M.C.Q

MCQ 1521 Mark

In the given figure, O is the centre of a circle. If $\angle\text{OAB}=40^\circ$ and C is a point on the circle, then $\angle\text{ACB}=?$

A
40°
B
50°
C
80°
D
100°

Answer

50°
Solution:
In $\triangle\text{OAB},$
$\text{OA}=\text{OB}$ [Radii of the same circle]
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$ [Angle opposite equal sides are equal]
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 40^\circ+40^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\ \angle\text{AOB}=100^\circ$
We know that, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
So, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$=\frac{1}{2}(100^\circ)$
$=50^\circ$

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MCQ 1531 Mark

In the given figure, O is the centre of a circle in which $\angle\text{OBA}=20^\circ$ and $\angle\text{OCA}=30^\circ.$ Then, $\angle\text{BOC}=?$

A
130º
B
90º
C
50º
D
100º

Answer

100º
Solution:
In $\triangle\text{OAB},$ we have:
OA = OB (Radii of a circle)
$\Rightarrow\angle\text{OAB}=\angle\text{OBA}=20^\circ$
In $\triangle\text{OAC},$ we have:
OA = OC (Radii of a circle)
$\Rightarrow\angle\text{OAC}=\angle\text{OCA}=30^\circ$
Now, $\angle\text{BAC}=(20^\circ+30^\circ)=50^\circ$
$\therefore\angle\text{BOC}=(2\times\angle\text{BAC})=(2\times50^\circ)=100^\circ$
$\Rightarrow\angle\text{BOC}=100^\circ$

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MCQ 1541 Mark

In the given figure, AD is a diameter of the circle with centre O. Chords AB, BC and CD are equal. If $\angle\text{DEF}=110^\circ,$ then $\angle\text{FAB}$ is equal to:

A
130º
B
110º
C
140º
D
120º

Answer

130º
Solution:

Here, given AB = BC = CD
Now, equal chords subtend equal angles at centre. So, $\angle\text{AOC}=\angle\text{BOC}=\angle\text{COD}$
Also, they lie in straight line so, $\angle\text{AOC}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}=60^\circ$
In $\triangle\text{AOB}$
$\text{AO}=\text{OB},\angle\text{OAB}=\angle\text{OBA}$
Since, $\angle\text{AOB}=60^\circ,\angle\text{OAB}=\angle\text{OBA}=60^\circ$
Now, $\angle\text{DOE}=\angle\text{AOB}=60^\circ$ (vertically opposite angle)
In $\triangle\text{DOE},\text{OD}=\text{OE}$ (radius)
so, $\angle\text{ODE}=\angle\text{OED}$
$\triangle\text{DOE},\angle\text{DOE}+\angle\text{ODE}+\angle\text{OED}=180^\circ$
$2\angle\text{ODE}=\angle\text{OED}=180-60=120^\circ$
$\angle\text{ODE}=\angle\text{OED}=60^\circ$
given was, $\angle\text{DEF}=110^\circ,$ so, $\angle\text{OEF}=110-60=50^\circ$
Now, in $\triangle\text{EOF}\ \text{OE}=\text{OF}$
so, $\angle\text{OEF}=\angle\text{OFE}=50^\circ$
In, $\triangle\text{EOF}\ \angle\text{FOE}=180-(50+50)=80^\circ$
Now, $\angle\text{DOE}+\angle\text{FOE}+\angle\text{AOF}=180^\circ$ (All lie on same straight line)
So, $\angle\text{AOF}=180-(80+60)=40^\circ$
Now, in $\triangle\text{AOF}\ \text{AO}=\text{FO}$
SO, $\angle\text{OFA}=\angle\text{OAF}$
In $\triangle\text{AOF},2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
$2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
$\angle\text{OAF}=90-20=70^\circ$
So, $\angle\text{FAB}=\angle\text{FAO}+\angle\text{OAB}$
$=70^\circ+60^\circ=130^\circ$

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MCQ 1551 Mark

In a circle of radius 17cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23cm. If the length of one chord is 16cm, then the length of the other is:

A
15cm
B
30cm
C
34cm
D
23cm

Answer

30cm
Solution:
Given that: Radius of the circle is 17cm, distance between two parallel chords AB and CD is 23cm, where AB = 16cm. We have to find the length of CD.

We know that the perpendicular drawn from the centre of the circle to any chord divides it into two equal parts.
AM = MB = 8cm
Let OM = x cm ⇒ ON = 23 - x
In right angled triangle OMB,
$\text{x}=\sqrt{172-82=15}$
Now, in triangle OND, ON = (23 - x)cm = (23 - 15)cm = 8cm
$\text{ND}=\sqrt{\text{OD}2-\text{ON}2}$
$\Rightarrow\text{ND}=\sqrt{172-82=15}$
Therefore, the length of the other chord is
CD = 2 × 15 = 30cm

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MCQ 1561 Mark

In the given figure, O is the centre of the circle. $\angle\text{OAB}$ and $\angle\text{OCB}$ are 40º and 30º respectively. Then, the measure of $\angle\text{AOC}$ is:

A
120º
B
170º
C
110º
D
140º

Answer

140º
Solution:

OA = OB = OC = Radius
So, $\angle\text{OAB}\angle\text{ABO}=40^\circ$
and, $\angle\text{OCB}=\angle\text{ABO}=30^\circ$
Thus, $\angle\text{ABC}=30+40=70^\circ$
So, $\angle\text{AOC}=70\times2=140^\circ$ {Angle subtended by arc at centre is twice of angle subtended at circumference}

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MCQ 1571 Mark

In the given figure, M, A, B and N are points on a circle having centre O. AN and MB cut at Y. If $\angle\text{NYB}=50^\circ$ and $\angle\text{YNB}=20^\circ,$ then reflex $\angle\text{MON}$ is equal to:

A
260º
B
200º
C
220º
D
240º

Answer

220º
Solution:

In triangle NYB,
$\angle\text{N}+\angle\text{Y}+\angle\text{B}=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-50^\circ-20^\circ=110^\circ$
Complete the cyclic quadrilateral, MBNX, where X being any point on the circumference in the major segment, we have:-
$\angle\text{MXN}=80^\circ-110^\circ=70^\circ$
So, minor $\angle\text{MON}=70^\circ\times2=140^\circ$
Hence, reflex $\angle\text{MON}=360^\circ-140^\circ=220^\circ$

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MCQ 1581 Mark

An equilateral triangle of side 9cm is inscribed in a circle. The radius of the circle is:

A
$3\text{cm}$
B
$3\sqrt{2}\text{cm}$
C
$3\sqrt{3}\text{cm}$
D
$6\text{cm}$

Answer

$3\sqrt{3}\text{cm}$
Solution:

Let $\triangle\text{ABC}$ be an equilateral triangle.
Let AD be one of its medians.
Then, $\text{AD}\perp\text{BC}$ and BD = 4.5cm
In right $\triangle\text{ADB},$
$\therefore\ \text{AD}=\sqrt{\text{AB}^2-\text{BD}^2}$ [By pythagoras theorem]
$=\sqrt{9^2-\frac{9^2}{2}}$
$=\sqrt{81-\frac{81}{2}}$
$=\sqrt{\frac{243}{2}}$
$=\frac{9\sqrt{3}}{2}\text{cm}$
Let G be the centroid of $\triangle\text{ABC}.$
Then, AG : GD = 2 : 1
$\therefore$ radius $=\text{AG}=\frac{2}{3}\text{AD}=\Big(\frac{2}{3}\times\frac{9\sqrt{3}}{2}\Big)=3\sqrt{3}\text{cm}$

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MCQ 1591 Mark

Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11cm, BE = 3cm and DE = 3.5cm, then CD = ?

A
10.5cm
B
9.5cm
C
8.5cm
D
7.5cm

Answer

8.5cm
Solution:
Construction: Join AC.
$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
⇒ AE × BE = DE × CE ...(i)
Then,
AE = AB + BE = 11 + 3 = 14cm, BE = 3cm, CE = (x + 3.5)cm and DE = 3.5cm
So, from (i), we get
14 × 3 = 3.5 × (CD + 3.5)
$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
⇒ 12 = CD + 3.5
⇒ CD = 8.5cm

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MCQ 1601 Mark

In the given figure, AB || || CD and O is the centre of the circle. If $\angle\text{ADC}=25^\circ,$ then the measure of $\angle\text{AEB}$ is:

A
40º
B
25º
C
80º
D
60º

Answer

40º
Solution:

Here, AB || || CD and $\angle\text{ADC}=25^\circ$
So, $\angle\text{DAB}=25^\circ$ (opposite interior angles are equal)
Now, $\angle\text{ADC}=25^\circ,$ so, $\angle\text{AOC}=50^\circ$ (Angle subtended by arc AC at centre is twice the angle subtended at circumference)
Similarly, $\angle\text{DAB}=25^\circ,$ So, $\angle\text{DOB}=50^\circ$(Angle subtended by arc BD at centre is twice the angle subtended at circumference)
$\angle\text{AOB}+\angle\text{DOB}+\angle\text{AOC}=180^\circ$ (All lie in straight line)
$\angle\text{AOB}=180-50=80^\circ$
Now, $\angle\text{AEB}=40^\circ$ (Angle subtended by arc AB at centre is twice the angle subtended at circumference)

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MCQ 1611 Mark

Write the correct answer in the following: In Fig. if $\angle\text{OAB}=40^\circ,$ then $\angle\text{ACB}$ is equal to:

A
50°.
B
40°.
C
60°.
D
70°.

Answer

50°.
Solution:
In $\triangle\text{QAB, OA} = \text{OB}$ [both are the radius of a circle]
$\angle\text{OAB} = \angle\text{OBA}\Rightarrow \angle\text{OBA} = 40^\circ$
[angles opposite to equal sides are equal]
Also, $\angle\text{AOB} = \angle\text{OBA}\Rightarrow \angle\text{BAO} = 180^\circ$
[by angle sum property of a triangle]
$\angle\text{AOB} + 40^\circ + 40^\circ = 180^\circ$
$\Rightarrow\ \angle\text{AOB} = 180^\circ – 80^\circ = 100^\circ$
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\angle\text{AOB} = 2 \angle\text{ACB} \Rightarrow 100^\circ =2 \angle\text{ACB}$
$\angle\text{ACB} = \frac{100}{2} = 50^\circ$

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MCQ 1621 Mark

If AB is a chord of a circle, P and Q are the two points on the circle different from A and B, then:

A
$\angle\text{APB}+\angle\text{AQB}=90^\circ$
B
$\angle\text{APB}+\angle\text{AQB}=180^\circ$ or $\angle\text{APB}+\angle\text{AQB}$
C
$\angle\text{APB}+\angle\text{AQB}=180^\circ$
D
$\angle\text{APB}=\angle\text{AQB}$

Answer

$\angle\text{APB}+\angle\text{AQB}=180^\circ$ or $\angle\text{APB}+\angle\text{AQB}$
Solution:
We are given AB is a chord of the circle; P and Q are two points on the circle different from A and B.
We have the following figure.
Case 1: Consider P and Q are on the same side of AB.

We know that angles in the same segment are equal.
Hence, $\angle\text{APB}=\angle\text{AQB}$
Case 2: Now consider P and Q are on the opposite sides of AB.
In this case, we have the following figure.

Since quadrilateral APBQ is a cyclic quadrilateral.
Therefore,
$\angle\text{APB}+\angle\text{AQB}=180^\circ$ (Sum of the pair of opposite angles of a cyclic quadrilateral is 180º)
Therefore,from case 1 and case 2, $\angle\text{APB}=\angle\text{AQB}$ or $\angle\text{APB}+\angle\text{AQB}=180^\circ$

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MCQ 1631 Mark

In the given figure, BOC is a diameter of a circle with centre O. If $\angle\text{BCA}=30^\circ,$ then $\angle\text{CDA}=?$

A
50º
B
45º
C
30º
D
60º

Answer

60º
Solution:
Angles in a semi-circle measure 90º.
$\therefore\angle\text{BAC}=90^\circ$
In $\triangle\text{ABC},$ we have:
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Angle sum property of a triangle)
$\therefore90^\circ+\angle\text{ABC}+30^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-120^\circ)=60^\circ$
$\therefore\angle\text{CDA}=\angle\text{ABC}=60^\circ$ (Angles in the same segment of a circle)
$\Rightarrow\angle\text{CDA}=60^\circ$

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MCQ 1641 Mark

The radius of a circle is 13cm and the length of one of its chords is 10cm. The distance of the chord from the centre is:

A
$11.5\text{cm}$
B
$12\text{cm}$
C
$\sqrt{69}\text{cm}$
D
$23\text{cm}$

Answer

$12\text{cm}$
Solution:

Let O be the centre of the circle with radius OA = 13cm.
AB is given to be 10cm.
Distance of a point to a line is always perpendicular to the line.
So, $\text{OL}\perp\text{AB}.$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
⇒ AL = LB = 5cm
In right $\triangle\text{OLA},$
OL² = AO² - AL² [By pythagoras theorem]
⇒ OL² = 13² - 5²⇒ OL² = 169 - 25
⇒ OL² = 144
⇒ OL = 12cm

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MCQ 1651 Mark

ABC is a triangle with B as right angle, AC = 5cm and AB = 4cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is:

A
4cm
B
5cm
C
3cm
D
6cm

Answer

6cm
Solution:

In the circle produce CB to P. Here PC is the required chord.
We know that perpendicular drawn from the centre to the chord divide the chord into two equal parts.
So, PC = 2BC
Now in $\triangle\text{ABC}$ apply Pythagoras theorem
AB² + BC² = AC²⇒ BC² = AC² - AB²⇒ BC² = 5² - 4²⇒ BC² = 25 - 16
⇒ BC² = 9
⇒ BC = 3cm
So, PC = 2 × BC
= 2 × 3
PC = 6cm

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MCQ 1661 Mark

How many circle can pass through three non-collinear points?

A
Three
B
One
C
Two
D
Four

Answer

One
Solution:
Only one circle can be drawn from three distinct points. Join any two sets of points. Make their respective perpendicular bisectors. The point where the two bisectors meet act as the centre of such a circle. Taking the distance between the centre and any of the given points as radius, we can draw the circle. Such a circle would definitely pass through the remaining two points as well.

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MCQ 1671 Mark

O is the centre of the given circle. If $\angle\text{APB}=120^\circ$ and $\angle\text{DBC}=25^\circ,$ then the measure of $\angle\text{ADB}$ is equal to:

A
60º
B
120º
C
95º
D
100º

Answer

95º
Solution:

Now, $\angle\text{APB}+\angle\text{CPB}=180^\circ$ (Linear Pair)
$120^\circ+\angle\text{CPB}=180^\circ$
$\angle\text{CPB}=60^\circ$
Now from angle sum property, we can calculate the values of $\angle\text{CPB}$ and we find that $\angle\text{CPB}=95^\circ$
Since, $\angle\text{PCB}=\angle\text{ADB}=95^\circ$

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MCQ 1681 Mark

If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than $\angle\text{AOB} =$

A
60°
B
90°
C
120°
D
None of these.

Answer

60°
Solution:

Chord AB = Chord BC = Chord CD
$\Rightarrow\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$ (equal chords subtend equal angles at the center)
Now, $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{AOB}+\angle\text{AOB}=180^\circ$
$\Rightarrow3\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=60^\circ$

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MCQ 1691 Mark

If two diameters of a circle intersect each other at right angles, then quadrilateral formed by joining their end points is a:

A
Rhombus.
B
Rectangle.
C
Parallelogram.
D
Square.

Answer

Square.
Solution:

AB and CD are diameters of a circle and diameter makes 90° at any point on circle.
$\Rightarrow\angle\text{CAD}=\angle\text{CBD}=\angle\text{BCA}=\angle\text{ADB}=90^\circ$
Also, diagonals AB and CD are perpendicular to each other.
Thus, ABCD is a square.

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MCQ 1701 Mark

In the figure, O is the center of the circle. If $\angle\text{OAB}=40^\circ,$ then $\angle\text{ACB}$ is equal to:

A
60º
B
70º
C
40º
D
50º

Answer

50º
Solution:

In $\triangle\text{AOB}$
$\angle\text{A}=\angle\text{B}=40^\circ$
And $\angle\text{A}+\angle\text{B}+\angle\text{O}=180^\circ$
$\Rightarrow\angle\text{O}=100^\circ$
So, $\angle\text{ACB}=\frac{100^\circ}{2}=50^\circ$

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MCQ 1711 Mark

In the given figure, BOC is a diameter of a circle with centre O. If $\angle\text{BCA}=30^\circ$ then $\angle\text{CDA}=?$

A
30°
B
45°
C
60°
D
50°

Answer

60°
Solution:
Since BOC is a diameter, $\angle\text{BAC}=90^\circ.$
In $\triangle\text{BAC},$
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ [Angle sum property]
$\Rightarrow\ 90^\circ+\angle\text{ABC}+30^\circ=180^\circ$
$\Rightarrow\ \angle\text{ABC}=60^\circ$
Since angles in the same segment of a circle are equal.
$\angle\text{CDA}=\angle\text{ABC}=60^\circ.$

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MCQ 1721 Mark

AB and CD are two equal chords of a circle with centre O such that $\angle\text{AOB}=80^\circ,$ then $\angle\text{COD}=?$

A
100º
B
80º
C
120º
D
40º

Answer

80º
Solution:
Given: AB = CD
We know that equal chords of a circle subtend equal angles at the centre.
$\therefore\angle\text{COD}=\angle\text{AOB}=80^\circ$
$\Rightarrow\angle\text{COD}=80^\circ$

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MCQ 1731 Mark

In the given figure, a circle is centred at O. The value of x is:

A
70º
B
110º
C
125º
D
55º

Answer

110º
Solution:

$\angle\text{ACO}=\angle\text{CAO}=20^\circ$ (because OA = OC)
$\angle\text{OBC}=\angle\text{OCB}=35^\circ$ (because OB = OC)
$\angle\text{ACB}=55^\circ$
$\text{x}=2\angle\text{ACB}=2\times55^\circ=110^\circ$

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MCQ 1741 Mark

ABC is a triangle with B as right angle, AC = 5cm and AB = 4cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is:

A
3cm.
B
4cm.
C
5cm.
D
6cm.

Answer

6cm.
Solution:

AD and AC are radii of same circle and CD is a chord.
Consider $\triangle\text{ABC},$
BC² = (AC)² - (AB)²
=5² - 4² = 25 - 16 = 9
⇒ BC = 3cm
Chord CD = 2 × BC = 6cm

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MCQ 1751 Mark

In the given figure, if $\angle\text{ABC}=45^\circ,$ then $\angle\text{AOC}=$

A
75º
B
45º
C
90º
D
60º

Answer

90º
Solution:
The angle made by an arc at the centre is double the angle made by it on any other point on the circumfrence.

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MCQ 1761 Mark

The whole arc of a circle is called.

A
Semi-circle
B
Segment
C
Circumference
D
Sector

Answer

Circumference
Solution:
Circumference is the total length of the circle or in other words its a perimeter of the circle.

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MCQ 1771 Mark

In the given figure, A and B are the centres of two circles having radii 5cm and 3cm respectively and intersecting at points P and Q respectively. If AB = 4cm, then the length of common chord PQ is:

A
3cm
B
7.5cm
C
9cm
D
6cm

Answer

6cm
Solution:
We know that the line joining their centres is the perpendicular bisector of the common chord.
Join AP.
Then AP = 5cm; AB = 4cm
Also, AP² = BP² + AB² [using pythagoras theorem]
⇒ BP² = AP² - AB²⇒ BP² = 5² - 4²⇒ BP = 3cm
$\therefore$ triangle ABP is a right angled and PQ = 2 × BP = (2 × 3)cm = 6cm

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MCQ 1781 Mark

An angle in the semicircle is:

A
360º
B
None of these.
C
180º
D
90º

Answer

90º
Solution:
The angle in a semicircle is always 90º.

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MCQ 1791 Mark

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle\text{ADC}=140^\circ,$ then $\angle\text{BAC}$ is equal to:

A
75º
B
40º
C
30º
D
50º

Answer

50º
Solution:
In the given quadrilateral,
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$140^\circ+\angle\text{ABC}=180^\circ$
$\angle\text{ABC}=40^\circ$
Since, AB is diameter so ABCD lies in semi-circle.
Thus, $\angle\text{BCA}=90^\circ$
In triangle, ABC,
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{BAC}=180^\circ-40^\circ-90^\circ=180^\circ-130^\circ=50^\circ$
$\angle\text{BAC}=50^\circ$

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MCQ 1801 Mark

Write the correct answer in the following: In Fig. BC is a diameter of the circle and $\angle\text{BAO}=60^\circ.$ Then $\angle\text{ADC}$ is equal to:

A
30°.
B
45°.
C
60°.
D
120°.

Answer

60°.
Solution:
In $\triangle\text{OAB},$ we have
OA = OB [Radii of the same circle]
$\therefore\angle\text{ABO}=\angle\text{BAO}$ [Angles opp. To equal sides are equal]
$\therefore\angle\text{ABO}=\angle\text{BAO}=60^\circ$ [Given]
Now, $\angle\text{ADC}=\angle\text{ABC}=60^\circ$
[$\because\angle\text{ABC}$ and $\angle\text{ADC}$ are angles in the same segment of circle, are equal]
Hence, $\angle\text{ADC}=60^\circ$
So, (c) is the correct answer.

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MCQ 1811 Mark

In the given figure, AOB is a diameter of a circle with centre O such that AB = 34cm and CD is a chord of length 30cm. Then the distance of CD from AB is:

A
8cm
B
15cm
C
18cm
D
6cm

Answer

8cm
Solution:

Construction: Join OC.
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
So, $\text{CL}=\frac{1}{2}\text{CD}=\frac{1}{2}(30)=15\text{cm}$
AB is the diameter.
So, $\text{AO}=\frac{1}{2}\text{AB}=\frac{1}{2}(34)=17\text{cm}.$
In $\triangle\text{OLC},$
OL² = OC² - CL²⇒ OL² = 17²- 15²⇒ OL² = 289 - 225
⇒ OL² = 64
⇒ OL = 8cm

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MCQ 1821 Mark

In the given figure, ABCD is a cyclic quadrilateral in which $\angle\text{BAD}=75^\circ,$ $\angle\text{ABD}=58^\circ$ and $\angle\text{ADC}=77^\circ,$ AC and BD intersect at P. The measure of $\angle\text{DPC}$ is:

A
94º
B
105º
C
92º
D
90º

Answer

92º
Solution:

Since AD acts as a chord also, So, $\angle\text{ABD}=\angle\text{ACD}=58^\circ$
Again as CD also acts as a chord also, therefore,
$\angle\text{DBC}=\angle\text{DAC}$
Now, $\angle\text{ABC}=\angle\text{ABD}+\angle\text{DBC}$
Also, $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-77^\circ=103^\circ$
And therefore
$\angle\text{DBC}=103^\circ-58^\circ=45^\circ$
Hence, $\angle\text{DAC}=45^\circ$
Since,
$\angle\text{DAC}=45^\circ$
So, $\angle\text{CAB}=75^\circ-45^\circ=30^\circ$
But, $\angle\text{CAB}=\angle\text{BDC}$
$\Rightarrow\angle\text{BDC}=30^\circ$
Now, In triangle CPD,
$\angle\text{C}+\angle\text{P}+\angle\text{D}=180^\circ$
$\Rightarrow58^\circ+\angle\text{P}+30^\circ=180^\circ$
$\Rightarrow\angle\text{P}=180^\circ-30^\circ-58^\circ=92^\circ$

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MCQ 1831 Mark

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If $\angle\text{ACD}=25^\circ,$ then $\angle\text{AOD}=?$

A
50°
B
75°
C
90°
D
100°

Answer

75°
Solution:
OB = BC [Given]
$\Rightarrow\ \angle\text{OBC}=\angle\text{BCO}=25^\circ$ [Angles opposite equal sides are equal]
Now,
$\angle\text{OBC}=\angle\text{BOC}+\angle\text{BCO}=25^\circ+25^\circ=50^\circ$
OA = OB [Radii of the same circle]
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$
In $\triangle\text{AOC},$
$\angle\text{AOD}=\angle\text{OAC}+\angle\text{ACO}$
$=\angle\text{OAB}+\angle\text{BCO}$
$=50^\circ+25^\circ$
$=75^\circ$

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MCQ 1841 Mark

In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$

A
45º
B
90º
C
30º
D
60º

Answer

60º
Solution:
We have:
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=(\frac{1}{2}\times90^\circ)=45^\circ$
$\Rightarrow\angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=(2\times30^\circ)=60^\circ$
$\therefore\angle\text{COD}=180^\circ-\angle\text{COA}=(180^\circ-60^\circ)=120^\circ$
$\Rightarrow\angle\text{CAO}=\frac{1}{2}\angle\text{COD}=(\frac{1}{2}\times120^\circ)=60^\circ$
$\Rightarrow\angle\text{CAO}=60^\circ$

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MCQ 1851 Mark

In the given figure, AOB is a diameter of a circle and CD || AB. If $\angle\text{BAD}=30^\circ$ then $\angle\text{CAD}=?$

A
30°
B
60°
C
45°
D
50°

Answer

30°
Solution:
Since AB || CD, $\angle\text{BAD}=\angle\text{CDA}=30^\circ$ [Alternate angles]
Since AOB is a diameter, $\angle\text{ADB}=90^\circ$
$\angle\text{CDB}=\angle\text{CDA}+\angle\text{ADB}=30^\circ$
$\Rightarrow\ \angle\text{CDB}=30^\circ+90^\circ$
$\Rightarrow\ \angle\text{CDB}=120^\circ$
We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{CAB}+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CAD}+\angle\text{DAB}+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CAD}+30^\circ+120^\circ=180^\circ$
$\Rightarrow\ \angle\text{CAD}=30^\circ$

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MCQ 1861 Mark

Two circle are congruent if they have equal.

A
Diameter
B
Chord
C
Radii
D
Secant

Answer

Radii
Solution:

As per theorem,
two circles are congruent only if they have same radii.

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MCQ 1871 Mark

In the given figure, $\angle\text{OPQ}=30^\circ$ and $\angle\text{ORQ}=57^\circ.$ Then, the measure of $\angle\text{POR}$ is:

A
66º
B
33º
C
54º
D
36º

Answer

54º
Solution:
$\angle\text{OPQ}=\angle\text{OQP}=30^\circ$ (Angles of isosceles triangle OPQ)
Also, in triangle OPQ,
$\angle\text{O}+\angle\text{P}+\angle\text{Q}=180^\circ$
$\angle\text{O}=30^\circ+30^\circ=180^\circ$
$\angle\text{O}=180^\circ-60^\circ=120^\circ$
So, $\angle\text{POQ}=120^\circ\ ...(\text{i})$
Again, in triangle ORQ
$\angle\text{R}=\angle\text{Q}=57^\circ$
And $\angle\text{O}+\angle\text{R}+\angle\text{Q}=180^\circ$
$\angle\text{O}+57^\circ+57^\circ=180^\circ$
$\angle\text{O}+180^\circ-114^\circ=66^\circ$
So, $\angle\text{ROQ}=66^\circ\ ...(\text{ii})$
From(1) and (2), we get :-
$\angle\text{POR}=\angle\text{POQ}-\angle\text{ROQ}$
$\Rightarrow\angle\text{POR}=120^\circ-66^\circ=54^\circ$

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MCQ 1881 Mark

ABCD is a cyclic quadrilateral such that $\angle\text{ADB} = 30^\circ$ and $\angle\text{DCA} = 80^\circ,$ then $\angle\text{DAB} =$

A
70°
B
100°
C
125°
D
150°

Answer

70°
Solution:

ABCD is a cyclic Quadrilateral.
Consider $\triangle\text{ABD}$ and $\triangle\text{ABC}.$
Both are on the same base AB and $\angle\text{ADB}$ and $\angle\text{ACB}$ are the angles in the same segment AB.
$\Rightarrow\angle\text{ADB}=\angle\text{ACB}=30^\circ$
$\Rightarrow\angle\text{BCD}=80^\circ+30^\circ=110^\circ$
In a cyclic Quadrilateral, sum of opposite angles is 180°
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$

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MCQ 1891 Mark

In the given figure, AC is a diameter of the given circle and $\angle\text{BCD}=50^\circ$ Then, $\angle\text{EAF}-\angle\text{ABC}$ is equal to:

A
25º
B
15º
C
20º
D
10º

Answer

15º
Solution:

$\angle\text{B}=90^\circ$ (Angle in semicircle)
$\angle\text{BAD}=180^\circ-75^\circ=105^\circ$
$\angle\text{EAF}=\angle\text{BAD}=105^\circ$
$\angle\text{EAF}-\angle\text{ABC}=105^\circ-90^\circ=15^\circ$

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MCQ 1901 Mark

Number of circles that can be drawn through three non-collinear points is:

Answer

1
Solution:

Three non-collinear points make a triangle and there is only one circle that can pass through all three points,
i.e. circumcircle of that triangle.

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MCQ 1911 Mark

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ,$ then $\angle\text{BAC}=?$

A
20º
B
45º
C
60º
D
30º

Answer

30º
Solution:
We have:
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
$\Rightarrow\angle\text{ABC}+120^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\angle\text{ABC}=60^\circ$
Also, $\angle\text{ACB}=90^\circ$ (Angle in a semicircle)
In $\triangle\text{ABC},$ we have:
$\angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow\angle\text{BAC}+90^\circ+60^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=(180^\circ-150^\circ)=30^\circ$
$\Rightarrow\angle\text{BAC}=30^\circ$

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MCQ 1921 Mark

Two chords AB and CD intersect at right angles. If, $\angle\text{BAC}=40^\circ,$ then $\angle\text{ABD}$ is equal to:

A
50º
B
55º
C
45º
D
60º

Answer

50º
Solution:

Let AB and CD intersect at O. Then In the triangle ACO,
$\angle\text{A}+\angle\text{C}+\angle\text{D}=180^\circ$
$40^\circ+\angle\text{C}+90^\circ=180^\circ$
$130^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=50^\circ$
As per the theorem, Angles in the same segment of a circle are equal.
Hence, $\angle\text{ABD}=\angle\text{C}=50^\circ$

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MCQ 1931 Mark

For what value of x in the figure, points A, B, C and D are concyclic?

A
9º
B
10º
C
11º
D
12º

Answer

10º
Solution:

If the quadrilateral ABCD is concyclic, then,
$\angle\text{A}+\angle\text{C}=180^\circ$
80º + 81º + x = 180º
x = 10º

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MCQ 1941 Mark

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance $\frac{\text{r}}{2}$ from O, then $\angle\text{BAO}=$

A
15º
B
60º
C
45º
D
30º

Answer

30º
Solution:

Let OD = r
$\text{OC}=\frac{\text{r}}{2}$
In $\angle\text{OAC}$ and $\angle\text{DAC}$
$\text{SAS}-\angle\text{OAC}\cong\angle\text{DAC}$
Now, in $\angle\text{OAD}$ equilateral
$\angle\text{AOD}=60^\circ$
$\angle\text{CAO}=\angle\text{BAO}=30^\circ$
$\Rightarrow\sin\theta=\frac{\text{r}}{\frac{2}{\text{r}}}=\frac{1}{2}$
$\Rightarrow\theta=30^\circ$

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MCQ 1951 Mark

If A , B, C are three points on a circle with centre O such that $\angle\text{AOB} = 90^\circ$ and $\angle\text{BOC} = 120^\circ,$ then $\angle\text{ABC} =$

A
60°
B
75°
C
90°
D
135°

Answer

75°
Solution:

$\angle\text{AOC}=\angle\text{AOB}+\angle\text{BOC}$
$=90^\circ+120^\circ=210^\circ$
$\angle\text{COA}=360^\circ-210^\circ=150^\circ$
If arc $\widehat{\text{COA}}$ makes 150° at centre, then it will make half angle of the centre at circumference.
$\Rightarrow\angle\text{CBA}$ or $\angle\text{ABC}=\frac{150^\circ}{2}=75^\circ$

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MCQ 1961 Mark

The relation between diameter and radius of a circle is:

A
$\text{d}=2\pi\text{r}$
B
d = 2r
C
r = 2d
D
d = r

Answer

d = 2r
Solution:
Radius is half the length of the diameter, thus diameter is twice the length of the radius.

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MCQ 1971 Mark

Two point on a circle makes the:

A
Diameters
B
Diameter
C
Chord
D
Secant

Answer

Chord
Solution:
A chord is the line joining any two points on the circle.

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MCQ 1981 Mark

Write the correct answer in the following: In Fig. if AOB is a diameter of the circle and AC = BC, then $\angle\text{CAB}$ is equal to:

A
30°.
B
60°.
C
90°.
D
45°.

Answer

45°.
Solution:
As AOB is a diameter of the circle,
$\angle\text{C}=90^\circ$
[$\because$ Angles in a semi-circle is 90°]
Now, AC = BC
$\angle\text{A}=\angle\text{B}$
[$\because$ Angles opposite to equal sides of triangle are equal]
Using angle sum property of a triangle, we have
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\Rightarrow2\angle\text{A}+90^\circ=180^\circ$
$\Rightarrow2\angle\text{A}=90^\circ\Rightarrow\angle\text{A}=90^\circ\div2=45^\circ$
Hence, (d) is the correct answer.

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MCQ 1991 Mark

Angle inscribed in a semicircle is:

A
120º
B
90º
C
75º
D
60º

Answer

90º
Solution:
Angle inscribed in a semicircle is a right angle. Its a given theorem.

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MCQ 2001 Mark

In the given figure, ABCD and ABEF are two cyclic quadrilaterals. If $\angle\text{BCD}=110^\circ$ then $\angle\text{BEF}=?$

A
55°
B
70°
C
90°
D
110°

Answer

110°
Solution:
Since ABCD is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAD}=70^\circ$
Since ABEF is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BEF}=180^\circ$
$\Rightarrow\ 70^\circ+\angle\text{BEF}=180^\circ$
$\Rightarrow\ \angle\text{BEF}=110^\circ$

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Maths M.C.Q