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MATHS M.C.Q

Circle · Rajasthan Board (RBSE) STD 9 (Rajasthan - English Medium). 42 questions.

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M.C.Q

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42 questions · timed · auto-graded

MCQ 11 Mark
In the given figure$, CD$ is the diameter of a circle with centre $O$ and $CD$ is perpendicular to chord $AB.$ If $AB = 12\ cm$ and $CE = 3\ cm,$ then radius of the circles is:
  • A
    $6\ cm$
  • B
    $9\ cm$
  • ✓
    $7.5\ cm$
  • D
    $8\ cm$
Answer
Correct option: C.
$7.5\ cm$
$OA = OC$
$\Rightarrow OA = OE + CE$
$\Rightarrow OA = OE + 3$
$\Rightarrow OE = OA - 3 ...(i)$
$\text{AE}=\frac{1}{2}\text{AB}$ [Perpendicular drawn from the centre of a circle to the chord bisect the chord]
$=\frac{1}{2}(12)=6\text{ cm}$
In right $\triangle\text{OEA},$
$OA^2 = OE^2 + AE^2$
$\Rightarrow OA^2 = (OA - 3)^2 + AE^2 [$From $(i)]$
$\Rightarrow OA^2 = OA^2 - 6OA + 9 + AE^2$
$\Rightarrow 6OA = 9 + 6^2$
$\Rightarrow 6OA = 9 + 36$
$\Rightarrow\ \text{OA}=\frac{45}{6}=7.5\text{ cm}$
So, the radius of the circle is $7.5\ cm.$
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MCQ 21 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:
  • A
    $11.5\text{ cm}$
  • ✓
    $12\text{ cm}$
  • C
    $\sqrt{69}\text{ cm}$
  • D
    $23\text{ cm}$
Answer
Correct option: B.
$12\text{ cm}$

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 31 Mark
 In the given figure, BOC is a diameter of a circle with centre O. If $\angle\text{BCA}=30^\circ$ then $\angle\text{CDA}=?$
  1. 30°
  2. 45°
  3. 60°
  4. 50°
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 41 Mark
In the given figure ABCD is a cyclic quadrilateral in which AB || DC and $\angle\text{BAD}=100^\circ.$ Then, $\angle\text{ABC}=?$
  1. 80°
  2. 100°
  3. 50°
  4. 40°
Answer
  1. 100°
Solution:
Since AB || DC,
$\angle\text{ADC}+\angle\text{BAD}=180^\circ$
$\Rightarrow\ 100^\circ+\angle\text{BAD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}=80^\circ$
We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{BAD}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 80^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=100^\circ$
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Question 51 Mark
In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If $\angle\text{AEB}=110^\circ$ and $\angle\text{CBE}=30^\circ,$ then $\angle\text{ADB}=?$
  1. 70°
  2. 60°
  3. 80°
  4. 90°
Answer
  1. 80°
Solution:
$\angle\text{AED}=\angle\text{ECB}+\angle\text{EBC}$
$\Rightarrow\ 110^\circ=\angle\text{ECB}+30^\circ$
$\Rightarrow\ \angle\text{ECB}=80^\circ$
Since angles in the same segment are equal,
$\angle\text{ADB}=\angle\text{ECB}=80^\circ$
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Question 61 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 = ?
  1. 10.5cm
  2. 9.5cm
  3. 8.5cm
  4. 7.5cm
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 71 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=130^\circ.$ Then, $\angle\text{ACB}=?$
  1. 50°
  2. 65°
  3. 115°
  4. 155°
Answer
  1. 115°
Solution:
Minor $\angle\text{AOB}=130^\circ$
Major $\angle\text{AOB}=360^\circ-130^\circ$
⇒ Major $\angle\text{AOB}=230^\circ$
Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(230^\circ)$
$\Rightarrow\ \angle\text{ACB}=115^\circ$
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Question 81 Mark
In the given figure, O is the centre of a circle in which $\angle\text{AOC}=100^\circ.$ Side AB of quadrilateral OABC has been produced to D. Then, $\angle\text{CBD}=?$
  1. 50°
  2. 40°
  3. 25°
  4. 80°
Answer
  1. 50°
Solution:

Construction: Let E be a point on the reamaining part of the circumference of the circle.
Join AE and CE.
$\angle\text{AEC}=\frac{1}{2}\angle\text{AOC}$
$\Rightarrow\ \angle\text{AEC}=\frac{1}{2}(100^\circ)=50^\circ$
Since AECB forms a cyclic quadrilateral.
$\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$
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Question 91 Mark
In the given figure, O is the centre of a circle. If $\angle\text{OAC}=50^\circ$ then $\angle\text{ODB}=?$
  1. 40°
  2. 50°
  3. 60°
  4. 75°
Answer
  1. 50°
Solution:
Since angles in the same segment of a circle are equal.
$\angle\text{CDB}=\angle\text{BAC}$
That is , $\angle\text{ODB}=\angle\text{OAC}=50^\circ.$
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Question 101 Mark
In the given figure, O is the centre of a circle in which $\angle\text{OAB}=20^\circ$ and $\angle\text{OCB}=50^\circ.$ Then, $\angle\text{AOC}=?$
  1. 50°
  2. 70°
  3. 20°
  4. 60°
Answer
  1. 60°
Solution:
OA = OC [Radii of the same circle]
$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$
In $\triangle\text{OAB,}$
$\angle\text{OBA}+\angle\text{OAB}+\angle\text{AOB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 20^\circ+20^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\ \angle\text{AOB}=140^\circ$
Now,
OB = OC [Radii of the same circle]
$\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}=50^\circ$
In $\triangle\text{OCB},$
$\angle\text{OBC}+\angle\text{OCB}+\angle\text{COB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 50^\circ+50^\circ+\angle\text{COB}=180^\circ$
$\Rightarrow\ \angle\text{COB}=80^\circ$
So,
$\angle\text{AOB}=\angle\text{AOC}+\angle\text{COB}$
$\Rightarrow\ \angle\text{AOC}=\angle\text{AOB}-\angle\text{COB}$
$\Rightarrow\ \angle\text{AOC}=140^\circ-80^\circ$
$\Rightarrow\ \angle\text{AOC}=60^\circ$
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Question 111 Mark
In the given figure, O is the centre of a circle and $\angle\text{ACB}=30^\circ.$ Then, $\angle\text{AOB}=?$
  1. 30°
  2. 15°
  3. 60°
  4. 90°
Answer
  1. 60°
Solution:
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,
$\angle\text{AOB}=2\angle\text{ACB}$
$=2(30^\circ)$
$=60^\circ$
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Question 121 Mark
In the given figure, O is the centre of a circle. If $\angle\text{AOB}=100^\circ$ and $\angle\text{AOC}=90^\circ$ then $\angle\text{BAC}=?$
  1. 85°
  2. 80°
  3. 95°
  4. 75°
Answer
  1. 85°
Solution:
$\angle\text{BOA}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Angles around a point are 360°]
$\Rightarrow\ 100^\circ+90^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\ \angle\text{BOC}=170^\circ$
Now,
$\angle\text{BAC}=\frac{1}{2}(\angle\text{BOC})=\frac{1}{2}(170^\circ)=85^\circ$
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MCQ 131 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:
  • ✓
    $8\ cm$
  • B
    $15\ cm$
  • C
    $18\ cm$
  • D
    $6\ cm$
Answer
Correct option: A.
$8\ cm$

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 141 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 $\angle\text{AOD}=?$
  1. 50°
  2. 75°
  3. 90°
  4. 100°
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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MCQ 151 Mark
A chord is at a distance of $8\ cm$ from the centre of a circle of radius $17\ cm.$ The length of the chord is:
  • A
    $25\ cm$
  • B
    $12.5\ cm$
  • ✓
    $30\ cm$
  • D
    $9\ cm$
Answer
Correct option: C.
$30\ cm$

Let $O$ be the centre of the circle with radius $OA = 17\ cm.$
Since $\text{OC}\perp\text{AB}.$
In right $\triangle\text{OCA},$
$OA^2= OC^2 + AC^2 [$By pythagoras theorem$]$
$AC^2= OA^2 - OC^2$
$\Rightarrow AC^2 = 17^2 - 8^2$
$\Rightarrow AC^2 = 289 - 64$
$\Rightarrow AC^2 = 225$
$\Rightarrow AC = 15\ cm$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\Rightarrow AB = 2AC = 2(15) = 30\ cm.$
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Question 161 Mark
In the given figure, AOB is a diameter of a circle and CD || AB. If $\angle\text{BAD}=30^\circ$ then $\angle\text{CAD}=?$
  1. 30°
  2. 60°
  3. 45°
  4. 50°
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 171 Mark
In the given figure, BOC is a diameter of a circle and AB = AC. Then, $\angle\text{ABC}=?$
  1. 30°
  2. 45°
  3. 60°
  4. 90°
Answer
  1. 45°
Solution:
Since BOC is a diameter of a circle, $\angle\text{BAC}$ is 90°.
Given that AB = AC.
$\Rightarrow\ \angle\text{ABC}=\angle\text{ACB}$
In $\triangle\text{BAC},$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ [Angle sum property]
$\Rightarrow\ \angle\text{ABC}+\angle\text{ABC}+90^\circ=180^\circ$
$\Rightarrow\ 2\angle\text{ABC}=90^\circ$
$\Rightarrow\ \angle\text{ABC}=45^\circ$
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MCQ 181 Mark
In the given figure, $O$ is the centre of a circle and diameter $AB$ bisects the chord $CD$ at a point $E$ such that $CE = ED = 8\ cm$ and $EB = 4\ cm.$ The radius of the circle is:
  • ✓
    $10\ cm$
  • B
    $12\ cm$
  • C
    $6\ cm$
  • D
    $8\ cm$
Answer
Correct option: A.
$10\ cm$
Construction: Join $OD.$
$OB = OD$
$\Rightarrow OB = OE + EB$
$\Rightarrow OB = OE + 4$
$\Rightarrow OE = OB - 4 ...(i)$
In right $\triangle\text{OED},$
$OD^2 = OE^2 + DE^2$
$\Rightarrow OD^2 = (OB - 4)^2 + DE^2 [$From $(i)]$
$\Rightarrow OD^2 = OB^2 - 80A + 16 + 8^2$
$\Rightarrow 8OD = 16 + 64$
$\Rightarrow 8OD = 80$
$\Rightarrow\ \text{OD}=\frac{80}{8}=10\text{ cm}$
So, the radius of the circle is $10\ cm.$
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Question 191 Mark
In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$
  1. 30°
  2. 45°
  3. 60°
  4. 90°
Answer
  1. 60°
Solution:
$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
$\Rightarrow\ \angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
Since AOD is a straight line,
$\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$
$\therefore\ \angle\text{COD}+60^\circ=180^\circ$
$\therefore\ \angle\text{COD}=120^\circ$
$\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$
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Question 201 Mark
In the given figure, equilateral $\triangle\text{ABC}$ is inscribed in a circle and ABCD is a quadrilateral, as shown. Then, $\angle\text{BDC}=?$
  1. 90°
  2. 60°
  3. 120°
  4. 150°
Answer
  1. 120°
Solution:
Since $\triangle\text{BDC}$ is an equilateral traingle, $\angle\text{BAC}=60^\circ.$
Since ABCD is a cyclic equilateral,
$\angle\text{BAC}+\angle\text{BDC}=180^\circ$
$\Rightarrow\ 60^\circ+\angle\text{BDC}=180^\circ$
$\Rightarrow\ \angle\text{BDC}=120^\circ$
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Question 211 Mark
In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ$ then $\angle\text{BAC}=?$
  1. 60°
  2. 30°
  3. 20°
  4. 45°
Answer
  1. 30°
Solution:
We know that the opposite angles of a quadrilateral are supplementary.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=60^\circ$
 Since BOC is a diameter $\angle\text{ACB}=90^\circ.$
In $\triangle\text{CAB},$
$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ [Angle sum property]
$\Rightarrow\ 60^\circ+\angle\text{BAC}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAC}=30^\circ$
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Question 221 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 $\angle\text{ACB}=?$
  1. 40°
  2. 50°
  3. 80°
  4. 100°
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 231 Mark
In the give figure, AB and CD are two intersecting chords of a circle. If $\angle\text{CAB}=40^\circ$ and $\angle\text{BCD}=80^\circ$then $\angle\text{CBD}=?$
  1. 80°
  2. 60°
  3. 50°
  4. 70°
Answer
  1. 60°
Solution:
Since angles in the same segment are equal,
$\angle\text{BDC}=\angle\text{BAC}=40^\circ$
In $\triangle\text{BDC},$
$\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ [Angle sum property]
$\Rightarrow\ 40^\circ+80^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\ \angle\text{CBD}=60^\circ$
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Question 241 Mark
In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that $\angle\text{ADC}=95^\circ$ and $\angle\text{ECF}=20^\circ.$ Then, $\angle\text{EAD}=?$
  1. 95°
  2. 85°
  3. 105°
  4. 75°
Answer
  1. 105°
Solution:
Since CF || AB, $\angle\text{ABC}=\angle\text{BCF}=85^\circ$
$\angle\text{BAD}=\angle\text{BCE}$
$\Rightarrow\ \angle\text{BAD}=\angle\text{BCF}+\angle\text{ECF}$
$\Rightarrow\ \angle\text{BAD}=105^\circ$
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Question 251 Mark
In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If $\angle\text{CBE}=100^\circ$ then $\angle\text{CDF}=?$
  1. 100°
  2. 80°
  3. 130°
  4. 90°
Answer
  1. 80°
Solution:
Since ABCD is a cyclic equilateral,
$\angle\text{CBE}=\angle\text{ADC}=100^\circ$
Since ADF is a straight line,
$\angle\text{CDF}+\angle\text{ADC}=180^\circ$
$\Rightarrow\ \angle\text{CDF}+100^\circ=180^\circ$
$\Rightarrow\ \angle\text{CDF}=80^\circ$
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Question 261 Mark
In the given figure, O is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{BOD}=?$
  1. 130°
  2. 50°
  3. 100°
  4. 80°
Answer
  1. 100°
Solution:
OA = OB [Radii of the same circle]
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$ [Angle opposite equal sides are equal]
$\angle\text{BOD}=\angle\text{OAB}+\angle\text{OBA}$
$\Rightarrow\ \angle\text{BOD}=50^\circ+50^\circ$
$\Rightarrow\ \angle\text{BOD}=100^\circ$
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Question 271 Mark
In the given figure, ABCD and ABEF are two cyclic quadrilaterals. If $\angle\text{BCD}=110^\circ$ then $\angle\text{BEF}=?$
  1. 55°
  2. 70°
  3. 90°
  4. 110°
Answer
  1. 110°
Solution:
Since ABCD is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$
$\Rightarrow\ \angle\text{BAD}=70^\circ$
Since ABEF is a cyclic qyadrilateral, we have:
$\angle\text{BAD}+\angle\text{BEF}=180^\circ$
$\Rightarrow\ 70^\circ+\angle\text{BEF}=180^\circ$
$\Rightarrow\ \angle\text{BEF}=110^\circ$
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Question 281 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOC}=130^\circ.$ Then, $\angle\text{ABC}=?$
  1. 50°
  2. 65°
  3. 115°
  4. 130°
Answer
  1. 115°
Solution:
Minor $\angle\text{AOC}=130^\circ$
Major $\angle\text{AOC}=360^\circ-130^\circ$
⇒ Major $\angle\text{AOC}=230^\circ$
Since $\angle\text{ABC}=\frac{1}{2}\text{major}\angle\text{AOC}$
$\Rightarrow\ \angle\text{ABC}=\frac{1}{2}(230^\circ)$
$\Rightarrow\ \angle\text{ABC}=115^\circ$
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Question 291 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, $\angle\text{BOC}=?$
  1. 50°
  2. 90°
  3. 100°
  4. 130°
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 301 Mark
In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and $\angle\text{CBD}=35^\circ.$ Then, $\angle\text{BAD}=?$
  1. 65°
  2. 70°
  3. 110°
  4. 90°
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 311 Mark
AB and CD are two equal chords of a circle with centre O such that $\angle\text{AOB}=80^\circ$ then $\angle\text{COD}=?$
  1. 100°
  2. 80°
  3. 120°
  4. 40°
Answer
  1. 80°
Solution:
Given that AB = CD.
Since equal chord, subtend equal angles at the centre,
$\angle\text{COD}=\angle\text{AOB}=80^\circ.$
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Question 321 Mark
In the given figure, O is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{CDA}=?$
  1. 40°
  2. 50°
  3. 75°
  4. 25°
Answer
  1. 50°
Solution:
OA = OB [Radii of the same circle]
$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=50^\circ$
Since angles in the same segment are equal, $\angle\text{ABC}=\angle\text{CDA}.$
That is, $\angle\text{ABO}=\angle\text{CDA}=50^\circ$
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Question 331 Mark
In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. If AB = 10cm, then CD = ?
  1. 5cm
  2. 12.5cm
  3. 15cm
  4. 10cm
Answer
  1. 10cm
Solution:
In $\triangle\text{BEO}$ and $\triangle\text{CFO},$
OB = OC [Radii of the same circle]
$\angle\text{OBE}=\angle\text{OCF}$ [Alternate angles since AB || CD]
$\angle\text{BOE}=\angle\text{COF}$ [Vertically angles]
$\Rightarrow\ \triangle\text{BEO}\cong\triangle\text{CFO}$ [ASA congruence criterion]
⇒ OE = OF [C.P.C.T.]
Since chord are equidistant from the centre are equal, AB = CD = 10 cm.
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Question 341 Mark
In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If $\text{OD}\perp\text{AB}$ such that OD = 6cm, then AC = ?
  1. 9cm
  2. 12cm
  3. 15cm
  4. 7.5cm
Answer
  1. 12cm
Solution:
In $\triangle\text{BOD}$ and $\triangle\text{CAB},$
Since BOC is the diameter, $\angle\text{CAB}=90^\circ.$
Also, $\angle\text{ODB}=90^\circ.$
So, $\angle\text{DBO}=\angle\text{ABC}$ [Common angles]
$\Rightarrow\ \triangle\text{BOD}\sim\triangle\text{BCA}$ [AA congruence criterion]
$\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{\text{BO}}{\text{BA}}$
$\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{1}{2}$ [Since radius = 2 diameter]
$\Rightarrow\ \frac{6}{\text{CA}}=\frac{1}{2}$
⇒ CA = 12cm that is, AC = 12cm.
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Question 351 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOC}=120^\circ.$ Then, $\angle\text{BDC}=?$
  1. 60°
  2. 45°
  3. 30°
  4. 15°
Answer
  1. 30°
Solution:
Since BOA is a diameter.
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\ \angle\text{BOC}=60^\circ$
So, $\angle\text{BDC}=\frac{1}{2}\angle\text{BOC}=\frac{1}{2}(60^\circ)=30^\circ$
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MCQ 361 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:
  • A
    $3\ cm$
  • ✓
    $6\ cm$
  • C
    $7.5\ cm$
  • D
    $9\ cm$
Answer
Correct option: B.
$6\ cm$
We know that, the line joining centres is the perpendicular bisector of the common chord.
Then,
$AP = 5\ cm, BP = 3\ cm$ and $AB = 4\ cm$
$AP^2 = 5^2 = 25$
$BP^2 + AB^2= 3^2 + 4^2 = 25$
In $\triangle\text{ABP},$
Since $AP^2 = BP^2 + AB^2$
$\triangle\text{ABP},$ is a right$-$angled triangle and $PQ = 2BP = 2(3) = 6\ cm.$
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Question 371 Mark
In the given figure, O is the centre of a circle. Then, $\angle\text{OAB}=?$
  1. 50°
  2. 60°
  3. 55°
  4. 65°
Answer
  1. 65°
Solution:
OA = OB [Radii of the same circle]
$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$
In $\triangle\text{OAB},$
$\angle\text{BOA}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ [Angle sum property]
$\Rightarrow\ 50^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$
$\Rightarrow\ 2\angle\text{OAB}=130^\circ$
$\Rightarrow\ \angle\text{OAB}=65^\circ$
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Question 381 Mark
An equilateral triangle of side 9cm is inscribed in a circle. The radius of the circle is:
  1. $3\text{cm}$
  2. $3\sqrt{2}\text{cm}$
  3. $3\sqrt{3}\text{cm}$
  4. $6\text{cm}$
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 391 Mark
In the given figure, O is the centre of a circle and $\angle\text{AOB}=140^\circ.$ Then, $\angle\text{ACB}=?$
  1. 70°
  2. 80°
  3. 110°
  4. 40°
Answer
  1. 110°
Solution:
Minor $\angle\text{AOB}=140^\circ$
Major $\angle\text{AOB}=360^\circ-140^\circ$
⇒ Major $\angle\text{AOB}=220^\circ$
Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(220^\circ)$
$\Rightarrow\ \angle\text{ACB}=110^\circ$
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Question 401 Mark
The angle in a semicircle measures:
  1. 45°
  2. 60°
  3. 90°
  4. 36°
Answer
  1. 90°
Solution:
The angle in a semicircle measures 90°.
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Question 411 Mark
Angles in the same segment of a circle area are:
  1. Equal
  2. Complementary
  3. Supplementary
  4. None of these
Answer
  1. Equal
Solution:
Angles in a the same segment of a circle area are equal.
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Question 421 Mark
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are inscribed in a circle such that $\angle\text{BAC}=60^\circ$ and $\angle\text{DBC}=50^\circ.$ Then $\angle\text{BCD}=?$
  1. 50°
  2. 60°
  3. 70°
  4. 80°
Answer
  1. 70°
Solution:
Since angles in the same segment of a circle are equal.
$\angle\text{BAC}=\angle\text{BDC}=60^\circ.$
In $\triangle\text{BDC},$
$\angle\text{BDC}+\angle\text{DBC}+\angle\text{BCD}=180^\circ.$
$\Rightarrow\ 60^\circ+50^\circ+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BCD}=70^\circ$
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M.C.Q - MATHS STD 9 Questions - Vidyadip