Circles Test

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Circles Test - 7

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Weekly Quiz Competition

Question 1
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 2
1 / -0

In the given figure, AB and AC are two tangents to a circle with centre O.

What is the perimeter of the figure OCAB, if AB = 7 cm and OC = 3 cm?

A
10 cm

B
40 cm

C
14 cm

D
20 cm

Solution

The lengths of the tangents drawn from an external point to a circle are equal.

So, OB = OC = Radius of the circle as O is the centre of the circle and AB = AC
Thus, perimeter of OCAB = OC + CA + AB + BO = (3 + 7 + 7 + 3) cm = 20 cm [1]
Question 3
1 / -0

In the given figure, AB is the diameter of the circle and points C and D are on the circumference, such that = 30° and = 70°. What is the measure of ?

A
40°

B
50°

C
30°

D
90°

Solution

As ABCD is a cyclic quadrilateral.
+ = 180°
= 180° -
= 180° - 70° = 110°
So, in
= 180° - (30° + 110°)
= 180° - 140° = 40°
Answer: (1)
Question 4
1 / -0

In the given figure, O is the centre of the circle, RPM = 30° and TPN = 60°. Then find the measure of PMN.

A
30°

B
60°

C
45°

D
90°

Solution

TPN = PMN (Alternate segment theorem)

So, TPN = 60° = PMN
60° = PMN [1]
Question 5
1 / -0

In the given figure, if PM and PN are tangents to the circle with centre Q, radius = 7 cm and PM = 7 cm, then what is the length of PQ?

A
7 cm

B
7 cm

C
14 cm

D
2 cm

Solution

PM is a tangent and MQ is the radius.
⇒ QM perpendicular to PM
⇒ PMQ is a right-angled Δ.
PQ² = PM² + MQ² (By Pythagoras Theorem)
PQ² = (7)² + (7)² = 2(7)²
PQ² = 98
PQ = 7 cm
Question 6
1 / -0

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

A
50°

B
205°

C
105°

D
52.5°

Solution

By angle sum property of a triangle, in triangle BOC, measure of angle BOC = 100°.
Measure of angle BDC = (Measure of angle BOC)/2 = 100°/2 = 50° (Angle subtended by a chord on the circumference is half the angle subtended by it at the centre of the circle)
Hence, option 1 is correct.
Question 7
1 / -0

In the figure (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T, such that CT becomes a tangent to the circle at point C. If ATC = 30° and ACT = 50°, find the measure of angle ∠BOA.

A
100°

B
110°

C
120°

D
None of these

Solution

According to the given information: CAT = 100° and BAC = 80°

ABC = 50° = ACT (Angles are in the alternate segment)
BCA = 180° - (80° + 50°) = 180° - 130° = 50°
Hence, BOA = 2 × BCA = 100°
Question 8
1 / -0

In the figure given below, AB is a diameter of the circle, TD is a tangent to the circle and AB = 2AD. If AHD = 36°and DBA = 30°, then what is the measure of CDT?

A
100°

B
110°

C
116°

D
126°

Solution

BDA = 90°, AHD = 36°, AB = 2AD
DBA = 30°
So, DCA = 30°(because angles by the same segment at the circumference are equal)
Now, we know that triangle ABD is a right triangle.
So, DAB = 60° and DBA = 30°
TDA = DBA (Angles in alternate segment)
DCA = 30°CDH = 6°(Exterior angle = Sum of interior opposite triangles for triangle DCH)
CDT = 6 + 90 + 30 = 126°
Hence, 126° is the correct answer.
Question 9
1 / -0

In a circle of radius 'a' and centre 'O', an isosceles triangle ABC is inscribed as shown below. Then the length of the equal side of the triangle is

A

B

C

D

Solution

Given: AB = AC
Hence, the perpendicular drawn through A will bisect the base.
Therefore, E is the midpoint.

Join OB

In triangle OEB,

OE = , OB = a

Applying Pythagoras theorem, we get

In triangle AEB,

AE =

Applying Pythagoras theorem, we get

AB =

=

=

=

=
Question 10
1 / -0

In the given figure, O is the centre of the circle and PQ is a tangent to the circle at P.

If OP = 3 cm and PQ = 12 cm, then what is the length of OQ?

A
3 cm

B
21 cm

C
9 cm

D
cm

Solution

The angle between the radius and the tangent of a circle is always 90°.
So, OPQ = 90°
POQ is a right-angled triangle.
By Pythagoras theorem,
OQ² = OP² + PQ²
OQ² = = 45 cm² + 144 cm² OQ = = 3cm
So, the length of OQ is 3cm.
Question 11
1 / -0

PA and PB are two tangents to a circle with centre O. If chord AB intersects OP at a point R, then which of the following relations is true?

A
AO = OR

B
OA² + AR² = OR²

C
OR² + OA² = AR²

D
OR² + AR² = OA²

Solution

A line drawn through the centre of a circle, which bisects a chord, is perpendicular to the chord.

Thus, OR AB
In the given figure, ARO = 90^o
ARO is a right-angled triangle.
Hence, OR² + AR² = OA² (by Pythagoras theorem)
Question 12
1 / -0

From an external point A, AB and AC are two tangents drawn to a circle with centre O. If BAC : BOC = 2 : 3, what is the measure of BOC?

A
45°

B
90°

C
108°

D
135°

Solution

If from an external point A, AB and AC are two tangents drawn to a circle with centre O, then OA bisects A.
Also, the angle between the radius of the circle and the tangent to the circle is 90°.

Thus, in the given figure,
ABO = 90°and ACO = 90°
Let BAC = 2x and BOC = 3x
Thus, in quadrilateral ABOC,
BAC + ABO + ACO + BOC = 360°
2x + 90° + 90° + 3x = 360°
5x = 360° - 180° = 180°
x = = 36°
BOC = 3 × 36° = 108°
Question 13
1 / -0

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes tangent to the circle at point C. If = 30° and = 50°, then measures

A
100°

B
150°

C
130°

D
140°

Solution

In triangle ACT,
= 50°, = 30° = 100°
Applying tangent secant theorem,
∠ABC = ∠ACT = 50°; and since is the external angle of the triangle ABC, ∠BCA = ∠CAT - ∠ABC = 50°.
= 100° (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 14
1 / -0

From a point P, a tangent is drawn to a circle. If the length of the tangent is 3 cm and the distance of point P from the centre of the circle is 5 cm, find the radius of the circle.

A
3 cm

B
4 cm

C
5 cm

D
6 cm

Solution

Let the tangent drawn from point P to the circle with centre O, meets the circle at point A.
AO is the radius of the circle.

Tangent to the circle makes a right angle with the radius of the circle.
So, APO is right angled triangle
Now, by applying Pythagoras theorem, we get
(OP)² = (PA)² + (OA)²
25 = 9 + (OA)²
= OA
OA = 4 cm
Question 15
1 / -0

In the given figure, if POS = x°, ROQ = y°, POQ = a°, ROS = b°, PS = RQ and RS = PQ, then which of the following relations is definitely true?

A
x = b

B
y = b

C
a = y

D
x = y

Solution

POS = x, ROQ = y, POQ = a, ROS = b, PS = RQ and RS = PQ (Given)

Now, equal chords of a circle subtend equal angles at the centre.
Therefore, x = y.
Hence, answer option 4 is correct.
Question 16
1 / -0

The number of tangents drawn from a point outside a circle is

A
2

B
3

C
4

D
1

Solution

We can draw only two tangents from a point outside a circle.
Question 17
1 / -0

PA and PB are two tangents to a circle with centre O. If AOP : OPB = 13 : 5, find the measure of OPA.

A
20°

B
25°

C
45°

D
15°

Solution

If AP and PB are tangents to the circle with centre O, then OP bisects P, and the angle between the tangents to the circle and the radius of the circle is 90°.

In the given figure, AOP : OPB = 13 : 5
So, AOP = 13x and OPB = 5x
PAO = 90° [Angle between the radius of the circle and the tangent to the circle]
Also, APO = OPB = 5x [ OP bisects P]
Now, in ΔAPO,
APO + AOP + PAO = 180°
5x + 13x + 90° = 180°
18x = 180° - 90° = 90°
x =
∴ APO = 5 5° = 25°
Question 18
1 / -0

Two circles 'a' and 'b' having equal radius 'k' intersect each other in such a way that each circle passes through the centre of the other circle. What is the ratio of the length of the common chord to the total circumference of both the circles?

A

B

C

D

Solution

Let AB be the common chord.
Here, radius of each circle = k
O' and O are the centres of both circles.
O'AO is an equilateral triangle with side k.
Its height will be k.
So, length of the chord = = k
So, required ratio = =
Question 19
1 / -0

In the given figure, OTQ = 36°,what is the measure of TPQ?

A
90°

B
72°

C
18°

D
144°

Solution

OT PT and OQ PQ (Radius is perpendicular to tangent at the point of contact)
By angle sum property of a quadrilateral, TOQ + TPQ = 180°......(i)
In triangle OTQ, which is an isosceles triangle,
2OTQ + TOQ = 180° ....... (ii)
Using (i) and (ii)
TPQ = 2OTQ
TPQ = 2 36° = 72°
Question 20
1 / -0

Using the given figure, find the following:

(i) SNR
(ii) SPQ

A
(i) 40°, (ii) 35°

B
(i) 25°, (ii) 45°

C
(i) 35°, (ii) 40°

D
(i) 45°, (ii) 35°

Solution

In the given figure,
PSR + PQR = 180° (Sum of the opposite angles of a cyclic quadrilateral is 180°.)
75° + PQR = 180°
PQR = 105°
In PSM,
PSM + SPM + PMS = 180°
75° + SPM + 65° = 180°
SPM = 180° – 140°
SPQ = 40°
SPQ + SRQ = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
40° + SRQ = 180°
SRQ = 140°
SRQ + SRN = 180° (Linear pair)
140° + SRN = 180°
SRN = 40°
Similarly, NSR = 180° – 75° = 105°
In NSR,
NSR + SRN + SNR = 180°
105° + 40° + SNR = 180°
SNR = 180 – 145°
SNR = 35°
Question 21
1 / -0

In the given figure, ∠DBC = 38° and O is the centre of the circle. Find the measure of angle x.

A
128°

B
132°

C
136°

D
142°

Solution

AC is a diameter of the circle.
∠ABC = 90° (Angle in the semicircle)
∠ABD + ∠DBC = ∠ABC
∠ABD + 38° = 90°
∠ABD = 52°
∠ABD + x = 180° (Interior opposite angles of a cyclic quadrilateral have sum 180°.)
x = 180° – 52° = 128°
Question 22
1 / -0

The length of a chord QR is 16 cm and tangents QT and RT are 17 cm long. Find the diameter of the circle if the distance of point T from the centre of the circle is 21 cm.

A
15 cm

B
18 cm

C
20 cm

D
24 cm

Solution

Given: Chord QR = 16 cm
Draw OS ⊥QR.
⇒ RS = SQ =× 16 = 8 cm (Perpendicular from the centre on a chord bisects the chord.)
Distance of point T from the centre of the circle, OT = 21 cm
Length of tangent QT = RT = 17 cm
In ΔQST,
QT²= QS²+ ST²
⇒ ST² = QT² – QS²
⇒ ST²= 17² – 8²=> ST²= 289 – 64
⇒ ST² = 225
⇒ ST = 15 cm
Now, OT = OS + ST
⇒ OS = OT – ST
⇒ OS = 21 – 15
⇒ OS = 6 cm
Now, in ΔOSR,
OR² = OS² + SR²
⇒ OR² = 6² + 8²⇒ OR²= 36 + 64
⇒ OR²= 100
⇒ OR = 10 cm
Diameter of the circle = 2 × radius
⇒ Diameter = 2 × 10 = 20 cm

Question 23

1 / -0

Match the columns.

Column I	Column II
(a) In the given figure, O is the centre of the circle and tangent PQ touches the circle at point P. PQ = 8 cm, find the length of the chord AB.	(i) cm
(b) In the given figure, O is the centre of the circle. Tangent XY touches the circle at point Z. Find the length of XZ.	(ii) cm
(c) In the given circle with centre O, YZ is a tangent to the circle which touches the circle at Z. AB is a chord such that area of triangle OAZ = cm². YZ = 6 cm and OY = 10 cm. Find the length of the chord AB.	(iii) cm

a - ii, b - i, c - iii

a - ii, b - iii, c - i

a - iii, b - ii, c - i

a - iii, b - i, c - ii

Solution

Part a:

∠OPQ = 90° (radius is perpendicular to tangent at the point of contact)
OP² = OQ² - QP²
OP² = 10² - 8²
OP² = 100 - 64 = 36
OP = 6 cm
OC = 6 - 2 = 4 cm
AC² = OA² - OC²
AC² = 6² - 4²
AC² = 36 - 16 = 20
AC = cm
C is the midpoint of the chord AB as ∠C = 90° (radius is perpendicular to the midpoint of the chord)
AB = cm

Part b:
MP² = OM² - OP²
MP² = 5² - 3²
MP² = 25 - 9
MP² = 16
MP = 4 cm

∠Z = 90° (radius is perpendicular to tangent at the point of contact)
So MN is parallel to XY
Triangle OMP is similar to triangle OXZ as ∠O is common, ∠Z = ∠P
Therefore:

XZ = = cm

Part c:
YZ is a tangent , so ∠YZO = 90°
OZ² = OY² - YZ²
OZ² = 10² - 6²
OZ² = 100 - 36
OZ = 8 cm
Area of triangle OAZ = 1/2 × OZ × AX = cm²AX = cmAs OX is perpendicular to AB, X is the midpoint of AB. (radius is perpendicular to the midpoint of the chord)AB = 2 × AX = cm

Question 24
1 / -0

In the following figure, there are two concentric circles. If MN is the radius of the smaller circle and AB is a chord in the bigger circle which touches the smaller circle at N, then what will be the length of AB given that the radii of the two circles are 7.5 cm and 4.5 cm?

A
8 cm

B
10 cm

C
12 cm

D
14 cm

Solution

We can see that the chord AB is tangent to the bigger circle.
Also, MN is the radius of the smaller circle.
Therefore, MN ⊥ AB (Radius of a circle is always perpendicular to the tangent at the point of contact.)
Now we can join MA and MB as shown below.

We can see that △MNA is a right-angled triangle.
According to Pythagoras theorem in △MNA,
MN² + NA² = AM²
4.5² + NA² = 7.5²
NA² = 56.25 - 20.25
NA² = 36
NA = 6 cm
In △MAB, MN ⊥ AB
AN = NB (Perpendicular from the centre of the circle bisects the chord.)
AB = AN + NB
AB = 6 + 6 = 12 cm
Question 25
1 / -0

The given circle with centre O has three tangents AB, BC and MN. Find the measure of ∠ABC.

A
20°

B
10°

C
28°

D
30°

Solution

In the given figure,
∠BRO = ∠OTP = 90° (Radius of a circle is perpendicular to the tangent at the point of contact.)
∠RPT = ∠MPB = 180° - 80° = 100° ( Vertically opposite angles)
In quadrilateral ORPT,
∠ROT = 360° - 90° - 90° - 100° = 80°
In triangle ROB,
∠RBO = 180° - 90° - 80° = 10°
As B is the common point of tangents CB and AB, OB bisects ∠ABC.
∠ABC = 2 × ∠RBO = 2 × 10° = 20°

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Re-Attempt Weekly Quiz Competition

Circles Test - 7

Result

Circles Test - 7

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SHARING IS CARING

Question 1

5 units

2.5 units

10 units

15 units

Solution

Question 2

10 cm

40 cm

14 cm

20 cm

Solution

Question 3

40°

50°

30°

90°

Solution

Question 4

30°

60°

45°

90°

Solution

Question 5

7 cm

7 cm

14 cm

2 cm

Solution

Question 6

50°

205°

105°

52.5°

Solution

Question 7

100°

110°

120°

None of these

Solution

Question 8

100°

110°

116°

126°

Solution

Question 9

Solution

Question 10

3 cm

21 cm

9 cm

cm

Solution

Question 11

AO = OR

OA2 + AR2 = OR2

OR2 + OA2 = AR2

OR2 + AR2 = OA2

Solution

Question 12

45°

90°

108°

135°

Solution

Question 13

100°

150°

130°

OA² + AR² = OR²

OR² + OA² = AR²

OR² + AR² = OA²