Class 9 Exam  >  Class 9 Notes  >  RD Sharma Solutions for Class 9 Mathematics  >  RD Sharma Solutions -Ex-16.2, Circles, Class 9, Maths

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics PDF Download

Q1) The radius of a circle is 8 cm and the length of one of its chords is 12 cm. Find the distance of the chord from the centre.

Solution:

Given that,

Radius of circle (OA) = 8cm

Chord (AB) = 12cm

Draw OC⊥AB

We know that

The perpendicular from centre to chord bisects the chord

∴ AC = BC = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 6cm

Now in ΔOCA, by Pythagoras theorem

AC+ OC = OA2

⇒ 62 + OC2  = 82

⇒ 36 + OC2  = 64

⇒ OC2  = 64-36

⇒ OC2  = 28

⇒ OC =  Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ OC = 5.291cm

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q2) Find the length of a chord which is at a distance of 5 cm from the centre of a circle of radius 10 cm.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given that,

Distance (OC) = 5cm

Radius of the circle (OA) = 10cm

In ΔOCA, by Pythagoras theorem

OC+ AC = OA2

⇒ 52 + AC2  = 102

⇒ 25 + AC2  = 100

⇒ AC2  = 100 – 25

⇒ AC2  = 75

⇒ AC = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ AC = 8.66cm

We know that, the perpendicular from the centre to chord bisects the chord

Therefore, AC = BC = 8.66cm

Then the chord AB = 8.66 + 8.66

= 17.32cm

 

Q3) Find the length of a chord which is at a distance of 4 cm from the centre of a circle of radius 6 cm.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given that,

Radius of the circle (OA) = 6cm

Distance (OC) = 4cm

In ΔOCA, by Pythagoras theorem

AC+ OC = OA2

⇒ AC2 + 42  = 62

⇒ AC2  = 36 – 16

⇒ AC2  = 20

⇒ AC =  Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒ AC = 4.47cm

We know that the perpendicular distance from centre to chord bisects the chord.

AC = BC = 4.47cm

Then AB = 4.47 + 4.47

= 8.94cm

 

Q4) Two chords AB, CD of lengths 5 cm, 11 cm respectively of a circle are parallel. If the distance between AB and CD is 3 cm, find the radius of the circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Construction: Draw OP⊥CD

Chord AB = 5cm

Chord CD = 11cm

Distance PQ = 3cm

Let OP = x cm

And OC = OA = r cm

We know that the perpendicular from centre to chord bisects it.

∴ CP = PD = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

And AQ = BQ = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematicscm

In ΔOCP, by Pythagoras theorem

OC2 = OP2+CP2

⇒r2 = x2+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics…..(i)

In ΔOQA, by Pythagoras theorem

OA2 = OQ2+AQ2

⇒r2 = (x+3)2+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics…..(ii)

Compare equation (i) and (ii)

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒x2−x2+6x = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics−9

⇒6x = 15

⇒x = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q5) Give a method to find the centre of a given circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

(1) Take three points A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of the chord AB and BC which intersect each other at O.

(4) Point O will give the required circle because we know that, the Perpendicular bisectors of chord always pass through the centre.

 

Q6) Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given:

C is the mid-point of chord AB.

To prove: D is the mid-point of arc AB.

Proof:

In ΔOACandΔOBC

OA = OB [Radius of circle]

OC = OC [Common]

AC = BC [C is the mid-point of AB]

ThenΔOAC≅ΔOBC[BySSScondition]

∴ ∠AOC = ∠BOC

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics
Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Hence, D is the mid-point of arc AB.

 

Q7) Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Given:

PQ is a diameter of circle which bisects the chord AB at C.

To Prove: PQ bisects ∠AOB

Proof:

In ∠AOCand∠BOC

OA = OB [Radius of circle]

OC = OC [Common]

AC = BC [Given]

Then ΔAOC≅ΔBOC          [By SSS  condition]

∠AOC = ∠BOC[C.P.C.T]

Hence PQ bisects ∠AOB.

 

Q8) Prove that two different circles cannot intersect each other at more than two points.

Solution:

Suppose two circles intersect in three points A, B, C.

Then A, B, C are non-collinear so a unique circle passes through these three points. This is contradiction to the face that two given circles are passing through A, B, C. Hence, two circles cannot intersect each other at more than two points.

 

Q9) A line segment AB is of length 5 cm. Draw a circle of radius 4 cm passing through A and B. Can you draw a circle of radius 2 cm passing through A and B? Give reason in support of your answer.

Solution:

(1) Draw a line segment AB of 5cm.

(2) Draw the perpendicular bisectors of AB.

(3) With centre A and radius of 4cm, draw an arc which intersects the perpendicular bisector at point O. The point O will be the required centre.

(4) Join OA.

(5) With centre O and radius OA, draw a circle.

No, we cannot draw a circle of radius 2cm passing through A and B because when we draw an arc of radius 2cm with centre A, the arc will not intersect the perpendicular bisector and we will not find the centre.

Q10) An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Let ABC be an equilateral triangle of side 9cm and let AD is one of its median.

Let G be the centroid of ΔABC. Then AG : GD = 2 : 1

We know that in an equilateral triangle, centroid coincides with the circum centre.

Therefore, G is the centre of the circumference with circum radius GA.

Also G is the centre and GD is perpendicular to BC.

Therefore, In right triangle ADB, we have

AB2  = AD2 + DB2

⇒ 92  = AD2 + DB2

⇒AD =  Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

∴ Radius = AG = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q11) Given an arc of a circle, complete the circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

(1) Take three points A, B and C on the given arc.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chords AB and BC which intersect each other at point O. Then O will be the required centre of the required circle.

(4) Join OA.

(5) With centre O and radius OA, complete the circle.

 

Q12) Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:

Each pair of circles have 0, 1 or 2 points in common.

The maximum number of points in common is ‘2’.

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

 

Q13) Suppose you are given a circle. Give a construction to find its centre.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Steps of Construction:

(1) Take three points A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which intersect each other at O.

(4) Point O will be the required centre of the circle because we know that the perpendicular bisector of the chord always passes through the centre.

 

Q14) Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Draw OM⊥ABandON⊥CD.

Join OB and OD.

BM = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics[Perpendicular from the centre bisects the chord]

ND = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Let ON be x, so OM will be 6 – x.

ΔMOB

OM2 + MB2  = OB2

(6−x)2+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = OB2

36+x2−12x+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = OB2…..(i)

In ΔNOD

ON2 + ND2  = OD2

x2+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = OD2

x2+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = OD2…..(ii)

We have OB = OD. [Radii of same circle]

So, from equation (i) and (ii).

36+x2−12x+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

⇒12x = 36+Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

x = 1

From equation (ii)

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

OD2 = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

OD = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

So, the radius of the circle is found to be 55√2cm

 

Q15) The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre?

Solution:

Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

Distance of smaller chord AB from centre of circle = 4cm, OM = 4cm

MB = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 3cm

InΔOMB

OM2+MB2 = OB2

42+92 = OB2

16+9 = OB2

OB = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

OB = 5cm

InΔOND

OD = OB = 5cm              [Radii of same circle]

ND = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics = 4cm

ON2+ND2 = OD2

ON2+42 = 52

ON2 = 25−16

ON = Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics

ON = 3cm

So, the distance of bigger chord from the circle is 3cm.

The document Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions | RD Sharma Solutions for Class 9 Mathematics is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.
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FAQs on Ex-16.2, Circles, Class 9, Maths RD Sharma Solutions - RD Sharma Solutions for Class 9 Mathematics

1. What is the difference between a circle and a sphere?
Ans. A circle is a two-dimensional figure that lies in a plane and is formed by all the points equidistant from a fixed point called the center. On the other hand, a sphere is a three-dimensional figure formed by all the points equidistant from a fixed point called the center. In simple terms, a circle is a flat shape, whereas a sphere is a solid shape.
2. How can I find the area of a circle?
Ans. The area of a circle can be found using the formula A = πr^2, where A represents the area and r represents the radius of the circle. To calculate the area, simply square the radius and multiply it by π (pi), which is approximately equal to 3.14.
3. How do I find the circumference of a circle?
Ans. The circumference of a circle can be calculated using the formula C = 2πr, where C represents the circumference and r represents the radius of the circle. To find the circumference, simply multiply 2π (pi) by the radius of the circle.
4. What is the relationship between the diameter and radius of a circle?
Ans. The diameter of a circle is the distance across the circle passing through the center, while the radius is the distance from the center to any point on the circle. The relationship between the diameter and radius is that the diameter is always twice the length of the radius. In other words, if r is the radius of a circle, then the diameter can be represented as 2r.
5. Can the radius of a circle be negative?
Ans. No, the radius of a circle cannot be negative. The radius is a measure of distance, and distance cannot be negative. It is always a positive value or zero if the circle has a radius of 0.
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