RD Sharma 2020 solution class 9 chapter 15 Circles Exercise 15.2

Exercise 15.2

Page-15.28

Question 1:

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.

Answer 1:

Let AB be a chord of a circle with centre O and radius 8 cm such that
AB = 12 cm
We draw and join OA.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have


Hence the distance of chord from the centre .

Question 2:

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

Answer 2:

Given that OA = 10 cm and OL = 5 cm, we have to find the length of chord AB.
Let AB be a chord of a circle with centre O and radius 10 cm such that AO = 10 cm
We draw and join OA.

Since, the perpendiculars from the centre of a circle to a chord bisect the chord.
Now in we have


Hence the length of chord

Question 3:

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

Answer 3:

Given that and , find the length of chord AB.
Let AB be a chord of a circle with centre O and radius 6 cm such that

We draw and join OA.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.
Now in we have

$\Rightarrow AL=\sqrt{20}=4.47$
$$\begin{gathered} AB= 2\times AL \\ =2\times 4.47 \\ =8.94 \mathrm{cm} \end{gathered}$$

Hence the length of the chord is 8.94 cm.

Question 4:

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

Answer 4:

Let A, B and C are three distinct points on a circle .
Now join AB and BC and draw their perpendicular bisectors.
The point of intersection of the perpendicular bisectors is the centre of given circle.
Hence O is the centre of circle.

Question 5:

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.

Answer 5:

Let MN is the diameter and chord AB of circle C(O, r) then according to the question
AP = BP.
Then we have to prove that .
Join OA and OB.

       
In ΔAOP and ΔBOP
          (Radii of the same circle)
AP = BP         (P is the mid point of chord AB)
OP = OP         (Common)
Therefore,
                      (by cpct)

Hence, proved.

Page-15.29

Question 6:

A line segment AB is of length 5cm. 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.

Answer 6:

Given that a line AB = 5 cm, one circle having radius of which is passing through point A and B and other circle of radius.

As we know that the largest chord of any circle is equal to the diameter of that circle.

So,

There is no possibility to draw a circle whose diameter is smaller than the length of the chord.

Question 7:

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

Answer 7:

Let ABC be an equilateral triangle of side 9 cm and let AD be one of its medians. Let G be the centroid of. Then

We know that in an equilateral triangle centroid coincides with the circumcentre. Therefore, G is the centre of the circumcircle with circumradius GA.

As per theorem, G is the centre and . Therefore,

In we have

Therefore radius AG = $\frac{2}{3}AD=3\sqrt{3} \mathrm{cm}$

 

Question 8:

Given an arc of a circle, complete the circle.

Answer 8:

Let PQ be an arc of the circle.

In order to complete the circle. First of all we have to find out its centre and radius.

Now take a point R on the arc PQ and join PR and QR.

Draw the perpendicular bisectors of PR and QR respectively.

Let these perpendicular bisectors intersect at point O.

Then OP = OQ, draw a circle with centre O and radius OP = OQ to get the required circle.

Question 9:

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

Answer 9:

Given that two different pairs of circles in the figure.

As we see that only two points A, B of first pair of circle and C, D of the second pair of circles are common points.

Thus only two points are common in each pair of circle.

Question 10:

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

Answer 10:

Given a circle C(O, r).

We take three points A, B and C on the circle.

Join AB and BC.

Draw the perpendicular bisector of chord AB and BC.

Let these bisectors intersect at point O.

Hence, O is the centre of circle.

Question 11:

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?

Answer 11:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 6 cm, CD = 8 cm and OP = 4 cm. let the radius of the circle be cm.

According to the question, we have to find OQ

Draw and as well as point O, Q, and P are collinear.

Let

Join OA and OC, then

OA = OC = r

Now and

So, AP = 3 cm and CQ = 4 cm

In we have

And in

Question 12:

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.

Answer 12:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 5 cm and CD = 11 cm. let the radius of the circle be cm.

Draw and as well as point O, Q and P are collinear.

Clearly, PQ = 3 cm

Let then

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

$$\begin{gathered} \Rightarrow 6x+\frac{61}{4}=\frac{121}{4} \\ \Rightarrow 6x=\frac{121-61}{4} \\ \Rightarrow 6x=\frac{60}{4} \\ \Rightarrow x=\frac{5}{2} \end{gathered}$$

Putting the value of x in (2) we get,

Question 13:

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.

Answer 13:

Let P is the mid point of chord AB of circle C(O, r) then according to question, line OQ passes through the point P.

Then prove that OQ bisect the arc AB.

Join OA and OB.

In $△AOP \mathrm{and} △BOP$

                     (Radii of the same circle)

                     (P is the mid point of chord AB)

                     (Common)

Therefore,

                    (by cpct)

Thus

Arc AQ = arc BQ

Therefore,

Hence Proved.

Question 14:

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

Answer 14:

We have to prove that two different circles cannot intersect each other at more than two points.

Let the two circles intersect in three points A, B and C.

Then as we know that these three points A, B and C are non-collinear. So, a unique circle passes through these three points.

This is a contradiction to the fact that two given circles are passing through A, B, C.

Hence, two circles cannot intersect each other at more than two points.

Hence, proved.

 

Question 15:

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.

Answer 15:

Let AB and CD be two parallel chord of the circle with centre O such that AB = 5 cm, CD = 11 cm and PQ = 6 cm. Let the radius of the circle be cm.

Draw and as well as point O, Q, and P are collinear.

Clearly, PQ = 6 cm

Let OQ = x cm then

Join OA and OC, then

OA = OC = r

Nowand

So, and

In we have

…… (1)

And

…… (2)

From (1) and (2) we get

Putting the value of x in (1) we get,