R.d Sharma 2022 Mcqs Solutions for Class 9 Maths Chapter 15 - Circles

Answer:

(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).

Answer:

(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).

Answer:

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.

Answer:

(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

Now, since ABCD is a cyclic quadrilateral

So, ∠DAB + ∠BCD = 180°

Hence the correct answer is option (a).

Answer:

(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).

Answer:

(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)

 

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).
 

Answer:

(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).

Answer:

(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).

Answer:

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).

Answer:

(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).

Answer:

(c) diameter

The greatest chord in a circle is the diameter of the circle.

Hence the correct answer is option (c).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

(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).

Answer:

Answer:

In ABCD, AB âˆ¥ DC and AD is the transversal.

∠DAB + ∠ADC = 180°          (Co-interior angle)

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 100°

Thus, in cyclic quadrilateral ABCD

∠ADC + ∠ABC = 180°          (∵ In a cyclic quadrilateral opposite angles are supplementary)

⇒ 100° + ∠ABC = 180°

⇒ ∠ABC = 80°

Hence, the correct answer is option (b).
 

Answer:

ABCD is a cyclic quadrilateral.

∴ ∠B + ∠D = 180° 

⇒ ∠B + 120° = 180°

⇒ ∠B = 60°                         .....(1)

Now, in â–³ABC, ∠ACB is the angle made by the diameter on the circumference.

∴ ∠ACB = 90°                        .....(2)

Now, using angle sum property in â–³ABC, we have

∠ACB + ∠BAC + ∠ABC = 180°

⇒ 90° + ∠BAC + 60° = 180°                 [From (1) and (2)]

⇒ ∠BAC = 30°

Hence, the correct answer is option (b).

Answer:

â–³ABC is an equilateral triangle.

∴ ∠ABC = ∠BCA = ∠CAB = 60°

Now ABDC is a cyclic quadrilateral as all its vertices lie on the circumference of the given circle.

∴ ∠CAB + ∠BDC = 180°

⇒ 60° + ∠BDC = 180°

⇒ ∠BDC = 120°

Hence, the correct answer is option (c).

Answer:

Since angles in the same segment of a circle are equal.

∴ ∠DAC = ∠DBC = 40º

Now, in â–³DCB, using the angle sum property of triangle, we have

∠DBC + ∠BDC + ∠DCB = 180º

⇒ 40º + 45º + ∠DCB = 180º

⇒ ∠DCB = 95º

Hence, the correct answer is option (b).

Answer:

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

∴ ∠BOD = 2 × ∠BAD

⇒ ∠BOD = 2 × 75°

⇒ ∠BOD = 150°             .....(1)

Construction: Draw a line from O to C.



Angles subtended by equal chords at the centre of the circle are equal.
∴ ∠BOC = ∠COD             .....(2)

Form (1), we have

∠BOC + ∠COD = 150°

⇒ ∠BOC + ∠BOC = 150°

⇒ ∠BOC = 75°

Hence, the correct answer is option (c).

Answer:

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

∴ ∠BOD = 2 × ∠BAD

⇒ ∠BOD = 2 × 75°

⇒ ∠BOD = 150°

Construction: Draw a line from B to D.



Angles subtended by radius on the chord are equal.
∴ ∠OBD = ∠ODB

Now, in â–³BOD using the angle sum property, we get

∠BOD + ∠ODB + ∠OBD = 180°

⇒ 150° + ∠OBD + ∠OBD = 180°

⇒ ∠OBD = 15°

Hence, the correct answer is option (b).

Answer:

ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180º

⇒ 75º + ∠BCD = 180º

⇒ ∠BCD = 105º

Hence, the correct answer is option (d).

Answer:

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

∴ ∠BOD = 2 × ∠BAD

⇒ ∠BOD = 2 × 75°

⇒ ∠BOD = 150°

Hence, the correct answer is option (a).

Answer:

Construction: Draw two lines from A to C and from C to B.



Now, ABCD is a cyclic quadrilateral.

∴ ∠ADC + ∠ABC = 180°

⇒ 120° + ∠ABC = 180°

⇒ ∠ABC = 60°

The angle subtended by the diameter to any point on the circle is a right angle.

∴ ∠BCA = 90°

Now, in â–³ABC, using the angle sum property, we have

∠BCA + ∠ABC + ∠CAB = 180°

⇒ 90° + 60° + ∠CAB = 180°

⇒ ∠CAB = 30°

Hence, the correct answer is option (c).

Answer:

Statement-2: Angle in a semi-circle is a right angle.

The angle subtended by the diameter to any point on the circle is a right angle.



Statement-2 is true.

Statement-1: Given: PQ is a diameter of a circle and R is a point on the circle.



From statement-2, "The angle subtended by the diameter to any point on the circle is a right angle".

$$\begin{gathered} \mathrm{ar}\left(\mathrm{PQR}\right)=\frac{1}{2}\times \mathrm{Height}\times \mathrm{Base} \\ \Rightarrow \mathrm{ar}\left(\mathrm{PQR}\right)=\frac{1}{2}\times \mathrm{PR}\times \mathrm{QR} \end{gathered}$$

Thus, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Statement-1: If P, Q, R, and S are four points such that ∠QPR = 40º and ∠QSR = 80º, then S is the centre of the circle through Q, R and P.
Let the location of points P, Q, R, and S can be shown below:



We know that the angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

So, if ∠QPR = 40º then ∠QSR = 80º.

Thus, S is the centre of the circle through Q, R and P.

So, Statement-1 is true.

Statement-2: If A, B, C and D are four points such that ∠BAC = 60º and ∠BDC = 60º, then A, B, C, D are concyclic points.

We know that all angles inscribed in a circle and subtended by the same chord are equal.



Here ∠BAC and ∠BDC are subtended by the same chord BC, and their measurements are equal.

Thus, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Hence, the correct answer is option (b).

Answer:

Statement-2: The sum of each pair of opposite angles of a cyclic quadrilateral is 180º.

For a cyclic quadrilateral ABCD, ∠A + ∠C = ∠B + ∠D = 180°.

Statement-1: In the given figure, ABCD is a cyclic quadrilateral. If AB âˆ¥ DC and ∠A = 80º, then ∠B = 80º, ∠C = ∠D = 100º.



In ABCD, AB âˆ¥ DC and AD is the transversal.

∠A + ∠D = 180°          (Co-interior angle)

⇒ 80° + ∠D = 180°

⇒ ∠D = 100°

Thus, in cyclic quadrilateral ABCD.

∠D + ∠B = 180°          (∵ In a cyclic quadrilateral opposite angles are supplementary)

⇒ 100° + ∠B = 180°

⇒ ∠B = 80°

Thus, in cyclic quadrilateral ABCD.

∠A + ∠C = 180°          (∵ In a cyclic quadrilateral opposite angles are supplementary)

⇒ 80° + ∠C = 180°

⇒ ∠C = 100°

Thus, Statement-1 is true.

Thus, Statement-2 is true and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

Construct: Take any point D on the circumference of the circle and make the quadrilateral ABCD.



We know that the angle subtended by the chord at the centre is double the angle subtended by it at any point on the circle.

∴ ∠ADC = 70°

Now, ABCD is a cyclic quadrilateral.

∴ ∠ADC + ∠ABC = 180°

⇒ 70° + ∠ABC = 180°

⇒ ∠ABC = 110°

Thus, Statement-1 is true.

Statement-2: 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.

Statement-2 is true and Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

Answer:

It is given that, AB and CD are two equal chords of a circle with centre O and OP ⊥ AB and OQ ⊥ CD and ∠POQ = 110º.

Equal chords of a circle are equally away from the centre of the circle.(Statement-2)

∴ OP = OQ

Angles made by equal sides are always equal.

∴ ∠OPQ = ∠OQP

Now, in â–³OPQ, using angle sum property

∠OPQ + ∠OQP + ∠POQ = 180°

⇒ ∠OPQ = 35°

Now, ∠OPQ + ∠OPB + ∠APQ = 180°

⇒ 35° + 90° + ∠APQ = 180°

⇒ ∠APQ = 55°

Thus, ​Statement-1 is false and Statement-2 is true.

Hence, the correct answer is option (d).

Answer:

Given: ∠AOB = 100º and ∠BOC = 120º

Sum of all the angles at the centre of the circle is 360º. Therefore

∠AOB + ∠BOC + ∠AOC = 360º

⇒ ∠AOC = 140º

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. (Statement-2)

∠ABC = 70º

Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Hence, the correct answer is option (a).