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Maths 4 Marks Questions

Circles · CBSE STD 9 (CBSE - English Medium). 83 questions.

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Question 14 Marks
Answer
Let us draw a perpendicular OM on line AD.

It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
$\therefore$ BM = MC ...(1)
And, AM = MD ...(2)
On subtracting equation (2) from (1), we obtain
AM − BM = MD − MC
⇒ AB = CD
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Question 24 Marks
A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and $\angle\text{ADC}=130^\circ.$ Find $\angle\text{BAC}.$
Answer


Draw a quadrilateral ABCD inscribed in a circle having centre O.Given, $\angle\text{ADC}=130^\circ$
Since, ABCD is a quadrilateral. inscribed in a circle, therefore ABCD becomes a cyclic quadrilateral.
$\because$ Since, the sum of opposite angle of a cyclic quadrilateral is 180°.
$\therefore\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=50^\circ$
Since, AB is a diameter of a circle, then AB subtends an angle to the circle is right angle.
$\therefore\angle\text{ACB}=90^\circ$
In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
$\Rightarrow\angle\text{BAC}+90^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-(90^\circ+50^\circ)$
$=180^\circ-140^\circ=40^\circ$
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Question 34 Marks
AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.
Answer
Given: AS and AC are two equal chords whose centre is O.

To prove: Centre O lies on the bisector of $\angle\text{BAC}.$
Construction: Join SC, draw bisector AD of $\angle\text{BAC}.$
Proof In $\angle\text{SAM}$ and $\angle\text{CAM},$
AS = AC [given]
$\angle\text{BAM} = \angle\text{CAM}$ [given]
and AM = AM [common side]
$\therefore\triangle\text{BAM}=\triangle\text{CAM}$ [by SAS congruence rule]
$\Rightarrow\text{BM}=\text{CM}$ [by CPCT]
and $\angle\text{MBA}=\angle\text{CMA}$ [by CPCT]
So, BM = CM and $\angle\text{BMA}=\angle\text{CMA}=90^\circ$
So, AM is the perpendicular bisector of the chord BC.
Hence, bisector of $\angle\text{BAC}$ i.e., AM passes through the centre O.
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Question 44 Marks
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
Answer
ABCD is a parallelogram. A circle through A, B is so drawn that intersects AD at P and BC at Q. We have to prove that P, Q, C and D are concyclic. Join PQ.

Now, side AP of the cyclic quadrilateral APQB produced to D.
$\therefore\ \text{Ext. }\angle1=\text{int.opp. }\angle\text{B}$
$\because$ BA || CD and BC cuts them
$\therefore\angle\text{B}+\angle\text{C}=180^\circ$
[ $\because$ sum of int. $\angle\text{s}$ on the same side of the transversal is 180°]
or $\angle1+\angle\text{C}=180^\circ\ [\therefore\angle1=\angle\text{B}\ (\text{proved})]$
$\therefore$ PDCQ is cyclic quadrilateral.
Hence, the point P, Q, and are concyclic.
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Question 54 Marks
Prove that the line joining the mid-point of a chord to the centre of the circle passes through the mid-point of the corresponding minor arc.
Answer
Given:

C is the mid-point of chord AB.
To prove: D is the mid-point of arc AB.
Proof:
In $\triangle\text{OAC}$ and $\triangle\text{OBC}$
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [C is the mid-point of AB]
Then $\triangle\text{OAC}\cong\triangle\text{OBC}$ [By SSS condition]
$\therefore\angle\text{AOC}=\angle\text{BOC}$
$\Rightarrow\text{m}\overline{\text{A}}\text{D}\cong\text{m}\overline{\text{B}}\text{D}$
$\Rightarrow\overline{\text{A}}\text{D}\cong\overline{\text{B}}\text{D}$
Hence, D is the mid-point of arc AB.
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Question 64 Marks
If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.
Answer
Given: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at points L and M.
To prove: AB || CD
Proof: AB and CD are two chords of a circle with centre O. Diameter POQ bisects them at L and M.

Then $\text{OL}\perp\text{AB}$
Also, $\text{OM}\perp\text{CD}$
$\therefore\ \angle\text{ALM}=\angle\text{LMD}=90^\circ$
Since alternate angles are equal, we have:
AB || CD
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Question 74 Marks
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
Answer
Given: $\triangle\text{ABC}$ is an isosceles triangle such that AB = AC and also DE || SC.
To prove: Quadrilateral BCDE is a cyclic quadrilateral.
Construction: Draw a circle passes through the points B, C, D and E.

Proof: In $\triangle\text{ABC},\ \ \text{AB}=\text{AC}$ [equal sides of an isosceles triangle]
$\Rightarrow\angle\text{ACB}=\angle\text{ABC}\ \ ...(\text{i})$
[angles opposite to the equal sides are equal]
Since, DE || BC
$\Rightarrow\angle\text{ADE}=\angle\text{ACB}$ [corresponding angles] ....(ii)
On adding both sides by $\angle\text{EDC}$ in Eq. (ii), we get,
$\angle\text{ADE}+\angle\text{EDC}=\angle\text{ACB}+\angle\text{EDC}$
$\Rightarrow180^\circ=\angle\text{ACB}+\angle\text{EDC}$ [$\angle\text{ADE}$ and $\angle\text{EDC}$ from linear pair aniom]
$\Rightarrow\angle\text{EDC}+\angle\text{ABC}=180^\circ$ [from Eq. (i)]
Hence, BCDE is cyclic quadrilateral, because sum opposite angles is 180°.
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Question 84 Marks
In the given figure, O is the centre of the circle. if $\angle\text{PBC}=25^\circ$ and $\angle\text{APB}=110^\circ,$find the value of $\angle\text{ADB}.$
Answer
From the given diagram, we have:

$\angle\text{ACB}=\angle\text{PCB}$
$\angle\text{BPC}=(180^\circ-110^\circ)=70^\circ$ [Linear pair]
Considering $\triangle\text{PCB},$ we have:
$\angle\text{PCB}+\angle\text{BPC}+\angle\text{PBC}=180^\circ$ [Angle sum property]
$\Rightarrow\ \angle\text{PCB}+70^\circ+25^\circ=180^\circ$
$\Rightarrow\ \angle\text{PCB}=(180^\circ-95^\circ)=85^\circ$
$\Rightarrow\ \angle\text{ACB}=\angle\text{PCB}=85^\circ$
We know that the angles in the same segment of a circle are equal.
$\therefore\ \angle\text{ADB}=\angle\text{ACB}=85^\circ$
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Question 94 Marks
In Fig. O is the centre of the circle, BD = OD and $\text{CD}\bot\text{AB}.$ Find $\angle\text{CAB}.$
Answer
Given, in the figure $\text{BD}=\text{OD},\ \text{CD}\bot\text{AB}$

In $\triangle\text{OBD},\ \ \ \text{BD}=\text{OD}$ [given]
$\text{OD}=\text{OB}$ [both are the radius of circle]
$\therefore\text{OB}=\text{OD}=\text{BD}$
Thus, $\triangle\text{ODB}$ is an equilateral triangle.
$\therefore\angle\text{BOD}=\angle\text{OBD}=\angle\text{ODB}=60^\circ$
In $\triangle\text{MBC}$ and $\triangle\text{MBD},\ \ \text{MB}=\text{MB}$ [common side]
$\angle\text{CMB}=\angle\text{BMD}=90^\circ$ and CM = MD [in a circle, any perpendicular draw on a chord also bisects the chord]
$\therefore\triangle\text{MBC}=\triangle\text{MBD}$ [by SAS congruence rile]
$\therefore\angle\text{MBC}=\angle\text{MBD}$ [by CPCT]
$\Rightarrow\angle\text{MBC}=\angle\text{OBD}=60^\circ\ \ \ [\because\angle\text{OBD}=60^\circ]$
Since, AB is a diameter of the circle,
$\therefore\angle\text{ACB}=90^\circ$
In $\triangle\text{ACB},\ \ \ \angle\text{CAB}+\angle\text{CBA}+\angle\text{ACB}=180^\circ$ [by angle sum property of triangle]
$\Rightarrow\angle\text{CAB}+60^\circ+90^\circ=180^\circ$
$\Rightarrow\angle\text{CAB}=180^\circ-(60^\circ+90^\circ)=30^\circ$
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Question 104 Marks
Two chords AB, CD of lengths 5cm, 11cm respectively of a circle are parallel. If the distance between AB and CD is 3cm, find the radius of the circle.
Answer
Let AB and CD be two parallel chord of the circle with center o such that AB = 5cm and CD = 11cm.let the radius of the circle be r cm.

Draw $\text{OP}\perp\text{AB}$ and $\text{OQ}\perp\text{CD}$ as well as point O, Q and P are collinear.
Clearly, PQ = 3cm
Let OQ = x then OP = x + 3
In $\triangle\text{OAP}$ and $\triangle\text{OCQ}$ we have
$\text{OA}^2=\text{OP}^2+\text{AP}^2$
$\Rightarrow\text{r}^2=(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2\dots(1)$
And
$\text{OC}^2=\text{OQ}^2+\text{CQ}^2$
$\Rightarrow\text{r}^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2\dots(2)$
From (1) and (2) we get
$(\text{x}+3)^2+\Big(\frac{5}{2}\Big)^2=\text{x}^2+\Big(\frac{11}{2}\Big)^2$
$\Rightarrow\text{x}^2+\text{6x}+9+\frac{25}{4}=\text{x}^2+\frac{121}{4}$
$\Rightarrow\text{6x}+\frac{61}{4}=\frac{121}{4}$
$\Rightarrow\text{6x}=\frac{121-61}{4}$ $\Rightarrow\text{6x}=\frac{60}{4}$
$\Rightarrow\text{x}=\frac{5}{2}$
Putting the value of x in (2) we get,
$\text{r}^2=\Big(\frac{5}{2}\Big)^2+\Big(\frac{11}{2}\Big)^2$
$=\frac{25}{4}+\frac{121}{4}$
$=\frac{146}{4}$
$\Rightarrow\text{r}=\sqrt{\frac{146}{4}}$
$\text{r}=\sqrt{\frac{146}{4}}\text{cm}$
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Question 114 Marks
In Fig. O is the centre of the circle, $\angle\text{BCO}=30^\circ.$ Find x and y.
Answer
Given, Ois the centre of the circle and $\angle\text{BCO}=30^\circ.$ In the given figure join OB and AC.

In $\triangle\text{BOC},\ \ \ \text{CO}=\text{BO}$ [both are the radius of circle]
$\therefore\angle\text{OBC}=\angle\text{OCB}=30^\circ$ [angles opposite to equal sides are equal]
$\therefore\angle\text{BOC}=180^\circ-(\angle\text{OBC}+\angle\text{OCE})$ [by angle sum property of a triangle]
$=180^\circ-(30^\circ+30^\circ)=120^\circ$ $\angle\text{BOC}=2\angle\text{BAC}$
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore\angle\text{BAC}=\frac{120^\circ}{2}=60^\circ$
Also, $\angle\text{BAE}=\angle\text{CAE}=30^\circ$ [AE is an angle bisector of angle A]
$\Rightarrow\angle\text{BAE}=\text{x}=30^\circ$
In $\triangle\text{ABE},\ \ \ \angle\text{BAE}+\angle\text{EBA}+\angle\text{AEB}=180^\circ$ [by angle sum property of triangle]
$\Rightarrow30^\circ+\angle\text{EBA}+90^\circ=180^\circ$
$\therefore\angle\text{EBA}=180^\circ-(90^\circ+30^\circ)=180^\circ-120^\circ=60^\circ$
Now, $\angle\text{EBA}=60^\circ$
$\Rightarrow\angle\text{ABD}+\text{y}=60^\circ$
$\Rightarrow\frac{1}{2}\times\angle\text{AOD}+\text{y}=60^\circ$
[in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle]
$\Rightarrow\frac{90^\circ}{2}+\text{y}=60^\circ$
$[\because\angle\text{AOD}=90^\circ,\text{ given}]$
$\Rightarrow45^\circ+\text{y}=60^\circ$
$\Rightarrow\text{y}=60^\circ-45^\circ$
$\therefore\text{y}=15^\circ$
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Question 124 Marks
In a cyclic quadrilateral ABCD, if $\angle\text{A}-\angle\text{C}=60^\circ,$ prove that the smaller of two is 60º
Answer
We have
$\angle\text{A}-\angle\text{C}=60^\circ\dots(1)$
Since, ABCD is a cyclic quadrilateral
Then $\angle\text{A}+\angle\text{C}=180^\circ\dots(2)$
Add equations (1) and (2)
$\angle\text{A}-\angle\text{C}+\angle\text{A}+\angle\text{C}=60^\circ+180^\circ$
$\Rightarrow2\angle\text{A}=240^\circ$
$\Rightarrow\angle\text{A}=\frac{240^\circ}{2}=120^\circ$
Put value of $\angle\text{A}$ in equation (2)
$120^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-120^\circ=60^\circ$
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Question 134 Marks
In Fig. $\angle\text{ADC}=130^\circ$ and chord BC = chord BE. Find $\angle\text{CBE}.$
Answer
In the given figure, we have $\angle\text{ADC}=130^\circ$ and chord BC = BE. We have to find $\angle\text{CBE}.$ Since ABCD is a cyclic quadrilateral and the opposite angles of a cyclic quadrilateral are supplementary.
$\therefore\angle\text{D}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\angle\text{OBC}=50^\circ....(1)$
In $\triangle\text{OBC}$ and $\triangle\text{OBE},$ we have
BC = BE [Given]
OC = OE [Radii of same circle]
OB = OB [Common side]
$\therefore\triangle\text{OBC}\cong\triangle\text{OBE}$ [By SSS cong. Rule]
$\angle\text{OBC}+\angle\text{OBE}=50^\circ$ $\big[$By C.P.C.T. and by (1) $\angle\text{OBC}=50^\circ\big]$
$\therefore\angle\text{OBC}+\angle\text{OBE}=50^\circ+50^\circ=100^\circ$
Hence, $\angle\text{CBE}=100^\circ$
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Question 154 Marks
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that $\angle\text{AED} = 95^\circ$ and $\angle\text{OBA} = 30^\circ$ Find $\angle\text{OAC.}$
Answer
We are given ABCD is a quadrilateral with center O, $\angle\text{ADE}=95^\circ$ and $\angle\text{OBA}=30^\circ$
We need to find $\angle\text{OAC}$
We are given the following figure:

Since $\angle\text{ADE}=95^\circ$
$\Rightarrow\angle\text{ADC}=180^\circ-95^\circ=85^\circ$
Since squo; ABCD is cyclic quadrilateral This means
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABO}+\angle\text{OBC}+\angle\text{ADC}=180^\circ$
$\Rightarrow30^\circ+\angle\text{OBC}+85^\circ=180^\circ$
$\Rightarrow\angle\text{OBC}=180^\circ-115^\circ=65^\circ$ Since OB = OC (radius)
$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=65^\circ$ In $\triangle\text{OBC}$
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\angle\text{OBC}=180^\circ$
$\angle\text{BOC}+2\times65^\circ=180^\circ$
$\angle\text{BOC}=180^\circ-130^\circ$ $\angle\text{BOC}=50^\circ$
Since DBAC and DBOC are formed on the same base which is chord.
So,
$\angle\text{BAC}=\frac{\angle\text{BOC}}{2}$
$=\frac{50^\circ}{2}$ $\angle\text{BAC}=25^\circ$
Consider $\triangle\text{BOA}$ which is isosceles triangle.
$\angle\text{OAB}=30^\circ$ $\Rightarrow\angle\text{OAC}+\angle\text{BAC}=30^\circ$
$\Rightarrow\angle\text{OAC}+25^\circ=30^\circ$
$\Rightarrow\angle\text{OAC}=5^\circ$
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Question 164 Marks
A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent.
Answer
Given A circle passing through three points A, B and C.
Construction: Draw perpendicular bisectors of AB and AC and they meet at a point O. Join OA, OB and OC.
To prove: Perpendicular bisector of BC, also passes through O i.e., LO, ON and OM are concurrent.
Proof: In $\triangle\text{QEA}$ and $\triangle\text{OEB},$

AE = BE [OL is the perpendicular bisector of AB]
$\angle\text{AEO}=\angle\text{BEO}\ \ \ [\text{each } 90^\circ]$ and $\text{OE}=\text{OE}$ [common side]
$\therefore\triangle\text{OEA}\cong\text{OEB}$ [by SAS congruence rule]
$\therefore\text{OA}=\text{OB}$
Similarly $\triangle\text{OFA}=\triangle\text{OFC}$ [by SAS congruence rule]
$\therefore\text{OA}=\text{OC}$ [by CPCT]
$\therefore\text{OA}=\text{OB}=\text{OC}=\text{r}$ [say]
Now, we draw a perpendicular from O to the BC and join them.
In $\triangle\text{OMB}$ and $\triangle\text{OMC},$
$\text{OB}=\text{OC}$ [proved above]
$\text{OM}=\text{OM}$ [common side]
and $\angle\text{OMB}=\angle\text{OMC}$ [each 90°]
$\therefore\triangle\text{OMB}\equiv\triangle\text{OMC}$ [by RHS congruence rule]
$\Rightarrow\text{BM}=\text{MC}$ [by CPCT]
Hence, OM is the perpendicular bisector of BC.
Hence, OL, ON and OM are concurret.
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Question 174 Marks
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD
Answer
Given: Two equal chords AB and CD of a circle intersecting at a point P.

To prove: PB = PD
Construction: Join OP, draw $\text{OL}\bot\text{AB}$ and $\text{OM}\bot\text{CD}$
Proof: We have, AB = CD
$\Rightarrow\text{OL}=\text{OM}$ [equal chords are equidistant from the centre]
In $\triangle\text{OLP}$ and $\triangle\text{OMP}\ \ \text{OL}=\text{OM}$ [proved above]
$\angle\text{OLP}=\angle\text{OMP}$ [each 90°] and OP = OP [common side]
$\therefore\triangle\text{OLP}=\triangle\text{OMP}$ [by RHS congruence rule]
$\Rightarrow\text{LP}=\text{MP}$ [by CPCT] ...(i)
Now, AB = CD
$\Rightarrow\frac{1}{2}(\text{AB})=\frac{1}{2}(\text{CD})$ [dividing both sides by 2]
$\Rightarrow\text{BL}=\text{DM}\ \ ...(\text{ii})$
[perpendicular drawn from centre to the circle bisects the chord i.e., AL = LB and CM = MD]
On subtracting Eq. (ii) from Eq. (ii) we get LP - BL = MP - DM ⇒ PB = PD
Hence proved.
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Question 184 Marks
Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha.
Answer
Let R, S and M be the position of Ishita, Isha and Nisha respectively.

$\text{AR}=\text{AS}=\frac{24}{2}=12\text{cm}$
OR = OS = OM = 20 cm [Radii of circle]
In $\triangle\text{OAR,}$
OA2 + AR2 = OR2
OA2 + 122 = 202
OA2 = 400 - 144 = 256m2
OA = 16m
We know that, in an isosceles triangle altitude divides the base.
So in $\triangle\text{RSM},\angle\text{RCS}=90^\circ$ and RC = CM
Area of $\triangle\text{ORS}=\frac{1}{2}\times\text{OA}\times\text{RS}$
$\Rightarrow\frac{1}{2}\times\text{RC}\times\text{OS}$
$=\frac{1}{2}\times16\times24$
$\Rightarrow\text{RC}\times20=16\times24$
$\Rightarrow\text{RC}=19.2$
$\Rightarrow\text{RM}=2(19.2)=38.4\text{m}$
So, the distance between Ishita and Nisha is 38.4m.
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Question 194 Marks
Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.
Answer
Given: $\triangle\text{ABC}$ and l is perpendicular bisector of BC.

To prove: Angles bisector of $\angle\text{A}$ and perpendicular bisector of BC intersect on the circumcircle of $\triangle\text{ABC}.$
Proof: Let the angle bisector of $\angle\text{A}$ intersect circumcircle of $\triangle\text{ABC}$ at P. Join BP and CP.
$\Rightarrow\angle\text{BAP}=\angle\text{BCP}$ [Angles in the same segment are equal]
$\Rightarrow\angle\text{BAP}=\angle\text{BCP}=\frac{1}{2}\angle\text{A}\ ...(1)$ [AP is bisector of $\angle\text{A}$]
Similary, we have $\angle\text{PAC}=\angle\text{PBC}=\frac{1}{2}\angle\text{A}\ \ ...(2)$
From equation (1) and (2), we have
$\angle\text{BCP}=\angle\text{PBC}$
$\Rightarrow\text{BP}=\text{CP}$
[$\because$ If the angles subtended by two chords of a circle at the centre are equal, the chords are equal]
⇒ P is on perpendicular bisector of BC.
Hence, angle bisector of $\angle\text{A}$ and perpendicular bisector of BC intersect on the circumcircle of $\triangle\text{ABC}.$
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Question 204 Marks
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer
Given: ABCD is a trapezium whose non-parallel sides AD and BC are equal.

To prove: Trapezium ABCD is a cyclic
Join BE, where BE || AD.
Proof: Since, AB || DE and AD || BE
Since, the quadrilateral ABCD is a parallelogram,
$\therefore\angle\text{BAD}=\angle\text{BED}\ \ ...(\text{i})$ [opposite angles of a parallelogram are equal] and AD = BE ...(ii) [opposite sides of a parallelogram are equal]
But, AD = BC [given] ...(iii) From Eqs. (ii) and (iii), BC = BE
$\Rightarrow\angle\text{BEC}=\angle\text{BCE}\ \ ...\text{(iv)}$ [angle opposite to equal sides are equal]
Also, $\angle\text{BEC}+\angle\text{BED}=180^\circ$ [linear pair axiom]
$\therefore\angle\text{BCE}+\angle\text{BAD}=180^\circ$ [from Eqs. (i) and (iv)]
If sum of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic.
Hence, trapezium ABCD is a cyclic.
Hence proved.
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Question 214 Marks
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If $\angle\text{DBC}=55^\circ$ and $\angle\text{BAC}=45^\circ,$ find $\angle\text{BCD}.$
Answer
Since angles in the same segment of a circle are equal.
$\therefore\angle\text{CAD}=\angle\text{DBC}=55^\circ$
$\therefore\angle\text{DAB}=\angle\text{CAD} +\angle\text{BAC}=55^\circ +45^\circ=100^\circ$
But, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$ [Opposite angles of a cyclic quadrilateral]
$\therefore\angle\text{BCD}=180^\circ-100^\circ=80^\circ$
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Question 224 Marks
In a cyclic quadrilateral ABCD, if $(\angle\text{B}-\angle\text{D})=60^\circ,$ show that the smaller of the two is 60°
Answer
ABCD is a cyclic quadrilateral
$\angle\text{B}-\angle\text{D}=60^\circ\dots(\text{i})$
And
$\angle\text{B}+\angle\text{D}=180^\circ\dots(\text{ii})$
Adding (i) and (ii) we get,
$2\angle\text{B}=240^\circ$
$\therefore\ \angle\text{B}=\frac{240}{2}=120^\circ$
Substituting the value of $\angle\text{B}=120^\circ$ in (i) we get
$120^\circ-\angle\text{D}=60^\circ$
$\Rightarrow\ \angle\text{D}=120^\circ-60^\circ=60^\circ$
The smaller of the two angles i.e. $\angle\text{D}=60^\circ.$
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Question 234 Marks
Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.
Answer


Let $\triangle\text{ABC}$ be a right triangle right angled at B. Let P be the mid-point of hypotenuse AC.
Draw a circle with centre at P and AC as a diameter.
Since, $\angle\text{ABC}=90^\circ.$ Therefore, the circle passes through B.
$\therefore$ BP = Radius
Also, AP = CP = Radius
$\therefore$ AP = BP = CP
Hence, $\text{BP}=\frac{1}2{}\text{AC}$
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Question 244 Marks
ABCD is a cyclic qudrilateral in which:
$\text{BC}\parallel\text{AD},\ \angle\text{ADC}=110^\circ$ and $\angle\text{BAC}=50^\circ.$ Find $\angle\text{DAC}.$
Answer


Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\angle\text{ABC}+110^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-110^\circ=70^\circ$
Since, AD || BC
Then, $\angle\text{DAB}+\angle\text{ABC}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{DAC}+50^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{DAC}=180^\circ-50^\circ-70^\circ=60^\circ$
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Question 254 Marks
In the given figure, $\angle\text{BAD}=75^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of x and y.
Answer


We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle
$\text{i.e.},\angle\text{BAD}=\angle\text{DCF}=75^\circ$
$\Rightarrow\ \angle\text{DCF}=\text{x}=75^\circ$
Again, the sum of opposite angles in a cyclic quadrilateral is 180°.
Thus, $\angle\text{DCF}=\angle\text{DEF}=180^\circ$
$\Rightarrow\ 75^\circ+\text{y}=180^\circ$
$\Rightarrow\ \text{y}=(180^\circ-75^\circ)=105^\circ$
Hence, $\text{x}=75^\circ$ and $\text{y}=105^\circ.$
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Question 264 Marks
In the given figure, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of x and y.
Answer
We have, $\angle\text{BAD}=78^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$
Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow78^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow​​\angle\text{BCD}=180^\circ-78^\circ=102^\circ$
Now, $\angle\text{BCD}+\angle\text{DCF}=180^\circ$ [Linear pair of angles]
$\Rightarrow102^\circ=\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-102^\circ=78^\circ$
Since, DCEF is a cyclic quadrilateral
Then, x + y = 180º
⇒ 78º + y = 180º
⇒ y = 180º - 78º = 102º
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Question 274 Marks
In the given figure, if O is the circumcentre of $\angle\text{ABC},$ then find the value of $\angle\text{OBC} + \angle\text{BAC.}$
Answer


Since, O is the circumcentre of $\triangle\text{ABC,}$ So, O would be centre of the circle passing through points A, B and C.
$\angle\text{ABC}=90^\circ$ (Angle in the semicircle is 90°)
$\Rightarrow\angle\text{OAB}+\angle\text{OBC}=90^\circ\dots(1)$
As OA = OB (Radii of the same circle)
$\therefore\angle\text{OAB}=\angle\text{OBA}$ (Angle opposite to equal sides are equal)
or,
$\angle\text{BAC}=\angle\text{OBA}$
From (1)
$\angle\text{BAC}+\angle\text{OBC}=90^\circ$
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Question 284 Marks
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
Answer
Given: Let ABCD be a cyclic quadrilateral and AD = BC.

Join AC and BD.
To prove: AC = BD
Proof: In $\triangle\text{AOD}$ and $\triangle\text{BOC},$
$\triangle\text{OAD} = \angle\text{OBC}$ and $\angle\text{ODA} = \angle\text{OCB}$ [since, same segments subtends equal angle to the circle]
AB = BC [given]
$\triangle\text{AOD} = \triangle\text{BOC}$ [by ASA congruence rule]
Adding is DOC on both sides, we get,
$\triangle\text{AOD}+ \triangle\text{DOC} \cong \triangle\text{BOC} + \triangle\text{DOC}$
$\Rightarrow \triangle\text{ADC} \cong \triangle\text{BCD}$
AC = BD [by CPCT]
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Question 294 Marks
If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of $\angle\text{BPC}.$
Answer
Since equal choeds of a circle subtends equal angle at the centre, so we have chord AB = chord AC [Given]
So, $\angle\text{AOB}=\angle\text{AOC}\ \ ...(1)$
Since the angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle,

$\therefore\angle\text{APC}=\frac{1}{2}\angle\text{AOC}\ ...(2)$ and $\angle\text{APB}=\frac{1}{2}\angle\text{AOB}\ ...(3)$
$\therefore\angle\text{APC}=\angle\text{APB}$ [From (1), (2) and (3)]
Hence, PA is the bisector of $\angle\text{BPC}.$
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Question 304 Marks
ABCD is a cyclic qudrilateral in which:
$\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ$ find $\angle\text{ADB}.$
Answer


Since, ABCD is a cyclic quadrilateral.
Then, $\angle\text{BAD}+\angle\text{BCD}=180^\circ$ 
$\Rightarrow\angle\text{BAD}+100^\circ=180^\circ$
$\Rightarrow\angle\text{BAD}=180^\circ-100^\circ=80^\circ$
In  by angle sum property
$\angle\text{ABD}+\angle\text{ADB}+\angle\text{BAD}=180^\circ$
$\Rightarrow70^\circ+\angle\text{ADB}+80^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-70^\circ-80^\circ=30^\circ$
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Question 314 Marks
A line segment AB is of length 5cm. Draw a circle of radius 4cm passing through A and B. Can you draw a circle of radius 2cm passing through A and B? Give reason in support of your answer.
Answer
  1. Draw a line segment AB of 5cm.
  2. Draw the perpendicular bisectors of AB.
  3. With centre A and radius of 4cm, draw an arc which intersects the perpendicular bisector at point O. The point O will be the required centre.
  4. Join OA.
  5. With centre O and radius OA, draw a circle.

No, we cannot draw a circle of radius 2cm passing through A and B because when we draw an arc of radius 2cm with centre A, the arc will not intersect the perpendicular bisector and we will not find the centre.

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Question 324 Marks
ABCD ia a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that:
  1. AD || BC.
  2. EB = EC.
Answer


Given ABCD is a cyclic quadrilateral in which EA = ED To prove:
  1. AD || BC.
  2. EB = EC.
Proof:
  1. Since EA = ED
    Then, $\angle\text{EAD}=\angle\text{EDA}\dots(1)$ [Oppo. angles to equal sides]
    Since, ABCD is a cyclic quadrilateral
    Then, $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
    But $\angle\text{ABC}+\angle\text{EBC}=180^\circ$ [Linear pair of angles]
    Then, $\angle\text{ADC}=\angle\text{EBC}\dots(2)$
    Compare equations (1) and (2)
    $\angle\text{EAD}=\angle\text{EBC}\dots(3)$
    Since, corresponding angles are equal
    Then, BC || AD
  1. From equation (3)
    $\angle\text{EAD}=\angle\text{EBC}\dots(3)$
    Similarly $\angle\text{EDA}=\angle\text{ECB}\dots(4)$
    Compare equations (1)(3) and (4)
    $\angle\text{EBC}=\angle\text{ECB}$
    $\Rightarrow \text{EB}=\text{EC}$ [Opposite angles to equal sides]

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Question 334 Marks
Two chords AB and CD of lengths 5cm and 11cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6cm, find the radius of the circle.
Answer
Draw $\text{OM}\perp\text{AB}$ and $\text{ON}\perp\text{CD}.$

Join OB and OD. $\text{BM}=\frac{\text{AB}}{2}=\frac{5}{2}$ [Perpendicular from the centre bisects the chord]
$\text{ND}=\frac{\text{CD}}{2}=\frac{11}{2}$ Let ON be x, so OM will be 6 - x.
$\triangle\text{MOB}$
$\text{OM}^2+\text{MB}^2=\text{OB}^2$
$(6-\text{x})^2+\Big(\frac{5}{2}\Big)^2+\text{OB}^2$
$36+\text{x}^2-12\text{x}+\frac{25}{4}=\text{OB}^2\dots(\text{i})$
In $\triangle\text{NOD}$
$\text{ON}^2+\text{ND}^2=\text{OD}^2$
$\text{x}^2+\Big(\frac{11}{2}\Big)^2=\text{OD}^2$
$\text{x}^2+\frac{121}{4}=\text{OD}^2\dots(\text{ii})$
We have OB = OD. [Radii of same circle] So, from equation (i) and (ii).
$36+\text{x}^2-\text{12x}+\frac{25}{4}=\text{x}^2+\frac{121}{2}$
$\Rightarrow\text{12x}=36+\frac{25}{4}-\frac{121}{4}$
$=\frac{144+25-121}{4}$
$=\frac{48}{4}=12$
$\text{x}=1$
From equation (ii)
$(1)^2+\Big(\frac{121}{4}\Big)=\text{OD}^2$
$\text{OD}^2=1+\frac{121}{4}=\frac{125}{4}$
$\text{OD}=\frac{5\sqrt{5}}2{}$
So, the radius of the circle is found to be $55\sqrt{2}\text{cm}.$
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Question 344 Marks
If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.
Answer
We have to prove that R, D, P and Q are concyclic.

Join RD, QD, PR and PQ. $\because$ RP joints R and P, the mid-point of AB and BC.
$\therefore$ RP || AC [Mid-point theorem]
Similarly, PQ || AB.
$\therefore$ ARPQ is a || gm
$\angle\text{RAQ}=\angle\text{RPQ}$ [opposite angle of a || gm] .... (1)
$\because$ ABD is a rt. $\angle\text{D}\triangle$ and DR is a median,
$\therefore$ RA = DR and $\angle1=\angle2\ ....(2)$
Similarly $\angle3=\angle4\ ....(3)$
Adding (2) and (3), we get
$\angle1+\angle3=\angle2+\angle4$
$\Rightarrow\angle\text{RDQ}=\angle\text{RAQ}$
$\angle\text{RPQ}$ [Proved above] Hence, R, D, P and Q are concyclic.
[$\because\angle\text{D}$ and $\angle\text{P}$ are subtended by RQ on the same side of it.]
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Question 354 Marks
O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that $\angle\text{BOD}=\angle\text{A}.$
Answer
Given: In a $\triangle\text{ABC}$ a circle is circumscribed having centre O.

Also, D is the mid-point of BC.
To prove: $\angle\text{BOD}=\angle\text{A}\ \ \text{or}\ \ \angle\text{BOD}=\angle\text{BAC}$
Construction: Join OB, OD and OC.
Proof: In $\triangle\text{BOD}$ and $\triangle\text{COD}$,
OB = OC [both are the radius of circle]
BD = DC [D is the mid-point of BC]
and OD = OD [common side]
$\therefore\triangle\text{BOD}\cong\triangle\text{COD}$ [by SSS congruence rule]
$\therefore\angle\text{BOD}=\angle\text{COD}$ [by CPCT] ....(i)
We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore2\angle\text{BAC}=\angle\text{BOC}$
$\Rightarrow\angle\text{BAC}=\frac{2}{2}\angle\text{BOD}$
$[\because\angle\text{BOC}=2\angle\text{BOD}]$ [from Eq. (i)]
$\Rightarrow\angle\text{BAC}=\angle\text{BOD}$
Hence proved.
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Question 364 Marks
In the given figure, O is the center of the circle. Find $\angle\text{CBD}.$
Answer
We have, $\angle\text{AOC}=100^\circ$
By degree measure theorem
$\angle\text{AOC}=2\angle\text{APC}$
$\Rightarrow100^\circ=2\angle\text{APC}$
$\Rightarrow\angle\text{APC}=\frac{100^\circ}{2}=50^\circ$
$\therefore\angle\text{APC}+\angle\text{ABC}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow50^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-50^\circ=130^\circ$
$\therefore\angle\text{ABC}+\angle\text{CBD}=180^\circ$ [Linear pair of angles]
$\Rightarrow130^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\angle\text{CBD}=180^\circ-130^\circ=50^\circ$
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Question 374 Marks
In the given figure, BD = DC and $\angle\text{CBD}=30^\circ,$ find $\angle\text{BAC}.$
Answer


BD = DC
$\Rightarrow\ \angle\text{BCD}=\angle\text{CBD}=30^\circ$ In $\triangle\text{BCD},$ we have:
$\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 30^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CDB}=(180^\circ-60^\circ)=120^\circ$
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, $\angle\text{CDB}+\angle\text{BAC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\ \angle\text{BAC}=(180^\circ-120^\circ)=60^\circ$
$\therefore\ \angle\text{BAC}=60^\circ$
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Question 384 Marks
Answer


AB and CD are two chords of a circle which intersect each other at P outside the circle.
AB = 6cm, BP = 2cm and PD = 2.5cm
$\therefore$ AP × BP = CP × DP
⇒ 8 × 2 = (CD + 2.5) × 2.5 $[\because$ CP = CD + DP$]$
Let CD = xcm
Thus, 8 × 2 = (CD + 2.5) × 2.5
⇒ 16 = 2.5x + 6.25
⇒ 2.5x = (16 - 6.25) = 9.75
$\Rightarrow\ \text{x}=\frac{9.75}{2.5}=3.9$
Hence, CD = 3.9cm
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Question 394 Marks
ABCD is a cyclic trapezium with AD || BC. If $\angle\text{B}=70^\circ,$determine other three angles of the trapezium.
Answer

We have  ABCD is a cyclic trapezium with AD || BC and $\angle\text{B}=70^\circ.$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, AD || BC
Then, $\angle\text{A}+\angle\text{B}=180^\circ$ [Co-interior angles]
$\Rightarrow\angle\text{A}+70^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-70^\circ=110^\circ$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow110^\circ+\angle\text{C}=180^\circ$
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Question 414 Marks
Answer
Given: In a circle AYDZBWCX, two chords AB and CD intersect at right angles.
To prove: arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle
Construction: Draw a diameter EF parallel to CD having centre M.
Proof: Since, CD || EF arc EC = arc PD …(i) arc ECXA = arc EWB [symmetrical about diameter of a circle] arc AF = arc BF …(ii)
We know that, arc ECXAYDF = semi-circle

arc EA + arc AF = semi-circle ⇒ arc EC + arc CXA + arc FB = semi-circle [from Eq. (ii)]
⇒ arc DF + arc CXA + arc FB = semi-circle [from Eq. (i)]
⇒ arc DF + arc FB + arc CXA = semi-circle
⇒ arc DZB + arc C × A = semi-circle We know that, circle divides itself in two semi-circles, therefore the remaining portion of the circle is also equal to the semi-circle.
$\therefore$ arc AYD + arc BEC = semi-ciecle
Hence proved.
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Question 424 Marks
If the two sides of a pair of opposite sides of a cyclic quadrilateral ae equal, prove that its diagonals are equal.
Answer


Given ABCD is a cyclic quadrilateral in which AB = DC
To prove AC = BD
Proof In $\triangle\text{PAB}$ and $\triangle\text{PDC}$ $\text{AB}=\text{DC}$ [Given]
$\angle\text{BAP}=\angle\text{CDP}$ [Angles in the same segment]
$\angle\text{PBA}=\angle\text{PCD}$ [Angles in same segment]
Then, $\triangle\text{PAB}\cong\triangle\text{PDC}$ [By ASA condition]
$\therefore\text{PA}=\text{PD}\dots(1)$ [C.P.C.T.] and $\text{PC}=\text{PB}\dots(2)$ [C.P.C.T.]
Add equation (1) and (2)
PA + PC = PD + PB
⇒ AC = BD
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Question 434 Marks
In a cyclic quadrilateral ABCD if $\text{m}\angle\text{A}=\big(\text{m}\angle\text{C}\big).$ Find $\text{m}\angle\text{A}.$ 
Answer

We have, $\angle\text{A}=3\angle\text{C}$
Let $\angle\text{C}=\text{x}$
Then, $\angle\text{A}=\text{3x}$
$\therefore\angle\text{A}+\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\text{3x}+\text{x}=180^\circ$
$\Rightarrow\text{4x}=180^\circ$
$\Rightarrow\text{x}=\frac{180^\circ}{4}=45^\circ$
$\therefore\angle\text{A}=\text{3x}$
$=3\times45^\circ$
$=135^\circ$
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Question 454 Marks
In the given figure, O is the centre of the circle. If $\angle\text{ACB}=50^\circ,$ find $\angle\text{OAB}.$
Answer
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended

by the arc at any point on the circumference.
$\angle\text{AOB}=2\angle\text{ACB}$
$=2\times50^\circ$ [Given]
$\angle\text{AOB}=100^\circ\dots(\text{i})$
Let us consider the triangle $\triangle\text{OAB}.$
$\text{OA}=\text{OB}$ (Radii of a circle)
Thus, $\angle\text{OAB}=\angle\text{OBA}$
In $\triangle\text{OAB},$ we have:
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$
$\Rightarrow\ 100^\circ+\angle\text{OAB}+\angle\text{OBA}=180^\circ$
$\Rightarrow\ 100^\circ+2\angle\text{OAB}=180^\circ$
$\Rightarrow\ 2\angle\text{OAB}=180^\circ-100^\circ=80^\circ$
$\Rightarrow\ \angle\text{OAB}=40^\circ$
Hence, $\angle\text{OAB}=40^\circ$
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Question 464 Marks
In figure, O is the centre of the circle, then prove that $\angle\text{x}=\angle\text{y}+\angle\text{z}.$
Answer
We have,
$\angle3=\angle4$ [Angles in same segment]
$\therefore\angle\text{x}=2\angle3$ [By degree measure theorem]
$\Rightarrow\angle\text{X}=\angle3+\angle3$
$\Rightarrow\angle\text{X}=\angle3+\angle4\dots(\text{i})$ $[\angle3=\angle4]$
But $​​\angle\text{y}=\angle3+\angle1$ [By exterior angle property]
$\Rightarrow\angle3=\angle\text{y}-\angle1\dots(\text{ii})$
From (i) and (ii)
$\angle\text{x}=\angle\text{y}-\angle1+\angle4$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle4-\angle1$
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}+\angle1-\angle1$ [By exterior angle property]
$\Rightarrow\angle\text{x}=\angle\text{y}+\angle\text{z}$
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Question 474 Marks
In the given figure, O is the center of the circle. If $\angle\text{CEA}=30^\circ,$ find the value of x, y and z.
Answer
We have, $\angle\text{CEA}=30^\circ$
Since, quad. ABCE is a cyclic quadrilateral.
Then, $\angle\text{ABC}+\angle\text{CEA}=180^\circ$
$\Rightarrow\text{x}+30^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-30^\circ=150^\circ$
By degree measure theorem
$\angle\text{AOC}=2\angle\text{CEA}$
$\Rightarrow\text{y}=2\times30^\circ=60^\circ$
$\therefore\angle\text{ADC}=\angle\text{CEA}$ [Angle in same segment]
$\Rightarrow\text{z}=30^\circ$
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Question 484 Marks
In the given figure, $\triangle\text{PQR}$ is an isosceles triangle with PQ = PR and $\text{m}\angle\text{PQR}=35^\circ.$ Find $\text{m}\angle\text{QSR}$ and $\text{m}\angle\text{QTR}.$
Answer
We have, $\angle\text{PQR}=35^\circ$
Since, $\triangle\text{PQR}$ is an isosceles triangle with PQ = PR.
Then, $\angle\text{PQR}=\angle\text{PRQ}=35^\circ$
In $\triangle\text{PQR},$ by angle sum property
$\angle\text{p}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$
$\Rightarrow\angle\text{p}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{p}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{QSR}=\angle\text{p}=110^\circ$ [Angles in same segment]
Now, $\angle\text{QSR}+\angle\text{QTR}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow110^\circ+\angle\text{QTR}=180^\circ$
$\Rightarrow\angle\text{QTR}=180^\circ-110^\circ=70^\circ$
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Question 494 Marks
Find the length of a chord which is at a distance of 4cm from the centre of a circle of radius 6cm.
Answer
Given that,

Distance (OC) = 4cm
Radius of the circle (OA) = 6cm
In $\triangle\text{OCA,}$ by Pythagoras theorem
AC2 + OC2 = OA2
⇒ AC2 +  42 = 62
⇒ AC2 =  36 - 16
⇒ AC2 = 20
$\Rightarrow\text{AC}=\sqrt{20}$
⇒ AC = 4.47cm
We know that the perpendicular distance from centre to chord bisects the chord.
AC = BC = 4.47cm
Then AB = 4.47 + 4.47
= 8.94cm
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Question 504 Marks
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
Answer
Given: ABCD is a cyclic quadrilateral. DP and QB are the bisectors of $\angle\text{D}$ and $\angle\text{B},$ respectively.
To prove: PQ is the diameter of a circle.
Construction: Join QD and QC.

Proof: Since, ABCD is a cyclic quadrilateral.
$\therefore\angle\text{CDA}+\angle\text{CBA}=180^\circ$ [sum of opposite angles of cylic quadrilateral is 180°]
On dividing both sides by 2, we get,
$\frac{1}{2}\angle\text{CDA}+\frac{1}{2}\angle\text{CBA}=\frac{1}{2}\times180^\circ=90^\circ$
$\Rightarrow\angle1+\angle2=90^\circ\ ...(\text{i})$ $\big[\angle1=\frac{1}{2}\angle\text{CDA and}\ \angle2=\frac{1}{2}\angle\text{CBA}\big]$
But, $\angle2=\angle3$ [angles in the same segment QC are equal] ...(ii)
$\angle1+\angle3=90^\circ$
From Eqe. (i) and (ii), $\angle\text{PDQ}=90^\circ$
Hence, PQ is a diameter of a circle, because diameter of the circle.
Subtends a right angle at the circumference.
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4 Marks Questions - Maths STD 9 Questions - Vidyadip