RD Chapter 16- Circles Ex-16.1 |
RD Chapter 16- Circles Ex-16.3 |
RD Chapter 16- Circles Ex-16.4 |
RD Chapter 16- Circles Ex-16.5 |
RD Chapter 16- Circles Ex-VSAQS |

The radius of a circleis 8 cm and the length of one of its chords is 12 cm. Find the distance of thechord 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 .

**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

**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

Hence the length of the chord is 8.94 cm.

**Answer
4** :
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

…… (2)

From (1) and (2) we get

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

**Answer
5** :

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

.

**Answer
6** :

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 and △BOP△AOP 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.

**Answer
7** :

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.

**Answer
8** :

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.

**Answer
9** :

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.

**Answer
10** :
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 =

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