Circles Test

Result

Circles Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

AB is a chord of a circle with centre O and radius 41 cm. If OMAB and OM = 9 cm, Find the length of chord AB.

A
12 cm

B
15 cm

C
24 cm

D
80 cm

Solution

AB is a chord of circle will centre O.
Radius = 41 cm, OM AB, OM = 9 cm
OA = radius = 41 cm
(OA) ² = (OM) ² + (AM) ² (By applying Pythagoras theorem)
(41)² = (9)²+ (AM) ² => AM = 40 cm
AB = 2(AM) [The perpendicular which passes through the centre of circle bisects the chord.]
AB = 2(40) = 80 cm
Correct option is 4.
Question 2
1 / -0

In the following figure, find the value of x.

A
25^o

B
65^o

C
68^o

D
None of these

Solution

AOB = 130^o [Given]
AOB = 2x [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]
x = (AOB) = (130^o) = 65^o
Question 3
1 / -0

P, Q, R, S, T and U are points on the circle shown below and the length of PQR is 6 cm. The length of arc STU is

A
12 cm

B
6cm

C
6 cm

D
Cannot be determined

Solution

Length of PQR = 6 cm
x is not given as the centre of circle.
So, length of PQR = length of STU
[Cannot be proved]
Hence, cannot be determined from the above data.
Question 4
1 / -0

In the given figure, ABO = 60^o. If O is the centre, then ACB is equal to

A
40^o

B
60^o

C
50^o

D
30^o

Solution

Given:ABO = 60^o
AO = OB (radius of the circle)
OAB = ABO [angles opposite to the equal sides are equal]
In triangle AOB,
AOB + OBA + OAB = 180^o
AOB + 60^o + 60^o = 180^o
AOB = 180^o - 120^o = 60^o
AOB = 2ACB [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]
ACB = 60^o/2 = 30^o
Question 5
1 / -0

AB is a chord of length 10 cm of a circle with centre O and radius 13 cm. Find the distance of the chord from the centre.

A
12 cm

B
6 cm

C
cm

D
None of these

Solution

Given: Radius (AO) = 13 cm
AB = 10 cm
AB = 2 AM
[The perpendicular which passes through the centre of a circle bisects the chord.]
10 cm = 2 AM => AM = 5 cm
In triangle AMO, by using Pythagoras theorem, we get
OM² + AM² = OA²
OM² + (5 cm)² = (13 cm)²
OM² = 169 cm² - 25 cm²
OM² = 144 cm²
=> OM = 12 cm
Distance of chord from centre = 12 cm
Question 6
1 / -0

In the given figure, ΔABC is inscribed in a circle with centre O. If ACB = 65°, find ABC.

A
25°

B
35°

C
Cannot be determined

D
None of these

Solution

ACB = 65°
BAC = 90° (Angle in a semicircle is a right angle)
In ABC, by angle sum property,
ABC + BCA + CAB = 180°
ABC + 65° + 90° = 180°ABC = 180° - 155° = 25°
Question 7
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 8
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 9
1 / -0

14 cm long chord of a circle is at a distance of 24 cm from the centre. Find the radius of the circle.

A
5 cm

B
17 cm

C
25 cm

D
30 cm

Solution

OM AB
AB = 2 AM (Perpendicular OM bisects chord AB)
14 cm = 2 AM
AM = 7 cm
In AOM, by applying Pythagoras theorem,
OA² = OM² + AM²
r²=(24)² + (7)²
r = 25 cm
Radius = 25 cm
Question 10
1 / -0

In the following figure if O is the centre of the circle, find ACB

A
55⁰

B
208⁰

C
54⁰

D
None of these

Solution

AOB = 180^o
2 ACB = AOB
[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.]
2 ACB = 108^o
ACB = 54^o
Question 11
1 / -0

C is the centre of the circle as shown in the diagram below. Then x + y is equal to

A
100⁰

B
110⁰

C
130⁰

D
145⁰

Solution

C = 130^o (Vertically opposite angles)
130^o + x^o + C + y^o = 360^o (complete angle)
130^o + x^o + 130^o + y^o = 360^o
x ^o+ y^o = 360^o - 130^o - 130^o= 360^o - 260^o
x^o + y^o = 100^o
Question 12
1 / -0

In the given figure, AOB = 120⁰ and O is the centre of the circle. The value of BCE is

A
80⁰

B
50⁰

C
60⁰

D
40⁰

Solution

∠ADB = ∠AOB/2 = 60^o ( ∵ 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)
∠ADB + ACB = 180^o [Cyclic quadrilateral]
60^o + ACB = 180^o
ACB = 180^o - 60^o = 120^oACB + ECB = 180^o (Complementary angles)
120^o + ECB = 180^o
ECB = 180^o - 120^o = 60^o
Question 13
1 / -0

A chord of a circle is 12 cm in length and its distance from the centre is 7 cm. Find the length of the chord of the same circle which is at a distance of 6 cm from the centre.

A
14 cm

B
18 cm

C
24 cm

D
30 cm

Solution

AB = 12 cm
OA² = OM² + AM² [By Pythagoras theorem in AOM]

AB = 2AM [OM bisects the chord AB]

12 = 2AM
AM = 12/2 = 6 cm

OA² = (7)² + (6)²

=> r =

Now, ON = 6 cm
OP² = PN² + NO²
r² = (PN)² + (6)²

=> = (PN)² + (6)²

PN = = = = 7 cm

Length of chord = 2(PN) = 2 7 cm = 14 cm
Question 14
1 / -0

An equilateral PQR is inscribed in a circle with centre O. Find QOR.

A
60⁰

B
120⁰

C
30⁰

D
None of these

Solution

QPR = 60^o (Each angle of equilateral triangle is 60^o)
QOR = 2 QPR [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.]
QOR = 2 (60^o) = 120^o
Question 15
1 / -0

E and F are two points on a circle O. Point G is inside circle O and point H is outside circle O. Find the degree measure of EHF.

A
60⁰

B
Greater than 60⁰

C
Less than 60⁰

D
Cannot be determined from given data

Solution

Arc EF subtends an angle EGF in side E and another EHE outside the circle. So, there is no relationship.
We cannot determine EHF from the above data.
Question 16
1 / -0

In the given figure, O is the centre and AOB = 80⁰. Find the value of APB.

A
45°

B
90⁰

C
140⁰

D
150°

Solution

AOB = 80^o
AOB + 1 = 360^o
1 = 360 - AOB
1 = 360 - 80^o
1 = 280^o
So, APB = 1
APB = (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.) APB = 140^o
Question 17
1 / -0

Which of the following statement(s) is/are true?

A
Two chords of a circle, equidistant from the centre, are equal.

B
Equal chords in a circle subtend equal angles at the centre.

C
Angle in a semi circle is a right angle.

D
All of the above

Solution

(i) Two chords of a circle, equidistant from the centre, are equal. This is a true statement.

(ii) Equal chords in a circle subtend equal angles at the centre. This is also a true statement.
(iii) Angle in a semi circle is a right angle. This is also a true statement.

i.e. All of the above statements are true.
Question 18
1 / -0

In the given figure, XYZ is inscribed in a circle with centre O. If the length of the chord YZ is equal to the radius of the circle OY, then YXZ =

A
30⁰

B
60⁰

C
80⁰

D
100⁰

Solution

Chord YZ = OY = Radius
So, YZ = OY = OZ
OYZ is an equilateral triangle.

In OYZ,

O= Y = Z = 60^o (All angles are equal in equilateral triangle.)

XYZ = 30^o
Question 19
1 / -0

Triangle PQR is inscribed in a circle. If PQ = 4 cm, then QR is equal to

A
4 cm

B
6 cm

C
8 cm

D
cannot be determined from given data.

Solution

PQ = 4 cm
P = R = X^o
So, PQ = QR = 4 cm
(Sides opposite to equal angles are equal)
Question 20
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°