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MATHS 3 Marks Question

Circles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 25 questions.

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Question 13 Marks
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Answer

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. since $\angle\text{ABC}=90^\circ$ and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.
Hence, point of intersection of AC and BD is the center of the circle. 
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Question 23 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have
$\angle\text{OAB}=35^\circ$ then,
$\angle\text{OBA}=\angle\text{OAB}=35^\circ$ [Angles opposite to equal radii]
In $\triangle\text{AOB},$ by angle sum property
$\Rightarrow\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$
$\Rightarrow\angle\text{AOB}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{AOB}=180^\circ-35^\circ-35^\circ=110^\circ$
$\therefore\angle\text{AOB}+\text{reflex }\angle\text{AOB}=360^\circ$ [Complex angle]
$\Rightarrow110^\circ+\text{reflex }\angle\text{AOB}=360^\circ$
$\Rightarrow\text{reflex }\angle\text{AOB}=360^\circ-110^\circ=250^\circ$
By degree measure theorem reflex
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow250^\circ=\text{2x}$
$\text{x}=\frac{250^\circ}{2}=125^\circ$
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Question 33 Marks
In the given figure, If ABC is an equilateral triangle. Find $\angle\text{BDC}$ and $\angle\text{BEC.}$
Answer
Since, $\triangle\text{ABC}$ is an equilateral triangle
Then, $\angle\text{BAC}=60^\circ$
$\therefore\angle\text{BDC}=\angle\text{BAC}=60^\circ$ [Angles in same segment]
Since, quad. ABEC is a cyclic quadrilateral.
Then, $\angle\text{BAC}+\angle\text{BEC}=180^\circ$
$\Rightarrow60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\angle\text{BEC}=180^\circ-60^\circ=120^\circ$
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Question 43 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
In $\triangle\text{DAB},$ by angle sum property
$\angle\text{ADB}+\angle\text{DAB}+\angle\text{ABD}=180^\circ$
$\Rightarrow32^\circ+\angle\text{DAB}+50^\circ=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-32^\circ-50^\circ$
$\Rightarrow\angle\text{DAB}=98^\circ+50^\circ$
Now, $\angle\text{OAB}+\angle\text{DCB}=180^\circ$ [Opposite angles of cyclic quadrilateral]
$\Rightarrow98^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=180^\circ-98^\circ=82^\circ$
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Question 53 Marks
Prove that two different circles cannot intersect each other at more than two points.
Answer
Suppose two circles intersect in three points A, B, C.
Then A, B, C are non-collinear so a unique circle passes through these three points. This is contradiction to the face that two given circles are passing through A, B, C.
Hence, two circles cannot intersect each other at more than two points.
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Question 63 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have$\angle\text{BAC}=50^\circ$ and $\angle\text{DBC}=70^\circ$
$\therefore\angle\text{BDC}=\angle\text{BAC}$ [Angle in same segment]
In $\triangle\text{BDC},$ by angle sum property
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\Rightarrow50^\circ+\text{x}+70^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-50^\circ-70^\circ=60^\circ$
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Question 73 Marks
In the given figure, O is the center of the circle. If $\angle\text{BOD}=160^\circ,$ find the value of x and y.
Answer
We have, $\angle\text{BOD}=160^\circ$
By degree measure theorem
$\angle\text{BOD}=2\angle\text{BCD}$
$\Rightarrow160^\circ=\text{2x}$
$\Rightarrow\text{x}=\frac{160^\circ}{2}=80^\circ$
$\therefore\angle\text{BAD}+\angle\text{BCD}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\text{y}+\text{x}=180^\circ$
$\Rightarrow\text{y}+80^\circ=180^\circ$
$\Rightarrow\text{y}=180^\circ-80^\circ=100^\circ$
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Question 83 Marks
If O is the centre of the circle, find the value of x in the following figures:
Answer
$\angle\text{AOC}=135^\circ$ $\therefore\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [Linear pair of angles]$\Rightarrow135^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-135^\circ$
$\Rightarrow\angle\text{BOC}=45^\circ$
By degree measure theorem$\angle\text{BOC}=2\angle\text{CPB}$
$\Rightarrow45^\circ=\text{2x}$
$\Rightarrow\text{x}=\frac{45^\circ}{2}=22\frac{1}{2}^\circ$
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Question 93 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
Given that, $\angle\text{ABD} = 40^\circ$ Then $\angle\text{BDC}=\angle\text{BAC}=52^\circ$ [Angle in same segment]Since OD = OC
Then $\angle\text{ODC}=\angle\text{OCD}$ [Opposite angle to equal radii]
$\Rightarrow\text{x}=52^\circ$
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Question 103 Marks
In the given figure, AB and CD are diameters of a circle with center O. if $\angle\text{OBD}=50^\circ,$ find $\angle\text{AOC}.$
Answer
We have, $\angle\text{OBD}=50^\circ$
Since, AB and CD are diameters of circle then O is the center of the circle.
$\therefore​​\angle\text{DBC}=90^\circ$ [Angle in semicircle]
$\Rightarrow\angle\text{DOB}+\angle\text{OBC}=90^\circ$
$\Rightarrow50^\circ+\angle\text{OBC}=90^\circ$
$\Rightarrow\angle\text{OBC}=90^\circ-50^\circ=40^\circ$
By degree measure theorem
$\angle\text{AOC}=2\angle\text{ABC}$
$\Rightarrow\angle\text{AOC}=2\times40^\circ=80^\circ$
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Question 113 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have,
$\angle\text{ABD} = 40^\circ$
$\angle\text{ACD}=\angle\text{ABD}=40^\circ$ [Angle in same segment]
In $\triangle\text{PCD},$ by angle sum property
$\angle\text{PDC}+\angle\text{CPO}+\angle\text{PDC}=180^\circ$
$\Rightarrow40^\circ+110^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=180^\circ-150^\circ$
$\Rightarrow\text{x}=30^\circ$
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Question 123 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have,
$\angle\text{BAC} = 35°$
$\angle\text{BDC}=\angle\text{BAC}=35^\circ$ [Angle in same segment]
In $\triangle\text{BCD},$ by angle sum property
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{DBC}=180^\circ$
$\Rightarrow35^\circ+\text{x}+65^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-35^\circ-65^\circ=80^\circ$
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Question 133 Marks
In the given figure, O is the center of the circle and $\angle\text{DAB}=50^\circ.$ Calculate the values of x and y.
Answer
We have, $\angle\text{DAB}=50^\circ$
By degree measure theorem
$\angle\text{BOD}=2\angle\text{BAD}$
$\Rightarrow\text{x}=2\times50^\circ=100^\circ$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow50^\circ+\text{y}=180^\circ$
$\Rightarrow\text{y}=180^\circ-50^\circ=130^\circ$
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Question 143 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have
$\angle\text{AOB}=60^\circ$ By degree measure theorem reflex
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow60^\circ=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{60^\circ}{2}=30^\circ$ [Angles opposite to equal radii]
$\Rightarrow\text{x}=30^\circ$
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Question 153 Marks
If ABCD is a cyclic quadrilateral in which AD || BC. Prove that $\angle\text{B}=\angle\text{C}.$
 
Answer
Since, ABCD is a cyclic quadrilateral with AD || BC
Then, $\angle\text{A}+\angle\text{C}=180^\circ\dots(1)$ [Opposite angles of cyclic quad.]
And, $\angle\text{A}+\angle\text{B}=180^\circ\dots(2)$ [Co-interior angles]
Compare equations (1) and (2)
$\angle\text{B}=\angle\text{C}$
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Question 163 Marks
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
Answer
Let ABCD be a cyclic quadrilateral, and let O be the centre of the corresponding circle.
Then, each side of quadrilateral ABCD is a chord of the circle and the perpendicular bisector of a chord always passes through the centre of the circle.
So, right bisectors of the sides of quadrilateral ABCD will pass through the center O of the corresponding circle. 
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Question 173 Marks
ABCD is a cyclic qudrilateral in which:
$ \angle\text{DBC}=80^\circ$ and $\angle\text{BAC}=40^\circ.$ Find $\angle\text{BCD}.$
Answer

$\angle\text{BAC}=\angle\text{BDC}=40^\circ$ [Angle in same segment]
In  by angle sum property
$\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$
$\Rightarrow80^\circ+\angle\text{BCD}+40^\circ=180^\circ$
$\Rightarrow\angle\text{BCD}=180^\circ-80^\circ-40^\circ=60^\circ$
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Question 183 Marks
In figure, it is given that O is the centre of the circle and $\angle\text{AOC} = 150^\circ.$ Find $\angle\text{ABC}.$
Answer
$\angle\text{AOC}=150^\circ$
$\therefore\angle\text{AOC}+\text{reflex }\angle\text{AOC}=360^\circ$ [complex angle]
$\Rightarrow150^\circ+\text{reflex }\angle\text{AOC}=360^\circ$
$\Rightarrow\text{reflex }\angle\text{AOC}=360^\circ-150^\circ$
$\Rightarrow\text{reflex }\angle\text{AOC}=210^\circ$
$\Rightarrow2\angle\text{ABC}=210^\circ$ [By degree measure theorem]
$\Rightarrow\angle\text{ABC}=\frac{210^\circ}{2}=105^\circ$
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Question 193 Marks
In the given figure ABCD is a cyclic quadrilateral. If $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ,$ find $\angle\text{ADB}.$
Answer
We have, $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=70^\circ$
$\therefore\angle\text{DAB}+\angle\text{BCD}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\angle\text{DAB}+100^\circ=180^\circ$
$\Rightarrow\angle\text{DAB}+180^\circ-100^\circ=80^\circ$
In $\triangle\text{DAB},$ by angle sum property
$\angle\text{ADB}+\angle\text{DAB}+\angle\text{ABD}=180^\circ$
$\Rightarrow\angle\text{ADB}+80^\circ+70^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-80^\circ-70^\circ=30^\circ$
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Question 203 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have$\angle\text{CBD}=65^\circ$
$\therefore\angle\text{ACB}+\angle\text{CBD}=180^\circ$ [Linear pair of angles]
$\Rightarrow\angle\text{ABC}=65^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-65^\circ=115^\circ$
$\therefore\text{reflex}\angle\text{AOC}=2\angle\text{ABC}$ [By degree measure theorem]
$\Rightarrow\text{x}=2\times115^\circ$
$\Rightarrow\text{x}=230^\circ$
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Question 213 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have$\angle\text{DOB}=40^\circ$ and $\angle\text{DBC}=90^\circ$ [Angle in a semicircle]
$\Rightarrow\angle\text{DOB}+\angle\text{OBC}=90^\circ$
$\Rightarrow40^\circ+\angle\text{OBC}=90^\circ$
$\Rightarrow\angle\text{OBC}=90^\circ-40^\circ=50^\circ$ By degree measure theorem
$\angle\text{AOC}=2\angle\text{OBC}$
$\Rightarrow\text{x}=2\times50^\circ=100^\circ$
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Question 223 Marks
If O is the centre of the circle, find the value of x in the following figures:
Answer
We have $\angle\text{ABC}=40^\circ$ $\angle\text{ACB}=90^\circ$ [Angle in semi circle] In $\triangle\text{ABC,}$ by angle sum property$\angle\text{CAB}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{CAB}+90^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{CAB}=180^\circ-90^\circ-40^\circ$
$\Rightarrow\angle\text{CAB}=50^\circ$
Now, $\angle\text{CDB}=\angle\text{CAB}$ [Angle is same in segment]
$\Rightarrow\text{x}=50^\circ$
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Question 233 Marks
In figure, if $\angle\text{ACB} = 40^\circ, \angle\text{DPB} = 120^\circ,$ find $\angle\text{CBD}.$
Answer
We have,
$\angle\text{ACB} = 40^\circ, \angle\text{DPB} = 120^\circ$
$\therefore\angle\text{APB}=\angle\text{DCB}=40^\circ$ [Angle in same segment]
In $\triangle\text{POB},$ by angle sum property
$\angle\text{PDB}+\angle\text{PBD}+\angle\text{BPD}=180^\circ$
$\Rightarrow40^\circ +\angle\text{PBD}+120^\circ=180^\circ$
$\Rightarrow\angle\text{PBD}=180^\circ-40^\circ-120^\circ$
$\Rightarrow\angle\text{PBD}=20^\circ$
$\therefore\angle\text{CBD}=20^\circ$
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Question 243 Marks
If O is the centre of the circle, find the value of x in the following figure:
Answer
We have$\angle\text{AOC}=120^\circ$ By degree measure theorem.
$\angle\text{AOC}=2\angle\text{APC}$
$\Rightarrow120^\circ=2\angle\text{APC}$
$\Rightarrow120^\circ=2\angle\text{APC}$
$\Rightarrow\angle\text{APC}=\frac{120^\circ}{2}=60^\circ$
$\angle\text{APC}+\angle\text{ABC}=180^\circ$ [Opposite angles of cyclic quadrilaterals]
$\Rightarrow60^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-60^\circ$
$\Rightarrow\angle\text{ABC}=120^\circ$
$\therefore\angle\text{ABC}+\angle\text{DBC}=180^\circ$ [Linear pair of angles]
$\Rightarrow120^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=180^\circ-120^\circ=60^\circ$
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Question 253 Marks
If the given figure, AOC is a diameter of the circle and arc $\text{AXB}=\frac{1}{2}$ arc BYC. Find $\angle\text{BOC.}$
Answer
We need to find $\angle\text{BOC}$
$\text{arc}\ \text{AXB}=\frac{1}{2} \text{arc}\ \text{BYC},$
$\angle\text{AOB}=\frac{1}{2}\angle\text{BOC}$
Also $\angle\text{AOB}+\angle\text{BOC}=180^\circ$
Therefore, $\frac{1}{2}\angle\text{BOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=\frac{2}{3}\times180^\circ=120^\circ$
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