RD Sharma 2019 Solutions for Class 9 Math Chapter 15 Circles are provided here with simple step-by-step explanations. These solutions for Circles are extremely popular among class 9 students for Math Circles Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RD Sharma 2019 Book of class 9 Math Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RD Sharma 2019 Solutions. All RD Sharma 2019 Solutions for class 9 Math are prepared by experts and are 100% accurate.

Page No 15.100:

Question 1:

In the given figure, ΔABC is an equilateral triangle. Find mBEC.
 

Answer:

It is given that, is an equilateral triangle

We have to find

Since is an equilateral triangle.

So

And

…… (1)

Since, quadrilateral BACE  is a cyclic qualdrilateral

So ,                                         (Sum of opposite angles of cyclic quadrilateral is.)

Hence

Page No 15.100:

Question 2:

In the given figure, ΔPQR is an isosceles triangle with PQ = PR and mPQR = 35°. Find mQSR and m QTR.

Answer:

Disclaimer: Figure given in the book was showing mPQR as mSQR. 


It is given that ΔPQR is an isosceles triangle with PQ = PR and mPQR = 35°

 

We have to find the m∠QSR and mQTR

Since ΔPQR is an isosceles triangle

So ∠PQR = ∠PRQ = 35° 

Then

Since PQTR is a cyclic quadrilateral

So

In cyclic quadrilateral QSRT we have

Hence,

and  



Page No 15.101:

Question 3:

In the given figure, O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Answer:

It is given that O is centre of the circle and BOD = 160°

             

We have to find the values of x and y.

As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Therefore,

Since, quadrilateral ABCD is a cyclic quadrilateral.

So,

x + y = 180°              (Sum of opposite angles of a cyclic quadrilateral is 180°.)

Hence and

Page No 15.101:

Question 4:

In the given figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100° and ∠ABD = 70°, find ∠ADB.

Answer:

It is given that ∠BCD = 100° and ABD = 70°

We have to find the ∠ADB

We have

A + C = 180°                     (Opposite pair of angle of cyclic quadrilateral)

So,

Now in is and

Therefore,

Hence,

Page No 15.101:

Question 5:

If ABCD is a cyclic quadrilateral in which AD || BC (In the given figure). Prove that ∠B = ∠C.
 

Answer:

It is given that, ABCD is cyclic quadrilateral in which AD || BC

We have to prove

Since, ABCD is a cyclic quadrilateral

So,

and               ..… (1)

and              (Sum of pair of consecutive interior angles is 180°) …… (2)

From equation (1) and (2) we have

…… (3)

…… (4)

Hence Proved

Page No 15.101:

Question 6:

In the given figure, O is the centre of the circle. Find ∠CBD.
 

Answer:

It is given that,

We have to find

Since,    (Given)

So,

                     (The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)

Now,

               (Opposite pair of angle of cyclic quadrilateral)

So,

…… (1)

      (Linear pair)

         ()

Hence

Page No 15.101:

Question 7:

In the given figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.
 

Answer:

It is given that, AB and CD are diameter with center O and

We have to find  

Construction: Join the point A and D to form line AD

Clearly arc AD subtends at B and at the centre.

Therefore,  AOD=2ABD=100° …… (1)

Since CD is a straight line then
        
DOA+AOC=180°        Linear pair

     

Hence

Page No 15.101:

Question 8:

On a semi-circle with AB as diameter, a point C is taken, so that m (∠CAB) = 30°. Find m (∠ACB) and m (∠ABC).

Answer:

It is given that, as diameter, is centre and

We have to find and

Since angle in a semi-circle is a right angle therefore

In we have

(Given)

    (Angle in semi-circle is right angle)

Now in we have

                                        

Hence and  

Page No 15.101:

Question 9:

In a cyclic quadrilateral ABCD if AB || CD and ∠B = 70°, find the remaining angles.

Answer:

It is given that, ABCD is a cyclic quadrilateral such that AB || CD and

Sum of pair of opposite angles of cyclic quadrilateral is 180°.

                  (
given)

So,

Also AB || CD and BC transversal

So,

Now

               

Page No 15.101:

Question 10:

In a cyclic quadrilateral ABCD, if m A = 3 (mC). Find mA.

Answer:

It is given that

ABCD is cyclic quadrilateral and

We have to find

Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.

So

And

Therefore

Hence



Page No 15.102:

Question 11:

In the given figure, O is the centre of the circle and ∠DAB = 50° . Calculate the values of x and y.

Answer:

It is given that, O is the centre of the circle and DAB=50°.

   

We have to find  the values of x and y.

ABCD is a cyclic quadrilateral and

So,

50° + y = 180°
y = 180° − 50°
y = 130°

 

Clearly is an isosceles triangle with OA = OB and

Then,

                        (Since)

So,

x + AOB = 180°     (Linear pair)

Therefore, x =  180° − 80° = 100°

Hence,



and

 

Page No 15.102:

Question 12:

In the given figure, if ∠BAC = 60° and ∠BCA = 20°, find ∠ADC.

Answer:

It is given that, and

We have to find the

In given we have

ABC+BCA+BAC=180°     Angle sum propertyABC=180°-60°+20°=100°

In cyclic quadrilateral we have

         (Sum of pair of opposite angles of a cyclic quadilateral is 180º)

Then,

Hence

Page No 15.102:

Question 13:

In the given figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Answer:

It is given that, ABC is an equilateral triangle

We have to find and

Since is an equilateral triangle

So,

And is cyclic quadrilateral

So      (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)

Then,

Similarly BECD is also cyclic quadrilateral

So,

Hence, and  .

Page No 15.102:

Question 14:

In the given figure, O is the centre of the circle. If ∠CEA = 30°, Find the values of x, y and z.

Answer:

It is given that, O is the centre of the circle and

We have to find the value of x, y and z.

Since, angle in the same segment are equal

So

And z = 30°               

As angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Since

Then,

y = 2z
   = 2 × 30°
   = 60°

Since, the sum of opposite pair of angles of a cyclic quadrilateral is 180°.

z + x = 180°
x = 180° − 30°
   = 150°


Hence,
x = 150°, y = 60° and z = 30°

Page No 15.102:

Question 15:

In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.

Answer:

It is given that, and, are cyclic quadrilateral

We have to find the value of x and y.

Since , is a cyclic quadrilateral

So      (Opposite angle of a cyclic quadrilateral are supplementary)

        ()

     ..… (1)


x = 180° − 102°
   = 78°

Now in cyclic quadrilateral DCFE
x + y =
180°    (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180°
− 78°
   = 102°

Hence, x = 78° and y = 102°

 

Page No 15.102:

Question 16:

In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°

Answer:

It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since ABCD is a cyclic quadrilateral

So ∠A + ∠C = 180°                 (Sum of opposite pair of angles of cyclic quadrilateral is 180°)   ..… (1)

And,

A – ∠C = 60°               (Given)            ..… (2)

Adding equation (1) and (2) we have

So, ∠C = 60°

Hence, smaller of two is 60°.

 



Page No 15.103:

Question 17:

In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

Answer:

Here, ABCD is a cyclic quadrilateral, we need to find x.

In cyclic quadrilateral the sum of opposite angles is equal to 180°.

Therefore,

 ADC+ABC=180°180°-80°+180°-x=180°x=100°

Hence, the value of x is 100°.

Page No 15.103:

Question 18:

ABCD is a cyclic quadrilateral in which:

(i) BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.

(ii) ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.

(iii) ∠BCD = 100° and ∠ABD = 70° find ∠ADB.

Answer:

(i) It is given that , and

   

We have to find

In cyclic quadrilateral ABCD

      ..… (1)

      ..… (2)

Since, 

So,

Therefore in ,

So ,                   ..… (3)

Now,                   ( and is transversal)

(ii) It is given that , and

We have to find

                    (Angle in the same segment are equal)

Hence,

(iii) It is given that, ∠BCD = 100° and ABD = 70°

As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.

DAB+BCD=180°DAB=180°-100°                  =80°

In ΔABD we have,

DAB+ABD+BDA=180°BDA=180°-150°=30°

Hence, 

Page No 15.103:

Question 19:

Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.

Answer:

Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Let the diagonals AC and BD intersect at O.

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

Now, means that circle described on AB as diameter passes through O.

Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.

Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Page No 15.103:

Question 20:

If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.

Answer:



To prove: AC = BD

Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.

AOD=BOC         O is the centre of the circleAOD=2ACD     and BOC=2BDCSince, AOD=BOCACD=BDC      .....1    ACB=ADB           .....2  Angle in the same segment are equalAdding 1 and 2BCD=ADC        .....3In ACD and BDCCD=CD    commonBCD=ADC    Using3AD=BC   givenHence, ACDBDC     SAS congruency criterionAC=BD    cpctHence Proved

Page No 15.103:

Question 21:

Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).

Answer:

ADB=90°    Angle in a semicircleADC=90°     Angle in a semicircleSo, ADB+ADC=90°+90°=180°Therefore, BDC is a line.Hence, the point of intersection of two circles lie on the third side.



 

Page No 15.103:

Question 22:

ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.

Answer:

If in cyclic quadrilateral , then we have to find the other three angles.

Since, AD is parallel to BC, So,
  (Alternate interior angles)

Now, since ABCD is cyclic quadrilateral, so

And,

Hence, A=110°, C=70° and D=110°.

Page No 15.103:

Question 23:

In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Answer:

It is given that is a cyclic quadrilateral with and as its diagonals.

We have to find

Since angles in the same segment of a circle are equal

So

Since (Opposite angle of cyclic quadrilateral)

Hence

Page No 15.103:

Question 24:

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Answer:




To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.

Proof:

We know that the perpendicular bisector of every chord of a circle always passes through the centre.

Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.

Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.

Page No 15.103:

Question 25:

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Answer:

Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.

Let O be the centre of the circle.

We know that the angle formed in the semicircle is 90°.

Since, ABCD is a rectangle, So

Therefore, AC and BD are diameter of the circle.

We also know that the intersection of any two diameter is the centre of the circle.

Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Page No 15.103:

Question 26:

ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:

(i) AD || BC

(ii) EB = EC.

Answer:

(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EAED, then we have to prove the following, AD || BC

(ii) EB = EC

 

 

(i) It is given that EA = ED, so

EAD=EDA=x

Since, ABCD is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence, AD || BC

(ii) Since, AD and BC are parallel to each other, so,

ECB=EDA   Corresponding anglesEBC=EAD    Corresponding anglesBut, EDA=EADTherefore, ECB=EBC EC=EBTherefore, ECB is an isosceles triangle.

 



Page No 15.104:

Question 27:

Prove that the angle in a segment shorter than a semicircle is greater than a right angle.

Answer:



QP is a major arc and PSQ is the angle formed by it in the alternate segment.We know that the angle subtended by an arc at the centre is twice the angle subtended by it at any point of the alternate segment of the circle.2PSQ=mQP2PSQ=360°-mPQ2PSQ=360°-POQ2PSQ=360°-180°      POQ<180°2PSQ>180°PSQ>90°

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.

Page No 15.104:

Question 28:

Prove that the angle in a segment greater than a semi-circle is less than a right angle.

Answer:



To prove: ABC is an acute angleProof:AD being the diameter of the given circleACD=90°    Angle in a semicircle is a right angleNow, in ACD, ACD=90° which means that ADC is an acute angle.    .....1Again, ABC=ADC     Angle in a same segment are always equalABC is also an acute angle.  Using1Hence proved

Page No 15.104:

Question 29:

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half the hypotenuse.

Answer:

We have to prove that

Let be a right angle at B and P be midpoint of AC

Draw a circle with center at P and AC diameter

Since therefore circle passing through B

So

Hence

Proved.



Page No 15.107:

Question 1:

In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If ∠APB = 70°, find ∠ACB.
 

Answer:

Consider the smaller circle whose centre is given as ‘O’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

Hence ,the measure of is.



Page No 15.108:

Question 2:

In the given figure, two congruent circles with centres O and O' intersect at A and B. If ∠AOB = 50°, then find ∠APB.
 

Answer:

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.

So, from the given figure we have,

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.

One of the properties of a rhombus is that the opposite angles are equal to each other.

So, since it is given that, we can say that the angle opposite it, that is to say that should also have the same value.

Hence we get

Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is.

A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

This means that,

Hence the measure of is

Page No 15.108:

Question 3:

In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

Answer:

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .

Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.

Since in a cyclic quadrilateral the opposite angles are supplementary, here

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment. The angle that the chord forms with any point on the circumference of a particular segment is always the same.

Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.

So,

Now,

In any triangle the sum of the interior angles need to be equal to .

Consider the triangle ΔABP,

PAB+ABP+APB=180°APB=180°-30°-58°APB=92°

From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,

Hence the measure of is.

Page No 15.108:

Question 4:

In the given figure, if ∠AOB = 80° and ∠ABC = 30°, then find ∠CAO.

Answer:

Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

In any triangle the sum of the interior angles need to be equal to 180°.

Consider the triangle

Since,  , we have. So the above equation now changes to

Considering the triangle ΔABC now,

Hence, the measure of is.

Page No 15.108:

Question 5:

In the given figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find ∠BCD : ∠ABE.
 

Answer:

It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.

So, we have.

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘ABCD’ is a rhombus.

Let.

We know that in a circle the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

By this property we have

In a rhombus the opposite angles are always equal to each other.

So,

Since the sum of all the internal angles in any triangle sums up to in triangle , we have

In the rhombus ‘ABCD’ since one pair of opposite angles are ‘’ the other pair of opposite angles have to be

From the figure we see that,

So now we can write the required ratio as,

Hence the ratio between the given two angles is .

Page No 15.108:

Question 6:

In the given figure, AB is a diameter of the circle such that ∠A = 35° and ∠Q = 25°, find ∠PBR.
 

Answer:

Let us first consider the triangle ΔABQ.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle ΔABQ we have,

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.

So, we have

Now considering the triangle we have,

From the given figure it can be seen that,

Now, we can also say that,

Hence the measure of the angle is 115°.

Page No 15.108:

Question 7:

In the given figure, P and Q are centres of two circles intersecting at B and C. ACD is a straight line. Then, ∠BQD =
 

Answer:

Consider the circle with the centre ‘P’.

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have

Since ‘ACD’ is a straight line, we have

Now let us consider the circle with centre ‘Q’. Here let ‘E’ be any point on the circumference along the major arc ‘BD’. Now ‘CBED’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, now we have

Hence, the measure of is .



Page No 15.109:

Question 8:

In the given figure, if O is the circumcentre of ∠ABC, then find the value of ∠OBC + ∠BAC.

Answer:



Since, O is the circumcentre of ABC, So, O would be centre of the circle passing through points A, B and C.

ABC = 90°             (Angle in the semicircle is 90°.)

OAB+OBC=90°                .....(1)

As OA = OB    (Radii of the same circle)

OAB=OBA     Angle opposite to equal sides are equalor, BAC= OBAFrom 1BAC+OBC=90°

Page No 15.109:

Question 9:

If the given figure, AOC is a diameter of the circle and arc AXB = 12 arc BYC. Find ∠BOC.

Answer:

We need to find

  
 

arc AXB=12arc BYC, AOB=12BOCAlso AOB+BOC=180°Therefore, 12BOC+BOC=180°BOC=23×180°=120°

Page No 15.109:

Question 10:

In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

Answer:

We are given ABCD is a quadrilateral with center O, ADE = 95° and OBA = 30°

We need to find OAC

We are given the following figure

Since ADE = 95°

⇒ ∠ADC = 180 ° − 95° = 85°

Since squo;ABCD is cyclic quadrilateral

This means

ABC + ADC = 180°

Since OB = OC (radius)

⇒ ∠OBC = OCB = 65°

In ΔOBC

Since ÐBAC and ÐBOC are formed on the same base which is chord.

So

Consider ΔBOA which is isosceles triangle.

OAB = 30°

Page No 15.109:

Question 1:

If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is

(a) 15 cm

(b) 16 cm

(c) 17 cm

(d) 34 cm

Answer:

(c) 17 cm

We will represent the given data in the figure

   

In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.

We know that perpendicular from the centre to any chord divides it into two equal parts.

So,  AM = MB = = 8 cm.

Now consider right triangle OMA and by using Pythagoras theorem

        

Hence, correct answer is option (c).

Page No 15.109:

Question 2:

The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

(a) 5 cm

(b) 25 cm

(c) 27 cm

(d) 7 cm

Answer:

(b) 25 cm

We will represent the given data in the figure

We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.

So , AM = MB = AB2=82 = 4 cm.

Using Pythagoras theorem in the ΔAMO,

         

Hence, the correct answer is option (b).



Page No 15.110:

Question 3:

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =

(a) 60°

(b) 45°

(c) 30°

(d) 15°

Answer:

We will associate the given information in the following figure.

Since AO = r (radius of circle)

AM = (given)

Extended OM to D where MD =

Consider the triangles AOM and triangle AMD

So by SSS property

So AD = AO = r and OD=OM+MD=r

Hence ΔAOD is equilateral triangle

So

We know that in equilateral triangle altitudes divide the vertex angles

Hence option (c) is correct.

Page No 15.110:

Question 4:

ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =

(a) 70°

(b) 100°

(c) 125°

(d) 150°

Answer:

(a) 70°

It is given that ABCD is cyclic quadrilateral
ADB = 90° and DCA = 80°. We have to find DAB

We have the following figure regarding the given information

BDA = BCA = 30°      (Angle in the same segment are equal)

Now, since ABCD is a cyclic quadrilateral

So, DAB + BCD = 180°

Hence the correct answer is option (a).

Page No 15.110:

Question 5:

A chord of length 14 cm is at  a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is

(a) 12 cm

(b) 14 cm

(c) 16 cm

(d) 18 cm

Answer:

(d) 18 cm

We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the centre.

We have the following figure

We are given AB = 14 cm, OD = 6 cm, MO = 2 cm, PQ = ?

Since, perpendicular from centre to the chord divide the chord into two equal parts

Therefore

Now consider the ΔOPQ in which OM = 2 cm

So using Pythagoras Theorem in ΔOPM

Hence, the correct answer is option (d).

Page No 15.110:

Question 6:

One chord of a circle is known to be 10 cm. The radius of this circle must be

(a) 5 cm

(b) greater than 5 cm

(c) greater than or equal to 5 cm

(d) less than 5 cm

Answer:

(b) greater than 5 cm

We are given length of a chord to be 10 cm and we have to give information about the radius of the circle.

Since in any circle, diameter of the circle is greater then any chord.

So diameter > 10

2r > 10

r > 5 cm

Hence, the correct answer is option (b)

 

Page No 15.110:

Question 7:

ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Answer:

(d) 6 cm

We are given a right triangle ABC such that, AC = 5 cm, AB = 4 cm. A circle is drawn with A as centre and AC as radius. We have to find the length of the chord of this circle passing through C and B. We have the following figure regarding the given information.

In the circle produce CB to P. Here PC is the required chord.

We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.

So,  PC = 2BC

Now in ΔABC apply Pythagoras theorem

So,  PC = 2 × BC

            = 2 × 3
            = 6 cm

Hence, the correct answer is option (d).

Page No 15.110:

Question 8:

If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than ∠AOB =

(a) 60°

(b) 90°

(c) 120°

(d) none of these

Answer:

(a) 60°



As we know that equal chords make equal angle at the centre.

Therefore,

AOB=BOC=CODAOB+BOC+COD=180°     Linear pair3AOB=180°AOB=60°


Hence, the correct answer is option (a).

Page No 15.110:

Question 9:

Let C be the mid-point of an arc AB of a circle such that m AB = 183°. If the region bounded by the arc ACB and the line segment AB is denoted by S, then the centre O of the circle lies

(a) in the interior of S

(b) in the exertior of S

(c) on the segment AB

(d) on AB and bisects AB

Answer:

(a) in the interior of S

Given:
m AB = 183° and C is mid-point of arc ABO is the centre.

With the given information the corresponding figure will look like the following

So the center of the circle lies inside the shaded region S.

Hence, the correct answer is option (a).

Page No 15.110:

Question 10:

In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are

(a) 90° and 270°

(b) 90° and 90°

(c) 270° and 90°

(d) 60° and 210°

Answer:

(c) 270° and 90°

We are given the major arc is 3 times the minor arc. We are asked to find the corresponding central angle.

See the corresponding figure.

We know that angle formed by the circumference at the centre is 360°.

Since the circumference of the circle is divided into two parts such that the angle formed by major and minor arcs at the centre are 3x and x respectively.

So = 90° and = 3x = 270°

Hence, the correct answer is option (c).

Page No 15.110:

Question 11:

If A and B are two points on a circle such that m (AB) = 260°. A possible value for the angle subtended by arc BA at a point on the circle is

(a) 100°

(b) 75°

(c) 50°

(d) 25°

Answer:

(c) 50°

We are given

Suppose point P is on the circle.

Since

So, = 360° − 260° = 100°

We know that angle subtended by chord AB at the centre is twice that of subtended at the point P

So, = = 50°

Hence, the correct answer is option (c).

Page No 15.110:

Question 12:

An equilateral triangle ABC is inscribed in a circle with centre O. The measures of ∠BOC is

(a) 30°

(b) 60°

(c) 90°

(d) 120°

Answer:

(d) 120°

We are given that an equilateral
ΔABC is inscribed in a circle with centre O. We need to find BOC

We have the following corresponding figure:

We are given AB = BC = AC

Since the sides AB, BC, and AC are these equal chords of the circle.

So, the angle subtended by these chords at the centre will be equal.

Hence

Hence, the correct answer is option (d).

Page No 15.110:

Question 13:

If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a

(a) rhombus

(b) rectangle

(c) parallelogram

(d) square

Answer:

(d) square

The given information in the form of the following figure is as follows:

Since, four sides of the quadrilateral ACBD are four chords which subtend equal angles at the centre. Therefore,

    (Since AB and CD are perpendicular diameters)

So sides AC, BC, BD and AD are equal, as equal chords subtend equal angle at the centre.

So , AC = CB = BD = DA    …… (1)

We know that diameters subtend an angle of measure 90° on the circle.

So,  …… (2)

From (1) and (2) we can say that is a square.

Hence, the correct answer is option (d).

 

Page No 15.110:

Question 14:

If ABC is an arc of a circle and ∠ABC = 135°, then the ratio of arc ABC  to the circumference is

(a) 1 : 4

(b) 3 : 4

(c) 3 : 8

(d) 1 : 2

Answer:

(c) 3 : 8

The length of an arc subtending an angle ‘’ in a circle of radius ‘r is given by the formula,

Length of the arc =

Here, it is given that the arc subtends an angle of with its centre. So the length of the given arc in a circle with radius ‘r’ is given as

Length of the arc =

The circumference of the same circle with radius ‘r’ is given as .

The ratio between the lengths of the arc and the circumference of the circle will be,

Hence, the correct answer is option (c).

Page No 15.110:

Question 15:

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is

(a) 60°

(b) 75°

(c) 120°

(d) 150°

Answer:

(d) 150°

We are given that the chord is equal to its radius.

We have to find the angle subtended by this chord at the minor arc.

We have the corresponding figure as follows:

We are given that

AO = OB = AB

So , AOB is an equilateral triangle.

Therefore, we have

AOB = 60°

Since, the angle subtended by any chord at the centre is twice of the angle subtended at any point on the circle.

Take a point P on the minor arc.

Since is a cyclic quadrilateral

So, opposite angles are supplementary. That is

Hence, the correct answer is option (d).



Page No 15.111:

Question 16:

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =

(a) 41°

(b) 23°

(c) 67°

(d) 18°

Answer:

Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be ‘O’.

We are given that and. This is shown in fig (2).

So we see that,

QPS=QPR+RPS             =67°+72°             =139°

(a) 41°

In a cyclic quadrilateral it is known that the opposite angles are supplementary.

Hence the correct answer is option (a).

Page No 15.111:

Question 17:

If A , B, C are three points on a circle with centre O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =

(a) 60°

(b) 75°

(c) 90°

(d) 135°

Answer:

(b) 75°

To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

Here we are given that ‘A’, ‘B’, ‘C’ are three points on a circle with centre ‘O’ such that and .

From the figure we see that,

Now, as seen earlier, the angle made by the arc ‘AC’ with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.

Since the point ‘C’ lies on the remaining part of the circle, the angle the arc ‘AC’ makes with this point has to be half of the angle ‘AC’ makes with the centre.Therefore we have,

Hence the correct answer is option (b).

Page No 15.111:

Question 18:

The greatest chord of a circle is called its

(a) radius

(b) secant

(c) diameter

(d) none of these

Answer:

(c) diameter

The greatest chord in a circle is the diameter of the circle.

Hence the correct answer is option (c).

Page No 15.111:

Question 19:

Angle formed in minor segment of a circle is

(a) acute

(b) obtuse

(c) right angle

(d) none of these

Answer:

(b) obtuse

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment.

The angle formed by the chord in the minor segment will always be obtuse.

Hence the correct answer is option (b).

Page No 15.111:

Question 20:

Number of circles that can be drawn through three non-collinear points is

(a) 1

(b) 0

(c) 2

(d) 3

Answer:

(a) 1

Suppose we are given three non-collinear points as A, B and C

1. Join A and B.

2. Join B and C.

3. Draw perpendicular bisector of AB and BC which meet at O as centre of the circle.

So basically we can only draw one circle passing through three non-collinear points A, B and C.

Hence, the correct answer is option (a).

Page No 15.111:

Question 21:

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =

(a) 45°

(b) 60°

(c) 75°

(d) 90°
 

Answer:

(d) 90°

We are given the following figure

 ACD = ABD     (Angle in the same segment are equal)

⇒ ∠ACD = y

Consider the ΔACM in which

Hence, the correct answer is option (d).

Page No 15.111:

Question 22:

In the given figure, if ∠ABC = 45°, then ∠AOC =

(a) 45°

(b) 60°

(c) 75°

(d) 90°

Answer:

(d) 90°

We have to find AOC.

 As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Hence, the correct answer is option (d).

Page No 15.111:

Question 23:

In the given figure, chords AD and BC intersect each other at right angles at a point P. If ∠DAB = 35°, then
ADC =
(a) 35°
(b) 45°
(c) 55°
(d) 65°

Answer:

(c) 55°

   (Angle in the same segment are equal.)

Also, since the chords ‘AD’ and ‘BC’ intersect perpendicularly we have,

Consider the triangle ,

Hence, the correct answer is option (c).

Page No 15.111:

Question 24:

In the given figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is

(a) 42°

(b) 48°

(c) 58°

(d) 52°


 

Answer:

(b) 48°

Construction: Join A and D.

Since AC is the diameter. So ADC will be 90°.

Therefore,

 ACB = ADB = 48°      (Angle in the same segment are equal.)

Hence, the correct answer is option (b).



Page No 15.112:

Question 25:

In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

(a) 2AB

(b) 2

(c) 12AB

(d) 12AB

Answer:

(d) 12AB

We are given a circle with centre at O and two perpendicular diameters AB and CD.

We need to find the length of AC.

We have the following corresponding figure:

Since, AB = CD                   (Diameter of the same circle)

Also, AOC = 90°

And,  AO =

Here,  AO = OC (radius)

In ΔAOC

Hence,  the correct answer is option (d).

Page No 15.112:

Question 26:

Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circle is

(a) r

(b) 2r AB

(c) 3r

(d) 32

Answer:

(c) 3r

We are given two circles of equal radius intersect each other such that each passes through the centre of the other.

We need to find the common chord.

We have the corresponding figure as follows:

  

 AO = AO = r (radius)

And OO = r

So, ΔOAO is an equilateral triangle.

We know that the attitude of an equilateral triangle with side r is given by

That is AM =

We know that the line joining centre of the circles divides the common chord into two equal parts.

So we have

Hence, the correct answer is option (c).

Page No 15.112:

Question 27:

If AB is a chord of a circle, P and Q are the two points on the circle different from A and B, then

(a) ∠APB = ∠AQB

(b) ∠APB + ∠AQB = 180° or ∠APB = ∠AQB

(c) ∠APB + ∠AQB = 90°

(d) ∠APB + ∠AQB = 180°

Answer:

(b)  APB + AQB = 180° or APB = AQB

We are given AB is a chord of the circle; P and Q are two points on the circle different from A and B.

We have following figure.

Case 1: Consider P and Q are on the same side of AB

We know that angle in the same segment are equal.

Hence, APB = AQB

Case 2: Now consider P and Q are on the opposite sides of AB

In this case we have the following figure:

Since quadrilateral APBQ is a cyclic quadrilateral.

Therefore,

APB + AQB = 180°    (Sum of the pair of opposite angles of cyclic quadrilateral is 180°.)

Therefore, APB = AQB or APB + AQB = 180°

Hence, the correct answer is option (b).

Page No 15.112:

Question 28:

AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is

(a) 6 cm

(b) 52 cm

(c) 7 cm

(d) 35 cm

Answer:

(d) 35 cm

Let distance between the centre and the chord CD be x cm and the radius of the circle is r cm.

We have to find the radius of the following circle:

 

In triangle OND,

…… (1)

Now, in triangle AOM,

…… (2)

From (1) and (2), we have,

Hence, the correct answer is option (d).

Page No 15.112:

Question 29:

In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

(a) 34 cm

(b) 15 cm

(c) 23 cm

(d) 30 cm

Answer:

(d) 30 cm

Given that: Radius of the circle is 17 cm, distance between two parallel chords AB and CD is 23 cm, where AB= 16 cm. We have to find the length of CD.

    

We know that the perpendicular drawn from the centre of the circle to any chord divides it into two equal parts.

So, AM = MB = 8 cm

Let OM = x cm

In triangle OMB,

Now, in triangle OND, ON = (23 − x) cm = (23 − 15) cm = 8 cm

Therefore, the length of the other chord is



Hence, the correct answer is option (d).

Page No 15.112:

Question 30:

In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

(a) 130°

(b) 115°

(c) 65°

(d) 165°



Answer:

(b) 115°

We have the following information in the following figure. Take a point P on the circle in the given figure and join AP and CP.

Since, the angle subtended by a chord at the centre is twice that of subtended atany part of the circle.

So, 

Since is a cyclic quadrilateral and we known that opposite angles are supplementary.

Therefore,

ABC+APC=180°ABC+65°=180°ABC=180°-65°ABC=115°

Hence, the correct answer is optoon (b).



Page No 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:

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 .

Page No 15.28:

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:

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

Page No 15.28:

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:

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

AL=20=4.47 

AB= 2×AL=2×4.47=8.94 cm

Hence the length of the chord is 8.94 cm.

Page No 15.28:

Question 4:

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

Answer:

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.

Page No 15.28:

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:

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 No 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:

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.

Page No 15.29:

Question 7:

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

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

Therefore radius AG = 23AD=33 cm

 

Page No 15.29:

Question 8:

Given an arc of a circle, complete the circle.

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 required circle.

Page No 15.29:

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:

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.

Page No 15.29:

Question 10:

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

Answer:

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.

Page No 15.29:

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:

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

Page No 15.29:

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:

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

6x+614=12146x=121-6146x=604x=52

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

Page No 15.29:

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:

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

                     (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.

Page No 15.29:

Question 14:

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

Answer:

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.

 

Page No 15.29:

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:

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,



Page No 15.47:

Question 1:

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha.

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 = 2x m.

This is shown in the fig (2).

Now, consider the triangle ΔBOD, 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

Hence, the distance between Ishita and Nisha is .

Page No 15.47:

Question 2:

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and hands to talk to each other. Find the length of the string of each phone.

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.

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

AB = BC = CA
Therefore,  ABC is an equilateral triangle.
OA = 40 m
Medians of equilateral triangle pass through the circumcentre (O)  of the equilateral triangle ABC. We know that medians intersect each other in the ratio 2 : 1. As AD is the median of equilateral triangle ABC, we can write

OAOD=2140OD=21OD=20 mAD=AO+OD=40+20 m=60 mIn ADC,AC2=AD2+DC2AC2=602+AC24         AC = BC, DC=12BCDC=12AC3AC24=3600AC2=4800AC=403 m

Hence the length of the string of each phone is



Page No 15.5:

Question 1:

Fill in the blanks:

(i) All points lying inside/outside a circle are called ........ points /.......points.

(ii) Circles having the same centre and different radii are called ...... circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in ......of the circle.

(iv) A continuous piece of a circle is ....... of the circle.

(v) The longest chord of a circle is a ....... of the circle.

(vi) An arc is a ..... when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and ....of the circle.

(viii) A circle divides the plane, on which it lies, in ...... parts.

Answer:

(i) interior/exterior
(ii) concentric
(iii) the exterior
(iv) arc
(v) diameter
(vi) semi-circle
(vii) centre
(viii) three



Page No 15.6:

Question 2:

Write the truth value (T/F) fo the following with suitable reasons:

(i) A circle is a plane figure.

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

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

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

(vi) Sector is the region between the chord and its corresponding arc.

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

(viii) The degree measure of a semi-circle is 180°

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 a fixed 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 θ 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 No 15.72:

Question 1:

In the given figure, O is the centre of the circle. If APB= 50°, find ∠AOB and ∠OAB.

Answer:

This question seems to be incorrect.

Page No 15.72:

Question 2:

In the given figure, O is the centre of the circle. Find ∠BAC.
 

Answer:

It is given that

And (given)

We have to find

In given triangle

(Given)

 OB = OA               (Radii of the same circle)

Therefore, is an isosceles triangle.

So, OBA=OAB                ..… (1)

                       (Given)

                    [From (1)]

So

Again from figure, is given triangle and

Now in

                  (Radii of the same circle)

OAC=OCA    

               (Given that)

Then,

Since

Hence

Page No 15.72:

Question 3:

If O is the centre of the circle, find the value of x in each of the following figures.

(i)


(ii)



(iii)



(iv)



(v)



(vi)



(vii)



(viii)



(ix)



(x)



(xi)



(xii)

Answer:

We have to find in each figure.

(i) It is given that

AOC+COB=180°   Linear pair

 
As we know the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Now ,x=12COB=2212°

Hence

(ii) As we know that = x                 [Angles in the same segment]

line is diameter passing through centre,

So,

BCA= 90°       Angle inscribed in a semicircle is a right angle 

  

 

CAB+ABC+BCA=180°    Angle sum propertyx+40°+90°=180°x=50°

(iii) It is given that



ABC=12Reflex AOC

 

So

And

Then

Hence

(iv)

  (Linear pair)

   

And
x =
 
 
Hence,

(v) It is given that

is an isosceles triangle.
  

Therefore

And,

In AOB,AOB+OBA+BAO=180°70°+BAO=180°BAO=110°

AOB=2ReflexACB
                     

Hence,

(vi) It is given that

 

And

COA+AOB=180°COA=180°-60°COA=120°

OCA is an isosceles triangle.

So

Hence,

(vii)                    (Angle in the same segment)

 

In we have

Hence

(viii)

 

As    (Radius of circle)

Therefore, is an isosceles triangle.

So          (Vertically opposite angles)

Hence,

(ix) It is given that

   

…… (1)          (Angle in the same segment)

ADB=ACB=32°  ......(2)           (Angle in the same segment)

Because and are on the same segment of the circle.

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

Hence,

(x) It is given that

  
BAC=BDC=35°                (Angle in the same segment)

 Now in BDC we have

BDC+DCB+CBD=180°35°+65°+CBD=180°CBD=180°-100°=80°

Hence,

(xi)

 

                    (Angle in the same segment)

In we have

Hence

(xii)
 

          (Angle in the same segment)

is an isosceles triangle

So,                   (Radius of the same circle)

Then

Hence



Page No 15.73:

Question 4:

O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A

Answer:

We have to prove that



Since, circumcenter is the intersection of perpendicular bisectors of each side of the triangle.

Now according to figure A, B, C are the vertices of ΔABC

In , is perpendicular bisector of BC

So, BD = CD                 

OB = OC             (Radius of the same circle)

And,

OD = OD         (Common)

Therefore,
BDOCDO    SSS congruency criterion

BOD=COD    by cpct

We know that angle formed any chord of the circle at the center is twice of the angle formed at the circumference by same chord

Therefore,

Therefore,



Hence proved

Page No 15.73:

Question 5:

In the given figure, O is the centre of the circle, BO is the bisector of ∠ABC. Show that AB = AC.

Answer:

It is given that,ABC is on circumference of circle BD is passing through centre.

Construction: Join A and C to form AC and extend BO to D such that BD be the perpendicular bisector of AC.

Now in BDA and BDC we have

AD = CD           (BD is the perpendicular bisector) 

So BDA=BDC=90°

  (Common)

BDABDC    SAS congruency criterion

Hence     (by cpct)

Page No 15.73:

Question 6:

In the given figure, O and O' are centres of two circles intersecting at B and C. ACD is a straight line, find x.

Answer:

It is given that

Two circles having center O and O' and ∠AOB = 130°

And AC is diameter of circle having center O


 

We have

So

Now, reflex

So

x°=360°-230°=130°

Hence,



Page No 15.74:

Question 7:

In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

Answer:

It is given that ∠ACB = 40° and ∠DPB = 120°

Construction: Join the point A and B

           (Angle in the same segment)

Now in BDP we have

DPB+PBD+BDP=180°120°+PBD+40°=180°PBD=20°

Hence

Page No 15.74:

Question 8:

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer:

We have to find and

Construction: - O is centre and r is radius and given that chord is equal to radius of circle

Now in we have

AO = OB = BA       ( It is given that chord is equal to radius of circle)

So, is an equilateral triangle

So, AOB=2ADB                (The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

Then

So

AEB=12ReflexAOB=12360°-60°=150°

Therefore,

and

Hence, the angle subtended by the chord at a point on the minor arc is 150° and also at a point on the major arc is 30°.

 

Page No 15.74:

Question 9:

In the given figure, it is given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.
 

Answer:

It is given that O is the centre of circle and A, B and C are points on circumference.

  (Given)

We have to find ABC

The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

ABC=12reflex AOC=12360°-150°=12×210°=105°

Hence,

Page No 15.74:

Question 10:

In the given figure, O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Answer:

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

  

We have to prove that x = ∠y + ∠z

∠4
and ∠3 are on same segment

So, ∠4 = ∠3           

               (Angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle)

…… (1)

               (Exterior angle is equal to the sum of two opposite interior angles)         …… (2)

4=z+1               (Exterior angle is equal to the sum of two opposite interior angles)  

                 …… (3)

Adding (2) and (3)

……(4)

From equation (1) and (4) we have 

Page No 15.74:

Question 11:

In the given figure, O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find ∠RTS.
 

Answer:

It is given that O is the centre and ROS=40°

We have

In right angled triangle RQT we have

RQT+QTR+TRQ=180°20°+QTR+90°=180°QTR=70°

Hence, RTS=70°



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