Circles - Class 9 Maths - Extra Questions

Let ABCD be a cyclic quadrilateral as shown in the figure.

Let us join OB and OD.

Then, $$\angle BOD = 2 \angle BCD$$

$$\angle BCD = \dfrac{1}{2}\angle BOD$$

Similarly, $$\angle BAD = \dfrac{1}{2}\angle DOB$$

$$\therefore  \angle BAD + \angle BCD= \dfrac{1}{2}\angle BOD + \dfrac{1}{2}\angle DOB$$

$$= \dfrac{1}{2} (\angle BOD + \angle DOB)$$

$$= \dfrac{1}{2}\times 360^{\circ}$$

$$= 180^{\circ}$$

Similarly $$\angle B + \angle D = 180^{\circ}$$

ABCD be a quadrilateral circumscribing the circle with centre 0.

            ABCD touches the circle at points P, Q, R and S.

            To prove: Opposite sider subtend supplementary angles at centre

$$\angle AOB + \angle COD = 180^\circ$$   $${\text{and}} \angle AOD + \angle BOC = 180^\circ$$

Proof:  In AOP and AOS.

AP = AS (length of tangent drawn from ext. pt are equal)

AO = AO (common)

OP = OS (both radius)

$$\therefore \,\,\,\,\Delta AOP \cong \Delta AOS\left( {{\text{SSS congruence}}} \right)$$

$$\angle AOP = \angle AOS\left( {CPCT} \right)$$

$$\Rightarrow$$   $$\angle 1 = \angle 8$$    $$....\left( 1 \right)$$

Similarly   $$\angle 2 = \angle 3 =$$   $$....\left( 2 \right)$$

$$\angle 3 = \angle 4$$ $$....\left( 3 \right)$$

$$\angle 6 = \angle 7$$   $$....\left( 4 \right)$$

$$\angle AOB + \angle COD = 180^\circ$$

Now, $$\angle 1 + \angle 2 + \angle 3 + \angle 4 + \angle 5 + \angle 6 + \angle 7 + \angle 8 = 360^\circ$$

$$\left( {\angle 1 + \angle 2} \right) + \left( {\angle 5 + \angle 6} \right) = 180^\circ$$

Similarly, $$\angle BOC + \angle AOD = 180^\circ$$

Hence proved.

Here, $$PQRS$$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$\angle P=\angle QPS=73^o$$ and $$\angle R=\angle QRS$$ are opposite angles.

Then, $$\angle P+\angle R=180^o$$

$$\implies$$   $$73^o+\angle R=180^o$$

$$\implies$$   $$\angle R=180^o-73^o=107^o$$ .

$$\angle QPR=?$$

$$\angle PRS=?$$

$$\angle PSR=?$$

$$\angle PQT=?$$

In $$\Delta PRQ$$ ,

$$\angle P+\angle R+\angle Q=180^{\circ}$$ ...(Angle sum property)

$$\implies$$ $$\angle QPR+90^{\circ}+65^{\circ}=180^{\circ}$$

$$\implies$$ $$\angle QPR=180^{\circ}-155^{\circ}=25^{\circ}$$ .

In $$\Delta PQT$$ ,

$$\angle P+\angle Q+\angle T=180^{\circ}$$   ...(Angle sum property)

$$\implies$$ $$\angle PQT + 90^o + 60^o = 180^o$$

$$\implies$$ $$\angle PQT=180^{\circ}-150^{\circ}=30^{\circ}$$ .

Now,

$$\angle PSR+\angle PQR=180^{\circ}$$ ...(Sum of opposite sides of cyclic quad= $$180^{\circ}$$ )

$$\implies$$ $$\angle PSR=180^{\circ}-65^{\circ}=115^{\circ}$$ .

In $$\Delta PSR$$ ,

$$\angle PSR+\angle P+\angle PRS=180^{\circ}$$   ...(Angle sum property)

$$\implies$$ $$\angle PSR=180^{\circ}-35^{\circ}-115^{\circ}=180^{\circ}-150^{\circ}$$

$$\implies$$ $$\angle PRS=30^{\circ}$$ .

Since $$\angle ABC = 90^o$$

$$AC$$ is the diameter because angle inscribed in semi-circle is a right a angle

Thus by pythagorean theorem, we have

$$AC^2 = AB^2 + BC^2 \\= 8^2 + 6^2 \\= 100$$

$$\therefore AC = 10 \\\ \ \ \ \ \ \ \ \ = d \\ \ \ \ \ \ \ \ \ \ =2r$$

Thus,

$$r = 5\ cm$$

1.   Let us consider any three points $$P, Q$$ and $$R$$ on the circumference of the circle. 

2.   Make chords $$PQ$$ and $$QR$$ by joining  $$P, Q$$ and $$R$$ .

3.   Draw $$AB$$ and $$CD$$ , the perpendicular bisectors of $$PQ$$ and $$QR$$ . They will intersect each other at $$O$$ . 

4. $$O$$ will be the center of the circle.

Let the two circles have centres $$O$$ and $$Q$$ and they intersect at two points $$A$$ and $$B.$$  

Then, $$AB$$ is the  common chord of the two circles.

$$OQ$$ is the line joining the centres of the two circles. Let $$OQ$$ cuts $$AB$$ at $$P$$ .
We need to prove  $$OQ$$ is the  perpendicular  bisector of $$AB$$ .
Draw line segment $$OA, OB, QA$$ and $$QB$$ .

In $$\triangle OAQ$$   and $$\triangle OBQ,$$  

$$OA = OB$$   ....(Radius of the circle)
$$QA = QB$$   ...(Radius of the circle)
and $$OQ = OQ$$   ....(common side)     
$$\triangle OAQ \cong \triangle OBQ$$     ....SSS test of  congruence

$$\Rightarrow \angle AOQ = \angle BOQ$$         ....c.a.c.t.     ....(1)

[Since $$\angle AOQ= \angle AOP$$ and $$\angle BOP= \angle BOQ$$ ]

In triangles $$AOP$$ and $$BOP$$ , 

$$OA = OB$$    ....(Radius of the circle)
$$\angle AOP= \angle BOP$$        .....From (1)

$$OP = OP$$       .....Common side
$$\triangle AOP \cong \triangle BOP$$    ....SAS test of congruence, 

$$\Rightarrow AP=BP$$ and $$\angle APO = \angle BPO$$

But,   $$\angle APO +\angle BPO= 180^o$$

$$2 \angle APO=180^o \Rightarrow \angle APO=90^o$$

$$AP = BP$$  and $$\angle APO =\angle BPO=90 ^o$$

Hence, $$OQ$$ is the perpendicular  bisector of $$AB$$ .

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius $$5$$ m at points $$R, S$$ and $$M$$ respectively.

Given: $$RS = SM = 6$$ m

We know that center $$O$$ lies on the bisector of $$\angle RSM$$ i.e. ray $$OS$$

Let $$OA \perp$$ chord $$RM$$

We know the perpendicular from center to the chord, bisects the chord.

Hence $$RA = AM$$

In $$\triangle SAR$$ , $$\angle A=90^o$$

$$SR^2=SA^2+RA^2$$               [By Pythagoras theorem]

$$\Rightarrow 36= SA^2+RA^2$$

$$\Rightarrow RA^2=36-SA^2$$        ... (1)

In the $$\triangle OAR$$ , 

$$RO^2=AO^2+RA^2$$              [By Pythagoras theorem]

$$\Rightarrow 25=(OS-AS)^2+RA^2$$

$$\Rightarrow RA^2=25-(OS-AS)^2$$

$$RA^2= 25- (5-AS)^2$$           ... (2)

From (1) and (2) , we get

$$\Rightarrow 36- SA^2=25-(5-SA)^2$$

$$\Rightarrow 11-SA^2+(5-SA)^2=0$$

$$\Rightarrow 11-SA^2+25-10SA+AS^2=0$$

$$\Rightarrow 10 SA=36$$

$$\Rightarrow SA=3.6$$

Putting $$SA = 3.6$$ in (1), we get

$$RA^2=3.6-(3.6)^2$$

          $$=36-12.96$$

          $$=36-12.96$$

          $$=23.04$$  m

$$\therefore RA = 4.8$$ m 

$$\Rightarrow RM= 2 RA= 2\times 4.8=9.6$$  m

Hence, the distance between Reshma and Mandip $$= 9.6$$ m

From the figure,

$$\Rightarrow \angle ADE +\angle ADC=180^o$$ (Linear pair) ....... (i)

Also, $$\Box ABCE$$ is a cyclic quadrilateral. 

So, $$\angle AED +\angle ABC=180^o$$  ....... (ii) (opposite angles of cyclic quadrilateral sum upto $$180^{o}$$ )

$$\angle ADC=\angle ABC$$ ....... (iii) (Opposite angles of a parallellogram are equal)

Using (i), (ii) and (iii), 

$$\angle AED+\angle ABC =\angle ADE+\angle ABC$$

$$\Rightarrow \angle AED = \angle ADE$$

Now, In $$\triangle AED$$ ,

$$\angle AED =\angle ADE$$

$$\Rightarrow AD=DE$$ (Sides opposite to equal angles)

In $$\triangle DEF$$ ,

$$\angle D= \angle EDF$$

But $$\angle EDF = \angle EDA +\angle FDA$$      ....angle addition property

Now, $$\angle EDA = \angle EBA$$ and $$\angle FDA = \angle FCA$$      .....Angles inscribed in the same arc

$$\therefore \angle EDF = \angle EBA + \angle FCA$$

$$=\dfrac 12 \angle B+ \dfrac 12 \angle C$$

[Since $$BE$$ is bisector of $$\angle B$$ and $$CF$$ is bisector $$\angle C$$ ]

$$\therefore \angle D= \dfrac {\angle B+\angle C}{2}$$        ....(1)

Similarly,  $$\angle E= \dfrac {\angle C+\angle A}{2}$$ and  $$\angle F= \dfrac {\angle A+ \angle B}{2}$$              ...(2)

Now, $$\angle A + \angle B + \angle C = 180^o$$        ....angle sum property of triangle

$$\therefore \angle B + \angle C = 180^{\circ}- \angle A$$          ...(3)

Similarly, $$\angle C + \angle A = 180^o - \angle B$$ and   $$\angle A + \angle B = 180^o - \angle C$$                   ...(4)

Substituting eq (3) in eq (1), we get

$$\therefore \angle D = \dfrac {180^o - \angle A}{2}$$

$$\angle D = 90^o - \dfrac 12 \angle A$$

Similarly, $$\angle E = 90^o- \dfrac 12 \angle B$$ and  $$\angle F = 90^o- \dfrac 12 \angle C$$

Join $$OA$$ and $$OC$$ .

Let the radius of the circle be $$r$$ cm and $$O$$ be the centre 

Draw $$OP \bot AB$$  and $$OQ \bot CD$$ . 

We know, $$OQ \bot CD$$ ,  $$OP \bot AB$$  and $$AB \parallel CD$$ . 

Therefore, points $$P, O$$ and $$Q$$ are collinear. So, $$PQ = 6$$ cm. 

Let $$OP = x$$ . 

Then, $$OQ = (6 – x)$$ cm. 
And $$OA = OC = r$$ . 

Also, $$AP = PB = 2.5$$ cm and $$CQ = QD = 5.5$$ cm.

(Perpendicular from the centre to a chord of the circle bisects the chord.)

  
In right triangles  $$QAP$$ and $$OCQ$$ , we have

$$OA^2=OP^2+AP^2$$ and $$OC^2=OQ^2+CQ^2$$

$$\therefore r^2=x^2+(2.5)^2$$                  ..... (1) 

and $$r^2=(6-x)^2+(5.5)^2$$      ..... (2) 

$$\Rightarrow x^2+(2.5)^2=(6-x)^2+(5.5)^2$$

$$\Rightarrow x^2+6.25=36-12x+x^2+30.25$$

$$12x=60$$

$$\therefore x=5$$

Putting $$x = 5$$ in (1), we get 

$$r^2=52+(2.5)^2=25+6.25=31.25$$

$$\Rightarrow r^2=31.25 \Rightarrow r=5.6$$
Hence, the radius of the circle is $$5.6$$ cm

$$\Rightarrow$$ In $$\triangle BDC$$ , 

$$\Rightarrow$$ $$\angle ADC$$ is the exterior angle 

$$\therefore \angle ADC=\angle DBC+\angle DCB$$      .... (1) 

(The measure of an exterior angle is equal to the sum of the remote interior angles)

$$\Rightarrow$$ By inscribed angle theorem,

$$\Rightarrow$$ $$\angle ADC= \dfrac 12\angle AOC$$ and $$\angle DCB=\dfrac 12 \angle DOE$$    ....... (2) 

From (1) and (2), we have

$$\Rightarrow$$ $$\dfrac 12 \angle AOC=\angle ABC+\dfrac 12 \angle DOE$$    ...[Since $$\angle DBC= \angle ABC$$ ] 

$$\Rightarrow \angle ABC=\dfrac 12 (\angle AOC- \angle DOE)$$

Hence,  $$\angle ABC$$  is equal to half the difference of angles subtended by the chords $$AC$$ and $$DE$$ at the centre.

Given:  Angle bisector of angle A and perpendicular bisector of BC intersect. 

Let the bisector of A meet the circumcircle at E.

In $$\triangle ABE$$ and $$\triangle AEC$$ ,

$$\angle BAE = \angle CAE$$         {AE Angle bisector of A}

$$\therefore BE = EC$$   .....(1)    {converse of the angle subtended by the chord of equal length at the same point are equal} 

Let D be the midpoint of BC and now connect ED.

In $$\triangle BDE$$  and $$\triangle CDE$$ ,  
$$BE = CE$$      ....from eq. (1)
$$BD = CD$$     ...{ $$D$$ is mid point of $$BC$$ }

$$DE = DE$$     ...common side

$$\therefore \triangle BDE \cong \triangle CDE$$      ...SSS test of congruence

$$\Rightarrow \angle BDE= \angle CDE$$     .....c.p.c.t.

Also, $$\angle BDE+\angle CDE=180^{\circ}$$      ...angles in linear pair

$$\therefore \angle BDE = \angle CDE = 90^{\circ}$$     

Therefore, $$DE$$ is the right angled bisector of $$BC$$ .

Hence, they intersect on the circumcircle of triangle ABC

Here, given a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$110^o$$ and $$x^o$$ are opposite angles.

Then, $$110^o+x^o=180^o$$

$$\implies$$   $$x^o=180^o-110^o$$

$$\implies$$   $$x^o=70^o$$ .

Also, $$85^o$$ and $$y^o$$ are opposite angles.

Then, $$85^o+y^o=180^o$$

$$\implies$$   $$y^o=180^o-85^o$$

$$\implies$$   $$y^o=95^o$$ .

Since $$CS$$ is perpendicular To $$AB$$ and co passes through center. $$CD$$ bisects $$AB$$

$$\implies AE = BE$$ where E is the point of intersection of $$AB$$ and $$CD$$

In $$\triangle ADE$$ and $$\triangle BDE$$

$$ED$$ is common

$$\angle DEA = \angle DEB = 90$$

$$AE = BE$$

By $$SAS$$ congruency the 2 triangles are congruent

$$AB = BD$$

Here, $$ABCD$$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$\angle B=2x^o$$ and $$\angle D=3x^o$$ are opposite angles.

Then, $$\angle B+\angle D=180^o$$

$$\implies$$   $$2x^o+3x^o=180^o$$

$$\implies$$   $$5x^o=180^o$$

$$\implies$$   $$x^o=\dfrac{180^o}{5}$$

$$\implies$$   $$x^o=36^o$$ .

Therefore,

$$\angle POQ \, is \, 180^\circ$$ since it is a straight line and angle subtended by arc of a circle at its center is twice the angle formed by the same arc of a point on the circle.

$$\angle POQ  = 2 \angle PRQ$$

$$\angle PRQ= \dfrac{180^o}{2} = 90^o$$

Where R is any point on the circle. 

$$\angle PRQ = \angle  PMQ = 90^{\circ}$$     (angle in the semi-circle)

In $$\triangle PRQ$$

(i) $$\angle QPR = 180 - \angle PRQ - \angle PQR = 180 – 90 – 55 = 35^\circ$$

In $$\triangle PMQ$$

(ii) $$\angle QPM = 180 – 50 – 90 = 40^\circ$$

$$\angle SPQ = \angle SPR + \angle RPQ = 25 + 35 = 60^{\circ}$$

(iii) PQRS is a cyclic quadrilateral. So opposite angles are supplementary.

$$\implies \angle SPQ + \angle SRQ = 180^{\circ}$$

$$\angle SRQ = 180 – 60 = 120^{\circ}$$

$$\angle SRP + \angle PRQ = 120^{\circ}$$

$$\implies \angle PRS = 120 – 90 = 30^{\circ}$$

Consider $$\angle CAD$$ and $$\angle CBD$$

They are both angles made by the same chord $$CD$$

$$\Rightarrow \angle CAD = \angle CBD = 30^\circ$$

Since $$ABCD$$ is a cyclic quadrilateral opposite angles are supplementary.

$$\Rightarrow \angle DAB + \angle DCB = 180^\circ$$

$$\angle DCB = 180^{\circ} – (\angle DAC + \angle CAB)$$

$$\angle DCB = 180^o – 30^o – 50^o = 100^\circ$$

As the radius of the given circles are equal so the common chord will bisect the line joining their centres

$$OO'=\sqrt { { (a-b) }^{ 2 }+{ (b-a) }^{ 2 } } \\ OO'=\sqrt { { (a-b) }^{ 2 }+{ (a-b) }^{ 2 } } \\ OO'=\sqrt { 2 } \left| a-b \right|$$

Now $$OO'=2AB$$

$$\Rightarrow OB=\dfrac { \sqrt { 2 } \left| a-b \right|  }{ 2 } =\dfrac { \left| a-b \right|  }{ \sqrt { 2 }  }$$

In $$\triangle AOB$$ $$,$$ $${ OA }^{ 2 }={ OB }^{ 2 }+{ AB }^{ 2 }$$

$$\Rightarrow { c }^{ 2 }={ \left( \dfrac { \left| a-b \right|  }{ \sqrt { 2 }  }  \right)  }^{ 2 }+{ AB }^{ 2 }\\ \Rightarrow { AB }^{ 2 }={ c }^{ 2 }-\dfrac { { (a-b) }^{ 2 } }{ 2 } \\ \Rightarrow { AB }^{ 2 }=\dfrac { 2{ c }^{ 2 }-{ (a-b) }^{ 2 } }{ 2 } \\ \Rightarrow AB=\sqrt { \dfrac { 2{ c }^{ 2 }-{ (a-b) }^{ 2 } }{ 2 }  } =\dfrac { \sqrt { 4{ c }^{ 2 }-{ 2(a-b) }^{ 2 } }  }{ 2 }$$

Length of common chord $$=AC=2AB$$

$$\Rightarrow AC=2\times \dfrac { \sqrt { 4{ c }^{ 2 }-{ 2(a-b) }^{ 2 } }  }{ 2 } =\sqrt { 4{ c }^{ 2 }-{ 2(a-b) }^{ 2 } }$$

Hence proved.

Equation of circle is   $${ x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }$$

Let the line $$\dfrac { x }{ a } +\dfrac { y }{ b } =1$$ intersect the circle at $$A$$ and $$B$$

Draw $$OC$$ perpendicular to $$AB$$

$$OC=\dfrac { |0+0-1| }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } }  }  } =\dfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }$$

In $$\triangle OAC$$ $$,$$ $${ OA }^{ 2 }={ AC }^{ 2 }+{ OC }^{ 2 }$$

$$\Rightarrow { r }^{ 2 }={ \left( \dfrac { ab }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 } }  }  \right)  }^{ 2 }+{ AC }^{ 2 }$$

$$\Rightarrow { AC }^{ 2 }={ r }^{ 2 }-\dfrac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } }$$

$$\Rightarrow { AC }^{ 2 }=\dfrac { { r }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })-{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } }$$

$$\Rightarrow AC=\sqrt { \dfrac { { r }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })-{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } }  }$$

length of chord $$=AB=2AC$$

$$\Rightarrow AB=2\sqrt { \dfrac { { r }^{ 2 }({ a }^{ 2 }+{ b }^{ 2 })-{ a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 } }  }$$

Here, $$PQRS$$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^\circ$$ .

Here, $$\angle PSR=82^\circ$$ and $$\angle PQR$$ are opposite angles.

Then, $$\angle PSR+\angle PQR=180^\circ$$

$$\Rightarrow 82^\circ+\angle PQR=180^\circ$$

$$\Rightarrow \angle PQR=180^\circ-82^\circ$$

$$\Rightarrow\angle PQR=98^\circ$$

Now,

$$\angle PQS+\angle RQS=\angle PQR$$

$$\Rightarrow$$   $$55^\circ+\angle RQS=98^\circ$$

$$\Rightarrow$$   $$\angle RQS=98^\circ-55^\circ$$

$$\Rightarrow\angle RQS=43^\circ$$ .

Hence, $$\angle RQS$$ is equal to $$43^\circ.$$

Given: Side of the square $$=a$$ cm

$$\therefore$$ Diameter of the circle $$=a$$ cm

Hence radius of the circle = $$\dfrac{a}{2}$$ cm

$$\therefore$$ Area of the circle $$=\pi r^2$$

                                  $$=\pi\times\left(\dfrac{a}{2}\right)^2$$

                                  $$=\pi\times\dfrac{a^2}{4}$$

                                  $$=\dfrac{22}{7}\times\dfrac{a^2}{4}$$

                                  $$=\dfrac{11a^2}{14}$$

Let (x,y) be a common point of chord y=mx+c, y=kx & 3x²−y²−2x+4y=0
First two give x & y which can be used in third to give (3−k²)c + (4k−2)(k−m) = 0
As a quadratic in k this is k²(4−c)−2k(2m+1)+(2m+3c) = 0
If the two values of k arising from this give perp lines thro O then $$\dfrac{(2m+3c)}{(4−c)} = −1$$
→ 2m+3c = −4+c → c=−m−2
Chord is thus y = mx–m−2 → y+2 = m(x−1) which always contains (1,−2)

Given: $$\angle PQR = 1.2$$

$$OQ = 20cm$$

To find: $$x = ?$$

Since OQ, OA and OB are radius:

OQ=20cm

OA=20cm

OB=20cm

Now, OQ=OC+CQ=OC+x

Also, CQ,CP,CR are radius , all equal to x

We know that OQ bisects

$$\angle AOB\,\,\,\omega \,\,OQ$$ passes

Through center C of smaller circle.

Then, $$\angle POC = {{1.2} \over 2} = 0.6$$

In $$\Delta OCP$$ , $$\sin \angle POC = {{PC} \over {OC}}$$

$$\Rightarrow \sin \left( {0.6} \right) = {x \over {OQ - x}}\left[ {OQ = OC + x} \right]$$

$$\Rightarrow \left( {20 - x} \right)\sin \left( {0.6} \right) = x$$

$$\Rightarrow x\left( {1 + \sin \left( {0.6} \right)} \right) = 20\sin \left( {0.6} \right)$$

$$\Rightarrow x = {{20\sin \left( {0.6} \right)} \over {1 + \sin \left( {0.6} \right)}}$$

Thus, $$x = {{20\sin \left( {0.6} \right)} \over {1 + \sin \left( {0.6} \right)}}$$

Here, $$AQST$$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$\angle A=\angle TAQ$$ and $$\angle S=\angle TSQ$$ are opposite angles.

Then clearly, $$\angle A+\angle S=180^o$$

$$\implies$$   $$\angle TAQ+\angle TSQ=180^o$$ .

That is the sum of $$\angle TAQ$$ and $$\angle TSQ$$ is $$180^o$$ .

Here, $$ABCD$$ is a cyclic quadrilateral.

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$\angle A=y$$ and $$\angle C=2y$$ are opposite angles.

Then, $$\angle A+\angle C=180^o$$

$$\implies$$   $$y+2y=180^o$$

$$\implies$$   $$3y=180^o$$

$$\implies$$   $$y=60^o$$ .

Here, $$\angle B=3x$$ and $$\angle D=2x$$ are opposite angles.

Then, $$\angle B+\angle D=180^o$$

$$\implies$$   $$3x+2x=180^o$$

$$\implies$$   $$5x=180^o$$

$$\implies$$   $$x=36^o$$ .

Hence, the four angles are:

$$\angle A=y=60^o$$ ,

$$\angle B=3x=3(36^o)=108^o$$ ,

$$\angle C=2y=2(60^o)=120^o$$

and $$\angle D=2x=2(36^o)=72^o$$ .

We have,

Circumfrance $$=2\pi r=44$$

$$\pi r=\dfrac{44}{2}$$

$$r=\dfrac{22}{\pi }=\dfrac{22}{\dfrac{22}{7}}$$

$$r=7$$

So, area of quadrant $$=\dfrac{1}{4}\pi {{r}^{2}}$$

$$\Rightarrow \dfrac{1}{4}\times \dfrac{22}{7}\times 7\times 7$$

$$\Rightarrow \dfrac{77}{2}$$

$$\Rightarrow 38.5\,\,c{{m}^{2}}$$

Hence, this is the answer.

Here, $$MNOP$$ is a cyclic quadrilateral.

Given, $$\angle P=4x+12, \angle N=3x+28$$ .

We know, in a cyclic quadrilateral, the sum of opposite angles are supplementary, i.e. $$180^o$$ .

Here, $$\angle P$$ and $$\angle N$$ are opposite angles.

Then, $$\angle P+\angle N=180^o$$

$$\implies$$   $$(4x+12)+(3x+28)=180^o$$

$$\implies$$   $$4x+3x+12+28=180^o$$

$$\implies$$   $$7x+40=180^o$$

$$\implies$$   $$7x=180^o-40^o$$

$$\implies$$   $$7x=140^o$$

$$\implies$$   $$x=20^o$$ .

Then, $$\angle P=4x+12=4(20^o)+12=80^o+12=92^o$$

and $$\angle N=3x+28=3(20^o)+28=60^o+28=88^o$$ .