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MATHS M.C.Q

Circles · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 249 questions.

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M.C.Q

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Question 11 Mark
AOB is the diameter of the circle. ABCD is a cyclic trapezium in which AB || DC. If $\angle\text{BED}=65^\circ,$ then $\angle\text{BDC}$ is equal to:
Answer
  1. 25º
    Solution:

    AB is diameter and ABCD is semi-circle, so $\angle\text{ADB}=90^\circ$
    Also, $\angle\text{BED}=\angle\text{BAD}=65^\circ$ {Angles in same segment}
    so, in $\triangle\text{ABD},\angle\text{BAD}+\angle\text{ABD}+\angle\text{ADB}=180^\circ$
    $\angle\text{ADB}=180-155=25^\circ$
    Now, since, AB || CD
    So, $\angle\text{ADB}=\angle\text{BDC}=25^\circ$ {alternate interior angles}
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Question 21 Mark
A chord of length $14\ cm$ is at a distance of $6\ cm$ from the centre of a circle. The length of another chord at a distance of $2\ cm$ from he centre of the circle is:
Answer
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
$AO^2 = AD^2 + OD^2 [$using paythagoras theorem$]$
$= 7^2 + 6^2$
$= 49 + 36$
$\text{AO}=\sqrt{85}$
Now consider the $\triangle\text{OPQ}$ in which $OM = 2\ cm$
So using Pythagoras theorem in $\triangle\text{OPM}$
$PM^2 = OP^2 - OM^2$
$=(\sqrt{85})^2-2^2 (\because OP = AO =$ radius$)$
$PM^2 = 81$
$PM = 9\ cm$
Hence $PQ = 2PM$
$= 2 \times 9$
$PQ = 18\ cm$
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Question 31 Mark
The given figure shows two intersecting circles. If $\angle\text{ABC}=75^\circ,$ then the measure of $\angle\text{PAD}$ is:
Answer
  1. 105º
    Solution:

    Since $\angle\text{PQC}+\angle\text{PBC}=180^\circ$ (Opposite angles of a cyclic quadrilateral))
    $\angle\text{PQC}=180^\circ-75^\circ=105^\circ$
    Again, $\angle\text{DPQ}+\angle\text{PQC}=180^\circ$ (Linear Pair)
    so, $\angle\text{DPQ}=75^\circ$
    Also, $\angle\text{PAD}+\angle\text{DQP}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
    $\angle\text{PAD}=105^\circ$
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Question 41 Mark
In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and $\angle\text{CBD}=35^\circ.$ Then, $\angle\text{BAD}=?$
Answer
  1. 70°
    Solution:
    BC = BD [Given]
    $\angle\text{BDC}=\angle\text{CBD}=35^\circ$ [Angle opposite equal sides are equal]
    In $\triangle\text{BCD},$
    $\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^\circ$ [Angle sum property]
    $\Rightarrow\ \angle\text{BCD}+35^\circ+35^\circ=180^\circ$
    $\Rightarrow\ \angle\text{BCD}=110^\circ$
    Since ABCD is a cyclic quadrilateral,
    $\angle\text{BAD}+\angle\text{BCD}=180^\circ$
    $\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
    $\Rightarrow\ \angle\text{BAD}=70^\circ$
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Question 51 Mark
In the given figure PQ and RS are two equal chords of a circle with centre O. OA and OB are perpendiculars on chords PQ and RS, respectively. If $\angle\text{AOB}=140^\circ,$ then $\angle\text{PAB}$ is equal to.
Answer
  1. 70º
    Solution:

    In triangle ABO, AO = BO
    So, $\angle\text{BAO}=\angle\text{ABO}=\text{x}$
    $\text{x}+\text{x}+140^\circ=180^\circ$
    $\Rightarrow2\text{x}=40^\circ$
    $\text{x}=20^\circ$
    Now $\angle\text{QAO}+\angle\text{BAO}+\angle\text{PAB}=180^\circ$
    Substituting the values we get:
    $90^\circ+20^\circ+\angle\text{PAB}=180^\circ$
    $\angle\text{PAB}=70^\circ$
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Question 61 Mark
In the given figure, O is the centre of a circle in which $\angle\text{OBA}=20^\circ$ and $\angle\text{OCA}=30^\circ.$ Then, $\angle\text{BOC}=?$
Answer
  1. 100°
    Solution:
    In $\triangle\text{OAB},$
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$ [Angle opposite equal sides are equal]
    In $\triangle\text{OAC},$
    OA = OC [Radii of the same circle]
    $\Rightarrow\ \angle\text{OCA}=\angle\text{OAC}=30^\circ$ [Angle opposite equal sides are equal]
    Now, $\angle\text{BAC}=\angle\text{BAO}+\angle\text{CAO}$
    $=20^\circ+30^\circ$
    $=50^\circ$
    $\angle\text{BOC}=2\angle\text{BAC}=2(50^\circ)=100^\circ.$
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Question 71 Mark
The chord of a circle is equal to its radius. The angle subtended by this chord at the minor arc of the circle, is:
Answer
  1. 150º
    Solution:
    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,
    $\triangle\text{AOB}$ is an equilateral triangle.
    Therefore, we have
    $\angle\text{AOB}=60^\circ$
    Since the angle subtended by any chord at the centre is twice the angle subtended at any point on the circle.
    So, $\angle\text{AQB}=\frac{\angle\text{AOB}}{2}=\frac{60}{2}=30^\circ$
    $\Rightarrow\angle\text{AQB}=30^\circ$
    Take a point P on the minor arc
    Since APBQ is a cyclic quadrilateral
    So, opposite angles are supplementary. That is
    $\angle\text{APB}+\angle\text{AQB}=180^\circ$
    $\angle\text{APB}+130^\circ=180^\circ\ [\angle\text{AQB}=30^\circ]$
    $\angle\text{APB}=180^\circ-30^\circ=150^\circ$
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Question 81 Mark
In a circle of radius $17\ cm,$ two parallel chords are drawn on opposite side of a diameter. The distance between the chords is $23\ cm$. If the length of one chord is $16\ cm,$ then the length of the other is:
Answer

$PQ = 23\ cm$
$AB = 16\ cm$
$\Rightarrow BP = AP = 8\ cm$
$r = 17\ cm$
$\Rightarrow EF =$ diameter $= 2r = 34\ cm$
Consider $\triangle\text{OPB},$
$r^2 = OP^2 + BP^2$
$\Rightarrow OP^2= (17)^2 - (8)^2 $
$= 289 - 64 $
$= 225$
$\Rightarrow OP = 15\ cm$
$\Rightarrow OQ = 23 - 15 = 8\ cm$
Consider $\triangle\text{OQD},$
$r^2 = OQ^2 + QD^2$
$\Rightarrow QD^2= r^2 - OQ^2 $
$= (17)^2 - (8)^2$
$= 225$
$\Rightarrow OD = 15\ cm$
$\Rightarrow CD = 2 \times QD = 30\ cm$
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Question 91 Mark
In the given figure, O is the centre of a circle. If $\angle\text{OAB}=40^\circ$ and C is a point on the circle, then $\angle\text{ACB}=?$
Answer
  1. 50°
    Solution:
    In $\triangle\text{OAB},$
    $\text{OA}=\text{OB}$ [Radii of the same circle]
    $\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$ [Angle opposite equal sides are equal]
    $\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 40^\circ+40^\circ+\angle\text{AOB}=180^\circ$
    $\Rightarrow\ \angle\text{AOB}=100^\circ$
    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.
    So, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
    $=\frac{1}{2}(100^\circ)$
    $=50^\circ$
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Question 101 Mark
One chord of a circle is known to be 10cm. The radius of this circle must be:
Answer
  1. Greater than 5cm.
    Solution:
    The longest chord of a circle is its diameter.
    ⇒ Diameter > 10cm
    ⇒ 2 × Radius > 10cm
    ⇒ Radius > 5cm
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Question 111 Mark
Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11cm, BE = 3cm and DE = 3.5cm, then CD = ?
Answer
  1. 8.5cm
    Solution:
    Construction: Join AC.
    $\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
    ⇒ AE × BE = DE × CE ...(i)
    Then,
    AE = AB + BE = 11 + 3 = 14cm, BE = 3cm, CE = (x + 3.5)cm and DE = 3.5cm
    So, from (i), we get
    14 × 3 = 3.5 × (CD + 3.5)
    $\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
    ⇒ 12 = CD + 3.5
    ⇒ CD = 8.5cm
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Question 121 Mark
In the given figure$, AB || \ || CD$ and $O$ is the centre of the circle. If $\angle\text{ADC}=25^\circ,$ then the measure of $\angle\text{AEB}$ is:
Answer

Here, $AB || \ || CD$ and $\angle\text{ADC}=25^\circ$
So, $\angle\text{DAB}=25^\circ ($opposite interior angles are equal$)$
Now, $\angle\text{ADC}=25^\circ,$ so, $\angle\text{AOC}=50^\circ ($Angle subtended by arc $AC$ at centre is twice the angle subtended at circumference$)$
Similarly, $\angle\text{DAB}=25^\circ,$ So, $\angle\text{DOB}=50^\circ($Angle subtended by arc $BD$ at centre is twice the angle subtended at circumference$)$
$\angle\text{AOB}+\angle\text{DOB}+\angle\text{AOC}=180^\circ ($All lie in straight line$)$
$\angle\text{AOB}=180-50=80^\circ$
Now, $\angle\text{AEB}=40^\circ ($Angle subtended by arc $AB$ at centre is twice the angle subtended at circumference$)$
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Question 131 Mark
Write the correct answer in the following: In Fig. if $\angle\text{OAB}=40^\circ,$ then $\angle\text{ACB}$ is equal to:
Answer
  1. 50°.
    Solution:
    In $\triangle\text{QAB, OA} = \text{OB}$ [both are the radius of a circle]
    $\angle\text{OAB} = \angle\text{OBA}\Rightarrow \angle\text{OBA} = 40^\circ$
    [angles opposite to equal sides are equal]
    Also, $\angle\text{AOB} = \angle\text{OBA}\Rightarrow \angle\text{BAO} = 180^\circ$
    [by angle sum property of a triangle]
    $\angle\text{AOB} + 40^\circ + 40^\circ = 180^\circ$
    $\Rightarrow\ \angle\text{AOB} = 180^\circ – 80^\circ = 100^\circ$
    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.
    $\angle\text{AOB} = 2 \angle\text{ACB} \Rightarrow 100^\circ =2 \angle\text{ACB}$
    $\angle\text{ACB} = \frac{100}{2} = 50^\circ$
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Question 141 Mark
If AB is a chord of a circle, P and Q are the two points on the circle different from A and B, then:
Answer
  1. $\angle\text{APB}+\angle\text{AQB}=180^\circ$ or $\angle\text{APB}+\angle\text{AQB}$
    Solution:
    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 the following figure.
    Case 1: Consider P and Q are on the same side of AB.

    We know that angles in the same segment are equal.
    Hence, $\angle\text{APB}=\angle\text{AQB}$
    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,
    $\angle\text{APB}+\angle\text{AQB}=180^\circ$ (Sum of the pair of opposite angles of a cyclic quadrilateral is 180º)
    Therefore,from case 1 and case 2, $\angle\text{APB}=\angle\text{AQB}$ or $\angle\text{APB}+\angle\text{AQB}=180^\circ$
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Question 161 Mark
In the given figure, AB is a diameter of the circle APBR. APQ and RBQ are straight lines. If $\angle\text{A}=35^\circ$ and then the measure of $\angle\text{PBR}$ is:
Answer
  1. 115º
    Solution:

    $\angle\text{APB}=\angle\text{BPQ}=90^\circ$
    Now,
    In $\triangle\text{APB},$
    $\angle\text{BAP}+\angle\text{APB}+\angle\text{ABP}=180^\circ$
    $35^\circ+90^\circ+\angle\text{ABP}=180^\circ$
    $\angle\text{ABP}=55^\circ$
    Again,
    In $\triangle\text{BPQ}$
    $\Rightarrow\angle\text{BPQ}+\angle\text{PQB}+\angle\text{PBQ}=180^\circ$
    $\Rightarrow90^\circ+25^\circ+\angle\text{PBQ}=180^\circ$
    $\Rightarrow\angle\text{PBQ}=65^\circ$
    Since, RBQ is a straight line,
    $\angle\text{RBA}+\angle\text{ABP}+\angle\text{PBQ}=180^\circ$
    $\angle\text{RBA}+55^\circ+65^\circ=180^\circ$
    $\angle\text{RBA}=60^\circ$
    Finally,
    $\angle\text{PBR}=\angle\text{ABP}+\angle\text{RBA}$
    $=55^\circ+60^\circ=115^\circ$
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Question 171 Mark
Write the correct answer in the following:
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle\text{ADC}=140^\circ,$ then $\angle\text{BAC}$ is equal to:
Answer
  1. 50°.
    Solution:
    Given, ABCD is a cyclic quadrilateral and $\angle\text{ADC}=140^\circ.$

    We know that, sum of the opposite angles in a cyclic quadrilateral is 180°.
    $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ 140^\circ+\angle\text{ABC}=180^\circ$
    $\Rightarrow\ \angle\text{ABC}=180^\circ-140^\circ$
    $\therefore\angle\text{ABC}=40^\circ$
    Since, $\angle\text{ACB}$ is an angle in a semi-circle.
    $\therefore\angle\text{ACB}=90^\circ$
    In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
    $\Rightarrow\angle\text{BAC}+90^\circ+40^\circ=180^\circ$
    $\Rightarrow\angle\text{BAC}=180^\circ-130^\circ=50^\circ$
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Question 181 Mark
In the given figure, O is the centre of a circle in which $\angle\text{OBA}=20^\circ$ and $\angle\text{OCA}=30^\circ.$ Then, $\angle\text{BOC}=?$
Answer
  1. 100º
    Solution:
    In $\triangle\text{OAB},$ we have:
    OA = OB (Radii of a circle)
    $\Rightarrow\angle\text{OAB}=\angle\text{OBA}=20^\circ$
    In $\triangle\text{OAC},$ we have:
    OA = OC (Radii of a circle)
    $\Rightarrow\angle\text{OAC}=\angle\text{OCA}=30^\circ$
    Now, $\angle\text{BAC}=(20^\circ+30^\circ)=50^\circ$
    $\therefore\angle\text{BOC}=(2\times\angle\text{BAC})=(2\times50^\circ)=100^\circ$
    $\Rightarrow\angle\text{BOC}=100^\circ$
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Question 191 Mark
In the given figure, AD is a diameter of the circle with centre O. Chords AB, BC and CD are equal. If $\angle\text{DEF}=110^\circ,$ then $\angle\text{FAB}$ is equal to:
Answer
  1. 130º
    Solution:

    Here, given AB = BC = CD
    Now, equal chords subtend equal angles at centre. So, $\angle\text{AOC}=\angle\text{BOC}=\angle\text{COD}$
    Also, they lie in straight line so, $\angle\text{AOC}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
    $\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}=60^\circ$
    In $\triangle\text{AOB}$
    $\text{AO}=\text{OB},\angle\text{OAB}=\angle\text{OBA}$
    Since, $\angle\text{AOB}=60^\circ,\angle\text{OAB}=\angle\text{OBA}=60^\circ$
    Now, $\angle\text{DOE}=\angle\text{AOB}=60^\circ$ (vertically opposite angle)
    In $\triangle\text{DOE},\text{OD}=\text{OE}$ (radius)
    so, $\angle\text{ODE}=\angle\text{OED}$
    $\triangle\text{DOE},\angle\text{DOE}+\angle\text{ODE}+\angle\text{OED}=180^\circ$
    $2\angle\text{ODE}=\angle\text{OED}=180-60=120^\circ$
    $\angle\text{ODE}=\angle\text{OED}=60^\circ$
    given was, $\angle\text{DEF}=110^\circ,$ so, $\angle\text{OEF}=110-60=50^\circ$
    Now, in $\triangle\text{EOF}\ \text{OE}=\text{OF}$
    so, $\angle\text{OEF}=\angle\text{OFE}=50^\circ$
    In, $\triangle\text{EOF}\ \angle\text{FOE}=180-(50+50)=80^\circ$
    Now, $\angle\text{DOE}+\angle\text{FOE}+\angle\text{AOF}=180^\circ$ (All lie on same straight line)
    So, $\angle\text{AOF}=180-(80+60)=40^\circ$
    Now, in  $\triangle\text{AOF}\ \text{AO}=\text{FO}$
    SO, $\angle\text{OFA}=\angle\text{OAF}$
    In $\triangle\text{AOF},2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
    $2\angle\text{OAF}+\angle\text{FOA}=180^\circ$
    $\angle\text{OAF}=90-20=70^\circ$
    So, $\angle\text{FAB}=\angle\text{FAO}+\angle\text{OAB}$
    $=70^\circ+60^\circ=130^\circ$
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Question 201 Mark
In a circle of radius 17cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23cm. If the length of one chord is 16cm, then the length of the other is:
Answer
  1. 30cm
    Solution:
    Given that: Radius of the circle is 17cm, distance between two parallel chords AB and CD is 23cm, where AB = 16cm. 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.
    AM = MB = 8cm
    Let OM = x cm ⇒ ON = 23 - x
    In right angled triangle OMB,
    $\text{x}=\sqrt{172-82=15}$
    Now, in triangle OND, ON = (23 - x)cm = (23 - 15)cm = 8cm
    $\text{ND}=\sqrt{\text{OD}2-\text{ON}2}$
    $\Rightarrow\text{ND}=\sqrt{172-82=15}$
    Therefore, the length of the other chord is
    CD = 2 × 15 = 30cm
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Question 211 Mark
In the given figure, O is the centre of the circle. $\angle\text{OAB}$ and $\angle\text{OCB}$ are  40º and 30º respectively. Then, the measure of $\angle\text{AOC}$ is:
Answer
  1. 140º
    Solution:

    OA = OB = OC = Radius
    So, $\angle\text{OAB}\angle\text{ABO}=40^\circ$
    and, $\angle\text{OCB}=\angle\text{ABO}=30^\circ$
    Thus, $\angle\text{ABC}=30+40=70^\circ$
    So, $\angle\text{AOC}=70\times2=140^\circ$ {Angle subtended by arc at centre is twice of angle subtended at circumference}
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Question 221 Mark
How many circle can pass through three non-collinear points?
Answer
  1. One
    Solution:
    Only one circle can be drawn from three distinct points. Join any two sets of points. Make their respective perpendicular bisectors. The point where the two bisectors meet act as the centre of such a circle. Taking the distance between the centre and any of the given points as radius, we can draw the circle. Such a circle would definitely pass through the remaining two points as well.
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Question 241 Mark
In the given figure, if $\angle\text{AOB}=80^\circ$ and $\angle\text{ABC}=30^\circ,$ then $\angle\text{CAO}$ is equal to:
Answer
  1. 60º
    Solution:

    In triangle AOB, OA = OB (Radii)
    $\Rightarrow\angle\text{OAB}=\angle\text{OBA}$
    Now,
    $\Rightarrow\angle\text{OAB}+\angle\text{ABO}+\angle\text{AOB}=180^\circ$
    $\Rightarrow2\angle\text{OAB}=80^\circ=180^\circ$
    $\Rightarrow\angle\text{OAB}=50^\circ$
    Now, $\angle\text{CAB}=\frac{1}{2}\angle\text{AOB}=\frac{80^\circ}{2}=40^\circ$
    Again, in triangle ABC,
    $\angle\text{CAB}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$
    $\angle\text{CAB}+30^\circ+40^\circ=180^\circ$
    $\angle\text{CAB}=110^\circ$
    Thus,
    $\angle\text{CAO}=\angle\text{CAB}-\angle\text{OAB }$
    $=110^\circ-50^\circ=60^\circ$
    Hence, proved.
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Question 251 Mark
In the given figure, M, A, B and N are points on a circle having centre O. AN and MB cut at Y. If $\angle\text{NYB}=50^\circ$ and $\angle\text{YNB}=20^\circ,$ then reflex $\angle\text{MON}$ is equal to:
Answer
  1. 220º
    Solution:

    In triangle NYB,
    $\angle\text{N}+\angle\text{Y}+\angle\text{B}=180^\circ$
    $\Rightarrow\angle\text{B}=180^\circ-50^\circ-20^\circ=110^\circ$
    Complete the cyclic quadrilateral, MBNX, where X being any point on the circumference in the major segment, we have:-
    $\angle\text{MXN}=80^\circ-110^\circ=70^\circ$
    So, minor $\angle\text{MON}=70^\circ\times2=140^\circ$
    Hence, reflex $\angle\text{MON}=360^\circ-140^\circ=220^\circ$
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Question 261 Mark
An equilateral triangle of side 9cm is inscribed in a circle. The radius of the circle is:
Answer
  1. $3\sqrt{3}\text{cm}$
    Solution:

    Let $\triangle\text{ABC}$ be an equilateral triangle.
    Let AD be one of its medians.
    Then, $\text{AD}\perp\text{BC}$ and BD = 4.5cm
    In right $\triangle\text{ADB},$
    $\therefore\ \text{AD}=\sqrt{\text{AB}^2-\text{BD}^2}$ [By pythagoras theorem]
    $=\sqrt{9^2-\frac{9^2}{2}}$
    $=\sqrt{81-\frac{81}{2}}$
    $=\sqrt{\frac{243}{2}}$
    $=\frac{9\sqrt{3}}{2}\text{cm}$
    Let G be the centroid of $\triangle\text{ABC}.$
    Then, AG : GD = 2 : 1
    $\therefore$ radius $=\text{AG}=\frac{2}{3}\text{AD}=\Big(\frac{2}{3}\times\frac{9\sqrt{3}}{2}\Big)=3\sqrt{3}\text{cm}$
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Question 271 Mark
In the given figure, BOC is a diameter of a circle with centre O. If $\angle\text{BCA}=30^\circ,$ then $\angle\text{CDA}=?$
Answer
  1. 60º
    Solution:
    Angles in a semi-circle measure 90º.
    $\therefore\angle\text{BAC}=90^\circ$
    In $\triangle\text{ABC},$ we have:
    $\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ (Angle sum property of a triangle)
    $\therefore90^\circ+\angle\text{ABC}+30^\circ=180^\circ$
    $\Rightarrow\angle\text{ABC}=(180^\circ-120^\circ)=60^\circ$
    $\therefore\angle\text{CDA}=\angle\text{ABC}=60^\circ$ (Angles in the same segment of a circle)
    $\Rightarrow\angle\text{CDA}=60^\circ$
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Question 281 Mark
The radius of a circle is $13\ cm$ and the length of one of its chords is $10\ cm$. The distance of the chord from the centre is:
Answer

Let $O$ be the centre of the circle with radius $OA = 13\ cm.$
$AB$ is given to be $10\ cm.$
Distance of a point to a line is always perpendicular to the line.
So, $\text{OL}\perp\text{AB}.$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\Rightarrow AL = LB = 5\ cm$
In right $\triangle\text{OLA},$
$OL^2 = AO^2 - AL^2 [$By pythagoras theorem$]$
$\Rightarrow OL^2 = 13^2 - 5^2$
$\Rightarrow OL^2 = 169 - 25$
$\Rightarrow OL^2 = 144$
$\Rightarrow OL = 12\ cm$
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Question 291 Mark
$\text{ABC}$ is a triangle with $B$ as right angle, $AC = 5\ cm$ and $AB = 4\ cm$. $A$ circle is drawn with $A$ as centre and $AC$ as radius. The length of the chord of this circle passing through $C$ and $B$ is:
Answer

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 $\triangle\text{ABC}$ apply Pythagoras theorem
$AB^2 + BC^2 = AC^2$
$\Rightarrow BC^2 = AC^2 - AB^2$
$\Rightarrow BC^2 = 5^2 - 4^2$
$\Rightarrow BC^2 = 25 - 16$
$\Rightarrow BC^2 = 9$
$\Rightarrow BC = 3\ cm$
So, $PC = 2 \times BC$
$= 2 \times 3$
$PC = 6\ cm$
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Question 301 Mark
If AB, BC and CD are equal chords of a circle with O as centre and AD diameter, than $\angle\text{AOB} =$
Answer
  1. 60°
    Solution:

    Chord AB = Chord BC = Chord CD
    $\Rightarrow\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$ (equal chords subtend equal angles at the center)
    Now, $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
    $\Rightarrow\angle\text{AOB}+\angle\text{AOB}+\angle\text{AOB}=180^\circ$
    $\Rightarrow3\angle\text{AOB}=180^\circ$
    $\Rightarrow\angle\text{AOB}=60^\circ$
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Question 331 Mark
Write the correct answer in the following: In Fig. if AOB is a diameter of the circle and AC = BC, then $\angle\text{CAB}$ is equal to:
Answer
  1. 45°.
    Solution:
    As AOB is a diameter of the circle,
    $\angle\text{C}=90^\circ$
    [$\because$ Angles in a semi-circle is 90°]
    Now, AC = BC
    $\angle\text{A}=\angle\text{B}$
    [$\because$ Angles opposite to equal sides of triangle are equal]
    Using angle sum property of a triangle, we have
    $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\Rightarrow2\angle\text{A}+90^\circ=180^\circ$
    $\Rightarrow2\angle\text{A}=90^\circ\Rightarrow\angle\text{A}=90^\circ\div2=45^\circ$
    Hence, (d) is the correct answer.
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Question 341 Mark
O is the centre of the given circle. If $\angle\text{APB}=120^\circ$ and $\angle\text{DBC}=25^\circ,$ then the measure of $\angle\text{ADB}$ is equal to:
Answer
  1. 95º
    Solution:

    Now, $\angle\text{APB}+\angle\text{CPB}=180^\circ$ (Linear Pair)
    $120^\circ+\angle\text{CPB}=180^\circ$
    $\angle\text{CPB}=60^\circ$
    Now from angle sum property, we can calculate the values of $\angle\text{CPB}$ and we find that $\angle\text{CPB}=95^\circ$
    Since, $\angle\text{PCB}=\angle\text{ADB}=95^\circ$
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Question 351 Mark
In the given figure, BOC is a diameter of a circle with centre O. If $\angle\text{BCA}=30^\circ$ then $\angle\text{CDA}=?$
Answer
  1. 60°
    Solution:
    Since BOC is a diameter, $\angle\text{BAC}=90^\circ.$
    In $\triangle\text{BAC},$
    $\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ [Angle sum property]
    $\Rightarrow\ 90^\circ+\angle\text{ABC}+30^\circ=180^\circ$
    $\Rightarrow\ \angle\text{ABC}=60^\circ$
    Since angles in the same segment of a circle are equal.
    $\angle\text{CDA}=\angle\text{ABC}=60^\circ.$
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Question 371 Mark
In the given figure, $A$ and $B$ are the centres of two circles having radii $5\ cm$ and $3\ cm$ respectively and intersecting at points $P$ and $Q$ respectively. If $AB = 4\ cm,$ then the length of common chord $PQ$ is:
Answer
We know that the line joining their centres is the perpendicular bisector of the common chord.
Join $AP.$
Then$ AP = 5\ cm; AB = 4\ cm$
Also$, AP^2 = BP^2 + AB^2 \ [$using pythagoras theorem$]$
$\Rightarrow BP^2 = AP^2 - AB^2$
$\Rightarrow BP^2 = 5^2 - 4^2$
$\Rightarrow BP = 3\ cm$
$\therefore$ triangle $\text{ABP}$ is a right angled and $PQ = 2 \times BP = (2 \times 3)cm = 6\ cm$
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Question 391 Mark
Write the correct answer in the following: In Fig. BC is a diameter of the circle and $\angle\text{BAO}=60^\circ.$ Then $\angle\text{ADC}$ is equal to:
Answer
  1. 60°.
    Solution:
    In $\triangle\text{OAB},$ we have
    OA = OB [Radii of the same circle]
    $\therefore\angle\text{ABO}=\angle\text{BAO}$ [Angles opp. To equal sides are equal]
    $\therefore\angle\text{ABO}=\angle\text{BAO}=60^\circ$ [Given]
    Now, $\angle\text{ADC}=\angle\text{ABC}=60^\circ$
    [$\because\angle\text{ABC}$ and $\angle\text{ADC}$ are angles in the same segment of circle, are equal]
    Hence, $\angle\text{ADC}=60^\circ$
    So, (c) is the correct answer.
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Question 401 Mark
In the given figure, $\text{AOB}$ is a diameter of a circle with centre $O$ such that $AB = 34\ cm$ and $CD$ is a chord of length $30\ cm.$ Then the distance of $CD$ from AB is:
Answer

Construction: Join $OC.$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
So, $\text{CL}=\frac{1}{2}\text{CD}=\frac{1}{2}(30)=15\text{cm}$
AB is the diameter.
So, $\text{AO}=\frac{1}{2}\text{AB}=\frac{1}{2}(34)=17\text{cm}.$
In $\triangle\text{OLC},$
$OL^2 = OC^2 - CL^2$
$\Rightarrow OL^2 = 17^2- 15^2$
$\Rightarrow OL^2 = 289 - 225$
$\Rightarrow OL^2 = 64$
$\Rightarrow OL = 8\ cm$
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Question 411 Mark
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance $\frac{\text{r}}{2}$ from O, then $\angle\text{BAO}=$
Answer
  1. 30º
    Solution:

    Let OD = r
    $\text{OC}=\frac{\text{r}}{2}$
    In $\angle\text{OAC}$ and $\angle\text{DAC}$
    $\text{SAS}-\angle\text{OAC}\cong\angle\text{DAC}$
    Now, in $\angle\text{OAD}$ equilateral
    $\angle\text{AOD}=60^\circ$
    $\angle\text{CAO}=\angle\text{BAO}=30^\circ$
    $\Rightarrow\sin\theta=\frac{\text{r}}{\frac{2}{\text{r}}}=\frac{1}{2}$
    $\Rightarrow\theta=30^\circ$
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Question 421 Mark
In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If $\angle\text{ACD}=25^\circ,$ then $\angle\text{AOD}=?$
Answer
  1. 75°
    Solution:
    OB = BC [Given]
    $\Rightarrow\ \angle\text{OBC}=\angle\text{BCO}=25^\circ$ [Angles opposite equal sides are equal]
    Now,
    $\angle\text{OBC}=\angle\text{BOC}+\angle\text{BCO}=25^\circ+25^\circ=50^\circ$
    OA = OB [Radii of the same circle]
    $\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$
    In $\triangle\text{AOC},$
    $\angle\text{AOD}=\angle\text{OAC}+\angle\text{ACO}$
    $=\angle\text{OAB}+\angle\text{BCO}$
    $=50^\circ+25^\circ$
    $=75^\circ$
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Question 431 Mark
In the given figure, ABCD is a cyclic quadrilateral in which $\angle\text{BAD}=75^\circ,$ $\angle\text{ABD}=58^\circ$ and $\angle\text{ADC}=77^\circ,$ AC and BD intersect at P. The measure of $\angle\text{DPC}$ is:
Answer
  1. 92º
    Solution:

    Since AD acts as a chord also, So, $\angle\text{ABD}=\angle\text{ACD}=58^\circ$
    Again as CD also acts as a chord also, therefore,
    $\angle\text{DBC}=\angle\text{DAC}$
    Now, $\angle\text{ABC}=\angle\text{ABD}+\angle\text{DBC}$
    Also, $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
    $\Rightarrow\angle\text{ABC}=180^\circ-77^\circ=103^\circ$
    And therefore
    $\angle\text{DBC}=103^\circ-58^\circ=45^\circ$
    Hence, $\angle\text{DAC}=45^\circ$
    Since,
    $\angle\text{DAC}=45^\circ$
    So, $\angle\text{CAB}=75^\circ-45^\circ=30^\circ$
    But, $\angle\text{CAB}=\angle\text{BDC}$
    $\Rightarrow\angle\text{BDC}=30^\circ$
    Now, In triangle CPD,
    $\angle\text{C}+\angle\text{P}+\angle\text{D}=180^\circ$
    $\Rightarrow58^\circ+\angle\text{P}+30^\circ=180^\circ$
    $\Rightarrow\angle\text{P}=180^\circ-30^\circ-58^\circ=92^\circ$
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Question 451 Mark
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$
Answer
  1. 60º
    Solution:
    We have:
    $\angle\text{AOB}=2\angle\text{ACB}$
    $\Rightarrow\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=(\frac{1}{2}\times90^\circ)=45^\circ$
    $\Rightarrow\angle\text{ACB}=45^\circ$
    $\angle\text{COA}=2\angle\text{CBA}=(2\times30^\circ)=60^\circ$
    $\therefore\angle\text{COD}=180^\circ-\angle\text{COA}=(180^\circ-60^\circ)=120^\circ$
    $\Rightarrow\angle\text{CAO}=\frac{1}{2}\angle\text{COD}=(\frac{1}{2}\times120^\circ)=60^\circ$
    $\Rightarrow\angle\text{CAO}=60^\circ$
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Question 461 Mark
In the given figure, AOB is a diameter of a circle and CD || AB. If $\angle\text{BAD}=30^\circ$ then $\angle\text{CAD}=?$
Answer
  1. 30°
    Solution:
    Since AB || CD, $\angle\text{BAD}=\angle\text{CDA}=30^\circ$ [Alternate angles]
    Since AOB is a diameter, $\angle\text{ADB}=90^\circ$
    $\angle\text{CDB}=\angle\text{CDA}+\angle\text{ADB}=30^\circ$
    $\Rightarrow\ \angle\text{CDB}=30^\circ+90^\circ$
    $\Rightarrow\ \angle\text{CDB}=120^\circ$
    We know that the opposite angles of a quadrilateral are supplementary.
    $\angle\text{CAB}+\angle\text{CDB}=180^\circ$
    $\Rightarrow\ \angle\text{CAD}+\angle\text{DAB}+\angle\text{CDB}=180^\circ$
    $\Rightarrow\ \angle\text{CAD}+30^\circ+120^\circ=180^\circ$
    $\Rightarrow\ \angle\text{CAD}=30^\circ$
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Question 481 Mark
$\text{ABC}$ is a triangle with $B$ as right angle, $AC = 5\ cm$ and $AB = 4\ cm \ A$. circle is drawn with$ A$ as centre and $AC$ as radius. The length of the chord of this circle passing through $C$ and $B$ is:
Answer

$AD$ and $AC$ are radii of same circle and $CD$ is a chord.
Consider $\triangle\text{ABC},$
$BC^2 = (AC)^2 - (AB)^2$
$=5^2 - 4^2 = 25 - 16 = 9$
$\Rightarrow BC = 3\ cm$
Chord $CD = 2 \times BC = 6\ cm$
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Question 501 Mark
The whole arc of a circle is called.
Answer
  1. Circumference
    Solution:
    Circumference is the total length of the circle or in other words its a perimeter of the circle.
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M.C.Q - MATHS STD 9 Questions - Vidyadip