Important Questions for CBSE Class 9 Maths Chapter 10

ur examination.

1 Marks Quetions

1. An angle in the semicircle is

(a) Right angle

(b)

(c)

(d) none of these

Ans. (a) Right angle

2. If the angles subtended by two chords of a circle at the centre are equal then the chords are

(a) not equal

(b) equal

(d) line equals

Ans. (b) equal

3. How many circle passing through three non-collinear points

(a) one

(b) two

(d) four

Ans. (a) one

4. The constant distance is called

(a) diameter

(b) radius

(d) circle

Ans. b) radius

5. PS and RS are two chord’s of a circle such that PQ=10cm and RS= 24cm and PQ||RS. The distance between PQ and RS is 17cm. Find the radius of circle

(a) 10cm

(b) 13cm

(d) none of these

Ans. (b) 13cm

6. A circle is drawn. It divides the plane into

(a) 3 Parts

(b) 4 Parts

(d) No Parts

Ans. (a) 3 Parts

7. The relation between diameter and radius of a circle is

(a) r=2d

(b) d=r

(d) d=2r

Ans. (c) d=2r

8. If P and Q are any two Points on a circle then PQ is called a

(a) diameter

(b) secant

(d) radius

Ans. c) chord

9. What is a diameter

(a)

(b)

(c)

(d)

Ans. (d)

10. Two point on a circle shows the

(a) radius

(b) chord

(d) diameters

Ans. b) chord

11. The whole arc of a circle is called

(a) circumference

(b) semi-circle

(d) segment

Ans. (a) circumference

12. One half of the whole arc of a circle

(a) semi-circle

(b) circumference

(d) sector

Ans. (a) semi-circle

13. Circle having same centre are said to be

(a) Concentric

(b) circle

(d) secant

Ans. (a) Concentric

14. The line which meet a circle in two points is called a

(a) chord of circle

(b) diameter

(d) secant of circle

Ans. (d) secant of circle

15. The sum of either pair of opposite angle of cyclic quadrilateral is

(a)

(b)

(c)

(d)

Ans. c)

16. Two circle are congruent if they have equal.

(a) diameter

(b) radius

(d) secant

Ans. (b) radius

17. Which equation is show the diameter of circle

(a)
(b)
(c)
(d)

Ans. (a)

18.of the whole circle shows
(a) semi-circle

(b) circumference

(d) segments

Ans. (a) semi-circle

19. Two circle are congruent if they have equal

(a) radius

(b) diameter

(d) secant

Ans. (a) radius

2 Marks Quetions

1. Fill in the blanks:

(i) The centre of a circle lies in _______________ of the circle.

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

(iii) The longest chord of a circle is a _______________ of the circle.

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

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

Ans. (i) Interior

(ii) Exterior

(iii) diameter

(iv) Semi-circle

(v) Chord

(vi) Three

2. Write True or False:

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

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

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

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

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

(vi) A circle is a plane figure.

Ans. (i) True

(ii) False

(iii) False

(iv) True

(v) False

(vi) True

3. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chord.

Ans. Given: AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined.

To prove: OEM = OEN

Construction: Draw OM AB and ON CD.

Proof: In right angled triangles OME and ONE,

OME = ONE [Each ]

OM = ON $� � � � � � ℎ � � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � �$

OE = OE $� � � � � �$

OMEONE $� � � � � � � � � � � � � � � � � � �$

OEM = OEN $� � � � � �$

4. In figure, A, B, C are three points on a circle with centre O such that BOC = AOB = If D is a point on the circle other than the arc ABC, find ADC.

Ans. AOC = AOB + BOC AOC =

Now AOC = 2ADC

[Angled subtended by an arc, at the centre of the circle is double the angle subtended by the same arc at any point in the remaining part of the circle]

ADC = AOC

ADC = =

5. In figure, PQR = where P, Q, R are points on a circle with centre O. Find OPR.

Ans. In the figure, Q is a point in the minor arc

= 2PQR ROP = 2PQR

ROP = =

Now POR + ROP =

POR + = POR = = …..(i)

Now OPR is an isosceles triangle.

OP = OR $� � � � � � � � ℎ � � � � � � �$

OPR = ORP $� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$ …..(ii)

Now in isosceles triangle OPR,

OPR + ORP + POR =

OPR + ORP + =

2OPR = $Misplaced &$

2OPR =

OPR =

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. DBC = BAC is findBCD. Further if AB = BC, find ECD.

Ans. Here, DBC = and BAC =

And DAC = DBC = $� � � � � � � � � � � � � � � � � �$

Now ABCD is a cyclic quadrilateral.

DAB + BCD =

$� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$

BCD =

7. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Ans. Given: A trapezium ABCD in which ABCD and AD = BC.

To prove: The points A, B, C, D are concyclic.

Construction: Draw DECB.

Proof: Since DECB and EBDC.

EBCD is a parallelogram.

DE = CB and DEB = DCB

Now AD = BC and DA = DE

DAE = DEB

But DEA + DEB =

DAE + DCB = [DEA = DAE and DEB = DCB]

DAB + DCB =

A + C =

Hence, ABCD is a cyclic trapezium.

8. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D, P, Q respectively (see figure). Prove that ACP = QCD.

Ans. In triangles ACD and QCP,

A = P and Q = D $� � � � � � � � � � � � � � � � � � �$

ACD = QCP $� ℎ � � � � � � � � �$ ……….(i)

Subtracting PCD from both the sides of eq. (i), we get,

ACD - PCD = QCP - PCD

ACPO = QCD

Hence proved.

9. Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Ans. Let two circles with respective centers A and B intersect each other at points C and D.

We have to prove ACB = ADB

Proof: In triangles ABC and ABD,

AC = AD =

BC = BD =

AB = AB $� � � � � �$

ABC ABD

$� � � � � � � � � � � � � � � � � � �$

ACB = ADB $� � � � � �$

10. Prove that the circle drawn with any drawn with any side of a rhombus as a diameter, passes through the point of intersection of its diagonals.

Ans. Let ABCD be a rhombus in which diagonals AC and BD intersect each other at point O.

As we know that diagonals of a rhombus bisect and perpendicular to each other.

AOB =

And if we draw a circle with side AB as diameter, it will definitely pass through point O (the point intersection of diagonals) because then AOB = will be the angle in a semi-circle.

11. AB = DC and diagonal AC and BD intersect at P in cyclic quadrilateral Prove that

Ans. In PAB and PDC

AB = DC

[Angle in the same segment

$� � � � � � � � � � � �$

12. Prove that if ABC and ADC are two right triangle with common hypotenuse AC.

Ans. [AC is the common hypotenuse of it and ]

Quadrilateral ABCD is cyclic

Now, chord CD subtends and

$� � � � � � � � ℎ � � � � � � � � � � � �$

13. Show that in isosceles triangle ABC, AB = AC and B,C intersects the sides AB and AC at D and E.

Ans. According to given: BCED forms a cyclic quadrilateral

From (i) and (ii) we get

But they form a Pair of corresponding angles

14. Prove that cyclic parallelogram is a rectangle.

Ans. Let ABCD be the given cyclic parallelogram

$� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$ ……..(ii)

From (i) and (ii)

is a rectangle.

15. A line is Passing through the centre of a circle. If it bisects chord AB and CD of the circle. Prove that AB||CD.

Ans. Line EF passes through the centre O and bisects chord AB at P and chord CD at Q

P, is the mid-Point of AB and Q is the mid-point of CD

But the line joining the mid-point of a chord to the centre of the circle is perpendicular to the chord.

And

16. AB and CB are two chords of circle. Prove that BO bisects .

Ans. Join OA and OC

In And

OA=OC (redii of circle)

OB=OB (common)

AB=AB (given)

Hence, BO bisects

17. If BC is diameter of circle with centre O and OD is to chord AB so prove CA=2 OD

Ans. Join AC
Given that
D is the mid-point of AB
O is the mid-Point of BC
Now in
OD is the line joining the mid points of sides BC and AB

Hence proved.

18. Given a method to find the centre of a circle

Ans. Take three distinct points (non-collinear) A, B and C on the circle. Join AB and BC Draw bisectors PQ and RS of AB and BC respectively, to intersect at O

Now, P, is the centre of the circle.

19. C point is taken so that =300 from a semi-circle with AB as diameter. So find and .

Ans. AB is a diameter and C is a point on the semi-circle

and

20. Two different circle can’t interact each other at more than two points so, prove it.

Ans. Let the two different circles intersect in three point A, B, C. Then these points A B and C one non-collinear

We know that through three non-collinear Points, one and only one circle can pass so, it contradicts the hypothesis.

21. O is the centre and , find the length of the chord AB

Ans. Perpendicular drawn from the centre to the chord bisects the chord.

In right angled triangle BPO,

22. If OA is perpendicular to CB, find the length of AB

Ans.

In right angles triangle OAB,

AB=4

23. Prove that ADE is an isosceles triangle if and

Ans. Given that AB and AC is two equal chords of the circle with centre O,

And

$� � � � � � ℎ � � � � � � � � � � � � � � � � � �$

Subtracting (i) from (ii)

AD=AE

is an isosceles

24. Prove that the exterior angle formed by producing a side of a cyclic quadrilateral is equal to the interior opposite angle. Prove

Ans. $� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$

25. Show that if AB and CD are two equal chord.

Ans. We know that equal chords of a circle are equidistant from the centre

26. From the above question. Show that

Ans. And

Adding equal to equals we get

27. Show that if OA and OC are radii of same circle. OB and OD are also radii of same circle.

Ans. In AOB and COD

OA=OC (radii of same circle)

OB=OD (radii of same circle)

AB=CD (given)

AOBCOD (by SSS)

$� � � �$

28. Prove that OM Bisect AB. If OMAB.

Ans. AB is a chord of the circle with centre O.

OMAB

OA=OB (radii of same circle)

OM=OM (common)

AOBCOD (by SSS)

$� � � ℎ 90 °$

OAMOBM (by SAS)

AB = BM

Hence OM bisects AB

29. Prove that OMAB if AB is chord of the circle with centre O. O is joined to the mid-point M and AB.

Ans. O is joined to the mid-point M to AB

OM=OM (common)

AM=MB (M is midpoint of AM)

OA=OB (radii of same circle)

AOBCOD (by SSS)

OMA=OMB (CPCT)

But OMA+OMB =180° (linear pair)

Thus, OMA=OMB =90°

Hence, OMAB

30. ABCD is a cyclic quadrilateral in a circle with centre O. Prove that

Ans. If we join OD and OB, we can get DOB=2C, DOB=2A

2(A+C) = DOB+ reflex DOB =

A+C =

3 Marks Quetions

1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Ans. I Part: Two circles are said to be congruent if and only if one of them can be superposed on the other so as to cover it exactly.

Let C (O, ) and C (O’, ) be two circles. Let us imagine that the circle C (O’, ) is superposed on C (O, ) so that O’ coincide with O. Then it can easily be seen that C (O’,) will cover C (O,) completely if and only if

Hence we can say that two circles are congruent, if and only if they have equal radii.

II Part: Given: In a circle (O, ), AB and CD are two equal chords, subtend AOB and COB at the centre.

To Prove: AOB = COD

Proof: In AOB and COD,

AB = CD $� � � � �$

AO = CO $� � � � � � � � ℎ � � � � � � � � � � �$

BO = DO $� � � � � � � � ℎ � � � � � � � � � � �$

AOB COD $� � � � � � � � � �$

AOB = COD $� � � � � �$

Hence Proved.

2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal

Ans. Given: In a circle (O, ), AB and CD subtend two angles at the centre such that AOB = COD

To Prove: AB = CD

Proof: : In AOB and COD,

AO = CO $� � � � � � � � ℎ � � � � � � � � � � �$

BO = DO $� � � � � � � � ℎ � � � � � � � � � � �$

AOB = COD $� � � � �$

AOB COD $� � � � � � � � � �$

AB = CD $� � � � � �$

Hence proved.

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

Ans. From the figure, we observe that when different pairs of circles are drawn, each pair have two points (say A and B) in common.

Maximum number of common points are two in number.

Suppose two circles C (O, ) and C (O’, ) intersect each other in three points, say A, B and C.

Then A, B and C are non-collinear points.

We know that:

There is one and only one circle passing through three non-collinear points.

Therefore, a unique circle passes through A, B and C.

O’ coincides with O and

A contradiction to the fact that C (O’, ) C (O, )

Our supposition is wrong.

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

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

Ans. Steps of construction:

(a) Take any three points A, B and C on the circle.

(b) Join AB and BC.

(d) Draw perpendicular bisector PQ of BC.

(e) Let LM and PQ intersect at the point O.

Then O is the centre of the circle.

Verification:

O lies on the perpendicular bisector of AB.

OA = OB……….(i)

O lies on the perpendicular bisector of BC.

OB = OC……….(ii)

From eq. (i) and (ii), we observe that

OA = OB = OC = (say)

Three non-collinear points A, B and C are at equal distance from the point O inside the circle.

Hence O is the centre of the circle.

5. If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Ans. Given: Let C (O, ) and C (O’, ) be two circles intersecting at A and B. AB is the common chord.

To prove: OO’ is the perpendicular bisector of the chord AB.

Construction: Join OA, OB, O’A, O’B.

Proof: In triangles OAO’ and OBO’,

OA = OB $� � � ℎ � � � � � �$

O’A = O’B $� � � ℎ � � � � � �$

OO’ = OO’ $� � � � � �$

OAO’ OBO’ $� � � � � � � � � � � � � � �$

AOO’ = BOO’ $� � � � � �$

AOM = BOM

Now in AOB,OA = OB

And AOB = OBA $� � � � � � � � � � � � �$

Also AOM = BOM

RemainingAMO = BMO

AMO = BMO = $� � � � � � � � � �$

OM AB

OO’ AB

Since OM AB

M is the mid-point of AB.

Hence OO’ is the perpendicular bisector of AB.

6. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Ans. Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.

Radius OA = 5 cm, Radius O’A = 3 cm,

Distance between their centers OO’ = 4 cm

In triangle AOO’,

25 = 16 + 9

25 = 25

Hence AOO’ is a right triangle, right angled at O’.

Since, perpendicular drawn from the center of the circle bisects the chord.

Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.

Therefore, length of chord AB = Diameter of circle II

Length of chord = 6 cm.

7. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD. (See figure)

Ans. Given: Line intersects two concentric circles with centre O at points A, B, C and D.

To prove: AB = CD

Construction: Draw OL

Proof: AD is a chord of outer circle and OL AD.

AL = LD ………(i) $� � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � ℎ � � ℎ � � �$

Now, BC is a chord of inner circle and OL BC

BL = LC ……(ii) $� � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � ℎ � � ℎ � � �$

Subtracting (ii) from (i), we get,

AL – BL = LD – LC

AB = CD

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

Ans. Let AB be the minor arc of circle.

Chord AB = Radius OA = Radius OB

AOB is an equilateral triangle.

AOB =

Now

AOB + BOA =

+ BOA =

BOA =

D is a point in the minor arc.

= 2BDA

BOA = 2BDA

BDA = BOA =

BDA =

Thus angle subtended by major arc, at any point D in the minor arc is

Let E be a point in the major arc

= 2AEB

AOB = 2AEB

AEB = AOB

AEB = =

9. In figure, ABC = ACB = find BDC.

Ans. In triangle ABC,

BAC + ABC + ACB =

BAC +

BAC =

BAC = ……….(i)

Since, A and D are the points in the same segment of the circle.

BDC = BAC

$� � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$

BDC = $� � � � � (�)$

10. In figure, A, B, C, D are four points on a circle. AC and BD intersect at a point E such that BEC = and ECD = Find BAC.

Ans. Given: BEC = and ECD =

DEC = BEC = $� � � � � � � � � �$

Now in DEC,

DEC + DCE + EDC = $� � � � � � � � � � � � � � � �$

EDC = EDC =

BAC = EDC = $� � � � � � � � � � � � � � � � � � �$

11. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Ans. Since AC is a diameter.

B = D = ……….(i)

$� � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � �$

Similarly A = C = ……….(ii)

Now AC = BD $� � � � � � � � � � � � � � � � � � � � �$

$� � � � � � � � � � � � � � � � � � � � ℎ � � � �$

AD = BC $� ℎ � � � � � � � � � � � � � � � � � � � � � � �$ ……….(iii)

Similarly AB = DC……….(iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angles of the quadrilateral is and opposite sides are equal.

Hence ABCD is a rectangle.

12. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Ans. Given: Two circles intersect each other at points A and B. AP and AQ be their respective diameters.

To prove: Point B lies on the third side PQ.

Construction: Join A and B.

Proof: AP is a diameter.

1 =

$� � � � � � � � � � � � � � � � �$

Also AQ is a diameter.

2 =

$� � � � � � � � � � � � � � � � �$

1 + 2 =

PBQ =

PBQ is a line.

Thus point B. i.e. point of intersection of these circles lies on the third side i.e., on PQ.

13. ABC and ADC are two right triangles with common hypotenuse AC. Prove that CAD = ABD.

Ans. We have ABC and ADC two right triangles, right angled at B and D respectively.

ABC = ADC[Each ]

If we draw a circle with AC (the common hypotenuse) as diameter, this circle will definitely passes through of an arc AC, Because B and D are the points in the alternate segment of an arc AC.

Now we have subtending CBD and CAD in the same segment.

CAD = CBD

Hence proved.

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

Ans. Let O be the centre of the circle.

Join OA and OC.

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

AE = EB = AB = = cm

And CF = FD = CD = = cm

Let OE =

OF =

Let radius of the circle be

In right angled triangle AEO,

$� � � � � � � � ℎ � � � � � � � ℎ � � � � �$

……….(i)

Again In right angled triangle CFO,

$� � � � � � � � ℎ � � � � � � � ℎ � � � � �$

……….(ii)

Equating eq. (i) and (ii),

Now from eq. (i),

Hence radius of the circle is cm.

15. 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 form the centre?

Ans. Let AB = 6 cm and CD = 8 cm are the chords of circle with centre O.

Join OA and OC.

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

AE = EB = AB = = 3 cm

And CF = FD = CD = = 4 cm

Perpendicular distance of chord AB from the centre O is OE.

OE = 4 cm

Now in right angled triangle AOE,

$� � � � � � � � ℎ � � � � � � � ℎ � � � � �$

= 9 + 16 = 25

= 5 cm

Perpendicular distance of chord CD from the center O is OF.

Now in right angled triangle OFC,

$� � � � � � � � ℎ � � � � � � � ℎ � � � � �$

OF = 3cm

Hence distance of other chord from the centre is 3 cm.

16. ABCD is a parallelogram. The circle through A, B and C intersect CD (produced it necessary) at E. Prove that AE = AD.

Ans. In figure (a),

ABCD is a parallelogram.

1 = 3……….(i)

ABCE is a cyclic quadrilateral.

1 + 6 = ……….(ii)

And 5 + 6 = ……….(iii) $� � � � � � � � � �$

From eq. (ii) and (iii),1 = 5……….(iv)

Now, from eq. (i) and (iv),

3 = 5 AE = AD $� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �$

In figure (b),

ABCD is a parallelogram.

1 = 3 and 2 = 4

Also ABCD and BC meets them.

1 + 2 = ……….(i)

And ADBC and EC meets them.

5 = 2 ……….(ii) $� � � � � � � � � � � � � � � � � � �$

Since ABCE is a cyclic quadrilateral.

1 + 6 = ……….(iii)

From eq. (i) and (iii),

1 + 2 = 1 + 6

2 = 6

But from eq. (ii),2 = 5

5 = 6

Now in triangle AED,

5 = 6

AE = AD $� � � � � � � � � � � � � � � � � � � � � � � � � �$

Hence in both the cases, AE = AD

17. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

Ans. Given: Two equal circles intersect in A and B.

A straight line through A meets the circles in P and Q.

To prove: BP = BQ

Construction: Join A and B.

Proof: AB is a common chord and the circles are equal.

Arc about the common chord are equal, i.e.,

Since equal arcs of two equal circles subtend equal angles at any point on the remaining part of the circle, then we have,

1 = 2

In triangle PBQ,

1 = 2 $� � � � � �$

Sides opposite to equal angles of a triangle are equal.

Then we have, BP = BQ

18. Pair of opposite sides of a cyclic quadrilateral are equal, Prove that the other two sides are parallel.

Ans. Given: A cyclic quadrilateral ABCD in which AD= BC

To Prove: AB|| CD

Construction: Join B and D

Proof: AD=BC

But these are alternate angles

AB||CD

19. Prove that the centre of the circle through A, B, C, D is the Point intersection of its diagonals.

Ans. Given: A cyclic rectangle ABCD in which diagonals AC and BD intersect at Point O

To Prove: O is the centre of the circle

Proof: ABCD is a rectangle

AC= BD

Now as the diagonals AC and BD are intersecting at O

AO=OC, OB=OD

AO=OC=OB=OD

A, B, C, D lie on the same circle.

20. In isosceles triangle ABC, AD = AE and D and E are equal on side AB and AC so prove that B,C,E and are cyclic

Ans. Given that

AD=AE

[Adding both side]

21. If two non – parallel sides of a trapezium are equal, prove that it is cyclic.

Ans. In it and

AD=BC

DK=CP [Distance between || sides

22. In circle bisector AD of of Passes through the center O of the circum circle of Prove AB=AC

Ans. Draw and

In and

AO=AO

OPAOQA

OP=OQ

Chords AB and AC are equidistant from centre O

AB=AC

23. Prove that the circle drawn with the equal sides as a diameter passes through the Point D. if D is the mid Point of BC of an isosceles triangle ABC with AB=AC

Ans. Join AD in ABD and ACD

AB=AC

AD=AD

BD=CD $� � � � � � � � � � � �$

24. If a Pair of opposite sides of a cyclic quadrilateral are equal, then the diagonals are also equal.

Ans. Given: A cyclic quadrilateral ABCD in which AB=DC

To Prove: diagonal AC=diagonal BD

Proof: (Angle in same segment of circle)

But these are the angles subtended by the diagonals AC and BD in the same circle

AC=BD

25. and find

Ans. (Angle in same segment)

26. AB is chord of a circle and AB Produced to C such that BC=OB and CO joined and produce the circle and meet to D if and prove that

Ans. Proof: In

BO=BC

In is produced to A, forming exterior

OB=OA [Radii of the same circle

Again in is produced to D, forming exterior

Hence proved.

27. Prove that if P is the centre of circle

Ans. Given: A circle with centre P, XY and YZ are two chords

To Prove:

Proof:

Similarly arc YZ subtends at centre and at remaining Part of the circle

Adding (i) and (ii)

28. Prove that OA is the perpendicular bisector of BC if

Ans. Let OA intersect BC in P produce AO to meet the circle at K

Now, AOK is the diameter

In ABP and ACP

AB=AC

AP=AP (Common)

(SAS)

BP=CP

Each =

29. Prove that the line joining the midpoint of the two parallel chords of a circle passes through the centre of the circle.

Ans. Let AB and CD be the two parallel chords of the circle with centre O P and Q are the mid-points of AB and CD join OP and OQ.

Draw or

And

And also

POQ is a straight line.

30. ABCD is a quadrilateral in which AD=BC and show A, B, C, D lie on a circle

Ans. Join AC and BD

In ACD and BDC

AD=BC

As these are two equal angles on the same side of a line segment CD.

The four points A, B, C and D are concyclic.

31. Prove that diagonal is also equal when pair of opposite sides of a cyclic quadrilateral are equal. Prove.

Ans. Given: A cyclic quadrilateral ABCD in which AB= DC

To Prove: diagonal AC= Diagonal BD

Proof: [Angle in the same segment

But these are the angle subtended by the diagonal AC and BD in the same circle.

AC=BD

32. In ABCD cyclic quadrilateral diagonal Intersect at and so find

Ans. $� � � � � � � � ℎ � � � � � � � � � � � �$

33. Find the value of if A, B, C, D are concylic points

Ans. $� � � � � � � � � �$

$� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � = � � � � � � � � � � � . � � � � �$

34. Calculate the measure of where O is the centre of the circle

Ans. ABP =

APB =

APB+ABP+PAB =

++PAB =

PAB = - -=

Subtends PAB and PQB in the same segment of the circle

PAB=PQB=48°

35. In the given Fig find .

Ans. Subtends PQR and PSR in the same segment of the circle

PQR = PSR =

SPR + PSR +SRP =(angle sum property of a triangle)

SRP = --

36. Find the length of AB, CD,AC and BD if two concentric circles with centre O have A,B,C,D as the Point of intersection with line l.

Ans. OMBC

BM=MC……..(i)

OMAD

AM = MD…….(ii)

From (i) and (ii)

AM - BM = MD - MC = 6-4

AB = CD = 2cm

AC = AB+BC

=2+8=10cm

BD=BC+CD

=8+2 = 10cm

37. If OPAB find the length of the chord AB.

Ans. AP = PB =AB

In right angled triangle BPO,

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4 Marks Quetions

1. If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.

Ans. Given: Let AB and CD are two equal chords of a circle of centers

O intersecting each other at point E within the circle.

To prove: (a) AE = CE (b) BE = DE

Construction: Draw OM AB, ON CD. Also join OE.

Proof: In right triangles OME and ONE,

OME = ONE =

OM = ON

$� � � � � � ℎ � � � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � �$

OE = OE $� � � � � �$

OMEONE $� � � � � � � � � � � � � � � � � � �$

ME = NE $� � � � � �$ ……….(i)

Now, O is the centre of circle and OM AB

AM = AB $� � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � ℎ � � ℎ � � �$ …..(ii)

Similarly, NC = CD ……….(iii)

But AB = CD $� � � � �$

From eq. (ii) and (iii), AM = NC ……….(iv)

Also MB = DN …….…(v)

Adding (i) and (iv), we get,

AM + ME = NC + NE

AE = CE $� � � � � � � � � � (�)$

Now AB = CD $� � � � �$

AE = CE $� � � � � �$

AB - AE = CD - CE

BE = DE $� � � � � � � � � � (�)$

2. Three girls Reshma, Salma and Mandip are standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?

Ans. Let Reshma, Salma and Mandip takes the position C, A and B on the circle.

Since AB = AC

The centre lies on the bisector of BAC.

Let M be the point of intersection of BC and OA.

Again, since AB = AC and AM bisects CAB.

AM CB and M is the mid-point of CB.

Let OM = then MA =

From right angled triangle OMB,

52 = + MB2 ……….(i)

Again, in right angled triangle AMB,

62 = + MB2 ……….(ii)

Equating the value of MB2 from eq. (i) and (ii),

Hence, from eq. (i),

MB2 = =

= =

MB = = 4.8 cm

BC = 2MB =

3. A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Ans. Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

A = B = C = $� � �$

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.

OD = OE = OF = cm $� � �$

Join OA, OB and OC.

Area of AOB = Area of BOC = Area of AOC

And Area of ABC

= Area of AOB + Area of BOC + Area of AOC

And Area of ABC =

= 3 ( )

……….(i)

Now, CE BC

BE = EC = BC [ Perpendicular drawn from the centre bisects the chord]

BE = EC =

BE = EC = $� � � � � � � . (�)$

BE = EC =

Now in right angled triangle BEO,

OE2 + BE2 = OB2 $� � � � � � � � ℎ � � � � � � � ℎ � � � � �$

= 10 m

And = = m

Thus distance between any two boys is m.

4. Let vertex of an angle ABC be located outside a circle and let the sides of the angle intersect chords AD and CE with the circle. Prove that ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Ans. Vertex B of ABC is located outside the circle with centre O.

Side AB intersects chord CE at point E and side BC intersects chord AD at point D with the circle.

We have to prove that

ABC = [AOC - DOE]

Join OA, OC, OE and OD.

Now AOC = 2AEC

$� � � � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � ℎ � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � �$

AOC = AEC ……….(i)

Similarly DOE = DCE ……….(ii)

Subtracting eq. (ii) from eq. (i),

[AOC - DOE] = AEC - DCE ……….(iii)

Now AEC = ADC $� � � � � � � � � � � � � � � � � � � � � � � � � � �$ ……….(iv)

Also DCE = DAE $� � � � � � � � � � � � � � � � � � � � � � � � � � �$ ……….(v)

Using eq. (iv) and (v) in eq. (iii),

[AOC - DOE] = DAE + ABD - DAE

[AOC - DOE] = ABD

Or [AOC - DOE] = ABC Hence proved.

5. AC and BD are chords of a circle which bisect each other. Prove that:

(i) AC and BD are diameters.

(ii) ABCD is a rectangle.

Ans. Given: AC and BD of a circle bisect each other at O.

Then OA = OC and OB = OD

To prove:

(i) AC and BD are the diameters. In other words, O is the centre of the circle.

(ii) ABCD is a rectangle.

Proof: (i) In triangles AOD and BOC,

AO = OC $� � � � �$

AOD = BOC $� � � � � � � � � � � � � .$

OD = OB $� � � � �$

AODCOB $� � � � � � � � � � � � �$

AD = CB $� � � � � �$

Similarly, AOBCOD

AB = CD

$� � � � � � � � � � � � � � � � � � � � ℎ � � � �$

AC = BD $� ℎ � � � � � � � � � � � � � � � � � � � � � � � �$

AC and BD are the diameters as only diameters can bisect each other as the chords of the circle.

(ii) Ac is the diameter. $� � � � � � � � (�)$

B = D = ……….(i) $� � � � � � � � � � � - � � � � � �$

Similarly, BD is the diameter.

A = C = ……….(ii) $� � � � � � � � � � � - � � � � � �$

Now diameters AC = BD

$� � � � � � � � � � � � � � � � � � � � ℎ � � � �$

AD = BC $� ℎ � � � � � � � � � � � � � � � � � � � � ℎ � � � � � � � � � �$ ……….(iii)

Similarly AB = DC ……….(iv)

From eq. (i), (ii), (iii) and (iv), we observe that each angle of the quadrilateral is and opposite sides are equal.

Hence ABCD is a rectangle.

6. Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that angles of the triangle are and respectively.