Circles Test

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Circles Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, EC is a tangent to the circle touching it at point B. Find the measure of ACB.

A
35°

B
45°

C
50°

D
55°

Solution

In ΔABD,
ADB + BAD + ABD = 180° [Angle sum property]
75° + 40° + ABD = 180°
ABD = 180° - 115°
ABD = 65°

Also, DBC = 40° [Alternate segment theorem]
ABC = ABD + DBC
ABC = 65° + 40°
ABC = 105°

In ΔABC,
ABC + BAC + ACB = 180° [Angle sum property]
105° + 40° + ACB = 180°
ACB = 180° - 145°
ACB = 35°
Question 2
1 / -0

In the given figure with center O, PQ, QR = 16 cm, RS = 18 cm and SP = 14 cm are tangents to the circle with radius 6 cm touching at points J, K, L and M, respectively. If PJ : QJ = 2 : 1, what will be the length of OP?

A
4 cm

B
7.21 cm

C
8 cm

D
10 cm

Solution

From the question figure we can make a new figure as follows:

Tangents drawn from an external point to a circle are equal.
So,
PJ = PM = w cm
SM = SL = x cm
RL = RK = y cm
QK = QJ = z cm
Also,
RS = x+ y = 18 cm ... (i)
QR = y + z = 16 cm ... (ii)
SP = x + w = 14 cm … (iii)
PQ = z + w … (iv)
So, from (i) and (ii),
x + y – (y + z) = 18 – 16
x – z = 2 cm
x = z + 2 cm ... (v)

From (iii) and (v),
(z + 2) + w = 14 cm
z + w = 14 – 2
PQ = z + w = 12 cm
As PJ : QJ = 2 : 1 = 2n : n
Also, w : z = 2n : n
And PJ + QJ = PQ = 12 cm
2n + n = 12 cm
3n = 12
n = 4
So, PJ = w = 2 × 4 = 8 cm
Now, △JOP is a right-angled triangle at ∠OJP.
OP² = PJ² + OJ²
OP² = 8² + 6²
OP² = 64 + 36
OP = 10 cm
Question 3
1 / -0

In the given figure, A and B are centres of the two circles where the area of the bigger circle is 4 times the area of the smaller circle. CD is a common tangent to both the circles at point E, and E is the midpoint of CD. If AC = 10 cm and CE = 8 cm, find the area of DBC.

A
88 cm²

B
96 cm²

C
112 cm²

D
144 cm²

Solution

AEC = BEC = 90° (The tangent at any point of a circle is perpendicular to the radius through the point of contact)
In ,
AE² = AC² – CE²
AE² = 10² – 8² = 100 – 64
AE² = 36
AE = 6 cm
The area of the bigger circle is 4 times the area of the smaller circle.
Hence, the radius of the bigger circle is double the radius of the smaller circle.
Radius of the smaller circle = AE = 6 cm
Radius of the bigger circle = BE = 12 cm
CD = 2 × CE = 16 cm
Area of =× CD × BE =× 16 × 12 = 96 cm²
Question 4
1 / -0

What will be the measure of ∠BAO in the given figure with O as centre?

A
16°

B
18°

C
28°

D
36°

Solution

∠BCD = 72°
So, ∠BOD = 2 × 72° [Angle subtended by any arc at the centre of the circle is double the angle subtended by that arc at the edge of the circle]
∠BOD = 144°
As the radius is always perpendicular to the tangent of the circle,
BO ⊥ AB and DO ⊥ AD
∠ABO = ∠ODA = 90°
In quadrilateral ABOD,
∠ABO + ∠BOD + ∠ODA + ∠DAB = 360°
90° + 144° + 90° + ∠DAB = 360°
∠DAB = 360° - 324°
∠DAB = 36°
Also, line OA bisects ∠DAB, as △ABO and △ADO are similar and OA is the common base.
So, ∠BAO = ∠DAB ÷ 2
∠BAO = 36 ÷ 2
∠BAO = 18°
Question 5
1 / -0

What will be the measure of ∠BAE in the given figure with centre O, if AB and AE are tangents and CD is the diameter of the semicircle?

A
60°

B
80°

C
100°

D
120°

Solution

In △BEO,
∠BEO = ∠EBO = 30° [Because OE = OB]
∠BEO + ∠EBO + ∠EOB = 180°
30° + 30° + ∠EOB = 180°
∠EOB = 180° - 60°
∠EOB = 120°
Also, ∠ABO = ∠AEO = 90° [Tangent is perpendicular to the radius of circle]
So, in quadrilateral ABOE,
∠ABO + ∠AEO + ∠EOB + ∠BAE = 360°
90° + 90° + 120° + ∠BAE = 360°
∠BAE = 360° - [90° + 90° + 120°]
∠BAE = 360° - 300°
∠BAE = 60°
Question 6
1 / -0

Two intersecting circles of radii cm and 50 cm have a common chord LM. The distance between the centres of the circles is 70 cm. The chord LM intersects the line OX joining the centres of circles in ratio 3 : 4. Find the length of chord LM provided O and X are centre of the respective circles.

A
60 cm

B
55 cm

C
50 cm

D
45 cm

Solution

N is the mid-point of LM.
So, LN = NM
Also, OX = 70 cm
Here, ON : NX = 3 : 4
3x + 4x = 70
x = 10cm
ON : NX = 30 : 40
In △LNX,
LX² = LN² + NX² (Pythagoras theorem)50² = LN² + 40²LN² = 2500 – 1600
LN² = 900
LN =
LN = 30 cm
As LN = NM
LM = LN + NM
LM = 30 + 30
LM = 60 cm So, length of the common chord = 60 cm
Question 7
1 / -0

DECF is a cyclic quadrilateral where CD is the diameter of the circle and AB is a tangent to the circle. If ∠BCE is 55°, then what will be the value of ∠CDE?

A
70°

B
65°

C
60°

D
55°

Solution

AB is a tangent and CD is the diameter of the circle.
CD ⊥ AB
∠DCB = 90°
∠DCB = ∠ECB + ∠ECD
90° = 55° + ∠ECD
∠ECD = 35°
Also, ∠CED = 90° (Angle subtended by the diameter is always a right angle)
In ΔCED,
∠CED + ∠ECD + ∠CDE = 180°(Angle sum property)
90° + 35° + ∠CDE = 180°
∠CDE = 180° – (90°+ 35°)
∠CDE = 180° – 125°
∠CDE = 55°
Question 8
1 / -0

In the following figure, the radius of the circle with centre O is extended to meet a tangent to the circle at a specific point T. Line OT is 3 times the radius of the same circle. What will be the length of the tangent if the radius of the circle is 15 cm?

A
44 cm

B
43.43 cm

C
42.43 cm

D
42 cm

Solution

Radius of the circle = 15 cm

Length of OT = 3 × 15 = 45 cm
If we join points O and R, we will have a right-angled triangle.

OR = 15 cm
And ∠ORT = 90° (Radius is always perpendicular to the tangent of circle)
So, by using Pythagoras theorem,
OT² = TR²+ OR²
45² = TR² + 15²
2025 = TR² + 225
TR² = 2025 – 225
TR =
TR = 42.43 cm
So, length of the tangent TR = 42.43 cm
Question 9
1 / -0

Three tangents of a circle are joined to make a triangle around the circle. The points where the circle is touching the tangents are P, Q and R; and the vertices where the tangents meet are L, M and N. What will be the value of , if the tangents LM, MN and NL measure 35 cm, 30 cm and 25 cm, respectively?

A
6.5 cm

B
7.5 cm

C
8.5 cm

D
4.5 cm

Solution

According to the question:

So, from the given information,

LP + PM = 35 cm … (i)
MQ + QN = 30 cm … (ii)
NR + RL = 25 cm … (iii)
Also, PM = MQ
QN = NR
RL = LP
Equations (ii) and (iii) can be written as:
PM + QN = 30 cm … (iv)
QN + LP = 25 cm … (v)
Adding equations (i), (iv) and (v),
2(LP + PM + QN) = 90 cm
LP + PM + QN = 45
Dividing both sides by 6,

Also, = = 7.5 cm
Question 10
1 / -0

What will be the measure of ∠QST in the given figure with centre O, if ∠TOQ = 140°?

A
110°

B
100°

C
90°

D
70°

Solution

∠TOQ = 2∠TPQ (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)
∠TPQ =
∠TPQ =
∠TPQ = 70°
Now, in cyclic quadrilateral PQST,
∠TPQ + ∠QST = 180° (Opposite angles of a cyclic quadrilateral are always supplementary, means they will add up to 180°)
70^o + ∠QST = 180°
∠QST = 180° - 70^°∠QST = 110°
Question 11
1 / -0

In the given figure, find the measures of ∠ADB and ∠EOF, if ∠ACH = ∠BCI = 15° and line CO is the angle bisector of ∠ACB. (Point O is the centre of the circle.)

A
60° and 120°

B
90° and 120°

C
90° and 180°

D
60° and 90°

Solution

Given: ∠ACH = ∠BCI = 15° and line CO is the angle bisector of ∠ACB.
Therefore,
∠ACO = ∠BCO =× ∠ACB ... (1)
We know that the angles in the semicircle are right angles.
Therefore, ∠ACB = ∠ADB = 90°
So, from equation (1), we get:

∠ACO = ∠BCO =× 90°
∠ACO = ∠BCO = 45°
Now, ∠HCO = ∠ACO - ∠ACH = 45° - 15°
∠HCO = 30°
Similarly, ∠ICO = 30°
∠HCI = ∠HCO + ∠ICO = 30° + 30° = 60°
∠HCI = ∠ECF = 60°
Now, ∠EOF = 2 × 60° = 120°
(The angle which an arc of the circle subtends at the centre is double the angle which it subtends at any point of the remaining part of the circumference)
Hence, option 2 is the answer.
Question 12
1 / -0

In the given figure, ∠BDC = 60° and ∠BEC measures three-quarters of ∠BOC. Point O and point E are the centres of the smaller and the bigger circle, respectively. Find the measure of ∠BFC.

A
120°

B
140°

C
135°

D
155°

Solution

Given: ∠BDC = 60° and ∠BEC measures three-quarters of ∠BOC.
Now,
∠BDC = 60°
∠BOC = 2 × 60° = 120°
(The angle which an arc of the circle subtends at the centre is double the angle which it subtends at any point of the remaining part of the circumference)
Now, ∠BEC measures three-quarters of ∠BOC.
∠BEC =× 120°
∠BEC = 90°
Therefore,
∠BAC =× ∠BEC
(The angle which an arc of the circle subtends at the centre is double the angle which it subtends at any point of the remaining part of the circumference)
∠BAC =× 90°
∠BAC = 45°
Now, we know that the sum of all the angles of a cyclic quadrilateral is 360° and also the sum of opposite angles of a cyclic quadrilateral is 180°.
∠BAC + ∠ABF + ∠ACF + ∠BFC = 360°
45° + (∠ABF + ∠ACF) + ∠BFC = 360°
45° + 180° + ∠BFC = 360°
∠BFC = 360° - 225° = 135°
Hence, option 3 is the answer.
Question 13
1 / -0

In the given figure, find the sum of ∠EBD and ∠COD, if ∠EBO = 15°, ∠BAC = 45° and OD is the median of △BOC. Point O is the centre of the given circle.

A
110°

B
135°

C
115°

D
105°

Solution

Given: ∠EBO = 15° and ∠BAC = 45°
Therefore,
∠BOC = 2 × 45° = 90°
Also, OD is the median of △BOC.
Now, it can be seen that △BOC is an isosceles triangle and OD is the median of △BOC. Then, for isosceles triangle, median, angle bisector and altitude all are the same.
Therefore, it can be concluded that line OD is the angle bisector of ∠BOC.
Therefore,
∠BOD = ∠COD =× 90°
∠BOD = ∠COD = 45°
As for isosceles triangle, median, angle bisector and altitude all are the same.
Therefore,
∠ODB = 90°
Now, we know that the sum of all the angles of a triangle is 180°.
Therefore, in △ODB,
∠OBD + ∠ODB + ∠BOD = 180°
∠OBD + 90° + 45° = 180°
∠OBD = 180° - 135°
∠OBD = 45°
Now,
∠EBD = ∠EBO + ∠OBD = 15° + 45° = 60°
Now, sum of ∠EBD and ∠COD = 60° + 45° = 105°
Hence, option 4 is the answer.
Question 14
1 / -0

From the given figure, find the measure of ∠GBI, if ∠DEA = 30° and ∠ACB = 60°. Point O is the centre of the circle.

A
120°

B
60°

C
30°

D
40°

Solution

Given,
∠DEA = 30°
∠ACB = 60°
Therefore,
∠AOB = 120°
(The angle which an arc of the circle subtends at the centre is double the one it subtends at any point of the remaining part of the circumference.)
In △AOB,
∠AOB + ∠OAB + ∠OBA = 180° (Sum of all the angles of the triangle is 180°.)
(∠OAB = ∠DEA = 30° because alternate interior angles.)
120° + 30° + ∠OBA = 180°
150° + ∠OBA = 180°
∠OBA = 30°
OB is the radius of the circle. A tangent is passed through the point B as shown in the given figure.
Line OB is perpendicular to the tangent which passes through point B.
∠OBF = 90°
∠OBF = ∠OBA + ∠ABF
90° = 30° + ∠ABF
∠ABF = 90° - 30° = 60°
∠GBI = 60° (Vertically opposite angles are equal.)
Hence, option 2 is the answer.
Question 15
1 / -0

In the given figure where point O is the centre of both the circles. If ∠LHK = 45° = ∠LIJ, find: ∠CGD + ∠JKO + ∠LIJ - (∠HKP + ∠COD)

A
5°

B
0°

C
10°

D
15°

Solution

Given,
∠LHK = 45° and
∠LIJ = 45°
Now,
∠HKP = ∠LHK = 45° (Because alternate interior angles are equal)
∠IJQ = ∠LIJ = 45° (Because alternate interior angles are equal)
∠GKO = ∠HKP = 45° (Vertically opposite angles are equal.)
∠GJO = ∠IJQ = 45° (Vertically opposite angles are equal.)
In △KOJ,
∠JKO + ∠KOJ + ∠OJK = 180° (Sum of all the angles of the triangle is 180°.)
45° + ∠KOJ + 45° = 180°
∠KOJ = 180° - 90° = 90°
∠COD = ∠KOJ = 90°
(Vertically opposite angles are equal.)
∠CGD = × ∠COD
= × 90° = 45°
Now,
∠CGD + ∠JKO + ∠LIJ - (∠HKP + ∠COD) = 45° + 45° + 45° - (45° + 90°)
= 135° - 135° = 0°
Hence, option 2 is the answer.
Question 16
1 / -0

In the figure given below, O is the centre of the circle and DE is a tangent to that circle which touches it at point C. Find the measure of ∠BAC.

A
40°

B
45°

C
50°

D
55°

Solution

∠BCE = 180° - 130° = 50° (Linear pair)
∠ACE = 90° (The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
∠ACB = ∠ACE - ∠BCE = 90° - 50° = 40°
∠ABC = 90° (Angle in the semicircle is of 90°.)
In Triangle ABC,
∠ABC + ∠BAC + ∠ACB = 180°
∠BAC = 180° - 90° - 40° = 50°
Question 17
1 / -0

In the given circle with centre O, XY and PQ are tangents which touch the circle at points A and B respectively. Find the measure of ∠PCY.

A
95°

B
100°

C
110°

D
125°

Solution

∠CBO = ∠CAO = 90°
(The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
Minor ∠AOB = 360° - 280° = 80°
In quadrilateral AOBC,
∠AOB + ∠OBC + ∠BCA + ∠CAO = 360°
80° + 90° + ∠BCA + 90° = 360°
∠BCA = 360 ° - 260° = 100°
∠BCA = ∠PCY = 100° (Vertically opposite angles)
Question 18
1 / -0

In the given circle with centre O, AB and CD are tangents to the circle at points Q and P respectively. Tangents intersect each other at point Z. Find the measure of x.

A
50°

B
60°

C
65°

D
75°

Solution

∠PZQ = ∠AZC = 130° (Vertically opposite angles)
PZ = QZ (The lengths of tangents drawn from an external point to a circle are equal.)
Hence, ∠ZPQ = ∠ZQP
In triangle ZPQ
∠ZPQ + ∠ZQP + ∠PZQ = 180°
2 × ∠ZPQ = 180° - 130° = 50°
∠ZPQ = 25°
∠ZPO = 90° (The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
∠QPO = x = ∠ZPO - ∠ZPQ = 90° - 25° = 65°
Question 19
1 / -0

In the given circle with centre O, AB and AD are tangents to the circle at points B and D respectively and DC = BC. Find the measure of ∠ODC.

A
32.5°

B
35°

C
38.5°

D
42°

Solution

∠ODA = ∠OBA = 90° (The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
In quadrilateral ABOD,
∠DOB = 360° - 90° - 90° - 50° = 130°
∠DCB = ∠DOB = 65°
CB = CD
AD = AB (The lengths of tangents drawn from an external point to a circle are equal.)
Therefore, ∠CDA = ∠ABC
In quadrilateral ABCD,
∠DAB + ∠ABC + ∠BCD + ∠CDA = 360°
2∠CDA = 360° - 50° - 65° = 245°
∠CDA = 122.5°
∠ODC = ∠CDA - ∠ODA = 122.5° - 90° = 32.5°
Question 20
1 / -0

In the given figure, O is the centre of the circle and AC and AD are tangents to the circle at points C and D respectively. DB = OC. Find the measure of ∠OCB and ∠COB.

A
9°, 165°

B
12°, 162°

C
16°, 158°

D
19°, 155°

Solution

OD = OC = DB
∠DOB = ∠DBO

∠ADO = ∠ACO = 90° (The tangent at any point of a circle is perpendicular to the radius through the point of contact.)
In triangle DOB,
∠DOB + ∠DBO + ∠ODB = 180°
2∠DBO = 180° - 90°
∠DBO = 45°
∠DBC = 45° + 6° = 51°
In triangle ABC,
∠CAB + ∠ABC + ∠ACB = 180°
∠ACB = 180° - 30° - 51° = 99°
∠OCB = ∠ACB - ∠ACO = 99° - 90° = 9°
In triangle OCB,
∠COB + ∠OCB + ∠OBC = 180°
∠COB = 180° - 6° - 9° = 165°
Question 21
1 / -0

In the given figure, AB and ED are two tangents drawn on two identical circles with centre I and G meeting at one point and ∠DEC and ∠ECD are equal. If ∠FDE = 40°, then find the measure of ∠AEC.

A
50°

B
55°

C
60°

D
65°

Solution

In the figure,
Given:
∠FDE = 40°
In triangle CFD,
∠CDF = 90° (Because angles formed by drawing lines from the ends of the diameter of a circle to its circumference form a right angle.)
Then, ∠CDE = 90° + 40° = 130°
Also, given that:
∠DEC = ∠ECD = x (Let it be x)
In triangle ECD,
∠CDE = 130° and ∠DEC = ∠ECD = x
=> x + x + 130° = 180° (Sum of interior angles of a triangle = 180°)
=> 2x = 180° - 130°
=> 2x = 50°
=> x = 25°
=> ∠DEC = ∠ECD = 25° ------ (1)
Now, we know that tangent is perpendicular to the radius.
So, ∠AED = ∠AEC + ∠DEC = 90°
=> ∠AEC + 25° = 90° (From (1))
=> ∠AEC = 90° - 25°
=> ∠AEC = 65°
Question 22
1 / -0

In the given figure with O as centre, diameter of the circle is 6 cm and AB = BC = 8 cm. Find the length of AO.

A
cm

B
cm

C
cm

D
cm

Solution

Given:
AB = BC = 8 cm
DC = 6 cm
In triangle ACD, using Pythagoras' theorem;
AC² = AD² + CD²
=> AD² = AC² – CD²
=> AD² = (AB + BC)² – CD²
=> AD² = (16)² – (6)²
AD² = 256 – 36
=> AD =
Now, in triangle AOD, using Pythagoras' theorem;
AO² = OD² + AD²
=> AO² = (3)² +
=> AO² = 9 + 220
=> AO² = 229
=> AO = cm

Question 23

1 / -0

Column I	Column II
(a) The measure of ∠PTS, of the given circle is	(i) 80°
(b) In the given figure, the measure of ∠CAD is	(ii) 115°
(c) In the given figure, the measure of angle CBX is	(iii) 65°

	(a)	(b)	(c)
1	(ii)	(iii)	(i)
2	(iii)	(ii)	(i)
3	(i)	(ii)	(iii)
4	(ii)	(i)	(iii)

1

2

3

4 Solution

(a)
In the given figure,

∠PRQ = 50°
=> ∠PSQ = 25° (Angle subtended by (minor arc/major arc) at the centre is twice the angle subtended at any point (on the major arc/minor arc).)
For ∠PTS
Sum of internal angles of a triangle is 180°.
=> ∠PSQ + ∠PTS + ∠SPT = 180°
=> 25° + ∠PTS + 40° = 180°
=> ∠PTS +65° = 180°
=> ∠PTS = 180° - 65°
=> ∠PTS = 115°
(b)

In quadrilateral BCDE,
∠BCD = 45°
∠BCD + ∠BED = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
=> 45° + ∠BED = 180°
=> ∠BED = 180° - 45°
=> ∠BED = 135°
Now, ∠BED + ∠AEB = 180° (Straight line angles)
=> 135° + ∠AEB = 180°
=> ∠AEB = 180° - 135°
=> ∠AEB = 45°
We have,
∠AEB + ∠EBA + ∠BAE = 180° (Sum of interior angles of a triangle is 180°.)
=> 45°+ 70° + ∠BAE = 180°
=> 115° + ∠BAE = 180°
=> ∠BAE = 65°
Since, ∠BAE = ∠CAD, ∠CAD = 45°

(c)
∠ADB = 30° also equals ∠ACB.

And ∠ACB also equals ∠XCB.
So, in triangle BXC we know,
∠BXC = 70°, and Angle XCB = 30°
Now sum of the angles of triangle is 180°.
∠CBX + ∠BXC + ∠XCB = 180°
∠CBX + 70° + 30° = 180°
∠CBX = 80°

Question 24
1 / -0

In the figure with O as centre, AB, BC and AC are three tangents drawn on a circle with centre O. The length BF = 2 units, length DC = 4 units, AB + AC = 50 units and AB = 26 units. Find the length of the perpendicular drawn from the centre of the circle to BC.

A
10 units

B
8 units

C
6 units

D
4 units

Solution

Given AB + AC = 50 units
AB = 26 units
=> AC = 24 units
Also, EC = CD = 4 units
And BE = BF = t (Length of tangents drawn from an external point to a circle are equal.)
AD = AC - CD
=> AD = 24 - 4 = 20 units

We know that,
26 - t = 20
=> 26 - 20 = t
=> t = 6 units
From triangle OBE,
OB² = BE² + OE²
(x + 2)² = 6² + x²
=> x² + 4 + 4x = 36 + x²
=> 4x = 36 - 4
=> 4x = 32
=> x = 8 units
=> OE = 8 units

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Circles Test - 6

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SHARING IS CARING

Question 1

35°

45°

50°

55°

Solution

Question 2

4 cm

7.21 cm

8 cm

10 cm

Solution

Question 3

88 cm2

96 cm2

112 cm2

144 cm2

Solution

Question 4

16°

18°

28°

36°

Solution

Question 5

60°

80°

100°

120°

Solution

Question 6

60 cm

55 cm

50 cm

45 cm

Solution

Question 7

70°

65°

60°

55°

Solution

Question 8

44 cm

43.43 cm

42.43 cm

42 cm

Solution

Question 9

6.5 cm

7.5 cm

8.5 cm

4.5 cm

Solution

Question 10

110°

100°

90°

70°

Solution

Question 11

60° and 120°

90° and 120°

90° and 180°

60° and 90°

Solution

Question 12

120°

140°

135°

155°

Solution

88 cm²

96 cm²

112 cm²

144 cm²