Exercise MCQ
Question 1
The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is
- 11.5 cm
- 12 cm
- 23 cm
Solution 1
Correct option: (b)


Question 2
A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
- 25 cm
- 12.5 cm
- 30 cm
- 9 cm
Solution 2
Correct option: (c)


Question 3
In the given figure, BOC is a diameter of a circle and AB = AC. Then ∠ABC =?
- 30°
- 45°
- 60°
- 90°

Solution 3

Question 4
In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB =?
- 30°
- 15°
- 60°
- 90°

Solution 4

Question 5
In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB =?
- 40°
- 50°
- 80°
- 100°

Solution 5

Question 6
In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then, the distance of CD from AB is
- 8 cm
- 15 cm
- 18 cm
- 6 cm

Solution 6


Question 7
AB and CD are two equal chords of a circle with centre O such that ∠AOB = 80°, then ∠COD =?
- 100°
- 80°
- 120°
- 40°

Solution 7

Question 8
In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circle is
- 6 cm
- 9 cm
- 7.5 cm
- 8 cm

Solution 8

Question 9
In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
- 10 cm
- 12 cm
- 6 cm
- 8 cm

Solution 9


Question 10
In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB ‖ CD. If AB = 10 cm, then CD =?
- 5 cm
- 12.5 cm
- 15 cm
- 10 cm

Solution 10

Question 11
In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = 25°, then ∠AOD =?
- 50°
- 75°
- 90°
- 100°

Solution 11

Question 12
In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If OD ⟘ AB such that OD = 6 cm, then AC =?
- 9 cm
- 12 cm
- 15 cm
- 7.5 cm

Solution 12

Question 13
An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is
- 3 cm
- 6 cm
Solution 13


Question 14
The angle in a semicircle measures
- 45°
- 60°
- 90°
- 36°
Solution 14

Question 15
Angles in the same segment of a circle area are
- equal
- complementary
- supplementary
- none of these
Solution 15

Question 16
In the given figures, ⧍ABC and ⧍DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°. Then, ∠BCD =?
- 50°
- 60°
- 70°
- 80°

Solution 16

Question 17
In the given figure, BOC is a diameter of a circle with centre O. If ∠BCA = 30°, then ∠CDA =?
- 30°
- 45°
- 60°
- 50°

Solution 17

Question 18
In the given figure, O is the centre of a circle. If ∠OAC = 50°, then ∠ODB =?
- 40°
- 50°
- 60°
- 75°

Solution 18

Question 19
In the given figure, O is the centre of a circle in which ∠OBA = 20° and ∠OCA = 30°. Then, ∠BOC =?
- 50°
- 90°
- 100°
- 130°

Solution 19

Question 20
In the given figure, O is the centre of a circle. If ∠AOB = 100° and ∠AOC = 90°, then ∠BAC =?
- 85°
- 80°
- 95°
- 75°

Solution 20

Question 21
In the given figure, O is the centre of a circle. Then, ∠OAB =?
- 50°
- 60°
- 55°
- 65°

Solution 21

Question 22
In the given figure, O is the centre of a circle and ∠AOC = 120°. Then, ∠BDC =?
- 60°
- 45°
- 30°
- 15°

Solution 22

Question 23
In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠CDA =?
- 40°
- 50°
- 75°
- 25°

Solution 23

Question 24
In the given figure, AB and CD are two intersecting chords of a circle. If ∠CAB = 40° and ∠BCD = 80°, then ∠CBD =?
- 80°
- 60°
- 50°
- 70°

Solution 24

Question 25
In the given figures, O is the centre of a circle and chords AC and BD intersect at E. If ∠AEB = 110° and ∠CBE = 30°, then ∠ADB =?
- 70°
- 60°
- 80°
- 90°
Solution 25

Question 26
In the given figure, O is the centre of a circle in which ∠OAB =20° and ∠OCB = 50°. Then, ∠AOC =?
- 50°
- 70°
- 20°
- 60°

Solution 26

Question 27
In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ∠ADC = 120°, then ∠BAC =?
- 60°
- 30°
- 20°
- 45°

Solution 27

Question 28
In the given figure ABCD is a cyclic quadrilateral in which AB ‖ DC and ∠BAD = 100°. Then ∠ABC =?
- 80°
- 100°
- 50°
- 40°

Solution 28

Question 29
In the given figure, O is the centre of a circle and ∠AOC =130°. Then, ∠ABC =?
- 50°
- 65°
- 115°
- 130°

Solution 29

Question 30
In the given figure, AOB is a diameter of a circle and CD AB. If ∠BAD = 30°, then ∠CAD =?
- 30°
- 60°
- 45°
- 50°

Solution 30

Question 31
In the given figure, O is the centre of a circle in which ∠AOC =100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD =?
- 50°
- 40°
- 25°
- 80°

Solution 31


Question 32
In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠BOD=?
- 130°
- 50°
- 100°
- 80°

Solution 32

Question 33
In the given figures, ABCD is a cyclic quadrilateral in which BC = CD and ∠CBD = 35°. Then, ∠BAD =?
- 65°
- 70°
- 110°
- 90°

Solution 33

Question 34
In the given figure, equilateral ⧍ ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then ∠BDC =?
- 90°
- 60°
- 120°
- 150°

Solution 34

Question 35
In the give figure, side Ab and AD of quad. ABCD are produced to E and F respectively. If ∠CBE = 100°, then ∠CDF =?
- 100°
- 80°
- 130°
- 90°

Solution 35

Question 36
In the given figure, O is the centre of a circle and ∠AOB = 140°. The, ∠ACB =?
- 70°
- 80°
- 110°
- 40°

Solution 36

Question 37
In the given figure, O is the centre of a circle and ∠AOB = 130°. Then, ∠ACB =?
- 50°
- 65°
- 115°
- 155°

Solution 37

Question 38
In the given figure, ABCD and ABEF are two cyclic quadrilaterals. If ∠BCD = 110°, then ∠BEF =?
- 55°
- 70°
- 90°
- 110°

Solution 38

Question 39
In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ∠ADC = 95° and ∠ECF =20°. Then, ∠BAD =?
- 95°
- 85°
- 105°
- 75°

Solution 39

Question 40
Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD =?
- 10.5 cm
- 9.5 cm
- 8.5 cm
- 7.5 cm

Solution 40

Question 41
In the given figure, A and B are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points P and Q respectively. If AB = 4 cm, then the length of common chord PQ is
- 3 cm
- 6 cm
- 7.5 cm
- 9 cm

Solution 41

Question 42
In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then ∠CAO =?
- 30°
- 45°
- 60°
- 90°

Solution 42

Ex. 12C
Question 1
In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that .
Find

Solution 1

Question 2
In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If

Solution 2



Question 3
In the given figure , O is the centre of the circle and arc ABC subtends an angle of at the centre . If AB is extended to P, find .

Solution 3



Question 4
In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced . If

Solution 4



Question 5
In the given figure, BD=DC and

Solution 5



Question 6
In the given figure, O is the centre of the given circle and measure of arc ABC is Determine .

Solution 6
Question 7
In the given figure, is equilateral. Find

Solution 7
Question 8
In the adjoining figure, ABCD is a cyclic quadrilateral in which .

Solution 8



Question 9
In the given figure , O is the centre of a circle and Find the values of x and y.

Solution 9



Question 10
In the given figure, O is the centre of the circle and . Calculate the vales of x and y.

Solution 10

Question 11
In the given figure , sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If , find the value of x.

Solution 11


Question 12
In the given figure, AB is a diameter of a circle with centre O and DO||CB. If , calculate

Also , show that is an equilateral triangle.

Solution 12




Question 13Two chord AB and CD of a circle intersects each other at P outside the circle. If AB=6cm, BP=2cm and PD=2.5 cm, Find CD.
Solution 13
Question 14
In the given figure , O is the centre of a circle. If , calculate


Solution 14


Question 15
In the given figure, is an isosceles triangle in which AB=AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.

Solution 15


Question 16
In the given figure, AB and CD are two parallel chords of a circle . If BDE and ACE are straight lines , intersecting at E, prove that is isosceles.

Solution 16

Question 17
In the given figure, Find the values of x and y.

Solution 17
Question 18
In the given figure , ABCD is a quadrilateral in which AD=BC and . Show that the pints A, B, C, D lie on a circle.

Solution 18



Question 19
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.Solution 19


Question 20
Prove that the circles described with the four sides of a rhombus . as diameter , pass through the point of intersection of its diagonals.Solution 20





Question 21
ABCD is a rectangle . Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.Solution 21



Question 22
Give a geometrical construction for finding the fourth point lying on a circle passing through three given points , without finding the centre of the circle. Justify the construction.Solution 22




Question 23
In a cyclic quadrilateral ABCD, if , show that the smaller of the two is
Solution 23

Question 24
The diagonal s of a cyclic quadrilateral are at right angles . Prove that perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.Solution 24



Question 25
On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.Solution 25

AB is the common hypotenuse of ΔACB and ΔADB.
⇒ ∠ACB = 90° and ∠BDC = 90°
⇒ ∠ACB + ∠BDC = 180°
⇒ The opposite angles of quadrilateral ACBD are supplementary.
Thus, ACBD is a cyclic quadrilateral.
This means that a circle passes through the points A, C, B and D.
⇒ ∠BAC = ∠BDC (angles in the same segment) Question 26
ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that Solution 26

Construction: Take a point E on the circle. Join BE, DE and BD.
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠BAD = 2∠BED

Now, EBCD is a cyclic quadrilateral.
⇒ ∠BED + ∠BCD = 180°
⇒ ∠BCD = 180° – ∠BED

In ΔBCD, by angle sum property
∠CBD + ∠CDB + ∠BCD = 180°

Ex. 12A
Question 1
A chord of length 16 cm is drawn in a circle of radius 10 cm. Find the distance of a chord from the centre of the circle.Solution 1



Question 2
Find the length of the chord which is at the distance of 3 cm from the centre of a circle of radius 5 cm.Solution 2



Question 3
A chord of length 30 cm is drawn at a distance of 8 cm from the centre of a circle. Find the radius of the circle.Solution 3




Question 4
In a circle of radius 5 cm, AB and CD are two parallel chords of lengths 8 cm and 6cm respectively. Calculate the distance between the chords if they are
(i) On the same side of the centre
(ii) On the opposite side of the centre.Solution 4








Question 5
Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of a circle of a radius 17 cm. Find the distance between the chords.Solution 5




Question 6
In the given figure , the diameter CD of a circle with centre O is perpendicular to chord AB. If AB=12 cm and CE=3cm, calculate the radius of the circle.

Solution 6




Question 7
In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE=ED=8 cm and EB=4 cm. Find the radius of the circle.

Solution 7

Question 8
In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC||DO and AC=2xOD.

Solution 8



Question 9
In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects . Prove that AB=CD.

Solution 9



Question 10
Prove that the diameter of the circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.Solution 10



Question 11
Prove that two different circles cannot intersects other at more than two points.Solution 11

Question 12
Two circle of the radii 10 cm and 8 cm intersects each other , and the length of the common chord is 12 cm. Find the distance between their centres.

Solution 12

Question 13
Two equal circle intersects in P and Q. A straight line through P meets the circles in A and B. Prove that QA= QB.

Solution 13



Question 14
If a diameter of the circle bisects each of the two chords of a circle then prove that the chords are parallel.Solution 14



Question 15
In the adjoining figure , two circles with centres at A and B, and of radii 5 cm and 3 cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q . Find the length of PQ.

Solution 15
Question 16
In the given figure , AB is a chord of a circle with centre O and AB is produced to C such that BC=OB. Also CO is a joined and produced to meet the circle in D. If , prove that x=3y.

Solution 16

Question 17
AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre then prove that 4q2 = p2 + 3r2.Solution 17

Let O be the centre of a circle with radius r.
⇒ OB = OC = r
Let AC = x
Then, AB = 2x
Let OM ⊥ AB
⇒ OM = p
Let ON ⊥ AC
⇒ ON = q
In ΔOMB, by Pythagoras theorem,
OB2 = OM2 + BM2

In ΔONC, by Pythagoras theorem,
OC2 = ON2 + CN2

Question 18
In the adjoining figure , O is the centre of a circle . If AB and AC are chordsof a circlesuch thatAB=AC, , prove that PB=QC.

Solution 18


Question 19
In the adjoining figure, BC is a diameter of a circle with circle with centre O. If AB and CD are two chords such that AB|| CD, prove that AB=CD.

Solution 19
Question 20
An equilateral triangle of side 9 cm is inscribed in a circle . Find the radius of the circle.Solution 20



Question 21

Solution 21


Question 22
In the adjoining figure , OPQR is a square. A circle drawn with centre O cuts the square in X and Y. prove that QX =QY.
Solution 22
Question 23
Two circle with centres O and O’ intersect at two points A and B. A line PQ is drawn parallel to OO’ through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO’.Solution 23

Draw OM ⊥ PQ and O’N ⊥ PQ
⇒ OM ⊥ AP
⇒ AM = PM (perpendicular from the centre of a circle bisects the chord)
⇒ AP = 2AM ….(i)
And, O’N ⊥ PQ
⇒ O’N ⊥ AQ
⇒ AN = QN (perpendicular from the centre of a circle bisects the chord)
⇒ AQ = 2AN ….(ii)
Now,
PQ = AP + PQ
⇒ PQ = 2AM + 2AN
⇒ PQ = 2(AM + AN)
⇒ PQ = 2MN
⇒ PQ = 2OO’ (since MNO’O is a rectangle)
Ex. 12B
Question 1
(i) In figure (1) , O is the centreof the circle . If (ii) In figure(2), A, B and C are three points on the circlewithcentre O such that .


Solution 1

Question 2
In the given figure, O is the centre of the circle and .
Calculate the value of .

Solution 2
Question 3
In the given figure , O is the centre of the circle .If , find the value of
Solution 3
Question 4
In the given figure , O is the centre of the circle. If

Solution 4

Question 5
In the given figure, O is the centre of the circle .If find .
Solution 5
Question 6
In the given figure , , calculate

Solution 6

Question 7
In the adjoining figure , DE is a chord parallel to diameter AC of the circle with centre O. If , calculate .

Solution 7

Question 8
In the adjoining figure, O is the centre of the circle. Chord CD is parallel to diameter AB. If calculate

Solution 8

Question 9
In the given figure, AB and CD are straight lines through the centre O of a circle. If , find

Solution 9
Question 10
In the given figure , O is the centre of a circle, , find .

Solution 10

Question 11
In the adjoining figure , chords AC and BD of a circle with centre O, intersect at right angles at E. if , calculate .

Solution 11

Question 12
In the given figure , O is the centre of a circle in which . Find

Solution 12

Question 13
In the given figure, O is the centre of the circle and ∠BCO = 30°. Find x and y.

Solution 13
Given, ∠AOD = 90° and ∠OEC = 90°
⇒ ∠AOD = ∠OEC
But ∠AOD and ∠OEC are corresponding angles.
⇒ OD || BC and OC is the transversal.
∴ ∠DOC = ∠OCE (alternate angles)
⇒ ∠DOC = 30° (since ∠OCE = 30°)
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠DOC = 2∠DBC

Now, ∠ABE = ∠ABC = ∠ABD + ∠DBC = 45° + 15° = 60°
In ΔABE,
∠BAE + ∠AEB + ∠ABE = 180°
⇒ x + 90° + 60° = 180°
⇒ x + 150° = 180°
⇒ x = 30° Question 14
In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB

Solution 14
Construction: Join AC

Given, BD = OD
Now, OD = OB (radii of same circle)
⇒ BD = OD = OB
⇒ ΔODB is an equilateral triangle.
⇒ ∠ODB = 60°
We know that the altitude of an equilateral triangle bisects the vertical angle.

Now, ∠CAB = ∠BDC (angles in the same segment)
⇒ ∠CAB = ∠BDE = 30° Question 15
In the given figure, PQ is a diameter of a circle with centre O. If

Solution 15

Question 16
In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line.

Solution 16
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠APB = 2∠ACB

Now, ACD is a straight line.
⇒ ∠ACB + ∠DCB = 180°
⇒ 75° + ∠DCB = 180°
⇒ ∠DCB = 105°
Again,

Question 17
In the given figure , . Show that BC is equal to the radius of the circumcircle of whose centre is O.

Solution 17

Question 18
In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E. Prove that


Solution 18
Join AC and BC

We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
⇒ ∠AOC = 2∠ABC ….(i)
Similarly, ∠BOD = 2∠BCD ….(ii)
Adding (i) and (ii),
∠AOC + ∠BOD = 2∠ABC + 2∠BCD
⇒ ∠AOC + ∠BOD = 2(∠ABC + ∠BCD)
⇒ ∠AOC + ∠BOD = 2(∠EBC + ∠BCE)
⇒ ∠AOC + ∠BOD = 2(180° – ∠CEB)
⇒ ∠AOC + ∠BOD = 2(180° – [180° – ∠AEC])
⇒ ∠AOC + ∠BOD = 2∠AEC

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