RD Sharma 2020 solution class 9 chapter 15 Circles MCQS

MCQS

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Question 1:

If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is

(a) 15 cm

(b) 16 cm

(c) 17 cm

(d) 34 cm

Answer 1:

(c) 17 cm

We will represent the given data in the figure
   
In the diagram AB is the given chord of 16 cm length and OM is the perpendicular distance from the centre to AB.
We know that perpendicular from the centre to any chord divides it into two equal parts.
So,  AM = MB = = 8 cm.
Now consider right triangle OMA and by using Pythagoras theorem

        
Hence, correct answer is option (c).

Question 2:

The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is

(a) $\sqrt{5} \mathrm{cm}$

(b) $2\sqrt{5} \mathrm{cm}$

(c) $2\sqrt{7} \mathrm{cm}$

(d) $\sqrt{7} \mathrm{cm}$

Answer 2:

(b) $2\sqrt{5} \mathrm{cm}$

We will represent the given data in the figure

We know that perpendicular drawn from the centre to the chord divides the chord into two equal parts.
So , AM = MB = $\frac{AB}{2}=\frac{8}{2}$ = 4 cm.
Using Pythagoras theorem in the ΔAMO,

         
Hence, the correct answer is option (b).

Question 3:

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =

(a) 60°

(b) 45°

(c) 30°

(d) 15°

Answer 3:

We will associate the given information in the following figure.

Since AO = r (radius of circle)
AM = (given)
Extended OM to D where MD =
Consider the triangles AOM and triangle AMD

So by SSS property

So AD = AO = r and OD=OM+MD=r
Hence ΔAOD is equilateral triangle
So
We know that in equilateral triangle altitudes divide the vertex angles

Hence option (c) is correct.

Question 4:

ABCD is a cyclic quadrilateral such that ∠ADB = 30° and ∠DCA = 80°, then ∠DAB =

(a) 70°

(b) 100°

(c) 125°

(d) 150°

Answer 4:

(a) 70°

It is given that ABCD is cyclic quadrilateral ∠ADB = 90° and ∠DCA = 80°. We have to find ∠DAB
We have the following figure regarding the given information

∠BDA = ∠BCA = 30°      (Angle in the same segment are equal) Now, since ABCD is a cyclic quadrilateral
So, ∠DAB + ∠BCD = 180°

Hence the correct answer is option (a).

Question 5:

A chord of length 14 cm is at  a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is

(a) 12 cm

(b) 14 cm

(c) 16 cm

(d) 18 cm

Answer 5:

(d) 18 cm

We are given the chord of length 14 cm and perpendicular distance from the centre to the chord is 6 cm. We are asked to find the length of another chord at a distance of 2 cm from the centre.
We have the following figure

We are given AB = 14 cm, OD = 6 cm, MO = 2 cm, PQ = ?
Since, perpendicular from centre to the chord divide the chord into two equal parts
Therefore

Now consider the ΔOPQ in which OM = 2 cm
So using Pythagoras Theorem in ΔOPM


Hence, the correct answer is option (d).

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Question 6:

One chord of a circle is known to be 10 cm. The radius of this circle must be

(a) 5 cm

(b) greater than 5 cm

(c) greater than or equal to 5 cm

(d) less than 5 cm

Answer 6:

(b) greater than 5 cm

We are given length of a chord to be 10 cm and we have to give information about the radius of the circle.
Since in any circle, diameter of the circle is greater then any chord.
So diameter > 10
⇒ 2r > 10
⇒ r > 5 cm
Hence, the correct answer is option (b)

Question 7:

ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. 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) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Answer 7:

(d) 6 cm

We are given a right triangle ABC such that, AC = 5 cm, AB = 4 cm. A circle is drawn with A as centre and AC as radius. We have to find the length of the chord of this circle passing through C and B. We have the following figure regarding the given information.

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 ΔABC apply Pythagoras theorem

So,  PC = 2 × BC
            = 2 × 3
            = 6 cm
Hence, the correct answer is option (d).

Question 8:

If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than ∠AOB =

(a) 60°

(b) 90°

(c) 120°

(d) none of these

Answer 8:

(a) 60°



As we know that equal chords make equal angle at the centre.

Therefore,

$$\begin{gathered} \angle AOB=\angle BOC=\angle COD \\ \angle AOB+\angle BOC+\angle COD=180° \left[\mathrm{Linear} \mathrm{pair}\right] \\ \Rightarrow 3\angle AOB=180° \\ \Rightarrow \angle AOB=60° \end{gathered}$$

Hence, the correct answer is option (a).

Question 9:

Let C be the mid-point of an arc AB of a circle such that m $\stackrel{⏜}{AB}$ = 183°. If the region bounded by the arc ACB and the line segment AB is denoted by S, then the centre O of the circle lies

(a) in the interior of S

(b) in the exertior of S

(c) on the segment AB

(d) on AB and bisects AB

Answer 9:

(a) in the interior of S

Given: m $\stackrel{⏜}{AB}$ = 183° and C is mid-point of arc ABO is the centre.
With the given information the corresponding figure will look like the following

So the center of the circle lies inside the shaded region S.
Hence, the correct answer is option (a).

Question 10:

In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are

(a) 90° and 270°

(b) 90° and 90°

(c) 270° and 90°

(d) 60° and 210°

Answer 10:

(c) 270° and 90°

We are given the major arc is 3 times the minor arc. We are asked to find the corresponding central angle.
See the corresponding figure.

We know that angle formed by the circumference at the centre is 360°.
Since the circumference of the circle is divided into two parts such that the angle formed by major and minor arcs at the centre are 3x and x respectively.

So = 90° and = 3x = 270°
Hence, the correct answer is option (c).

Question 11:

If A and B are two points on a circle such that m $\stackrel{⏜}{\left(AB\right)}$ = 260°. A possible value for the angle subtended by arc BA at a point on the circle is

(a) 100°

(b) 75°

(c) 50°

(d) 25°

Answer 11:

(c) 50°

We are given

Suppose point P is on the circle.
Since
So, = 360° − 260° = 100°
We know that angle subtended by chord AB at the centre is twice that of subtended at the point P
So, = = 50°
Hence, the correct answer is option (c).

Question 12:

An equilateral triangle ABC is inscribed in a circle with centre O. The measures of ∠BOC is

(a) 30°

(b) 60°

(c) 90°

(d) 120°

Answer 12:

(d) 120°

We are given that an equilateral ΔABC is inscribed in a circle with centre O. We need to find ∠BOC
We have the following corresponding figure:

We are given AB = BC = AC
Since the sides AB, BC, and AC are these equal chords of the circle.
So, the angle subtended by these chords at the centre will be equal.
Hence

Hence, the correct answer is option (d).

Question 13:

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 13:

(d) square

The given information in the form of the following figure is as follows:

Since, four sides of the quadrilateral ACBD are four chords which subtend equal angles at the centre. Therefore,
    (Since AB and CD are perpendicular diameters)
So sides AC, BC, BD and AD are equal, as equal chords subtend equal angle at the centre.
So , AC = CB = BD = DA    …… (1)
We know that diameters subtend an angle of measure 90° on the circle.
So,  …… (2)
From (1) and (2) we can say that is a square.

Hence, the correct answer is option (d).

Question 14:

If ABC is an arc of a circle and ∠ABC = 135°, then the ratio of arc $\stackrel{⏜}{ABC}$  to the circumference is

(a) 1 : 4

(b) 3 : 4

(c) 3 : 8

(d) 1 : 2

Answer 14:

(c) 3 : 8

The length of an arc subtending an angle ‘’ in a circle of radius ‘r’ is given by the formula,
Length of the arc =
Here, it is given that the arc subtends an angle of with its centre. So the length of the given arc in a circle with radius ‘r’ is given as
Length of the arc =
The circumference of the same circle with radius ‘r’ is given as .
The ratio between the lengths of the arc and the circumference of the circle will be,

Hence, the correct answer is option (c).

Question 15:

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle is

(a) 60°

(b) 75°

(c) 120°

(d) 150°

Answer 15:

(d) 150°

We are given that the chord is equal to its radius.
We have to find the angle subtended by this chord at the minor arc.
We have the corresponding figure as follows:

We are given that
AO = OB = AB
So , $△$AOB is an equilateral triangle.
Therefore, we have
∠AOB = 60°
Since, the angle subtended by any chord at the centre is twice of the angle subtended at any point on the circle.

Take a point P on the minor arc.
Since is a cyclic quadrilateral
So, opposite angles are supplementary. That is

Hence, the correct answer is option (d).

Question 16:

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =

(a) 41°

(b) 23°

(c) 67°

(d) 18°

Answer 16:

Here we have a cyclic quadrilateral PQRS with PR being a diameter of the circle. Let the centre of this circle be ‘O’.
We are given that and. This is shown in fig (2).

So we see that,
$$\begin{gathered} \angle QPS=\angle QPR+\angle RPS \\ =67°+72° \\ =139° \end{gathered}$$
(a) 41°

In a cyclic quadrilateral it is known that the opposite angles are supplementary.

Hence the correct answer is option (a).

Question 17:

If A , B, C are three points on a circle with centre O such that ∠AOB = 90° and ∠BOC = 120°, then ∠ABC =

(a) 60°

(b) 75°

(c) 90°

(d) 135°

Answer 17:

(b) 75°

To solve this problem we need to know that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

Here we are given that ‘A’, ‘B’, ‘C’ are three points on a circle with centre ‘O’ such that and .
From the figure we see that,

Now, as seen earlier, the angle made by the arc ‘AC’ with the centre of the circle will be twice the angle it makes in any point in the remaining part of the circle.
Since the point ‘C’ lies on the remaining part of the circle, the angle the arc ‘AC’ makes with this point has to be half of the angle ‘AC’ makes with the centre.Therefore we have,

Hence the correct answer is option (b).

Question 18:

The greatest chord of a circle is called its

(a) radius

(b) secant

(c) diameter

(d) none of these

Answer 18:

(c) diameter

The greatest chord in a circle is the diameter of the circle.
Hence the correct answer is option (c).

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Question 19:

Angle formed in minor segment of a circle is

(a) acute

(b) obtuse

(c) right angle

(d) none of these

Answer 19:

(b) obtuse

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment.
The angle formed by the chord in the minor segment will always be obtuse.
Hence the correct answer is option (b).

Question 20:

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

(a) 1

(b) 0

(c) 2

(d) 3

Answer 20:

(a) 1

Suppose we are given three non-collinear points as A, B and C

1. Join A and B.
2. Join B and C.
3. Draw perpendicular bisector of AB and BC which meet at O as centre of the circle.
So basically we can only draw one circle passing through three non-collinear points A, B and C.
Hence, the correct answer is option (a).

Question 21:

In the given figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =

(a) 45°

(b) 60°

(c) 75°

(d) 90°
 

Answer 21:

(d) 90°

We are given the following figure

 ∠ACD = ∠ABD     (Angle in the same segment are equal)
⇒ ∠ACD = y
Consider the ΔACM in which

Hence, the correct answer is option (d).

Question 22:

In the given figure, if ∠ABC = 45°, then ∠AOC =

(a) 45°

(b) 60°

(c) 75°

(d) 90°

Answer 22:

(d) 90°

We have to find ∠AOC.

 As 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.

Hence, the correct answer is option (d).

Question 23:

In the given figure, chords AD and BC intersect each other at right angles at a point P. If ∠DAB = 35°, then
∠ADC =
(a) 35°
(b) 45°
(c) 55°
(d) 65°

Answer 23:

(c) 55°

   (Angle in the same segment are equal.)
Also, since the chords ‘AD’ and ‘BC’ intersect perpendicularly we have,

Consider the triangle ,


Hence, the correct answer is option (c).

Question 24:

In the given figure, O is the centre of the circle and ∠BDC = 42°. The measure of ∠ACB is

(a) 42°

(b) 48°

(c) 58°

(d) 52°

Answer 24:

(b) 48°

Construction: Join A and D.

Since AC is the diameter. So ∠ADC will be 90°.
Therefore,

 ∠ACB = ∠ADB = 48°      (Angle in the same segment are equal.)
Hence, the correct answer is option (b).

Question 25:

In a circle with centre O, AB and CD are two diameters perpendicular to each other. The length of chord AC is

(a) 2AB

(b) $\sqrt{2}$

(c) $\frac{1}{2}AB$

(d) $\frac{1}{\sqrt{2}}AB$

Answer 25:

(d) $\frac{1}{\sqrt{2}}AB$

We are given a circle with centre at O and two perpendicular diameters AB and CD.
We need to find the length of AC.
We have the following corresponding figure:

Since, AB = CD                   (Diameter of the same circle)
Also, ∠AOC = 90°
And,  AO =
Here,  AO = OC (radius)
In ΔAOC

Hence,  the correct answer is option (d).

Question 26:

Two equal circles of radius r intersect such that each passes through the centre of the other. The length of the common chord of the circle is

(a) $\sqrt{r}$

(b) $\sqrt{2}r AB$

(c) $\sqrt{3}r$

(d) $\frac{\sqrt{3}}{2}$

Answer 26:

(c) $\sqrt{3}r$

We are given two circles of equal radius intersect each other such that each passes through the centre of the other.
We need to find the common chord.
We have the corresponding figure as follows:
  
 AO = AO′ = r (radius)
And OO′ = r
So, ΔOAO′ is an equilateral triangle.
We know that the attitude of an equilateral triangle with side r is given by
That is AM =
We know that the line joining centre of the circles divides the common chord into two equal parts.
So we have

Hence, the correct answer is option (c).

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Question 27:

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) ∠APB = ∠AQB

(b) ∠APB + ∠AQB = 180° or ∠APB = ∠AQB

(c) ∠APB + ∠AQB = 90°

(d) ∠APB + ∠AQB = 180°

Answer 27:

(b)  ∠APB + ∠AQB = 180° or ∠APB = ∠AQB

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 following figure.
Case 1: Consider P and Q are on the same side of AB

We know that angle in the same segment are equal.
Hence, ∠APB = ∠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,
∠APB + ∠AQB = 180°    (Sum of the pair of opposite angles of cyclic quadrilateral is 180°.)
Therefore, ∠APB = ∠AQB or ∠APB + ∠AQB = 180°
Hence, the correct answer is option (b).

Question 28:

AB and CD are two parallel chords of a circle with centre O such that AB = 6 cm and CD = 12 cm. The chords are on the same side of the centre and the distance between them is 3 cm. The radius of the circle is

(a) 6 cm

(b) $5\sqrt{2 }\mathrm{cm}$

(c) 7 cm

(d) $3\sqrt{5 }\mathrm{cm}$

Answer 28:

(d) $3\sqrt{5 }\mathrm{cm}$

Let distance between the centre and the chord CD be x cm and the radius of the circle is r cm.

We have to find the radius of the following circle:


In triangle OND,
…… (1)
Now, in triangle AOM,
…… (2)
From (1) and (2), we have,

Hence, the correct answer is option (d).

Question 29:

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

(a) 34 cm

(b) 15 cm

(c) 23 cm

(d) 30 cm

Answer 29:

(d) 30 cm

Given that: Radius of the circle is 17 cm, distance between two parallel chords AB and CD is 23 cm, where AB= 16 cm. 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.
So, AM = MB = 8 cm
Let OM = x cm
In triangle OMB,

Now, in triangle OND, ON = (23 − x) cm = (23 − 15) cm = 8 cm

Therefore, the length of the other chord is


Hence, the correct answer is option (d).

Question 30:

In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

(a) 130°

(b) 115°

(c) 65°

(d) 165°

Answer 30:

(b) 115°

We have the following information in the following figure. Take a point P on the circle in the given figure and join AP and CP.

Since, the angle subtended by a chord at the centre is twice that of subtended atany part of the circle.
So, 
Since is a cyclic quadrilateral and we known that opposite angles are supplementary.
Therefore,
$$\begin{gathered} \angle ABC+\angle APC=180° \\ \Rightarrow \angle ABC+65°=180° \\ \Rightarrow \angle ABC=180°-65° \\ \Rightarrow \angle ABC=115° \end{gathered}$$
Hence, the correct answer is optoon (b).