that the ray PC is the bisector of ∠NPM.

**Answer:**

Given that PM = PN

We know that Congruent chords of a circle are equidistant from the center of the circle.

Therefore, AC = CB ………………(1)

Also,

A perpendicular drawn from the centre of a circle on its chord bisects

the chord.

CB bisects PN as PB = BN,

Similarly, CA bisects PM as PA = AM.

In Î”APC and Î”BPC,

∠CAP = ∠ CBP = 90°

PC = PC (common side)

AC = CB (From eq (1))

∴ Î”APC≅ Î”BPC (RHS congruence)

∴ ∠ APC = ∠ BPC (by CPCT)

Hence proved that PC is the bisector of ∠ NPM.

###### Practice Set 6.3

**Question 1.**

Construct Î”ABC such that ∠B = 100^{0}, BC = 6.4, ∠C = 50^{0} and construct its incircle.

**Answer:**

Steps of Construction:

1.Construct Î”ABC of given dimensions.

2.Draw bisectors of two angles, ∠A and ∠B.

3.Denote the point of intersection as O.

4.Draw perpendicular OP on AB.

5.Draw a circle with O as center and OP as radius.

**Question 2.**

Construct Î”PQR such that ∠70^{0}, ∠R = 50^{0}, QR = 7.3cm, and construct its circumcircle.

**Answer:**

Steps of Construction:

1.Construct Î”PQR of given dimensions.

2.Draw perpendicular bisectors of two sides, QR and PR.

3.Denote the point of intersection as O.

4.Draw a circle with O as center and OP as radius.

**Question 3.**

Construct Î”XYZ such that XY = 6.7 cm, YZ = 5.8 cm, XZ = 6.9 cm. Construct its incircle.

**Answer:**

Steps of Construction:

1.Construct Î”XYZ of given dimensions.

2.Draw bisectors of two angles, ∠X and ∠Y.

3.Denote the point of intersection as O.

4.Draw perpendicular OA on XY.

5.Draw a circle with O as center and OA as radius.

**Question 4.**

In Î”LMN, LM = 7.2cm, ∠M = 105^{0}, MN = 6.4cm, then draw Î”LMN and construct its circumcircle.

**Answer:**

Steps of Construction:

1.Construct Î”LMN of given dimensions.

2.Draw perpendicular bisectors of two sides, LM and MN.

3.Denote the point of intersection as O.

4.Draw a circle with O as center and OM as radius.