Rd Sharma 2021 for Class 9 Maths Chapter 15 - Circles

Answer:

It is given that, is an equilateral triangle

We have to find

Since is an equilateral triangle.

So

And

…… (1)

So ,                                         (Sum of opposite angles of cyclic quadrilateral is

Hence

Answer:

Disclaimer: Figure given in the book was showing m∠PQR as m∠SQR. 

It is given that ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°

We have to find the m∠QSR and m∠QTR

Since ΔPQR is an isosceles triangle

So ∠PQR = ∠PRQ = 35° 

Then

Since PQTR is a cyclic quadrilateral

So

In cyclic quadrilateral QSRT we have

Hence,

and  

Answer:

It is given that O is centre of the circle and ∠BOD = 160°

We have to find the values of x and y.

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.

Therefore,

Since, quadrilateral ABCD is a cyclic quadrilateral.

So,

x + y = 180°              (Sum of opposite angles of a cyclic quadrilateral is 180°.)

Hence and

Answer:

It is given that ∠BCD = 100° and ∠ABD = 70°

We have to find the ∠ADB

We have

∠A + ∠C = 180°                     (Opposite pair of angle of cyclic quadrilateral)

So,

Now in is and

Therefore,

Hence,

Answer:

It is given that, ABCD is cyclic quadrilateral in which AD || BC

We have to prove

Since, ABCD is a cyclic quadrilateral

So,

and               ..… (1)

and              (Sum of pair of consecutive interior angles is 180°) …… (2)

From equation (1) and (2) we have

…… (3)

…… (4)

Hence Proved

Answer:

It is given that,

We have to find

Since,    (Given)

So,

                     (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.)

Now,

               (Opposite pair of angle of cyclic quadrilateral)

So,

…… (1)

      (Linear pair)

         ()

Hence

Answer:

It is given that, AB and CD are diameter with center O and

We have to find  

Construction: Join the point A and D to form line AD

Clearly arc AD subtends at B and at the centre.

Therefore,  $\angle AOD=2\angle ABD=100°$ …… (1)

Since CD is a straight line then
        
$\angle DOA+\angle AOC=180° \left(\mathrm{Linear} \mathrm{pair}\right)$

     

Hence

Answer:

It is given that, as diameter, is centre and

We have to find and

Since angle in a semi-circle is a right angle therefore

In we have

(Given)

    (Angle in semi-circle is right angle)

Now in we have

Hence and  

Answer:

It is given that, ABCD is a cyclic quadrilateral such that AB || CD and

Sum of pair of opposite angles of cyclic quadrilateral is 180°.

                  ( given)

So,

Also AB || CD and BC transversal

So,

Now

Answer:

It is given that

ABCD is cyclic quadrilateral and

We have to find

Since ABCD is cyclic quadrilateral and sum of opposite pair of cyclic quadrilateral is 180°.

So

And

Therefore

Hence

Answer:

It is given that, O is the centre of the circle and $\angle DAB=50°$.

We have to find  the values of x and y.

ABCD is a cyclic quadrilateral and

So,

50° + y = 180°
y = 180° − 50°
y = 130°
 

Clearly is an isosceles triangle with OA = OB and

Then,

                        (Since)

So,

x + $\angle AOB$ = 180°     (Linear pair)

Therefore, x =  180° − 80° = 100°

Hence,

and

Answer:

It is given that, and

We have to find the

In given we have

$$\begin{gathered} \angle ABC+\angle BCA+\angle BAC=180° \left(\mathrm{Angle} \mathrm{sum} \mathrm{property}\right) \\ \Rightarrow \angle ABC=180°-\left(60°+20°\right)=100° \end{gathered}$$

In cyclic quadrilateral we have

         (Sum of pair of opposite angles of a cyclic quadilateral is 180º)

Then,

Hence

Answer:

It is given that, ABC is an equilateral triangle

We have to find and

Since is an equilateral triangle

So,

And is cyclic quadrilateral

So      (Sum of opposite pair of angles of a cyclic quadrilateral is 180°.)

Then,

Similarly BECD is also cyclic quadrilateral

So,

Hence, and  .

Answer:

It is given that, O is the centre of the circle and

We have to find the value of x, y and z.

Since, angle in the same segment are equal

So

And z = 30°               

As 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.
Since

Then,

y = 2z
   = 2 × 30°
   = 60°

Answer:

It is given that, and, are cyclic quadrilateral

We have to find the value of x and y.

Since , is a cyclic quadrilateral

So      (Opposite angle of a cyclic quadrilateral are supplementary)

        ()

     ..… (1)

x = 180° − 102°
   = 78°

Now in cyclic quadrilateral DCFE
x + y = 180°    (Opposite angles of a cyclic quadrilateral are supplementary)
y = 180° − 78°
   = 102°

Hence, x = 78° and y = 102°
 

Answer:

It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°

Since ABCD is a cyclic quadrilateral

So ∠A + ∠C = 180°                 (Sum of opposite pair of angles of cyclic quadrilateral is 180°)   ..… (1)

And,

∠A – ∠C = 60°               (Given)            ..… (2)

Adding equation (1) and (2) we have

So, ∠C = 60°

Hence, smaller of two is 60°.

 

Answer:

Here, ABCD is a cyclic quadrilateral, we need to find x.

In cyclic quadrilateral the sum of opposite angles is equal to 180°.

Therefore,

 $$\begin{gathered} \angle ADC+\angle ABC=180° \\ \Rightarrow 180°-80°+180°-x=180° \\ \Rightarrow x=100° \end{gathered}$$

Hence, the value of x is 100°.

Answer:

(i) It is given that , and

We have to find

In cyclic quadrilateral ABCD

      ..… (1)

      ..… (2)

Since, 

So,

Therefore in ,

So ,                   ..… (3)

Now,                   ( and is transversal)

(ii) It is given that , and

We have to find

                    _{(Angle in the same segment are equal)}

_Hence,

(iii) It is given that, ∠BCD = 100° and ∠ABD = 70°

As we know that sum of the opposite pair of angles of cyclic quadrilateral is 180°.

$$\begin{gathered} \angle DAB+\angle BCD=180° \\ \Rightarrow \angle DAB=180°-100° \\ =80° \end{gathered}$$

In ΔABD we have,

$$\begin{gathered} \angle DAB+\angle ABD+\angle BDA=180° \\ \Rightarrow \angle BDA=180°-150°=30° \end{gathered}$$

_Hence, 

Answer:

Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Let the diagonals AC and BD intersect at O.

We know that the diagonals of a rhombus intersect at right angle.

Therefore,

Now, means that circle described on AB as diameter passes through O.

Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.

Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.

Answer:

To prove: AC = BD

Proof: We know that equal chords subtend equal at the centre of circle and the angle subtended by a chord at the centre is twice the angle subtended by it at remaining part of the circle.

$$\begin{gathered} \angle AOD=\angle BOC \left(O \mathrm{is} \mathrm{the} \mathrm{centre} \mathrm{of} \mathrm{the} \mathrm{circle}\right) \\ \angle AOD=2\angle ACD \\ \mathrm{and} \angle BOC=2\angle BDC \\ \mathrm{Since}, \angle AOD=\angle BOC \\ \Rightarrow \angle ACD=\angle BDC .....\left(1\right) \\ \angle ACB=\angle ADB .....\left(2\right) \left(\mathrm{Angle} \mathrm{in} \mathrm{the} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{Adding} \left(1\right) \mathrm{and} \left(2\right) \\ \angle BCD=\angle ADC .....\left(3\right) \\ \mathrm{In} △ACD \mathrm{and} △BDC \\ CD=CD \left(\mathrm{common}\right) \\ \angle BCD=\angle ADC \left[\mathrm{Using}\left(3\right)\right] \\ AD=BC \left(\mathrm{given}\right) \\ \mathrm{Hence}, △ACD\cong BDC \left(\mathrm{SAS} \mathrm{congruency} \mathrm{criterion}\right) \\ \therefore AC=BD \left(\mathrm{cpct}\right) \\ \mathrm{Hence} \mathrm{Proved} \end{gathered}$$

Answer:

Answer:

If in cyclic quadrilateral , then we have to find the other three angles.

Since, AD is parallel to BC, So,
  (Alternate interior angles)

Now, since ABCD is cyclic quadrilateral, so

And,

$\mathrm{Hence}, \angle A=110°, \angle C=70° \mathrm{and} \angle D=110°$.

Answer:

It is given that is a cyclic quadrilateral with and as its diagonals.

We have to find

Since angles in the same segment of a circle are equal

So

Since (Opposite angle of cyclic quadrilateral)

Hence

Answer:

To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.

Proof:

We know that the perpendicular bisector of every chord of a circle always passes through the centre.

Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.

Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.

Answer:

Here, ABCD is a cyclic rectangle; we have to prove that the centre of the corresponding circle is the intersection of its diagonals.

Let O be the centre of the circle.

We know that the angle formed in the semicircle is 90°.

Since, ABCD is a rectangle, So

Therefore, AC and BD are diameter of the circle.

We also know that the intersection of any two diameter is the centre of the circle.

Hence, the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Answer:

(i) If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA =  ED, then we have to prove the following, AD || BC

(ii) EB = EC

(i) It is given that EA = ED, so

$\angle EAD=\angle EDA=x$

Since, ABCD is cyclic quadrilateral

Now,

Therefore, the adjacent angles andare supplementary

Hence, AD || BC

(ii) Since, AD and BC are parallel to each other, so,

$$\begin{gathered} \angle ECB=\angle EDA \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \angle EBC=\angle EAD \left(\mathrm{Corresponding} \mathrm{angles}\right) \\ \mathrm{But}, \angle EDA=\angle EAD \\ \mathrm{Therefore}, \angle ECB=\angle EBC \\ \Rightarrow EC=EB \\ \mathrm{Therefore}, △ECB \mathrm{is} \mathrm{an} \mathrm{isosceles} \mathrm{triangle}. \end{gathered}$$

Answer:

$$\begin{gathered} \stackrel{⏜}{QP} \mathrm{is} \mathrm{a} \mathrm{major} \mathrm{arc} \mathrm{and} \angle PSQ \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{formed} \mathrm{by} \mathrm{it} \mathrm{in} \mathrm{the} \mathrm{alternate} \mathrm{segment}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{an} \mathrm{arc} \mathrm{at} \mathrm{the} \mathrm{centre} \mathrm{is} \mathrm{twice} \mathrm{the} \mathrm{angle} \mathrm{subtended} \mathrm{by} \mathrm{it} \mathrm{at} \mathrm{any} \mathrm{point} \mathrm{of} \mathrm{the} \mathrm{alternate} \mathrm{segment} \mathrm{of} \mathrm{the} \mathrm{circle}. \\ \therefore 2\angle PSQ=m\left(\stackrel{⏜}{QP}\right) \\ \Rightarrow 2\angle PSQ=360°-m\left(\stackrel{⏜}{PQ}\right) \\ \Rightarrow 2\angle PSQ=360°-\angle POQ \\ \Rightarrow 2\angle PSQ=360°-180° \left(\because \angle POQ<180°\right) \\ \Rightarrow 2\angle PSQ>180° \\ \Rightarrow \angle PSQ>90° \end{gathered}$$

Thus, the angle in a segment shorter than a semi-circle is greater than a right angle.

Answer:

$$\begin{gathered} \mathrm{To} \mathrm{prove}: \angle ABC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle} \\ \mathrm{Proof}: \\ AD \mathrm{being} \mathrm{the} \mathrm{diameter} \mathrm{of} \mathrm{the} \mathrm{given} \mathrm{circle} \\ \Rightarrow \angle ACD=90° \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{semicircle} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angle}\right] \\ \mathrm{Now}, \mathrm{in} △ACD, \angle ACD=90° \mathrm{which} \mathrm{means} \mathrm{that} \angle ADC \mathrm{is} \mathrm{an} \mathrm{acute} \mathrm{angle}. .....\left(1\right) \\ \mathrm{Again}, \angle ABC=\angle ADC \left[\mathrm{Angle} \mathrm{in} \mathrm{a} \mathrm{same} \mathrm{segment} \mathrm{are} \mathrm{always} \mathrm{equal}\right] \\ \Rightarrow \angle ABC \mathrm{is} \mathrm{also} \mathrm{an} \mathrm{acute} \mathrm{angle}. \left[\mathrm{Using}\left(1\right)\right] \\ \mathrm{Hence} \mathrm{proved} \end{gathered}$$

Answer:

We have to prove that

Let be a right angle at B and P be midpoint of AC

Draw a circle with center at P and AC diameter

Since therefore circle passing through B

So

Hence

Proved.

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:

Given:
AD is a diameter
AB is chord
AD = 34 cm
AB = 30 cm


Let O be the centre of the circle.
AO = OD = 17 cm            ...(1)

Let OL is a line perpendicular to AB, where L is the point on AB.
Then, AL = LB = 15 cm              ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆ALO,
Using pythagoras theorem,
AL² + LO² = AO²
⇒ 15² + LO² = 17²        (From (1) and (2))
⇒ 225 + LO² = 289
⇒ LO² = 289 − 225
⇒ LO² = 64
⇒ LO = 8 cm

Now, In ∆ABD,
Using mid-point theorem: The line segment joining the mid-points of two sides of a triangle is parallel to the third side and is equal to the half of it.
Therefore, LO = $\frac{1}{2}$BD
⇒ BD = 2LO
⇒ BD = 2(8)
⇒ BD = 16 cm​


Hence, BD = 16 cm.

Answer:

Given:
AD is a diameter
AB is chord
AD = 34 cm
AB = 30 cm


Let O be the centre of the circle.
AO = OD = 17 cm            ...(1)

Let OL is a line perpendicular to AB, where L is the point on AB.
Then, AL = LB = 15 cm              ...(2) (∵ a perpendicular from the centre of the circle to the chord, bisects the chord)

In ∆ALO,
Using pythagoras theorem,
AL² + LO² = AO²
⇒ 15² + LO² = 17²        (From (1) and (2))
⇒ 225 + LO² = 289
⇒ LO² = 289 − 225
⇒ LO² = 64
⇒ LO = 8 cm

Thus, the distance of the chord from the centre is 8 cm.


Hence, the distance of AB from the centre of the circle is 8 cm.

Answer:

Given:
AB = 12 cm
BC = 16 cm
AB is perpendicular to BC


Since, AB is perpendicular to BC
Therefore, the circle formed by joining A, B and C is a circle with diameter AC.

In ∆ABC,
Using pythagoras theorem,
AB² + BC² = AC²
⇒ 12² + 16² = AC²       (given)
⇒ 144 + 256 =  AC²
⇒ AC²= 400
⇒ AC = 20

Thus, the diameter of the circle is 20 cm.
Therefore, the radius of the circle is half of the diameter of the circle.
Radius = $\frac{1}{2}\left(20\right)$ =10 cm

Hence, the radius of the circle passing through the points A, B and C is 10 cm.

Answer:

Given:
ABCD is a cyclic quadrilateral
AB is a diameter of the circle circumscribing ABCD
∠ADC = 140^∘


In a cyclic quadrilateral, the sum of opposite angles is 180^∘.


Thus, ∠ADC + ∠ABC = 180°
⇒ 140° + ∠ABC = 180°
⇒ ∠ABC = 180° − 140°
⇒ ∠ABC = 40°     ...(1)

Since AB is the diameter of the circle
Therefore, ∠ACB = 90°  (angle in the semi circle)    ...(2)

In ∆ABC,
∠BAC + ∠ACB + ∠ABC = 180° (angle sum property)
⇒ ∠BAC + 90° + 40° = 180° (From (1) and (2))
⇒ ∠BAC + 130° = 180°
⇒ ∠BAC = 180° − 130°
⇒ ∠BAC = 50°


Hence, ∠BAC =  50°.

Answer:

Given:
Two chords AB and CD of a circle are each at a distance of 6 cm from the centre.


The chords which are equidistant from the centre of the circle are of equal length.

Hence, the ratio of their length is 1 : 1.

Answer:

Given:
AB =AC
AB and AC lies on the opposite sides of OA


The chords of equal length, are equidistant from the centre of the circle.

In ∆OAB and ∆OAC,
AB = AC (given)
OA = OA (common)
OB = OC (radius of the circle)

By SSS property,
∆OAB ≅ ∆OAC

Therefore, ∠OAB = ∠OAC (by C.P.C.T.)

Hence, ∠OAB = ∠OAC.

Answer:

Given:
Two congruent circle with centres O and O' intersect at two points P and Q.


In ∆OPQ and ∆O'PQ,
OP = O'P (radius)
PQ = PQ (common)
OQ = O'Q (radius)

By SSS property,
∆OPQ ≅ ∆O'PQ

Therefore, ∠POQ = ∠PO'Q (by C.P.C.T.)

Hence, ∠POQ : ∠PO'Q = 1 : 1.

Answer:

Given:
AOB is a diameter of a circle
C is a point on the circle

Since, AOB is the diameter of the circle
Therefore, ∠ACB = 90°  (angle in the semi circle)    ...(2)

In right angled ∆ABC,
Using pythagoras theorem.
AC² + BC² = AB²


Hence, AC² + BC²  = AB².

Answer:

Given:
O is the circumcentre of ∆ABC
D is the mid-point of the base BC

In ∆BOD and ∆COD,
OB = OC (radius)
BD = CD (D is the mid-point of the base BC)
OD = OD (common)

By SSS property,
∆BOD ≅ ∆COD

Therefore, ∠BOD = ∠COD (by C.P.C.T.)
⇒ ∠BOC = ∠BOD + ∠COD
⇒ ∠BOC = 2∠BOD      ..(1)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠BOC = 2∠BAC
⇒ 2∠BOD = 2∠BAC
⇒ ∠BOD = ∠BAC


Hence, ∠BOD = ∠BAC.

Answer:

Given:
O is the circumcentre of ∆ABC


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠BOC = 2∠BAC     ...(1)


In ∆OBC ,
OB = OC (radius)
Thus, ∠OBC = ∠OCB       ...(2)

Also, ∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)
⇒ 2∠OBC + 2∠BAC = 180°     (from (1) and (2))
⇒ ∠OBC + ∠BAC = 90°

Hence, ∠OBC + ∠BAC = 90°.

Answer:

Given:
A chord of a circle is equal to its radius

Let AB is a chord and O is the centre of the circle.

AB = OA = OB (∵ Chord is equal to the radius)
⇒ ∆ABO is equilateral triangle

Thus, ∠AOB = 60°     ...(1)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ADB , where D is any point on the major segment of the circle
⇒ 2∠ADB = 60°     (from (1) and (2))
⇒ ∠ADB = 30°

Hence, the angle subtended by the chord at a point in major segment is 30°.

Answer:

Given:
A pair of opposite sides of a quadrilateral ABCD are equal.
i.e., AB = CD and AD = BD

Then, the quadrilateral can be a parallelogram, a rectangle, rhombus or a square.
In all the cases the diagonals bisects each other.

Hence, its diagonals are bisecting.

Answer:

Given:
Arcs AXB and CYD of a circle are congruent

We know, if any two arcs are congruent, then their corresponding chords are equal.
Thus, Chord AB = Chord CD

Hence, AB : CD = 1 : 1.

Answer:

Given:
A, B and C are three points on a circle


Let ABC be a triangle.
We know, a circumcentre is the point of intersection of the perpendicular bisectors of the triangle.

Thus, the perpendicular bisector of AB, BC and CA intersect at a point known as circumcentre.

Hence, the perpendicular bisector of AB, BC and CA are concurrent.

Answer:

Given:
AB and AC are equal chords of a circle


Let O be the centre of the circle.

In ∆OAB and ∆OAC,
AB = AC (given)
OA = OA (common)
OB = OC (radius of the circle)

By SSS property,
∆OAB ≅ ∆OAC

Therefore, ∠OAB = ∠OAC (by C.P.C.T.)

Thus, ∠BAC = 2∠OAB.

Hence, the bisector of ∠BAC passes through the centre.

Answer:

Given:
ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D
∠CBD + ∠CDB = k∠BAD


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠CAD = 2∠CBD      ...(1)
Also,  ∠CAB = 2∠CDB      ...(2)

Adding (1) and (2), we get
∠CAD + ∠CAB = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2∠CBD + 2∠CDB
⇒ ∠BAD = 2(∠CBD + ∠CDB)
⇒ ∠CBD + ∠CDB = $\frac{1}{2}$∠BAD

Hence, k = $\underline{\frac{1}{2}}$.

Answer:

Given:
AB and AC subtend angles equal to 90° and 150° respectively at the centre
i.e., ∠AOB = 90° and ∠AOC = 150°     ...(1)

In ∆AOB,
OA = OB (radius)
∴ ∠OBA = ∠OAB (angles opposite to equal sides are equal)    ...(2)

Now, ∠OBA + ∠OAB + ∠AOB = 180° (angle sum property)    
⇒ 2∠OAB +  90° = 180°           (From (1) and (2))
⇒ 2∠OAB = 180° − 90°
⇒ 2∠OAB = 90°
⇒ ∠OAB = 45°         ...(3)


In ∆AOC,
OA = OC (radius)
∴ ∠OCA = ∠OAC (angles opposite to equal sides are equal)    ...(4)

Now, ∠OCA + ∠OAC + ∠AOC = 180° (angle sum property)    
⇒ 2∠OAC +  150° = 180°           (From (1) and (4))
⇒ 2∠OAC = 180° − 150°
⇒ 2∠OAC = 30°
⇒ ∠OAC = 15°         ...(5)


Thus,
∠BAC = ∠OAC + ∠OAB 
⇒ ∠BAC = 15° + 45°               (From (3) and (5))
⇒ ∠BAC = 60°

Hence, ∠BAC = 60°.

Answer:

Given:
O and O' are the centres of two congruent circles
AXB of circle centred at O, subtends an angle of 75^∘ at the centre O 
arc PYQ of circle centred at O' ,subtends an angle 25^∘at the centre O'



Since, the circles are congruent
Therefore, they have same radius of measure r cm.      ...(1)

We know, Length of arc = $\frac{\theta }{360°}\times 2\mathrm{\pi }r$

Thus,
$$\begin{gathered} \mathrm{Length} \mathrm{of} \mathrm{arc} AXB=\frac{75°}{360°}\times 2\mathrm{\pi }r ...\left(2\right) \\ \mathrm{Length} \mathrm{of} \mathrm{arc} PYQ=\frac{25°}{360°}\times 2\mathrm{\pi }r ...\left(3\right) \\ \frac{\mathrm{Length} \mathrm{of} \mathrm{arc} AXB}{\mathrm{Length} \mathrm{of} \mathrm{arc} PYQ}=\frac{{\displaystyle \frac{75°}{360°}}\times 2\mathrm{\pi }r}{{\displaystyle \frac{25°}{360°}}\times 2\mathrm{\pi }r} \\ =\frac{{\displaystyle 75°}}{{\displaystyle 25°}} \\ =\frac{{\displaystyle 3}}{{\displaystyle 1}} \end{gathered}$$


Hence, the ratio of the arcs AXB and PYQ is 3 : 1.

Answer:

Given:
AB = CD
OP ⊥ AB and OQ ⊥ CD
∠POQ = 150°     ...(1)


In ∆POQ,
OP = OQ (equal chords are equidistant from the centre)
∴ ∠OPQ = ∠OQP (angles opposite to equal sides are equal)    ...(2)

Now, ∠OPQ + ∠OQP + ∠POQ = 180° (angle sum property)    
⇒ 2∠OPQ +  150° = 180°           (From (1) and (2))
⇒ 2∠OPQ = 180° − 150°
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15°         ...(3)


Since, OP ⊥ AB
Thus, ∠OPA = 90°     ....(4)


Now, ∠OPA = ∠OPQ + ∠APQ     
⇒ 90° = 15° + ∠APQ          (From (3) and (4))
⇒ ∠APQ = 90° − 15°
⇒ ∠APQ = 75°


Hence, ∠APQ = 75°.

Answer:

Given:
OA = 5cm        ...(1)
AB = 8 cm
OD is perpendicular to AB


We know, perpendicular from the centre to the chord bisects the chord.
Therefore, AC = CB = $\frac{1}{2}AB$
⇒ AC = 4 cm     ...(2)


In right angled ∆OAC,
Using pythagoras theorem
OA² = AC² + OC²
⇒ 5² = 4² + OC²    (From (1) and (2))
⇒ 25 = 16 + OC²
⇒ OC² = 25 − 16
⇒ OC² = 9
⇒ OC = 3 cm


OD = 5 cm (radius)
∴ CD = OD − OC
⇒ CD = 5 − 3
⇒ CD = 2 cm

​
Hence, CD is equal to 2 cm.

Answer:

Given:
∠ABC = 20°        ...(1)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ABC
⇒ ∠AOC = 2(20°)     (From (1))
⇒ ∠AOC = 40°

​
Hence, ∠AOC is equal to 40°.

Answer:

Given:
AOB is a diameter of the circle
AC = BC


We know, the diameter subtends a right angle to any point on the circle.
∴ ∠ACB = 90°      ...(1)

In ∆ACB,
AC = BC (given)
∴ ∠CAB = ∠CBA (angles opposite to equal sides are equal)    ...(2)

Now,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ 2∠CAB +  90° = 180°   (From (1) and (2))
⇒ 2∠CAB = 180° −  90°
⇒ 2∠CAB = 90°
⇒ ∠CAB = 45°

​
Hence, ∠CAB is equal to 45°.

Answer:

Given:
∠AOB = 90°      ...(1)
∠ABC = 30°      ...(2)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ 90° = 2∠ACB
⇒ ∠ACB = 45°    ...(3)


In ∆ACB,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ ∠CAB + 30° + 45° = 180°   (From (2) and (3))
⇒ ∠CAB + 75°= 180°
⇒ ∠CAB = 180° −  75°
⇒ ∠CAB = 105°    ...(4)

Also, in ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA (angles opposite to equal sides are equal)    ...(5)

Now,
∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 2∠OAB +  90° = 180°   (From (1) and (5))
⇒ 2∠OAB = 180° −  90°
⇒ 2∠OAB = 90°
⇒ ∠OAB = 45°  ...(6)


​∠CAO = ∠CAB − ∠OAB
           = 105° − 45° (From (4) and (6))
           = 60°

Hence, ∠CAO is equal to 60°.

Answer:

Given:
∠OAB = 40°      ...(1)


In ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA = 40° (angles opposite to equal sides are equal)    ...(2)

Now,
∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 40° + 40° + ∠AOB = 180°   (From (1) and (2))
⇒ ∠AOB = 180° −  80°
⇒ ∠AOB = 100°  ...(3)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ 100° = 2∠ACB    (From (3))
⇒ ∠ACB = 50°


Hence, ∠ACB = 50°.

Answer:

Given:
∠DAB = 60°      ...(1)
∠ABD = 50°      ...(2)


In ∆ADB,
∠DAB + ∠DBA + ∠ADB = 180° (angle sum property)
⇒ 60° + 50° + ∠ADB = 180°   (From (1) and (2))
⇒ ∠ADB = 180° −  110°
⇒ ∠ADB = 70°  ...(3)


We know, angles in the same segment of the circle are equal.
Thus, ∠ADB = ∠ACB
⇒ 70° = ∠ACB    (From (3))
⇒ ∠ACB = 70°


Hence, ∠ACB = 70°.

Answer:

Given:
BC is a diameter of circle
∠BAO = 60°      ...(1)

In ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA = 60° (angles opposite to equal sides are equal)    ...(2)

Also,
∠AOC = ∠OAB + ∠OBA (exterior angle)
⇒ ∠AOC = 60° + 60° (From (2))
⇒ ∠AOC = 120°    ...(3)

We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ADC
⇒ 120° = 2∠ADC    (From (3))
⇒ ∠ADC = 60°


Hence, ∠ADC = 60°.

Answer:

Given:
AOB is a diameter of circle
∠ADC = 120°

Quadrilateral ADCB is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^∘.


Thus, ∠ADC + ∠CBA = 180°
⇒ 120° + ∠CBA = 180°
⇒ ∠CBA = 180° − 120°
⇒ ∠CBA = 60°     ...(1)

We know, the diameter subtends a right angle to any point on the circle.
∴ ∠ACB = 90°      ...(2)


In ∆ACB,
∠CAB + ∠CBA + ∠ACB = 180° (angle sum property)
⇒ ∠CAB + 60° + 90° = 180°   (From (1) and (2))
⇒ ∠CAB = 180° −  150°
⇒ ∠CAB = 30°  ...(3)

Hence, ∠CAB = 30°.

Answer:

Given:
AOC is a diameter of circle
arc AXB = $\frac{1}{2}$ arc BYC
⇒  ∠BOA =  $\frac{1}{2}$∠BOC   ..(1)


Now, ∠BOA + ∠BOC = 180°      (Angles on a straight line)
⇒ $\frac{1}{2}$∠BOC + ∠BOC = 180°      (From (1))
⇒ $\frac{3}{2}$∠BOC = 180°  
⇒ ∠BOC = $\frac{2}{3}\times$180°
⇒ ∠BOC = 120°


Hence, ∠BOC = 120°.

Answer:

Given:
∠ABC = 45^∘   ..(1)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOC = 2∠ABC
⇒ ∠AOC = 2(45°)       (From (1))
⇒ ∠AOC = 90°


Hence, ∠AOC = 90°.

Answer:

Given:
∠ADC = 130^∘     ...(1)
chord BC = chord BE       ...(2)


Quadrilateral ADCB is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^∘.
Thus, ∠ADC + ∠CBA = 180°
⇒ 130° + ∠CBA = 180°
⇒ ∠CBA = 180° − 130°
⇒ ∠CBA = 50°     ...(3)


In ∆CBO and ∆EBO,
BC = BE (given)
OB = OB (common)
OC = OE (radius of the circle)

By SSS property,
∆OCB ≅ ∆OEB

Therefore, ∠OBC = ∠OBE = 50° (by C.P.C.T.)    ...(4)

Thus, ∠CBE = ∠OBE + ∠OBC
                     = 50° + 50°      (From (4))
                     = 100°


Hence, ∠CBE = 100°.

Answer:

Given:
∠ACB = 40^∘      ...(1)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ ∠AOB = 2(40°)       (From (1))
⇒ ∠AOB = 80°           ...(2)


In ∆AOB,
OA = OB (radius)
∴ ∠OAB = ∠OBA (angles opposite to equal sides are equal)         ...(3)


∠OAB + ∠OBA + ∠AOB = 180°  (angle sum property)
⇒ 2∠OAB + 80° = 180°      (From (2) and (3))
⇒ 2∠OAB = 180° − 80°
⇒ 2∠OAB = 100°
⇒ ∠OAB = 50°

​
Hence, ∠AOB = 80° and ​∠OAB = 50°.

Answer:

Given:
AOB is a diameter of the circle


We know, the diameter subtends a right angle to any point on the circle.
∴ ∠AEB = 90°      ...(1)


Quadrilateral ACDE is a cyclic quadrilateral.

In a cyclic quadrilateral, the sum of opposite angles is 180^∘.
Thus, ∠ACD + ∠DEA = 180°      ...(2)


Adding (1) and (2), we get
∠ACD + ∠DEA + ∠AEB = 90° + 180°
⇒ ​∠ACD + ∠DEB = 270°


Hence, ∠ACD + ∠BED = 270°.

Answer:

Given:
∠OAB = 30^∘        ...(1)
∠OCB = 57^∘        ...(2)


In ∆COB,
OC = OB (radius)
∴ ∠OCB = ∠OBC = 57^∘ (angles opposite to equal sides are equal)         ...(3)


∠OCB + ∠OBC + ∠COB = 180°  (angle sum property)
⇒ 57^∘ + 57^∘ + ∠COB = 180°      (From (3))
⇒ ∠COB = 180° − 114°
⇒ ∠COB = 66°                ...(4)


In ∆AOB,
OA = OB (radius)
∴ ∠OAB = ∠OBA = 30^∘ (angles opposite to equal sides are equal)         ...(5)


∠OAB + ∠OBA + ∠AOB = 180°  (angle sum property)
⇒ 30^∘ + 30^∘ + ∠AOB = 180°      (From (5))
⇒ ∠AOB = 180° − 60°
⇒ ∠AOB = 120°         ...(6)


Now,
∠AOB = ∠AOC + ∠COB
⇒ 120° = ∠AOC + 66°      (From (4) and (6))
⇒ ∠AOC = 120° − 66°
⇒ ∠AOC = 54°


Hence, ∠BOC = 66° and ∠AOC = 54°.

Answer:

Consider the smaller circle whose centre is given as ‘O’.

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.

So, here we have

Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

Hence ,the measure of is.

Answer:

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’.

So, from the given figure we have,

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.

One of the properties of a rhombus is that the opposite angles are equal to each other.

So, since it is given that, we can say that the angle opposite it, that is to say that should also have the same value.

Hence we get

Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is.

A property of a circle is 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.

This means that,

Hence the measure of is

Answer:

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to .

Here we have a cyclic quadrilateral ABCD. The centre of this circle is given as ‘O’.

Since in a cyclic quadrilateral the opposite angles are supplementary, here

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 that the chord forms with any point on the circumference of a particular segment is always the same.

Here, ‘CD’ is a chord and ‘A’ and ‘B’ are two points along the circumference on the major segment formed by the chord ‘CD’.

So,

Now,

In any triangle the sum of the interior angles need to be equal to .

Consider the triangle ΔABP,

$$\begin{gathered} \angle PAB+\angle ABP+\angle APB=180° \\ \Rightarrow \angle APB=180°-30°-58° \\ \Rightarrow \angle APB=92° \end{gathered}$$

From the figure, since ‘AC’ and ‘BD’ intersect at ‘P’ we have,

Hence the measure of is.

Answer:

Consider the given circle with the centre ‘O’. Let the radius of this circle be ‘r’. ‘AB’ forms a chord and it subtends an angle of 80° with its centre, that is.

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.

So, here we have

In any triangle the sum of the interior angles need to be equal to 180°.

Consider the triangle

Since,  , we have. So the above equation now changes to

Considering the triangle ΔABC now,

Hence, the measure of is.

Answer:

It is given that ‘ABCD’ is a parallelogram. But since ‘A’ is the centre of the circle, the lengths of ‘AB’ and ‘AD’ will both be equal to the radius of the circle.

So, we have.

Whenever a parallelogram has two adjacent sides equal then it is a rhombus.

So ‘ABCD’ is a rhombus.

Let.

We know that in a circle 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.

By this property we have

In a rhombus the opposite angles are always equal to each other.

So,

Since the sum of all the internal angles in any triangle sums up to in triangle , we have

In the rhombus ‘ABCD’ since one pair of opposite angles are ‘’ the other pair of opposite angles have to be

From the figure we see that,

So now we can write the required ratio as,

Hence the ratio between the given two angles is .

Answer:

Let us first consider the triangle ΔABQ.

It is known that in a triangle the sum of all the interior angles add up to 180°.

So here in our triangle ΔABQ we have,

By a property of the circle we know that an angle formed in a semi-circle will be 90°..

In the given circle since ‘AB’ is the diameter of the circle the angle which is formed in a semi-circle will have to be 90°.

So, we have

Now considering the triangle we have,

From the given figure it can be seen that,

Now, we can also say that,

Hence the measure of the angle is 115°.

Answer:

Consider the circle with the centre ‘P’.

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.

So, here we have

Since ‘ACD’ is a straight line, we have

Now let us consider the circle with centre ‘Q’. Here let ‘E’ be any point on the circumference along the major arc ‘BD’. Now ‘CBED’ forms a cyclic quadrilateral.

In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.

So here,

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.

So, now we have

Hence, the measure of is .

Answer:

Since, O is the circumcentre of $△ABC$, So, O would be centre of the circle passing through points A, B and C.

$\angle ABC$= 90°             (Angle in the semicircle is 90°.)

$\Rightarrow \angle OAB+\angle OBC=90°$                .....(1)

As OA = OB    (Radii of the same circle)

$$\begin{gathered} \therefore \angle OAB=\angle OBA \left(\mathrm{Angle} \mathrm{opposite} \mathrm{to} \mathrm{equal} \mathrm{sides} \mathrm{are} \mathrm{equal}\right) \\ \mathrm{or}, \angle BAC= \angle OBA \\ \mathrm{From} \left(1\right) \\ \mathit{\angle }BAC+\angle OBC=90° \end{gathered}$$

Answer:

We need to find

Answer:

We are given ABCD is a quadrilateral with center O, ∠ADE = 95° and ∠OBA = 30°

We need to find ∠OAC

We are given the following figure

Since ∠ADE = 95°

⇒ ∠ADC = 180 ° − 95° = 85°

Since squo;ABCD is cyclic quadrilateral

This means

∠ABC + ∠ADC = 180°

Since OB = OC (radius)

⇒ ∠OBC = ∠OCB = 65°

In ΔOBC

Since ÐBAC and ÐBOC are formed on the same base which is chord.

So

Consider ΔBOA which is isosceles triangle.

∠OAB = 30°

Answer:

Let AB be a chord of a circle with centre O and radius 8 cm such that

AB = 12 cm

We draw and join OA.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have

Hence the distance of chord from the centre .

Answer:

Given that OA = 10 cm and OL = 5 cm, we have to find the length of chord AB.

Let AB be a chord of a circle with centre O and radius 10 cm such that AO = 10 cm

We draw and join OA.

Since, the perpendiculars from the centre of a circle to a chord bisect the chord.

Now in we have

Hence the length of chord

Answer:

Given that and , find the length of chord AB.

Let AB be a chord of a circle with centre O and radius 6 cm such that

We draw and join OA.

Since, the perpendicular from the centre of a circle to a chord bisects the chord.

Now in we have

$\Rightarrow AL=\sqrt{20}=4.47$

$$\begin{gathered} AB= 2\times AL \\ =2\times 4.47 \\ =8.94 \mathrm{cm} \end{gathered}$$

Hence the length of the chord is 8.94 cm.

Answer:

Let A, B and C are three distinct points on a circle .

Now join AB and BC and draw their perpendicular bisectors.

The point of intersection of the perpendicular bisectors is the centre of given circle.

Hence O is the centre of circle.

Answer:

Let MN is the diameter and chord AB of circle C(O, r) then according to the question

AP = BP.

Then we have to prove that .

Join OA and OB.

       

In ΔAOP and ΔBOP

          (Radii of the same circle)

AP = BP         (P is the mid point of chord AB)

OP = OP         (Common)

Therefore,

                      (by cpct)

Hence, proved.