MCQ(1M) - Maths STD 9 Questions - Vidyadip

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MCQ(1M)

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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.

$7.5\ cm$

$OA=OC$
$\Rightarrow OA=OE+CE$
$\Rightarrow OA=OE+3$
$\Rightarrow OE=OA-3$
$AE=\frac{1}{2} AB [$ Perpendicular drawn from the centre of a circle to the chord bisect the chord $]$
$=\frac{1}{2}(12)=6 \ cm$
$\text { In right } \triangle OEA,$
$OA^2=O E^2+A E^2$
$\Rightarrow OA^2=(O A-3)^2+AE^2[\text { from (i) }]$
$\Rightarrow O A^2=O A^2-6 O A+9+AE^2$
$\Rightarrow 6 OA=9+6^2$
$\Rightarrow 6 OA=9+36$
$\Rightarrow OA=\frac{45}{6}=7.5 \ cm$
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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MCQ 31 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, $CL =\frac{1}{2} CD =\frac{1}{2}(30)=15 \ cm$
$A B$ is the diameter.
So, $AO =\frac{1}{2} AB =\frac{1}{2}(34)=17 \ cm$.
In $\triangle OLC$,
$OL ^2= OC ^2- CL ^2$
$\Rightarrow OL ^2=17^2-15^2$
$\Rightarrow OL ^2=289-225$
$\Rightarrow OL ^2=64$
$\Rightarrow OL =8 \ cm$

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MCQ 41 Mark

In the given figure $, \text{BOC}$ is a diameter of a circle with centre $O$. If $\angle\text{BCA}=30^\circ$ then $\angle\text{CDA}=?$

A
$30^\circ$
B
$45^\circ$
✓
$60^\circ$
D
$50^\circ$

Answer

Correct option: C.

$60^\circ$

Since $\text{BOC}$ 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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MCQ 51 Mark

The angle in a semicircle measures :

A
$45^\circ$
B
$60^\circ$
✓
$90^\circ$
D
$36^\circ$

Answer

Correct option: C.

$90^\circ$

$90^\circ$

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MCQ 61 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 $OC \perp AB$.
In right $\triangle OCA$,
$OA ^2= OC ^2+ AC ^2[ By$ pythagoras theorem $]$
$AC^2=OA^2-O O^2$
$\Rightarrow AC^2=17^2-8^2$
$\Rightarrow AC^2=289-64$
$\Rightarrow AC=225$
$\Rightarrow AC=15 \ cm$
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\Rightarrow A B=2 A C=2(15)=30 \ cm$

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MCQ 71 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, $\angle\text{AOC}=?$

A
$50^\circ$
B
$70^\circ$
C
$20^\circ$
✓
$60^\circ$

Answer

Correct option: D.

$60^\circ$

$OA = OC\ [$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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MCQ 81 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{ACB}=30^\circ.$ Then, $\angle\text{AOB}=?$

A
$30^\circ$
B
$15^\circ$
✓
$60^\circ$
D
$90^\circ$

Answer

Correct option: C.

$60^\circ$

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 $,\angle\text{AOB}=2\angle\text{ACB}$
$=2(30^\circ)$
$=60^\circ$

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MCQ 91 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 $\angle\text{BAC}=?$

✓
$85^\circ$
B
$80^\circ$
C
$95^\circ$
D
$75^\circ$

Answer

Correct option: A.

$85^\circ$

$\angle\text{BOA}+\angle\text{AOC}+\angle\text{BOC}=360^\circ \ [$Angles around a point are $360^\circ ]$
$\Rightarrow\ 100^\circ+90^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow\ \angle\text{BOC}=170^\circ$
Now,
$\angle\text{BAC}=\frac{1}{2}(\angle\text{BOC})$
$=\frac{1}{2}(170^\circ)=85^\circ$

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MCQ 101 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 $\angle\text{AOD}=?$

A
$50^\circ$
✓
$75^\circ$
C
$90^\circ$
D
$100^\circ$

Answer

Correct option: B.

$75^\circ$

$OB = BC \ [$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 111 Mark

In the given figure, $\text{AOB}$ is a diameter of a circle and $CD \| AB$. If $\angle\text{BAD}=30^\circ$ then $\angle\text{CAD}=?$

✓
$30^\circ$
B
$60^\circ$
C
$45^\circ$
D
$50^\circ$

Answer

Correct option: A.

$30^\circ$

Since $AB \| CD, \angle\text{BAD}=\angle\text{CDA}=30^\circ \ [$Alternate angles$]$
Since $\text{AOB}$ 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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MCQ 121 Mark

In the given figure, $\text{BOC}$ is a diameter of a circle and $AB = AC$. Then, $\angle\text{ABC}=?$

A
$30^\circ$
✓
$45^\circ$
C
$60^\circ$
D
$90^\circ$

Answer

Correct option: B.

$45^\circ$

Since $\text{BOC}$ is a diameter of a circle, $\angle\text{BAC}$ is $90^\circ .$
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 131 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 $OD.$

$O B=O D$
$\Rightarrow O B=O E+E B$
$\Rightarrow O B=O E+4$
$\Rightarrow O E=O B-4 \ldots \text { (i) }$
$\text { In right } \triangle O E D$
$O D^2=O E^2+D E^2$
$\Rightarrow O D^2=(O B-4)^2+D E^2[$ From $( i)]$
$\Rightarrow O D^2=O B^2-80 A+16+8^2$
$\Rightarrow 8 O D=16+64$
$\Rightarrow 8 O D=80$
$\Rightarrow O D=\frac{80}{8}=10 \ cm$
So, the radius of the circle is $10 \ cm .$

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MCQ 141 Mark

In the given figure, $\angle\text{AOB}=90^\circ$ and $\angle\text{ABC}=30^\circ.$ Then, $\angle\text{CAO}=?$

A
$30^\circ$
B
$45^\circ$
✓
$60^\circ$
D
$90^\circ$

Answer

Correct option: C.

$60^\circ$

$\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$
$\Rightarrow\ \angle\text{ACB}=45^\circ$
$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$
Since $\text{AOD}$ 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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MCQ 151 Mark

In the given figure $,O$ is the centre of a circle. If $\angle\text{OAC}=50^\circ$ then $\angle\text{ODB}=?$

A
$40^\circ$
✓
$50^\circ$
C
$60^\circ$
D
$75^\circ$

Answer

Correct option: B.

$50^\circ$

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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MCQ 161 Mark

Angles in the same segment of a circle area are:

✓
Equal
B
Complementary
C
Supplementary
D
None of these

Answer

Correct option: A.

Equal

Angles in a the same segment of a circle area are equal.

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MCQ 171 Mark

In the given figure, equilateral $\triangle\text{ABC}$ is inscribed in a circle and $\text{ABCD}$ is a quadrilateral, as shown. Then, $\angle\text{BDC}=?$

A
$90^\circ$
B
$60^\circ$
✓
$120^\circ$
D
$150^\circ$

Answer

Correct option: C.

$120^\circ$

Since $\triangle\text{BDC}$ is an equilateral traingle, $\angle\text{BAC}=60^\circ.$
Since $\text{ABCD}$ 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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MCQ 181 Mark

In the given figure $\text{AOB}$ is a diameter and $\text{ABCD}$ is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ$ then $\angle\text{BAC}=?$

A
$60^\circ$
✓
$30^\circ$
C
$20^\circ$
D
$45^\circ$

Answer

Correct option: B.

$30^\circ$

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 $\text{BOC}$ 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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MCQ 191 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 $\angle\text{ACB}=?$

A
$40^\circ$
✓
$50^\circ$
C
$80^\circ$
D
$100^\circ$

Answer

Correct option: B.

$50^\circ$

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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MCQ 201 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 $\angle\text{CBD}=?$

A
$80^\circ$
✓
$60^\circ$
C
$50^\circ$
D
$70^\circ$

Answer

Correct option: B.

$60^\circ$

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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MCQ 211 Mark

In the given figure, $\text{ABCD}$ 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, $\angle\text{EAD}=?$

A
$95^\circ$
B
$85^\circ$
✓
$105^\circ$
D
$75^\circ$

Answer

Correct option: C.

$105^\circ$

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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MCQ 221 Mark

In the given figure, sides $AB$ and $AD$ of quad. $\text{ABCD}$ are produced to $E$ and $F$ respectively. If $\angle\text{CBE}=100^\circ$ then $\angle\text{CDF}=?$

A
$100^\circ$
✓
$80^\circ$
C
$130^\circ$
D
$90^\circ$

Answer

Correct option: B.

$80^\circ$

Since $\text{ABCD}$ is a cyclic equilateral,
$\angle\text{CBE}=\angle\text{ADC}=100^\circ$
Since $\text{ADF}$ 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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MCQ 231 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{BOD}=?$

A
$130^\circ$
B
$50^\circ$
✓
$100^\circ$
D
$80^\circ$

Answer

Correct option: C.

$100^\circ$

$OA = OB\ [$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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MCQ 241 Mark

In the given figure, $\text{ABCD}$ and $\text{ABEF}$ are two cyclic quadrilaterals. If $\angle\text{BCD}=110^\circ$ then $\angle\text{BEF}=?$

A
$55^\circ$
B
$70^\circ$
C
$90^\circ$
✓
$110^\circ$

Answer

Correct option: D.

$110^\circ$

Since $\text{ABCD}$ 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 $\text{ABEF}$ 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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MCQ 251 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{AOC}=130^\circ.$ Then, $\angle\text{ABC}=?$

A
$50^\circ$
B
$65^\circ$
✓
$115^\circ$
D
$130^\circ$

Answer

Correct option: C.

$115^\circ$

Minor $\angle\text{AOC}=130^\circ$
Major $\angle\text{AOC}=360^\circ-130^\circ$
$\Rightarrow$ 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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MCQ 261 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, $\angle\text{BOC}=?$

A
$50^\circ$
B
$90^\circ$
✓
$100^\circ$
D
$130^\circ$

Answer

Correct option: C.

$100^\circ$

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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MCQ 271 Mark

In the give figure, $\text{ABCD}$ is a cyclic quadrilateral in which $BC = CD$ and $\angle\text{CBD}=35^\circ.$ Then, $\angle\text{BAD}=?$

A
$65^\circ$
✓
$70^\circ$
C
$110^\circ$
D
$90^\circ$

Answer

Correct option: B.

$70^\circ$

$BC = BD \ [$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 $\text{ABCD}$ 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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MCQ 281 Mark

$AB$ and $CD$ are two equal chords of a circle with centre $O$ such that $\angle\text{AOB}=80^\circ$ then $\angle\text{COD}=?$

A
$100^\circ$
✓
$80^\circ$
C
$120^\circ$
D
$40^\circ$

Answer

Correct option: B.

$80^\circ$

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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MCQ 291 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{OAB}=50^\circ.$ Then, $\angle\text{CDA}=?$

A
$40^\circ$
✓
$50^\circ$
C
$75^\circ$
D
$25^\circ$

Answer

Correct option: B.

$50^\circ$

$OA = OB \ [$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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MCQ 301 Mark

In the given figure, $\text{BOC}$ is a diameter of a circle with centre $O$. If $AB$ and $CD$ are two chords such that $AB \| CD$. If $AB = 10\ cm,$ then $CD =$ ?

A
$5\ cm$
B
$12.5\ cm$
C
$15\ cm$
✓
$10\ cm$

Answer

Correct option: D.

$10\ cm$

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$]$
$\Rightarrow\ \triangle\text{BEO}\cong\triangle\text{CFO} \ [\text{ASA}$ congruence criterion$]$
$\Rightarrow OE = OF \ [\text{C.P.C.T.]}$
Since chord are equidistant from the centre are equal, $AB = CD = 10 \ cm.$

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MCQ 311 Mark

In the given figure, $AB$ is a chord of a circle with centre $O$ and $\text{BOC}$ is a diameter. If $\text{OD}\perp\text{AB}$ such that $OD = 6\ cm,$ then $AC = $?

A
$9\ cm$
✓
$12\ cm$
C
$15\ cm$
D
$7.5\ cm$

Answer

Correct option: B.

$12\ cm$

In $\triangle\text{BOD}$ and $\triangle\text{CAB},$
Since $\text{BOC}$ 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$]$
$\Rightarrow\ \frac{6}{\text{CA}}=\frac{1}{2}$
$\Rightarrow CA = 12\ cm$ that is, $AC = 12\ cm.$

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MCQ 321 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{AOC}=120^\circ.$ Then, $\angle\text{BDC}=?$

A
$60^\circ$
B
$45^\circ$
✓
$30^\circ$
D
$15^\circ$

Answer

Correct option: C.

$30^\circ$

Since $\text{BOA}$ 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 331 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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MCQ 341 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 $\angle\text{BCD}=?$

A
$50^\circ$
B
$60^\circ$
✓
$70^\circ$
D
$80^\circ$

Answer

Correct option: C.

$70^\circ$

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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MCQ 351 Mark

Two chords $AB$ and $CD$ of a circle intersect each other at a point $E$ outside the circle. If $AB = 11\ cm, BE = 3\ cm$ and $DE = 3.5\ cm,$ then $CD = ?$

A
$10.5\ cm$
B
$9.5\ cm$
✓
$8.5\ cm$
D
$7.5\ cm$

Answer

Correct option: C.

$8.5\ cm$

Construction: Join $AC$.
$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
$\Rightarrow AE \times BE = DE \times CE ...(i)$
Then,
$AE = AB + BE $
$= 11 + 3 = 14\ cm, BE = 3\ cm, $
$CE = (x + 3.5)cm$ and $DE = 3.5\ cm$
So, from $(i),$ we get
$14 \times 3 = 3.5 \times (CD + 3.5)$
$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
$\Rightarrow 12 = CD + 3.5$
$\Rightarrow CD = 8.5\ cm$

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MCQ 361 Mark

In the given figure, $O$ is the centre of a circle. Then, $\angle\text{OAB}=?$

A
$50^\circ$
B
$60^\circ$
C
$55^\circ$
✓
$65^\circ$

Answer

Correct option: D.

$65^\circ$

$OA = OB\ [$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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MCQ 371 Mark

An equilateral triangle of side $9\ cm$ is inscribed in a circle. The radius of the circle is:

A
$3\text{ cm}$
B
$3\sqrt{2}\text{ cm}$
✓
$3\sqrt{3}\text{ cm}$
D
$6\text{ cm}$

Answer

Correct option: C.

$3\sqrt{3}\text{ cm}$

Let $\triangle\text{ABC}$ be an equilateral triangle.
Let $AD$ be one of its medians.
Then, $\text{AD}\perp\text{BC}$ and $BD = 4.5\ cm$
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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MCQ 381 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{AOB}=140^\circ.$ Then, $\angle\text{ACB}=?$

A
$70^\circ$
B
$80^\circ$
✓
$110^\circ$
D
$40^\circ$

Answer

Correct option: C.

$110^\circ$

Minor $\angle\text{AOB}=140^\circ$
Major $\angle\text{AOB}=360^\circ-140^\circ$
$\Rightarrow$ 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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MCQ 391 Mark

In the given figure $\text{ABCD}$ is a cyclic quadrilateral in which $AB \| DC$ and $\angle\text{BAD}=100^\circ.$ Then, $\angle\text{ABC}=?$

A
$80^\circ$
✓
$100^\circ$
C
$50^\circ$
D
$40^\circ$

Answer

Correct option: B.

$100^\circ$

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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MCQ 401 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 $\angle\text{ADB}=?$

A
$70^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$90^\circ$

Answer

Correct option: C.

$80^\circ$

$\angle\text{AED}=\angle\text{ECB}+\angle\text{EBC}$
$\Rightarrow\ 110^\circ=\angle\text{ECB}+30^\circ$
$\Rightarrow\ \angle\text{ECB}=80^\circ$
Since angles in the same segment are equal,
$\angle\text{ADB}=\angle\text{ECB}=80^\circ$

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MCQ 411 Mark

In the given figure, $O$ is the centre of a circle and $\angle\text{AOB}=130^\circ.$ Then, $\angle\text{ACB}=?$

A
$50^\circ$
B
$65^\circ$
✓
$115^\circ$
D
$155^\circ$

Answer

Correct option: C.

$115^\circ$

Minor $\angle\text{AOB}=130^\circ$
Major $\angle\text{AOB}=360^\circ-130^\circ$
$\Rightarrow $ 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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MCQ 421 Mark

In the given figure, $O$ is the centre of a circle in which $\angle\text{AOC}=100^\circ.$ Side $AB$ of quadrilateral $\text{OABC}$ has been produced to $D$. Then, $\angle\text{CBD}=?$

✓
$50^\circ$
B
$40^\circ$
C
$25^\circ$
D
$80^\circ$

Answer

Correct option: A.

$50^\circ$

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 $\text{AECB}$ forms a cyclic quadrilateral.
$\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$

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MCQ(1M) - Maths STD 9 Questions - Vidyadip