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Circles Test - 6

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Circles Test - 6

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Question 1
1 / -0

In the given figure with centre O, AOB = 30° and ∠BOC = 54°. Find ABC.

A
130°

B
132°

C
138°

D
140°

Solution

AOC = AOB + COB = 84°
Reflex AOC = 360° - 84° = 276°The angle subtended by arc at centre is double the angle subtended by it on the remaining part of the circle.
ABC = × (ReflexAOC)
ABC = 138°
Question 2
1 / -0

In the given figure, O is the centre of the circle. Find the measure of ABC.

A
105°

B
100°

C
80°

D
75°

Solution

According to the question,
DOC = 360° - 260° = 100°

Now,
OD = OC (Radius of circle)
ODC = OCD (As OD = OC)

In triangle ODC,
ODC + OCD + DOC = 180° (Sum of the interior angles of triangle)
2 ODC = 180° - 100° = 80°
ODC = 80° ÷ 2 = 40°
ADC = 40° + 35° = 75°
We know that opposite angles of cyclic quadrilateral are supplementary.
ABC = 180° - 75° = 105°
Question 3
1 / -0

In the given figure, a quadrilateral is inscribed in the circle. If ∠ADC = 55°, ∠ABD = 65°, ∠BAD = 85°, then find the value of ∠BDC & ∠BCD.

A
∠BDC = 25°, ∠BCD = 95°

B
∠BDC = 35°, ∠BDC = 75°

C
∠BDC = 45°, ∠BCD = 65°

D
∠BDC = 55°, ∠BCD = 55°

Solution

By angle-sum property of triangle ABD,
∠ADB = 180° - 85° - 65° = 30°
=> ∠BDC = ∠ADC - ∠ADB = 55° - 30° = 25°
∠BAD + ∠BCD = 180°
(Sum of opposite angles of a cyclic quadrilateral is 180°)
=> ∠BCD = 180° - 85° = 95°
Thus, ∠BDC = 25°and ∠BCD = 95°
Question 4
1 / -0

In the given figure, find the value of y when P is the centre of the circle.

A
12

B
15

C
28

D
36

Solution

We know that the measure of an inscribed angle is half the measure of the central angle with the same intercepted arc.

y + 60 = 6y
5y = 60
y = 12
Hence, the value of y = 12.
Question 5
1 / -0

What is the measure of angle PRS, if SQ is the diameter of the circle with centre O?

A
10°

B
15°

C
20°

D
30°

Solution

∠SPR = ∠SQR = 55° (Angles in the same segment are equal)
∠SRQ = 90° (Angle in the semicircle is right angle)
∠QSR = 180° - (∠SQR + ∠SRQ) ...............(Sum of the interior angles of triangle is 180°)
= 180° - (55° + 90°)
= 180° - 145° = 35°
Now, ∠PSR = ∠PSQ + ∠QSR = 75° + 35° = 110°
In triangle PSR,
∠PRS = 180° - (∠PSR + ∠SPR) (Sum of the interior angles of triangle is 180°)
= 180° - (110° + 55°)
= 180° - 165°
= 15°
Question 6
1 / -0

In the following figure, ABO = 43° and OA and OB are the radii of the cirlce. Find the measure of angle ADB.

A
143°

B
94°

C
133°

D
166°

Solution

Given, ABO = 43°
As OA = OB, then OAB = OBA = 43°.
In triangle OAB,
OAB + OBA + AOB = 180°
43° + 43° + ∠AOB = 180°
86° + AOB = 180°
AOB = 94°
And reflex AOB = 360° - 94° = 266°
An inscribed angle is half of the central angle.
So,
ADB = of reflex AOB

ADB = of 266 = 133°
Question 7
1 / -0

In the given figure, if ABCD is a cyclic quadrilateral with respective angles, then what will be the value of x - y?

A
8.5

B
4.5

C
6.5

D
5.5

Solution

Given, ABCD is cyclic quadrilateral.
DAB + BCD = 180°
2x + 16° = 180°
x = 82°
CBA + CDA = 180°
2y + 33° = 180°
y = 73.5°
x - y = 82° - 73.5° = 8.5°
Question 8
1 / -0

In the given figure, PQRS is a cyclic quadrilateral and ∠PQW is 145°. What is the measure of a° + b° + c° if ∠PRS is two fifth of ∠PSU?

A
145°

B
169°

C
215°

D
250°

Solution

Given, ∠PQW = 145°
∠PQW = ∠PSR = 145° (Exterior angle of cyclic quadrilateral is equal to interior opposite angle)
∠PSU = a° = 180 - 145° = 35°

In triangle PSR, ∠PRS is two fifth of ∠PSU (Given)
∠PRS = = 14°

Also, ∠PRS + ∠PSR + ∠SPR = 180° (Sum of the angle of triangles is 180°)
14° + 145° + ∠SPR = 180°
159° + ∠SPR = 180°
∠SPR = 180° - 159° = 21°
∠SPR = ∠STR = 21° (Angles on the same segment are equal)
Hence, c = 21°
∠SPV + ∠SPR = 180°
∠SPV + 21° = 180°
∠SPV = 159° or b = 159°
According to the question,
a + b + c = 35° + 159° + 21° = 215°
Question 9
1 / -0

What is the measure of ∠ADC in the given figure?

A
110°

B
120°

C
130°

D
140°

Solution

∠ACB = 90° (Angle in the semicircle)
In triangle ABC,
∠ABC = 180° - 90° - 40° = 50° (Angle sum property of triangle)
ABCD is a cyclic quadrilateral.
Hence, ∠ADC + ∠ABC = 180°
∠ADC = 180° - 50° = 130°
Question 10
1 / -0

There is a circle which has a diameter PQ = 34 cm. A chord which is equal to 16 cm, has been drawn from point P to point Y. Find the least distance between centre of circle and the chord of circle.

A
15 cm

B
16 cm

C
17 cm

D
18 cm

Solution

Given, PQ = 34 cm, PY = 16 cm

PO = = = 17 cm

Draw perpendicular OX on PY as perpendicular line gives the least distance between two points.
Also, perpendicular drawn from centre to the chord bisect the chord.

So, PX = = 8 cm

Now in right angle triangle ΔPXO,
OX² = PO²- PX²OX² = 17² - 8²
OX²= 289 - 64
OX² = 225
OX = 15 cm
So, the least distance between centre of circle and the chord of circle is 15 cm.
Question 11
1 / -0

In the given circle with centre O, ∠CBA = 50° and ∠ACB = 25°. Find ∠ODB.

A
90°

B
100°

C
115°

D
145°

Solution

∠ACB = (1/2) of ∠AOB (As the angle subtended by an arc at the centre is double the angle subtended by it on any other part of the circle)
∠AOB = 2 × 25 = 50°
In triangle OAB, OA = OB (Radii of the circle)
∠OAB = ∠OBA
∠OAB + ∠AOB + ∠OBA = 180° (Angle sum property of triangle)
2∠OAB = 180° - 50°
∠OAB = 130 ÷ 2 = 65°
∠ODB = ∠OAB + ∠CBA = 65° + 50° = 115° (As the external angle of a triangle is equal to the sum of other two interior angles)
Question 12
1 / -0

In the given figure, EFHG is a cyclic quadrilateral such that ∠EFG = 50° and ∠FHE = 60°. Find the measure of ∠FEG.

A
100°

B
90°

C
80°

D
70°

Solution

From the question, ∠GHE = ∠EFG = 50° [Angles in same segment of a circle are equal]
Now, ∠GHF = 50° + 60° = 110°
∠FEG + ∠GHF = 180°
∠FEG + 110° = 180°
∠FEG = 180° - 110°
∠FEG = 70°
Question 13
1 / -0

In the given figure, O is the centre of the circle with chords AP and BP being produced to R and Q respectively. If ∠QPR = 55°, find the measure of ∠AOB.

A
90°

B
100°

C
110°

D
120°

Solution

∠APB =∠RPQ = 55°
Now, ∠AOB and ∠APB are angles subtended by an arc AB at centre and at the remaining part of the circle.
∴ ∠AOB = 2∠APB = 2 x 55° = 110°
Question 14
1 / -0

In the given figure with centre O, if ∠OPQ = 58°, find the measure of x.

A
32°

B
36°

C
38°

D
40°

Solution

In ΔOPQ,
OP = OQ [radii of same circle]
∠OQP = ∠OPQ = 58° [Angles opposite to equal sides in a triangle are equal]
Now,∠OPQ + ∠POQ + ∠OQP = 180° [angle sum properly of triangle]
58° + ∠POQ + 58° = 180°
∠POQ = 180° - 116°
∠POQ = 64°
Therefore, ∠PRQ = x = 32° [∵ angle subtended by an arc at the centre is double the angle subtended by it at the remaining part of the circle]
Question 15
1 / -0

In figure, PQSR is a cyclic quadrilateral in which PQ is extended till F and QE || RS. If ∠FQE = 18° and ∠RPQ = 97°, find ∠PRS.

A
105°

B
104°

C
101°

D
100°

Solution

Sum of opposite angles of a cyclic quadrilateral is 180°.
∴ ∠RPQ + ∠QSR = 180°
97° + ∠QSR = 180°
∠QSR =180°- 97° = 83°
∵ QE || RS
∴ ∠SQE = ∠QSR = 83° [alternate interior angles]
∴ ∠SQF = ∠SQE + ∠FQE
83° + 18° = 101°
Now, ∠PQS + ∠SQF = 180° [linear pair]
and ∠PQS + ∠PRS = 180° [opposite angles of cyclic quadrilateral]
Thus, ∠PQS + ∠PRS = ∠PQS +∠SQF
∠PRS = ∠SQF
∠PRS = 101° [∵ ∠SQF = 101°]
Question 16
1 / -0

In the given figure, AB is the diameter and the chord CD is parallel to AB. Find the measure of CEB when BCD = 40°.

A
30°

B
40°

C
50°

D
60°

Solution

BCD = ABC = 40° (Alternate angles as AB is parallel to CD)
ACB = 90° (Angle in a semicircle is a right angled triangle)
In triangle ABC,
ACB + ABC + BAC = 180°
90° + 40° + BAC = 180°
130° + BAC = 180°
BAC = 50°
Now, CAB = CEB (Angles in the same segment are equal)
So, CEB = 50°
Question 17
1 / -0

If AP = PB and O is the centre of the circle with AC as a straight line, find the value of x + y.

A
200°

B
210°

C
220°

D
230°

Solution

Given: AP = PB
As the line drawn through the centre of the circle bisecting the chord is perpendicular to chord,
then, ∠APO = x = 90°
∠COP = y = ∠OAP + ∠APO (Exterior angle of triangle equals sum of two interior angles)
= 30° + 90° = 120°
x + y = 90° + 120° = 210°
Question 18
1 / -0

In the given figure, O is the centre of the circle and AOC = 120°. Find the value of 2x + y.

A
150°

B
300°

C
350°

D
240°

Solution

As we know that angle made by same arc at circle, i.e. on the circumference, is half of the angle made by same arc at centre,
then, x = × 120° = 60°
x + y = 180° (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180° - 60° = 120°
According to the question:
2x + y
= 2 × 60° + 120°
= 240°
Question 19
1 / -0

In the given figure with centre O, find ∠PQR, if AO = OB and ∠QPR = 40°.

A
40°

B
50°

C
70°

D
80°

Solution

PQ = PR (Equal chords are equidistant from centre and AO = OB)
Hence, PQR is an isosceles triangle where ∠PQR = ∠PRQ.
In triangle PQR,
∠PQR + ∠PRQ + ∠QPR = 180° (Sum of the interior angles of triangle is 180°)
2∠PQR = 180° - 40° = 140°
∴ ∠PQR = 140 ÷ 2 = 70°
Question 20
1 / -0

In the given figure, find the measure of ∠XOZ, if ΔXYZ is an equilateral triangle and XYZO is a cyclic quadrilateral.

A
60°

B
80°

C
90°

D
120°

Solution

ΔXYZ is an equilateral triangle.
Therefore, ∠XYZ = 60°
Now, ∠XOZ +∠XYZ = 180° [Sum of the opposite angles of a cyclic quadrilateral is 180°]
∠XOZ + 60° = 180°
∠XOZ =180° - 60°
∠XOZ = 120°
Question 21
1 / -0

Fill in the blanks.

(a) A chord that passes through the centre of a circle is called a ____X_____, and is the longest chord.
(b) A __Y____ of a circle is a straight line segment whose endpoints both lie on the circle.
(c) Central angles subtended by arcs of the same length are __Z___.

A
1

B
2

C
3

D
4

Solution

(a) A chord that passes through the centre of a circle is called a diameter, and is the longest chord.

(b) A chord of a circle is a straight line segment whose endpoints both lie on the circle.

(c) Central angles subtended by arcs of the same length are equal.
Angle DOC = Angle AOB.
Question 22
1 / -0

In the given figure, O is the centre of smaller circle with a chord AB where AC = CB. Find ∠SRQ and ∠RSP.

A
85°, 130°

B
90°, 100°

C
110°, 150°

D
160°, 120°

Solution

AC = CB
∠ACO = 90° (Line drawn through the centre of circle to bisect the chord is perpendicular to the chord)
∠OAC = 180° - ∠ACO - ∠AOC (Sum of interior angles of triangle is 180°)
∠OAC = 180° - 90° - 55° = 35°
In triangle APQ,
∠APQ = 180° - ∠PAQ - ∠PQA (Sum of interior angles of triangle is 180°)
∠APQ = 180° - 35° - 50° = 95°
PQRS is a cyclic quadrilateral.
∠SRQ = 180° - ∠APQ ( Opposite angles of a cyclic quadrilateral are supplementary)
∠SRQ = 180° - 95° = 85°
∠RSP = 180° - ∠ RQP ( Opposite angles of a cyclic quadrilateral are supplementary)
∠RSP = 180° - 50° = 130°
Question 23
1 / -0

In the given figure, O is the centre of the circle with AB as diameter. Which of the following statements is incorrect?

A
∠ACB is 65° more than ∠AED

B
∠AOD is 25° more than ∠AED

C
∠AED + ∠ACB = 105°

D
∠ABC = ∠AOD

Solution

∠ACB = 90° (Angle in the semicircle is a right angle)
∠ABC = 180° - ∠ACB - ∠CAB (Sum of interior angles of triangle is 180°)
∠ABC = 180° - 90° - 40° = 50°
∠AOD = 180° - 130° = 50° (Linear pair)
∠AOD = 2∠AED (Angle subtended by an arc at the centre of circle is double the angle subtended at any other part of the circle)
∠AED = 25°
Hence, ∠AED + ∠ACB = 25° + 90° = 115°
Correct answer is option 3.
Question 24
1 / -0

State 'T' for true and 'F' for false.
(i) The angle subtended by an arc at the centre is half the angle subtended by it at any point on the remaining part of the circle.
(ii) In a circle, an angle between a tangent and a chord from the point where they meet is double the angle in the alternate segment.
(iii) Two tangents drawn on a circle from a point outside are equal in length.

A
1

B
2

C
3

D
4

Solution

(i) ''The angle subtended by an arc at the centre is half the angle subtended by it at any point on the remaining part of the circle.'' This statement is FALSE because 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.

(ii) In a circle, an angle between a tangent and a chord from the point where they meet is double the angle in the alternate segment. This statement is FALSE because angles are equal, which implies angle CEA is equal to angle CDE:

(iii) Two tangents drawn on a circle from a point outside are equal in length. This statement is TRUE.

AP = PB
Question 25
1 / -0

In the given circle with centre O, find ∠ABC if ∠OCB = 55° and AB = AC.

A
70.5°

B
72.5°

C
68.5°

D
60.5°

Solution

OB = OC (Radius of the circle)
Therefore, ∠OBC = ∠OCB = 55°
In ΔOBC,
∠OBC + ∠OCB + ∠BOC = 180° (Sum of the interior angles of triangle is 180°)
55° + 55° + ∠BOC = 180°
∠BOC = 180° - 110°
∠BOC = 70°
Now,
∠BAC = 1/2 of ∠BOC (Angle subtended by arc at the centre of the circle is double the angle subtended at any other part of the circle)
∠BAC = 1/2 of 70° = 35°
In triangle ABC,
∠ABC = ∠ACB (As AB = AC)
∠ABC + ∠ACB + ∠BAC = 180° (Sum of the interior angles of triangle is 180°)
2∠ABC = 180° - 35°
∠ABC = 145 ÷ 2 = 72.5°

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