Circles Test

Result

Circles Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

In the given circle, O is the centre and length of chord XY = length of chord ZW. If XOY = 60°, then what is the measure of ZOW?

A
120°

B
60°

C
30°

D
90°

Solution

Given: O is the centre of the circle, length of chord XY = length of chord ZW, and XOY = 60°.

If XY = ZW, then ZOW = XOY(equal chords of a circle subtend equal angles at the centre)
Hence, ZOW = 60° (Given XOY = 60°)
Thus, answer option 2 is correct.
Question 2
1 / -0

In the given figure, AOB is a diameter of a circle with centre O. If BOD = 140°, then find the measure of ACD.

A
20°

B
40°

C
60°

D
90°

Solution

BOD = 140°(Given)
BOD + DOA = 180° (Sum of angles on a straight line)
DOA = 40°
DOA = 2ACD (Angle made by the chord at the centre is twice that subtended by it at the circumference)
40° = 2ACD
ACD = 20°
Option (1) is correct.
Question 3
1 / -0

O is the centre of the circle as shown in the diagram. The distance between P and Q is 4 cm such that the line through P and Q passes through the centre O. The measure of ROQ is

A
25°

B
35°

C
105°

D
90°

Solution

y° = 2∠QPR
(An angle subtended at the centre of a circle by an arc is twice the angle subtended at any point on the circumference by the same arc)
QPR = y°
y° = OPR + PRO [Exterior angle property in POR]
y° = + 45°
y° - y°/2 = 45°
y° = 2 45°
y°= 90°
Question 4
1 / -0

In the given circle, O is the centre, AB = BC and OM = 5 units. The length of ON is equal to ____.

A
5 units

B
2.5 units

C
10 units

D
15 units

Solution

It is given that O is the centre, AB = BC and OM = 5 units.

AB = BC
OM = ON [Equal chords of a circle are equidistant from the centre.]
ON = 5 units [OM = 5 units]
Question 5
1 / -0

In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P. If DBC = 60° and BAC = 40°, find the measure of BCD.

A
100°

B
120°

C
90°

D
80°

Solution

DAC = DBC = 60° [Angles in the same segment]
Now, DAB = DAC + CAB = 60° + 40° = 100°
We have:
BCD + DAB = 180° [Opposite angles of a cyclic quadrilateral]
BCD + 100° = 180°
BCD = 80°
Question 6
1 / -0

In the given figure, O is the centre of the circle. What is the measure of ACB?

A
260°

B
100°

C
50°

D
200°

Solution

^{Angle subtended by a chord of a circle at the centre of the circle is twice the angle subtended by the same chord on the circle.}
Question 7
1 / -0

In the given circle, O is the centre and AB = CD. What is the measure of AOB?

A
50°

B
60°

C
100°

D
80°

Solution

O is the centre of circle and AB = CD.

Now, in COD, COD + ODC + DCO = 180° [Sum of angles of a triangle is 180°]
COD + 50° + 50° = 180°
COD = 180° – 100° = 80°
Also, AB = CD
Angles subtended at the centre of a circle by equal chords. AB and CD are also equal.
AOB = COD = 80°
Question 8
1 / -0

In the given figure, O is the centre of the circle and AOC = 112°. Find the measure of ABC.

A
56°

B
112°

C
224°

D
124°

Solution

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Also, the sum of the opposite angles of a cyclic quadrilateral is 180°.
It is given that AOC = 112°
Also, ADC = AOC
ADC = = 56°
Now, ADC + ABC = 180°( ABCD is a cyclic quadrilateral.)
56° + ABC = 180°
ABC = 124°
Question 9
1 / -0

In the following figure, O is the centre of the circle and ∠AOB = 88°. What is the measure of ∠OCB?

A
88°

B
44°

C
45°

D
60°

Solution

In the given figure, join BC and AB.
∠AOB = 88° (Given)
∴ ∠BOC = 180° - 88° = 92°(Angle of linear pair)
Now, ΔOBC is an isosceles triangle.
∴ OB = OC = Radius
∴ ∠OCB = ∠OBC = x
Now, x + x + 92 = 180°
⇒ 2x = 88°
⇒ x = 44°
Question 10
1 / -0

In the given figure, O is the centre of the circle, AOB = POQ = 30° and XOY = ZOW = 40°. Also, XY = 4 units and AB = 3.5 units. The value of (AB + ZW + PQ + XY) is

A
11.25 units

B
22.5 units

C
15 units

D
7.8 units

Solution

It is given that AOB = POQ = 30°, XOY = ZOW = 40°, XY = 4 units and AB = 3.5 units.

If the angles subtended by two chords of a circle at the centre are equal, the chords are also equal.
Here, AOB = POQ
Therefore, AB = PQ = 3.5 units
And XOY = ZOW
So, XY = ZW = 4 units
Then, AB + ZW + PQ + XY = (3.5 + 4 + 3.5 + 4) units = 15 units
Question 11
1 / -0

In the given figure, O is the centre of the circle and CBD = 56°. Find the measure of AEC.

A
124°

B
117°

C
112°

D
100°

Solution

Join EC.
ABC = 180° - 56° = 124°
Now, ABC = AEC ( Angles are made by the same chord AC)
⇒ AEC = 124°
Question 12
1 / -0

If in the given figure, O is the centre of the circle and AEB + ACB + ADB = 3x, then find the measure of AOB.

A

B
x

C

D
2x

Solution

AEB + ACB + ADB = 3x (Given)

We have, AEB = ACB = ADB [Angles in the same segment of circle are equal]
AEB + ACB + ADB = 3x .... (1)
Or, AEB + AEB + AEB = 3x [From (1)]
Or, 3AEB = 3x
Or, AEB = x
Thus, AOB = 2AEB = 2x (The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle)
Question 13
1 / -0

In the given figure, O is the centre of the circle and AE is a diameter. If AB = BC and BFC = 25°, find the value of ABC.

A
130°

B
120°

C
115°

D
105°

Solution

Angles subtended by equal chords are equal.
AFB = 25° = BFC
In the cyclic quadrilateral ABCF,
AFC = 25° + 25° = 50°ABC = 180° - 50° = 130° (because the sum of opposite angles of a cyclic quadrilateral is 180°)
Question 14
1 / -0

Consider the figure:

If x = 60° and y = 120°, then (w - z) is equal to

A
60°

B
30°

C
100°

D
110°

Solution

x = 60°, y = 120°(Given)

Sum of opposite angles of a cyclic quadrilateral is 180°.
Therefore, x + w = 180° or w = 180° - x
= 180° - 60° = 120°
Also, y + z = 180° or z = 180 - y = 180° - 120° = 60°
So, w - z = 120° - 60° = 60°
Question 15
1 / -0

In the given figure, O is the centre of the circle.

If CBN = 74°, then AOC is equal to

A
64°

B
116°

C
148°

D
232°

Solution

CBN = 74°
CBA = 180 - 74°
CBA = 106°
CBA = 1 (An angle subtended at the centre of a circle by an arc is twice the angle subtended at any point on the circumference by the same arc.)
106° = 1
1 = 2 106°
1 = 212°
1 + 2 = 360°
2 + 212°= 360°
2 = 360° - 212°
2 = 148°
Question 16
1 / -0

In the given figure, O is the centre and m AOC = 130°. Then, m CBE is equal to

A
50°

B
100°

C
105°

D
115°

Solution

AOC = 130°
AOC = 2ABC [An angle subtended at the centre of a circle by an arc is twice than the angle subtended at any point on the circumference by the same arc]
130° = 2ABC
⇒ ABC = 130°/2 = 65°
ABC + CBE = 180° (Supplementary angles)
65° + CBE = 180°
CBE = 180° - 65° = 115°
Question 17
1 / -0

What is the length of PN, if AB = 30, PM = 8, DC = 16 and P is the centre of the circle?

A
15

B
16

C
17

D
None of these

Solution

PM AB
Any perpendicular from the centre to the chord bisects the chord.

In ΔAMP
AP² = MP² + AM²
AP² = 8² + 15² = 64 + 225 = 289
AP = = 17

In ΔCNP
CP² = NP² + CN²
17² = NP² + 8²
17² - 8² = NP²
NP = 15
Question 18
1 / -0

In the given figure, O is the centre of the circle, ST = UV, SV is a straight line and SOT = 85°. The measure of UOT is

A
85°

B
10°

C
20°

D
42.5°

Solution

O is the centre of the circle, ST = UV, SV is a straight line and SOT = 85°.

Now, ST = UV SOT = UOV [Equal chords subtend equal angles at the centre.]
Thus, UOV = 85°(SOT = 85°)
Now, the sum of adjacent angles on the same straight line is 180°.
So, SOT + UOT + UOV = 180°
85° + UOT + 85° = 180°
UOT = 180° - 170° = 10°
Question 19
1 / -0

In the following figure, AC is diameter of the circle. AB = BC and AED = 135°. Find the values of DEC and BAD.

A
35° and 55°

B
45° and 75°

C
45° and 90°

D
60° and 30°

Solution

Join EC.

AC is the diameter.
Here, AEC = 90° (Angle in semicircle)
DEC = AED - AEC
= 135° - 90°
DEC = 45°
Join AD.
DAC = DEC = 45° (Angles in the same segment are equal.)
ABC = 90° (Angle in semicircle)
BAC = BCA = = 45°
BAD = BAC + CAD
= 45° + 45° = 90°
Question 20
1 / -0

In the given figure, AB is the diameter of the circle. C and D lie on the semi-circle. If ABC = 65° and CAD = 45°, DCA equals

A
45°

B
25°

C
20°

D
15°

Solution

Since ABCD is cyclic,
D = 180° - 65°
= 115° [Sum of the opposite angles of cyclic quadrilateral is supplementary.]
In ADC,
DCA = 180° - (115° + 45°) (Angle sum property of triangle)
= 180° - 160° = 20°
Question 21
1 / -0

In the given figure, chord ED is parallel to the diameter AC of the circle. If CBE = 65°, then what is the value of DEC?

A
35°

B
55°

C
45°

D
25°

Solution

CBE = 65° (Given)
EBA = 25° (CBA = 90°)
Also, ACE = 25° (Angles subtended by the same arc at the circumference are equal.)
DEC = ACE = 25° (Alternate angles)

Question 22

1 / -0

Fill in the blanks with the help of the given table.

(i) If two arcs are P , then their corresponding chords are Q .
(ii) The minimum number of points required to draw a circle is R .
(iii) If one side of a cyclic quadrilateral is produced, then the exterior angle so formed is equal to the S .

	P	Q	R	S
A	congruent	unequal	2	angle in a semi-circle
B	not congruent	equal	3	angle in a semi-circle
C	not congruent	unequal	4	angle subtended by an arc at the centre
D	congruent	equal	3	interior opposite angle

A

B

C

D

Solution

(i) If two arcs are congruent, then their corresponding chords are equal.
(ii) The minimum number of points required to draw a circle is 3.
(iii) If one side of a cyclic quadrilateral is produced, then the exterior angle so formed is equal to the interior opposite angle.

Hence, option 4 is correct.

Question 23
1 / -0

State 'T' for true and 'F' for false for the following statements.

(i) The region between an arc and the two radii, joining the centre to the end points of the arc is called a segment.
(ii) The length of a complete circle is called diameter.
(iii) For a given chord, the area of the major segment is always equal to the area of the minor segment.
(iv) From any two given points, one can draw infinite number of circles passing through them.

A
(i) - F, (ii) - F, (iii) - F, (iv) - T

B
(i) - F (ii) - T, (iii) - T, (iv) -F

C
(i) - F, (ii) - F, (iii) - F, (iv) - F

D
(i) - F, (ii) - T, (iii) - F, (iv) - F

Solution

(i) The given statement is false as the region between an arc and the two radii, joining the centre to the end points of the arc is called a sector.
(ii) The given statement is false as the length of a complete circle is called circumference.
(iii) The given statement is false as major segment and minor segment are equal only when the chord is equal to the diameter.
(iv) The given statement is true as from any two given points, one can draw infinite number of circles passing through them.
Question 24
1 / -0

In the given figure, ABCD is a cyclic quadrilateral. A circle passing through the points A and B meets AD and BC at points E and F, respectively. Which of the given statements is/are true for the given problem?

(i) EF DC
(ii) ∠1 = ∠2
(iii) ∠2 = ∠3
(iv) DC AB

A
Only (i)

B
Only (ii) and (iv)

C
Only (i) and (iii)

D
None of the statements is true.

Solution

As ABCD is a cyclic quadrilateral, ∠1 + ∠3 = 180° (Sum of opposite angles of a cyclic quadrilateral)
Also, as AEFB is a cyclic quadrilateral;
∠1 + ∠3 = 180° (Sum of opposite angles of a cyclic quadrilateral)
∠1 + ∠3 = ∠1 + ∠2
∠3 = ∠2
∴ EF || DC (Corresponding angles are equal.)
Hence, option 3 is correct.
Question 25
1 / -0

In the given figure, AB is a diameter of the circle and CD is a chord equal to the radius of the circle. If AC and BD are extended and intersect at a point E, then which of the following steps is INCORRECT in order to prove that ∠AEB = 60°?

Step - 1: Join OC, OD and BC. Triangle ODC is an equilateral triangle.
Step - 2: ∠COD = 60°. Now, ∠CBD = 2∠COD (Angle subtended by an arc at the centre = 2 × Angle subtended by the arc at any point on the circle)
Step - 3: This gives ∠CBD = 30°.
Step - 4: ∠ACB = 90° (Angle subtended by a semi-circle)
Step - 5: So, ∠BCE = 180° – ∠ACB = 90°
Step - 6: ∠CEB = 90° – 30° = 60°, i.e. ∠AEB = 60°

A
Step - 3

B
Step - 2

C
Step - 4

D
Step - 5

Solution

Step - 1: Join OC, OD and BC. Triangle ODC is an equilateral triangle.
Step - 2: ∠COD = 60°. Now, ∠CBD = 2∠COD (Angle subtended by an arc at the centre = 2 × Angle subtended by the arc at any point on the circle)
Step - 3: This gives ∠CBD = 30°.
Step - 4: ∠ACB = 90° (Angle subtended by a semi-circle)
Step - 5: So, ∠BCE = 180° – ∠ ACB = 90°
Step - 6: ∠ CEB = 90° – 30° = 60°, i.e. ∠ AEB = 60°
Among the given steps, only step - 2 is incorrect as ∠CBD = ∠COD.