Maths MCQ Class XI Ch-11 | Conic Section

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 Mathematics MCQ Class XI Chapter-11

Conic Section


MCQ based on conic sections like circle Parabola, Ellipse, Hyperbola. Multiple Choice Questions on conic section chapter 11 class XI strictly according to the CBSE syllabus and pattern.

MCQ BASED ON CIRCLE CLASS XI

Question 1

Find the equation of circle with centre at origin and radius
5 units.
a) x2 + y2 = 25
b) x2 + y2 = 5
c) x2 = 25
d) y2 = 25

Answer a

Question 2

The point (6, 2) lie ___________ the circle x2 + y2 – 2x -4y – 36 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer c

Question 3

Find the equation of circle with centre at (2, 5) and radius
5 units.
a) x2 + y2   + 4x – 10y + 4 = 0
b) x2 + y2  – 4x – 10y +
4 = 0
c) x2 + y2  + 4x + 10y +
4 = 0
d) x2 + y2  + 4x – 10y –
4 = 0

Answer b

Question 4

Find the centre of the circle with equation x2 + y2 – 4x – 10y + 4 =
0.
a) (-2, 5)
b) (-2, -5)
c) (2, -5)
d) (2, 5)

Answer: d
Explanation: Comparing the equation with general
form x2 + y2 + 2gx + 2fy + c =
0, we get
2g = – 4 ⇒ g = – 2
2f = -10 ⇒ f = – 5
c = 4
Centre is at (-g, -f) i.e. (2, 5).

Question 5

Find the radius of the circle with equation x2 + y2 – 4x – 10y + 4 =
0.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 6

The centre of
the circle 4x2 + 4y2 – 8x + 12y – 25 = 0 is

a) (-2, 3)

b) (1, -3/2)

c) (-4, 6)

d) (4, -6)

Answer: (b)
(1, -3/2)

Explanation:

Given circle
equation: 4x2 + 4y2 – 8x + 12y – 25 = 0

⇒ x2 + y2 –
(8x/4) + (12y/4) – (25/4) = 0

⇒ x2 +y2 -2x
+3y -(25/4) = 0 …(1)

As we know that the
general equation of a circle is x2 + y2 + 2gx + 2fy + c = 0,
and the centre of the circle = (-g, -f)

Hence, by comparing
equation (1) and the general equation,

2g = -2,  ⇒ g = -1

2f = 3,    ⇒ f = 3/2

Now, substitute the
values in the centre of the circle (-g, -f), we get,

Centre = (1, -3/2).

Question 7

If a circle pass through (2, 0) and (0, 4) and centre at
x-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c
Explanation: Equation of circle with centre at
x-axis (a, 0) and radius r units is
(x – a)2 + (y)2 = r2
⇒ (2 – a)2 + (0)2 = r2
And (0 – a)2 + (4)2 = r2
⇒ (a – 2)2 = a2 + 42 ⇒ (- 2)(2a
– 2) =16 ⇒ a – 1 = – 4 ⇒ a = – 3


So, r2 = (2 + 3)2 = 5 2  ⇒ r = 5 units.

Question 8
The center of the circle 4x² + 4y² – 8x + 12y –
25 = 0 is?
(a) (2,-3)
(b) (-2,3)
(c) (-4,6)
(d) (4,-6)

Answer a

Question 9

If a circle pass through (4, 0) and (0, 2) and centre at
y-axis then find the radius of the circle.
a) 25 units
b) 20 units
c) 5 units
d) 10 units

Answer: c

Question 10

The point (1, 4) lie ___________ the circle x2 + y2 – 2x – 4y + 2 = 0.
a) inside circle
b) outside circle
c) on the circle
d) either inside or outside

Answer b

Question 11
The radius of the circle 4x² + 4y² – 8x + 12y –
25 = 0 is?
(a) √57/4
(b) √77/4
(c) √77/2
(d) √87/4

Answer c

MCQ BASED ON PARABOLA CLASS XI



Question 1

Find the focus of parabola with equation y2 = 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer c

Question 2

Find the focus of parabola with equation y2 = – 100x.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer d

Question 3

Find the focus of parabola with equation x2 = 100y.
a) (0, 25)
b) (0, -25)
c) (25, 0)
d) (-25, 0)

Answer a

Question 4

The focus of the
parabola y2 = 8x is

a) (0,
2)

b) (2,
0)

c) (0,
-2)

d) (-2,
0)

Answer: (b)
(2, 0)

Explanation:

Given parabola
equation y2 = 8x …(1)

Here, the
coefficient of x is positive and the standard form of parabola is y2 =
4ax …(2)

Comparing (1) and
(2), we get

4a = 8

a = 8/4 = 2

We know that the
focus of parabolic equation y2 = 4ax is (a, 0).

Therefore, the
focus of the parabola y2 =8x is (2, 0).

Question 5

The length of
the latus rectum of x2 = -9y is equal to

a)  3 units

b) – 3 units

c)  9/4 units

d)  9 units

Answer: (d) 9
units

Explanation:

Given parabola
equation: x2 = – 9y …(1)

Since the
coefficient of y is negative, the parabola opens downwards.

The general
equation of parabola is x2 = – 4ay…(2)

Comparing (1) and
(2), we get

-4a = -9

a = 9/4

We know that the
length of latus rectum = 4a = 4(9/4) = 9.

Therefore, the
length of the latus rectum of x2 = -9y is equal to 9 units.

Question 6

The parametric
equation of the parabola y2 = 4ax is

a) x = at; y = 2at

b) x = at2;
y = 2at

c) x = at2;
y2 = at3

d) x = at2;
y = 4at

Answer: (b) x
= at2; y = 2at

Question 7

Find the equation of latus rectum of parabola y2 = 100x.


a) x = 25
b) x = – 25
c) y = 25
d) y = – 25

Answer a

Question 8

Find the equation of latus rectum of parabola x2 = – 100y.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer c

Question 9

Find the equation of directrix of parabola y2=-100x.
a) x = 25
b) x = – 25
c) y = – 25
d) y = 25

Answer a

Question 10
If a parabolic reflector is 20 cm in diameter
and 5 cm deep then the focus of parabolic reflector is
(a) (0 0)
(b) (0, 5)
(c) (5, 0)
(d) (5, 5)

Answer c

Question 11

The equation of parabola with vertex at origin the axis is
along x-axis and passing through the point (2, 3) is
(a) y² = 9x
(b) y² = 9x/2
(c) y² = 2x
(d) y² = 2x/9

Answer b

Question 12

Find the vertex of the parabola y2 =
4ax.
a) (0, 4)
b) (0, 0)
c) (4, 0)
d) (0, -4)

Answer b

Question 13

Find the equation of axis of the parabola y2 = 24x.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer d

Question 14

Find the equation of axis of the parabola x2 = 24y.
a) x = 0
b) x = 6
c) y = 6
d) y = 0

Answer a

Question 15

Find the length of latus rectum of the parabola y2 = 40x.


a) 4 units
b) 10 units
c) 40 units
d) 80 units

Answer c

Question 16

At what point of the parabola x² = 9y is the abscissa three times that of ordinate
(a) (1, 1)
(b) (3, 1)
(c) (-3, 1)
(d) (-3, -3)

Answer b

MCQ BASED ON ELLIPSE CLASS – XI

Question 1

An ellipse has ___________ vertices and ____________ foci.
a) two, one
b) one, one
c) one, two
d) two, two

Answer d

Question 2

Find the coordinates of foci of ellipse (x/25)2 + (y/16)2 = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer a

Question 3
In an ellipse, the distance between its foci is
6 and its minor axis is 8 then its eccentricity is
(a) 4/5
(b) 1/√52
(c) 3/5
(d) 1/ 2

Answer c

Question 4
A rod of length 12 CM moves with its and always
touching the co-ordinate Axes. Then the equation of the locus of a point P on
the road which is 3 cm from the end in contact with the x-axis is
(a) x²/81 + y²/9 = 1
(b) x²/9 + y²/81 = 1
(c) x²/169 + y²/9 = 1
(d) x²/9 + y²/169 = 1

Answer a

Question 5

Find the coordinates of foci of ellipse (x/16)2 + (y/25)2 = 1.
a) (±3, 0)
b) (±4, 0)
c) (0, ±3)
d) (0, ±4)

Answer c

Question 6

What is major axis length for ellipse (x/25)2 + (y/16)2 = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer d

Question 7

For the ellipse
3x2 + 4y2 = 12, the length of the latus rectum is:

a)  2/5
b)  3/5
c)  3
d) 4

Answer: (c) 3

Explanation:

Given ellipse
equation: 3x2 + 4y2 = 12

The given equation
can be written as (x2/4) + (y2/3) = 1…(1)

Now, compare the
given equation with the standard ellipse equation: (x2/a2)
+ (y2/b2) = 1, we get

a = 2 and b = √3

Therefore, a >
b.

If a>b, then the
length of latus rectum is 2b2/a

Substituting the
values in the formula, we get

Length of latus
rectum = [2(√3)2] /2 = 3

Question 8

What is minor axis length for ellipse (x/25)2 + (y/16)2 = 1?
a) 5 units
b) 4 units
c) 8 units
d) 10 units

Answer c

Question 9

What is equation of latus rectums of ellipse (x/25)2 + (y/16)2 = 1?

a) x = ±3
b) y = ±3
c) x = ±2
d) y = ±2

Answer a

Question 10

In an ellipse,
the distance between its foci is 6 and the minor axis is 8, then its
eccentricity is

a) 1/2

b)  1/5

c)   3/5

d)   4/5

Answer: (c)
3/5

Explanation:

Given that the
minor axis of ellipse is 8.(i. e) 2b = 8. So, b=4.

Also, the distance
between its foci is 6. (i. e) 2ae = 6

Therefore, ae = 6/2
= 3

We know that b2 =
a2(1-e2)

b2 =
a2 – a2e2

b2 =
a2 – (ae)2

Now, substitute the
values to find the value of a.

(4)2 =
a2 -(3)2

16 = a2 –
9

a2 =
16+9 = 25.

So, a = 5.

The formula to
calculate the eccentricity of ellipse is e = √[1-(b2/a2)]

e = √[1-(42/52)]

e = √[(25-16)/25]

e = √(9/25) = 3/5.

Question 11

A man running a race course notes that the sum of the
distances from the two flag posts from him is always 10 meter and the distance
between the flag posts is 8 meter. The equation of posts traced by the man is
(a) x²/9 + y²/5 = 1
(b) x²/9 + y2 /25 = 1
(c) x²/5 + y²/9 = 1
(d) x²/25 + y²/9 = 1

Answer d

MCQ BASED ON HYPERBOLA CLASS XI

Question 1

A hyperbola has ___________ vertices and ____________ foci.

a) two, one

b) one, one

c) one, two

d) two, two

Answer d

Question 2

Find the coordinates of foci of hyperbola (x/9)2−(y/16)2=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer a

Question 3
The equation of a hyperbola with foci on the
x-axis is
(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y² = (a² + b²)

Answer b

Question 4

The eccentricity of hyperbola is
a. e =1
b. e > 1
c. e < 1
d. 0 < e < 1
Answer: (b) e > 1

Question 5

Find the coordinates of foci of hyperbola (y/16)2−(/x9)2=1.

a) (±5,0)

b) (±4,0)

c) (0,±5)

d) (0,±4)

Answer c

Question 6

What is transverse axis length for hyperbola (x/9)2−(y/16)2=1?

a) 5 units

b) 4 units

c) 8 units

d) 6 units

Answer d

Question 7

What is conjugate axis length for hyperbola (x/9)2−(y/16)2=1?

a) 5 units

b) 4 units

c) 8 units

d) 10 units

Answer c

Question 8

What is equation of latus rectums of hyperbola (x/9)2−(y/16)2=1?

a) x = ±5

b) y = ±5

c) x = ±2

d) y = ±2

Answer a 

Question 9

The length of
the transverse axis is the distance between the ____.

a)   Two vertices

b)   Two Foci

c)   Vertex and the
origin

d)   Focus and the
vertex

Answer: (a)
Two vertices



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Maths MCQ Class IX Ch-14 | Statistics

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 Mathematics MCQ | Class 09 | Chapter 14
STATISTICS

Multiple Choice Questions (MCQ)

  • MCQ Based on the Data.
  • MCQ  Based on the Mean of Data.
  • MCQ Based on the Median.
  • MCQ Based on the Mode.

Features

  • In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
  • Solution Hints are also given to some difficult problems. 
  • Each MCQ contains four options from which one option is correct. 

Action Plan

  • First of all students should Learn and write all basic points and Formulas related to the Chapter 14 Statistics.
  • Start solving  the NCERT Problems with examples.
  • Solve the important assignments on the Chapter 14 Class IX.
  • Then start solving the following MCQ.

MCQ |Chapter 14 | Statistics | Class IX

Question: 1

The collection of information, collected for a purpose is called:

a) Mean

b) Median

c) Mode

d) Data

Answer: d

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Maths MCQ Class IX Ch-15 | Probability

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       Mathematics MCQ | Class 09 | Chapter 15

PROBABILITY

Multiple Choice Questions (MCQ)

  • MCQ Based on the definition of probability.
  • MCQ  Based on the tossing of coins.
  • MCQ Based on the tossing of die.
  • MCQ Based on the Playing Cards.

Features

  • In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
  • Solution Hints are also given to some difficult problems. 
  • Each MCQ contains four options from which one option is correct. 

Action Plan

  • First of all students should Learn and write all basic points and Formulas related to the Chapter 15 Probability.
  • Start solving  the NCERT Problems with examples.
  • Solve the important assignments on the Chapter 15 Class IX.
  • Then start solving the following MCQ.

MCQ |Chapter 15 | Probability | Class IX

Question 1

1) The probability of
each event, when a coin is tossed for 1000 times with frequencies: Head: 455
& Tail: 545 is:

a)
0.455 & 0.545

b)
0.5 & 0.5

c)
0.45 & 0.55

d)
455 & 545

Answer: a

Question 2

Marks
obtained by a student in a test is shown in the table below.

Test no.

1

2

3

4

5

Marks

81

87

76

70

90

What is the probability that the student has scored more than
80?
a) 3/5
b) 4/5
c) 2/5
d) 1/2

Answer a

Question 3

When a die is thrown, what is
the probability of getting even number

a) 1/6

b) 1/3

c) 1/2

d) 2/3

Answer c

Question 4

The sum of all probabilities equal to:

a)
4

b)
1

c)
3

d)
2

Answer: b

Question 5

3) The probability of each
event lies between:

a)
1 & 2

b)
1 & 10

c)
0 & 1

d)
0 & 5

Answer: c

Question 6

Probability
of impossible event is 

a)
1

b)
0

c) Less
than 0

d)
Greater than 1

Answer
b

Question 7

If P(E) = 0.44, then P(not
E) will be:

a)
0.44

b)
0.55

c)
0.50

d)
0.56

Answer: d

Explanation: We know;

P(E) + P(not E) = 1

0.44 + P(not E) = 1

P(not E) = 1 – 0.44 = 0.56

Question 8

Probability
of certain (sure) event is

a) 1

b) 0

c)
Greater than 1

d)
Less than 0

Answer
a

Question 9

 If P(E) = 0.38, then
probability of event E, not occurring is:

a)
0.62

b)
0.38

c)
0.48

d)
1

Answer: a

Explanation: P(not E) = 1 – P(E) = 1-0.38 = 0.62

Question 10

Two coins are tossed simultaneously.
The probability of getting atmost one head is

a) 1/4

b) 3/4

c) 1/2

d) 1/4

Answer b

Question 11

The probability of drawing
an ace card from a deck of cards is:

a)
1/52

b)
1/26

c)
4/13

d)
1/13

Answer: d

Explanation: There are 4 aces in a deck of card.

Hence, the probability of taking one ace out of 52 cards = 4/52
= 1/13

Question 12

A coin is tossed 1000 times,
if the probability of getting a tail is 3/8, how many times head is obtained.

a) 525

b) 375

c) 625

d) 725

Answer b

Question 13

If the probability of an
event to happen is 0.3 and the probability of the event not happening is:

a)
0.7

b)
0.6

c)
0.5

d)
None of the above

Answer: a

Explanation: Probability of an event not happening = 1 – P(E)

P(not E) = 1 – 0.3 = 0.7

Question 14

A coin is tossed 1000 times,
if the probability of getting a tail is 3/8, how many times head is obtained.

a) 525

b) 375

c) 625

d) 725

Answer c

Question 15

 A dice is thrown. The
probability of getting 1 and 5 is:

a)
1/6

b)
2/3

c)
1/3

d)
1/2

Answer: c

Explanation: The probability of getting 1 and 5 = 2/6 = 1/3

Question 16

In a football match, Ronaldo
makes 4 goals from 10 panalty kicks. The probability of converting a penalty
kick into a goal by Ronaldo, is

a) 1/4

b) 1/6

c) 1/3

d) 2/5

Answer d

Question 17

A batsman hits boundaries
for 6 times out of 30 balls. Find the probability that he did not hit the boundaries.

a)
1/5

b)
2/5

c)
3/5

d)
4/5

Answer: d

Explanation: No. of boundaries = 6

No. of balls = 30

No. of balls without boundaries = 30 – 6 =24

Probability of no boundary = 24/30 = 4/5

Question 18

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a king or queen

a) 1/13

b) 2/13

c) 3/13

d 1/52

Answer b

Question 19

Three coins were tossed
200 times. The number of times 2 heads came up is 72. Then the probability of 2
heads coming up is:

a)
1/25

b)
2/25

c)
7/25

d)
9/25

Answer: d

Explanation: Probability = 72/200 = 9/25

Question 20

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a red face card

a) 1/13

b) 2/13

c) 3/26

d 1/26

Answer c

Question 21

What is the probability of getting an odd number less than 4, if a die is
thrown?

a) 1/6

b) 1/2

c) 1/3

d) 0

Answer: c

Explanation:
Sample space, S = {1, 2, 3, 4, 5, 6}

Favorable
outcomes = {1, 3}

Therefore, the
probability of getting an odd number less than 4 = 2/6 = 1/3.

Question 22

From a deck of 52 shuffled playing
cards, a card is drawn What is the probability of drawing a black king card

a) 1/13

b) 2/13

c) 3/26

d 1/26

Answer d

Question 23

What is the probability of impossible events?

a) 1

b) 0

c) More than 1

d) Less than 1

Answer: b

Explanation: The
probability of an impossible event is always 0.

Question 24

 Performing an event once is called

a) Sample

b) Trial

c) Error

d) None of the above

Answer: b

Explanation:
Performing an event once is called a trial.

Question 25

In a counting from 1 to 10,
what is the probability of finding a prime number

a) 1/2

b)  3/10

c) 2/5

d) 1/10

Answer c

Question 26

 A card is drawn from a well-shuffled deck of 52 cards. What is the probability
of getting a king of the red suits?

a) 3/36

b) 1/26

c) 3/26

d) 1/16

Answer: b

Explanation: In
a pack of 52 cards, there are a total of 4 king cards, out of which 2 are red
and 2 are black.

Therefore, in a
red suit, there are 2 king cards. 

Hence, the
probability of getting a king of red suits = 2/52 = 1/26.

Question 27

In a counting from 1 to 10,
what is the probability of finding a composite number

a) 1/2

b)  3/10

c) 2/5

d) 1/10

Answer a

Question 28

 Find the probability of a selected number is a multiple of 4 from the numbers
1, 2, 3, 4, 5, …15.

a) 1/5

b) 1/3

c) 4/12

d) 2/15

Answer: a

Explanation: S =
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Multiples of 4
from the sample space = {4, 8, 12}

Therefore, the
probability of the selected number is a multiple of 5 is 3/15 = 1/5.

Question 29

In the word MATHEMATICS, what
is the probability of finding a vowel

a) 1/11

b) 2/11

c) 3/11

d) 4/11

Answer d

Question 30 

What is the probability of drawing a queen from the deck of 52 cards?

a) 1/26

b) 1/52

c) 1/13

d) 3/52

Answer: c

Explanation:
Total cards = 52

Number of queens
in a pack of 52 cards = 4

Hence, the
probability of drawing a queen from a deck of 52 cards = 4/52 = 1/13

Question 31

In the word MATHEMATICS, what
is the probability of finding a consonant

a) 5/11

b) 6/11

c) 4/11

d) 7/11

Answer d

Question 32

Which of the following cannot be the probability of an event?

a) 1

b) 0

c) 0.75

d) 1.3

Answer: d

Explanation: The
probability of an event always lies between 0 and 1.

Question 33

There are 4 green and 2 red balls in a basket. What is the probability of
getting the red balls? 

a) 1/2

b) 1/3

c) 1/5

d) 1/6

Answer: b

Explanation:
Total balls = 4 green + 2 red = 6 balls

No. of red balls
= 2.

Hence, the
probability of getting the red balls = 2/6 = 1/3

Question 34

Empirical probability is also known as 

a) Classic probability

b) Subjective probability

c) Experimental probability

d) None of the above

Answer: c

Explanation: Empirical
probability is also known as experimental probability.

Question 35

If two coins are tossed simultaneously, then what is the probability of getting
exactly two tails?

a) 1/4

b) 1/2

c) 1/3

d) None of the above

Answer: a

Explanation: If
two coins are tossed, then the sample space, S = {HH, HT, TH, TT}

Favorable
outcome (Getting exactly two tails) = {TT} 

Therefore, the
probability of getting exactly two heads = 1/4

 


Maths MCQ Class IX Ch-10 | Circle

by

      Mathematics MCQ | Class 09 | Chapter 10

CIRCLE

Multiple Choice Questions (MCQ)

  • MCQ Based on the definition of circle.
  • MCQ  Based on the properties of circle.
  • MCQ Based on the cyclic quadrilateral.

Features

  • In the MCQ given below you find the important MCQ which are strictly according to the CBSE syllabus and are very useful for the CBSE Examinations. 
  • Solution Hints are also given to some difficult problems. 
  • Each MCQ contains four options from which one option is correct. 

Action Plan

  • First of all students should Learn and write all basic points and Formulas related to the Chapter 10 Circle.
  • Start solving  the NCERT Problems with examples.
  • Solve the important assignments on the Chapter 10 Class IX.
  • Then start solving the following MCQ.

MCQ |Chapter 10 | Circle | Class IX

Question 1

The collection of all the points in a plane, which are at a fixed distance from a fixed point in the plane, is called
a) square
b) triangle
c) rectangle
d) circle

Answer d

Question 2

The centre of the circle lies in______ of the circle.
a) Interior
b) Exterior
c) Circumference
d) None of the above
Answer: a

Question 3

A circle divides the plane into __________ parts.
a) two
b) three
c) four
d) five
Answer b

Question 4

The longest chord of the circle is:
a) Radius
b) Arc
c) Diameter
d) Segment
Answer: c

Question 5

From the diagram given below, the shaded area is __________

a) major segment
b) minor segment
c) major sector
d) minor segment
Answer b

Question 6

How many circles can pass through three points which are not collinear?
a) One
b) Two
c) Three
d) Four
Answer a

Question 7

Equal _____ of the congruent circles subtend equal angles at the centres.

a) Segments
b) Radii
c) Arcs
d) Chords
Answer: d
Explanation: See the figure below:

Let ΔAOB and ΔCOD are two triangles inside the circle.

OA = OC and OB = OD (radii of the circle)

AB = CD (Given)

So, ΔAOB ≅ ΔCOD (SSS congruency)

∴  ∠AOB = ∠COD ……….. By CPCT

Hence, this prove the statement.

Question 8

The region between an arc and the two radii, joining the centre to the end points of the arc is called __________
a) arc
b) chord
c) sector
d) area of circle
Answer c

Question 9

When does both segments and both sectors become equal?
a) When two arcs are equal
b) When radius = 2 x chord
c) When two arcs are unequal
d) When chord = radius
Answer a

Question 10

If chords AB and CD of congruent circles subtend equal angles at their centres, then:
a) AB = CD
b) AB > CD
c) AB < AD
d) None of the above
Answer: a
Explanation:
In triangles AOB and COD,
∠AOB = ∠COD (given)
OA = OC and OB = OD (radii of the circle)
So, ΔAOB ≅ ΔCOD. (SAS congruency)
∴ AB = CD (By CPCT)

Question 11
From the diagram given below, if O is the centre of the circle and OL is perpendicular to chord AB, then which of the following is true?

a) AL > LB
b) AL < LB
c) AL ≠ LB
d) AL = LB
Answer d

Question 12

If there are two separate circles drawn apart from each other, then the maximum number of common points they have:
a) 0
b) 1
c) 2
d) 3
Answer: a

Question 13

What is the value of AC if radius of circle is 5cm and OB = 3cm?

a) 5cm

b) 8cm
c) 4cm
d) 2cm

Question 14

The
angle subtended by the diameter of a semi-circle is:

a) 90o
b) 45o
c) 180o
d) 60o
Answer: c

Explanation: Angle in a semi circle is always right

Question 15

Find the value of AB if radius of circle is 10cm and PQ = 12cm and RS = 16cm?

a) 14cm
b) 16cm
c) 4cm
d) 2cm
Answer d

Question 16
If AB and CD are two chords of a circle intersecting at point E, as per the given figure. Then:



a) ∠BEQ > ∠CEQ
b) ∠BEQ = ∠CEQ
c) ∠BEQ < ∠CEQ
d) None of the above
Answer: b

Explanation:

OM = ON …………….. (Equal chords are always equidistant from the centre)

OE = OE  …………….. (Common)

∠OME = ∠ONE …… (Each = 90o)

So, ΔOEM ≅ ΔOEN ……… (by RHS congruence rule)

Hence, ∠MEO = ∠NEO …….. (by CPCT)

∴ ∠BEQ = ∠CEQ

Question 17

Find the radius of circle if AB = 7cm, RS = 6cm and PQ = 8cm.

a) 5cm

b) 8cm
c) 4cm
d) 20cm
Answer a

Explanation:

Let r be the radius of circle.
Since perpendicular from centre to a chord bisects the chord,
AR = RS/2 = 3cm and BP = PQ/2 = 4cm
In ΔBOP, ∠OBP = 90°
By Pythagoras theorem, OP2 = OB2 + BP2
⇒ r2 = x2 + (4)2
⇒ r2 = x2 + 16  ——————- (i)
Similarly, In ΔAOR, ∠OAR = 90°
By Pythagoras theorem, OR2 = OA2 + AR2
⇒ r2 = (7 – x)2 + (3)2
⇒ r2 = (7 – x)2 + 9  ———————— (ii)
From equation i and ii, x2 + 16 = (7 – x)2 + 9
⇒ x2 + 16 = 72 – 2 x 7 x x + x2 + 9
⇒ 2 x 7 x x = 49 – 16 + 9
⇒ x = 3
Substituting value of x in equation i, r2 = 32 + 16
⇒ r = 5cm.




Question 18
If a line intersects two concentric circles with centre O at A, B, C and D, then:
a) AB = CD
b) AB > CD
c) AB < CD
d) None of the above
Answer: a
Explanation: See the figure below:

From the above fig., OM ⊥ AD.

Therefore, AM = MD …….. (1)

Also, since OM ⊥ BC, OM bisects BC.

Therefore, BM = MC ……… (2)

From equation (1) and equation (2)

AM – BM = MD – MC

∴ AB = CD

Question 19
Two circles of radius 13cm and 17cm intersect at A and B. What is the distance between the centres of the circles if AB = 24cm?

a) 15cm
b) 25cm
c) 22cm
d) 14cm
Answer d

Explanation:
Since line joining the centres of the circles bisects the common chord, AC = BC = 12cm.
Also, In ΔAOC, ∠OCA = 90°
By Pythagoras theorem, OA2 = OC2 + AC2
⇒ 152 = OC2 + 122
⇒ OC2 = 225 – 144
⇒ OC = 9cm
Similarly, In ΔAO’C, ∠O’CA = 90°
By Pythagoras theorem, O’A2 = O’C2 + AC2
⇒ 132 = O’C2 + 122
⇒ OC2 = 169 – 144
⇒ OC = 5cm
Now, OO’ = OC + O’C
⇒ OO’ = 9 + 5 = 14cm.

Question 20

In the below figure, the value of ∠ADC is:

a) 60°
b) 30°
c) 45°
d) 55°
Answer: c

Explanation: ∠AOC = ∠AOB + ∠BOC

So, ∠AOC = 60° + 30°

∴ ∠AOC = 90°

An angle subtended by an arc at the centre of the circle is twice the angle subtended by that arc at any point on the rest part of the circle.

So, ∠ADC = 1/2∠AOC

= 1/2 × 90° = 45°

Question 21
In the given figure, find angle OPR.

a) 20°
b) 15°
c) 12°
d) 10°
Answer: d

Explanation: Major arc PR subtend Reflex ∠POR at the centre of the circle and ∠PQR in the remaining part of the circle

So, Reflex ∠POR = 2 × ∠PQR,

We know the values of angle PQR as 100°

So, Reflex ∠POR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160° 

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP = x

By angle sum property of triangle, we know:

∠POR + ∠OPR + ∠ORP = 180°

160° + x + x  = 180°

2x = 180° – 160°

2x = 20° ⇒ x = 10°

Question 22

In the given figure, ∠AOB = 90º and ∠ABC = 30º, then ∠CAO is equal to:

a) 30º
b) 45º
c) 60º
d) 90º
Answer: c
Explanation:

Given that ∠AOB = 90º and ∠ABC = 30º
OA = OB (Radii of the circle)
Let ∠OAB = ∠OBA = x
In the triangle OAB,
∠OAB + ∠OBA + ∠AOB = 180° (By using the angle sum property of triangle)
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = 90°/ 2 = 45°
Therefore, ∠OAB = 45° and ∠OBA = 45°
By using the theorem, “ the angles subtended by arcs at the centre of the circle double the angle subtended at the remaining part of the circle”, we can write
∠AOB = 2∠ACB
This can also be written as,
∠ACB = ½ ∠AOB = (½) × 90° = 45°
Now, apply the angle sum property of triangle on the triangle ABC,
∠ACB + ∠BAC + ∠CBA = 180°
45° + 30° + ∠BAC = 180°
75° + ∠BAC = 180° ⇒ ∠BAC = 180° – 75°
⇒ ∠BAC = 105°
∠CAO + ∠OAB = 105°
∠CAO + 45° = 105°
⇒ ∠CAO = 105° – 45° = 60°
Question 23

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:


a) 30º
b) 40º
c) 50º
d) 80º
Answer: c

Explanation:

Given that ABCD is a cyclic quadrilateral and ∠ADC = 140º.

We know that the sum of opposite angles of a cyclic quadrilateral is 180°.

Hence, ∠ADC + ∠ABC = 180° 

140° + ∠ABC = 180°

∠ABC = 180° – 140° = 40° 

Now  ∠ACB = 90° . . . . . . .  Angles in the semi circle is right

By using the angle sum property of triangle in the triangle, ABC, 

∠CAB + ∠ABC + ∠ACB = 180°

∠CAB + 40° + 90° = 180° 

∠CAB + 130°   = 180° 

∠CAB = 180° – 130°

∠CAB = 50° 

Therefore, ∠CAB or ∠BAC =50°.

Question 24

 In the given figure, if ∠OAB = 40º, then ∠ACB is equal to

a) 40º
b) 50º
c) 60º
d) 70º
Answer: b
Explanation:

Given that ∠OAB = 40º,

In  the triangle OAB, 

OA = OB (radii)

 ∠OAB = ∠OBA  = 40°  …… (∴ angles opposite to equal sides are equal)

Now, by using the angle sum property of triangle, we can write

∠AOB + ∠OBA + ∠BAO = 180° 

∠AOB + 40° + 40° = 180° 

∠AOB = 180 – 80° = 100° 

Since, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle

∠AOB = 2 ∠ACB 

∠ACB = ½ ∠AOB

Hence, ∠ACB = 100°/2 = 50°.

Question 25

In the given figure, if ∠ABC = 20º, then ∠AOC is equal to:


a) 10º
b) 20º
c) 40º
d) 60º
Answer: c
Explanation:
Given that, ∠ABC = 20º.
since the angle subtended by an arc at the centre of the circle is double the angle subtended at the remaining part.
∠AOC = 2∠ABC
∠AOC = 2 × 20° = 40°.


Question 26
In the given figure, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:

a) 2 cm
b) 3 cm
c) 4 cm
d) 5 cm
Answer: a

Explanation:

From the given diagram, we can observe that OC is perpendicular to chord AB. Therefore, OC bisects the chord AB

Hence. AC = CB

Also,  AC + CB = AB

AC + CB = 8

2AC = 8 (Since, AC = CB)

AC = 8/2 = 4 cm

As, the triangle OCA is a right-angled triangle, by using Pythagoras theorem, we can write 

AO2 = AC2 + OC2

52 = 42 + OC2

52 − 42 = OC2

OC2 = 9

OC = 3 cm

As, OD is the radius of the circle, OA = OD = 5cm

CD = OD – OC

CD = 5 – 3 = 2 cm

Hence, the value of CD is equal to 2cm.

Question 27

Find the value of x if ABCD is a cyclic quadrilateral if ∠1 : ∠2 = 3 : 6.

a) 90°
b) 45°
c) 60°
d) 20°
Answer: d

Explanation: Since ABCD is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°
⇒ ∠1 + ∠2 = 180°
⇒ 3k + 6k = 180° ⇒ k = 20°
Now, x = ∠1 [Exterior angle formed when one side of cyclic quadrilateral is produced is equal to the interior opposite angle]
⇒ x = 3k ⇒ x = 60°.

Question 28

If PQRS is a cyclic quadrilateral and PQ is diameter, find the value of ∠PQS.

a) 45°
b) 110°
c) 20°
d) 80°
Answer: c
Explanation: Since PQRS is a cyclic quadrilateral, Sum of either pair of opposite angles of cyclic quadrilateral is 180°. ⇒ ∠QRS + ∠QPS = 180°

 ⇒ ∠QPS = 70°  ————–(i)
Also, ∠PSQ = 90°  [Angle in a semicircle] ———-(ii)
In ΔPSQ,  ∠PSQ + ∠SPQ + ∠SQP = 180°  [Angle sum property of triangle]

⇒ 90° + 70° + ∠PQS = 180°  [from equation i and ii]
⇒ ∠PQS = 20°.

Question 29

In the given figure, BC is the diameter of the circle and ∠BAO = 60º. Then ∠ADC is equal to

a) 30º
b) 45º
c) 60º
d) 120º
Answer: c
Explanation: 

Given that ∠BAO = 60° 

Since OA = OB, ∠OBA = 60°

Then ∠ADC = 60° ………. (Angles in the same segment of a circle are equal).

 Question 30
If RS is equal to the radius of circle, find the value of ∠PTQ.

a) 90°
b) 60°
c) 45°
d) 30°
Answer: b
Explanation: Join OS, OR and RQ
From figure, ∠PRQ = 90° as angle in a semicircle is right angle.
Now, ∠QRT = 180° – ∠PRQ  [Linear Pair]
⇒ ∠QRT = 180° – 90° = 90° ————–(i)
In ΔROS, OS = OR = RS ⇒ ΔROS is an equilateral triangle.
⇒ ∠ROS = 60°
Now, ∠SQR = ½ ∠ROS  [angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference]
⇒ ∠SQR = 30° —————-(ii)
In ΔTQR, ∠TRQ + ∠QTR + ∠TQR = 180°  [Angle sum property of triangle]
⇒ 90° + ∠PQR + 30° = 180°  [from equation i and ii]
⇒ ∠PQR = 60°.
Question 31

In the given figure, if ∠DAB = 60º, ∠ABD = 50º, then ∠ACB is equal to:

a) 50º
b) 60º
c) 70º
d) 80º
Answer: c

Explanation:

∠DAB = 60º, ∠ABD = 50º

By using the angle sum property in the triangle ABD

∠DAB + ∠ABD + ∠ADB = 180º

60º + 50º + ∠ADB = 180º

∠ADB = 180º – 110º

∠ADB = 70º

∠ADB = ∠ACB ……… (Angles in the same segment of a circle are equal).

∠ACB =70º

Question 32

In the given figure, if AOB is a diameter of the circle and AC = BC, then ∠CAB is equal to:

a) 30º
b) 45º
c) 60º
d) 90º
Answer: b
Explanation:

We know that the angle in a semi circle is = 90°

Hence, ∠ACB = 90° 

Also, given that AC = BC 

∠CAB = ∠CBA = x …..  (let) (As, the angles opposite to equal sides are also equal) 

Now, by using the angle sum property of triangle in ∆ACB, we can write

 ∠CAB + ∠ABC + ∠BCA = 180°

x + x + 90° = 180° 

2x = 180° – 90° ⇒ 2x = 90° ⇒ x = 45°

∠CAB = x = 45°

Question 33

Find the value of ∠ABC if O is the centre of circle.

a) 60°
b) 120°
c) 240°
d) 180°
Answer: a
Explanation: Reflex ∠AOC = 240°

⇒ ∠AOC = 360° – Reflex ∠AOC = 360° – 240° = 120°
Since angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference, ∠AOC = 2∠ABC
⇒ ∠ABC = ½ ∠AOC = 60°.

Question 34 
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:

a) 6 cm
b) 8 cm
c) 10 cm
d) 12 cm
Answer: c

Explanation:

Given that AB = 12 cm, BC = 16 cm and AB is perpendicular to BC.

Hence, AC is the diameter of the circle passing through points A, B and C.

Hence, ABC is a right-angled triangle. 

Thus by using the Pythagoras theorem: 

AC2 = (CB)2 + (AB)2 

⇒ AC2 = (16)2 + (12)2 

⇒ AC2 = 256 + 144 

⇒ AC2 = 400 

Hence, the diameter of the circle, AC = 20 cm.

Thus, the radius of the circle is 10 cm.

Question 35

Find the value of ∠QOR if O is the centre of circle.

a) 50°
b) 65°
c) 130°
d) 30°
Answer c
Explanation: Join OP

In ΔPOQ, OP = OQ  [Radii of same circle]
⇒ ∠OPQ = ∠OQP = 20°  [Angles opposite to equal sides are equal]
In ΔPOR, OP = OR  [Radii of same circle]
⇒ ∠OPR = ∠ORP = 45°  [Angles opposite to equal sides are equal]
Now, ∠QPR = ∠OPR + ∠OPQ = 45° + 20° = 65°
Angle subtended by an arc of circle at the centre is twice the angle subtended by the arc on circumference ⇒ ∠ROQ = 2∠QPR = 2 x 65° = 130°.

Question 36

AD is the diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is

a) 4 cm
b) 8 cm
c) 15 cm
d) 17 cm
Answer: b

Explanation:

Given that, Diameter, AD = 34 cm.

Chord, AB = 30 cm.

Hence, the radius of the circle, OA = 17 cm

Now, consider the figure.

From the centre “O”. OM is perpendicular to the chord AB.

(i.e) OM ⊥ AM 

AM =  ½  AB 

AM =  ½ (30) = 15 cm

 Now by using the Pythagoras theorem in the right triangle AOM,

AO2 = OM2 + AM2 

OM2 = AO2– AM2 

OM2= 172 – 152 

OM2 = 64 

OM = √64 

OM = 8 cm

Download Math MCQ Worksheet For Class IX