### Page 16.5 Ex. 16.1

Q1.

**Answer :**

(i) All points lying inside or outside a circle are called interior points or exterior points.

Let the point *P*, *Q* and *R* are lie inside, outside or on the circle *C*(*O*, *r*) in such a way as given in the figure

Thus the answer is

(ii) Given that the circles having the same centre and different radii are called

As we know that the circles having the same centre and different radii are called concentric circle.

Thus the answer is

(iii) As we know that appoint whose distance from the centre of a circle is greater than its radius lies in the exterior of the circle as shown in the figure

Thus the answer is

(iv) As we know that a continuous piece of a circle is an arc of the circle.

Let *P* and *Q* be points on the circle *C* (*O*, *r*) then the piece are arcs of the circle *C* (*O*, *r*)

Thus the answer is arc.

(v) Given that the largest chord of the circle is a diameter of the circle.

As we know that a circle having so many diameters and a diameter of a given circle is one of the largest chords of the circle.

Thus the answer is diameter.

(vi) Given that an arc a semicircle when its ends are the ends of the diameter.

As we know that a diameter of a circle divides it into two equal parts which are arcs and each of two arcs is called a semicircle.

Thus the answer is semicircle.

(vii) As we know that segment of a circle is the region between arc and chord of the circle.

Let *PQ* be a chord of the circle *C* (*O*, *r*), then *PQ* divides the region enclosed by the circle into two parts. Each of the part is called segment of the circle.

Thus the answer is chord.

(viii) As we know that a circle divides the plane on which it lies in three parts as shown in the figure.

Thus the answer is three.

Q2.

**Answer :**

(i) Given that a circle is a plane figure.

As we know that a circle is a collection of those points in a plane that are at a given constant distance from affixed point in the plane.

Thus the given statement is

(ii) Given that line segment joining the centre to any point on the circle is a radius of the circle.

As we know that line segment joining the centre to any point on the circle is a radius of the circle.

Thus the given statement is

(iii) Given that if a circle is divided into three equal arcs each is a major arc.

As we know that if points *P*, *Q* and *R* lies on the given circle *C*(*O*, *r*)in such a way that

Then each arc is called major arc.

Thus the given statement is

(iv) It is given that a circle has only finite number of equal chords.

As we know that a circle having infinite number of unequal chords.

Thus the given statement is.

(v) Given that a chord of the circle, which is twice as long as its radius is diameter of the circle.

As we know that a chord of a circle which is largest to others and passing through the centre of the circle and twice as long as its radius is called diameter of the circle.

Thus the given statement is .

(vi) It is given that sector is the region between the chord and its corresponding arc.

As we know that the region between the chord and its corresponding arc is called sector.

Thus the given statement is.

(vii) Given that the degree measure of an arc is the complement of the central angle containing the arc.

As we know that the degree measure of a minor arc is the measure of the central angle containing the arc and that of a major arc is 360° minus the degree measure of the corresponding minor arc.

Let degree measure of an arc is ? of a given circle is denoted by

Thus the given statement is.

(viii) Given that the degree measure of a semi-circle is 180°.

As we know that the diameter of a circle divides into two equal parts and each of these two arcs are known as semi-circle.

and are semi circle

Hence,

Thus the given statement is.

### Page 16.24 Ex. 16.2

Q1.

**Answer :**

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

Q2.

**Answer :**

Given that *OA* = 10 cm and *OL* = 5 cm, 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

Q3.

**Answer :**

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 chord

Q4.

**Answer :**

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

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

Clearly,

Let then

In we have

…… (1)

And

…… (2)

From equation (1) and (2) we get

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

Q5.

**Answer :**

Let *A*, *B* and *C* are three distinct points on a circle whose radius is

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.

Q6.

**Answer :**

Let *P* is the mint 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

(Both are equal to radius).

(Mid point of chord *AB*)

(Common side)

Therefore,

Thus

Arc *AQ = *arc* BQ*

Hence Proved.

Q7.

**Answer :**

Let *MN *is the diameter and chord *AB* of circle *C*(*O, r*) then according to question

*AP* = *BP.*

Then prove that *.*

Join *OA *and* OB.*

In ?AOP and ?BOP

(Each equal to radius).

*AP* = *BP*

(Mid point of chord *AB*)

*OP* = *OP*

(Common side)

Therefore,

Hence, proved.

### Page 16.25 Ex. 16.2

Q8.

**Answer :**

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.

Q9.

**Answer :**

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 chord length.

Q10.

**Answer :**

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

Q11.

**Answer :**

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 require circle.

Q12.

**Answer :**

Given that two different pairs of circles in 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.

Q13.

**Answer :**

Given that 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.

Q14.

**Answer :**

Let *AB* and *CD *be two parallel chord of the circle with centre O such that *AB* = 5cm,

*CD* = 11cm 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* then

Join *OA* and *OC*, then

*OA* =* OC*

= *r*

Nowand

So, and

In we have

…… (1)

And

…… (2)

From equation (1) and (2) we get

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

Q15.

**Answer :**

Let *AB* and *CD *be two parallel chord of the circle with centre O such that *AB* = 6 cm,

*CD* = 8 cm and* O* *P* = 4 cm. let the radius of the circle be cm.

According to question, Find O*Q*

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

Let

Join *OA* and *OC*, then

*OA* =* OC*

= *r*

Nowand

So, *AP* = 3 cm and *CQ* = 4 cm

In we have

And in

### Page 16.40 Ex. 16.3

Q1.

**Answer :**

Using the data given in the question we can draw a diagram that looks like fig (1).

From the figure we see that it is an isosceles triangle that has been circumscribed in a circle of radius *R* = 20 m..

The equal sides of the isosceles triangle measure 24 m in length. The length of the base of the isosceles triangle is what we are required to find out.

Since it is an isosceles triangle the perpendicular dropped from the vertex *A* to the base will pass though the circumcentre of the triangle. Let ‘*h*’ be the height of the triangle.

Since the triangle has been circumscribed by a circle of radius ‘*R*’ the length of the distances from ‘*O*’ to any of the three persons would be ‘*R*’.

Let the positions of the persons Isha, Ishita and Nisha be replaced by ‘*A*’, ‘*B*’ and ‘*C*’ respectively. And let the length of the unknown base be, BC = 2*x*

This is shown in the fig (2).

Now, consider the triangle ?*BOD*, we have

At the same time consider, we have

Substitute this value in equation we got for ‘*R*’, we get

Now we have got the value of the height of the triangle as *h* = 14.4 m

Substituting the value of *h* in the below equation,

Now we have the value of *x*=19.2 m

We need the value of

We assumed

Hence the distance between Ishita and Nisha is

Q2.

**Answer :**

From the given data, we see that the given situation is equivalent to an equilateral triangle circumscribed by a circle.

Let the positions of the three boys Ankur, Amit and Anand be denoted by the points ‘*A*’,’*B*’ and ‘*C*’. Let ‘*O*’ be the centre of the circle, ‘*a*’ is the sides of the equilateral triangle and ‘*R*’ is its circumradius.

This figure would look like fig (1).

Now, in an equilateral triangle with side ‘*a*’, the height, ‘*h*’ of the equilateral triangle would be,

The circumradius, ‘*R*’ of the same equilateral triangle will be two-thirds the height of the equilateral triangle.

Substituting the value of the circumradius, *R *= 40 m we have

Hence the length of the string of each phone is

### Page 16.60 Ex. 16.4

Q1.

**Answer :**

It is given that *O* is the circle and ?*APB* = 50°

We have to find ?*COB* and ?*OBC*

According to the figure *O* is center of the circle and ?*ABC* is triangle

(Since *O* is center and *A* is on circumference)

(Since ?*CAB* = 50°)

So

Now in and

So (radius of circle)

Then (base angle)

Here sum of

Hence and

### Page 16.61 Ex. 16.4

Q2.

**Answer :**

It is given that *O* is the center of triangle and *A*, *B* and *C* are points on circumference

And (Given that)

We have to find ?*ABC*

From refection angle we have

Hence

Q3.

**Answer :**

It is given that

The circle have center *O* and *A*, *B* and *C* are points on circumference

And (given)

We have to find

In given triangle

(Given)

And (radius of circle)

Therefore is isosceles triangle.

So …… (1)

(Given)

(From equation (1))

So

Again from figure, is given triangle and

Now in side (radius of circle)

Similarly like first

(Given that)

Then

Since

Hence

Q4.

**Answer :**

We have to find in each figure.

(i) It is given that and

Now ,

Hence

(ii)

As we know that (angle form by same segment on circumference are equal)

Now are isosceles triangle and line is diameter passing through center.

So

Hence

(iii) It is given that

So …… (1)

And

…… (2)

Then

Hence

(iv)

(Linear angle)

And

Hence

(v) It is given that

Is isosceles triangle

Therefore

And

So reflection

Hence

(vi) It is given that

And

Is isosceles triangle

So

Hence

(vii) (Given that)

In we have

Hence

(viii) Is isosceles triangle

Because (radius of circle)

So (vertical angle)

Hence

(ix) It is given that

…… (1)

…… (2)

Because and are form on same segment of the circle.

Now from equation (1) and (2) we have

Hence

(x) It is given that

(Because form on same segment)

Now in we have

Hence

(xi)

In we have

Hence

(xii) (Angle form on same segment of triangle)

Is isosceles triangle

So (radius of triangle)

Then

Hence

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