M.C.Q - Maths STD 9 Questions - Vidyadip

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M.C.Q

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42 questions · auto-graded multiple-choice test.

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.

$7.5\ cm$

$\mathrm{OA}=\mathrm{OC}$
$\Rightarrow \mathrm{OA}=\mathrm{OE}+\mathrm{CE}$
$\Rightarrow \mathrm{OA}=\mathrm{OE}+3$
$\Rightarrow \mathrm{OE}=\mathrm{OA}-3 \ldots \text { (i) }$
$\mathrm{AE}=\frac{1}{2} \mathrm{AB}[\text { Perpendicular drawn from the centre of a circle to the chord bisect the chord] }$
$=\frac{1}{2}(12)=6 \mathrm{~cm}$
$\text { In right } \triangle \mathrm{OEA},$
$\mathrm{OA}^2=\mathrm{OE}+\mathrm{E}^2+\mathrm{AE}^2$
$\Rightarrow \mathrm{OA}^2=(\mathrm{OA}-3)^2+\mathrm{AE}^2[\text { From (i) }]$
$\Rightarrow \mathrm{OA}^2=\mathrm{OA}^2-6 \mathrm{OA}+9+\mathrm{AE}{ }^2$
$\Rightarrow 6 \mathrm{OA}=9+6^2$
$\Rightarrow 6 \mathrm{OA}=9+36$
$\Rightarrow \mathrm{OA}=\frac{45}{6}=7.5 \mathrm{~cm}$
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 $O A=13 \mathrm{~cm}$.
$A B$ is given to be $10\ cm.$
Distance of a point to a line is always perpendicular to the line.
So, $\mathrm{OL} \perp \mathrm{AB}$.
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\Rightarrow A L=L B=5 \mathrm{~cm}$
$\text { In right } \triangle \mathrm{OLA},$
$\mathrm{OL}^2=A O^2-\mathrm{AL}^2[\text { By pythagoras theorem }]$
$\Rightarrow \mathrm{OL}^2=13^2-5^2$
$\Rightarrow \mathrm{OL}^2=169-25$
$\Rightarrow \mathrm{OL}^2=144$
$\Rightarrow \mathrm{OL}=12 \mathrm{~cm}$

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MCQ 31 Mark

In the given figure, $A O B$ is a diameter of a circle with centre $O$ such that $A B=34 \mathrm{~cm}$ and $C D$ is a chord of length $30\ cm.$ Then the distance of $C D$ from $A B$ 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, $\mathrm{CL}=\frac{1}{2} \mathrm{CD}=\frac{1}{2}(30)=15 \mathrm{~cm}$
$A B$ is the diameter.
So, $\mathrm{AO}=\frac{1}{2} \mathrm{AB}=\frac{1}{2}(34)=17 \mathrm{~cm}$.
In $\triangle O L C$,
$\mathrm{OL}^2= \mathrm{OC}^2-\mathrm{CL}^2$
$\Rightarrow \mathrm{OL}^2=17^2-15^2$
$\Rightarrow \mathrm{OL}^2=289 \cdot-225$
$\Rightarrow O L^2=64$
$\Rightarrow \mathrm{OL}=8 \mathrm{~cm}$

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MCQ 41 Mark

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

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

Answer

Correct option: C.

$60^\circ$

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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MCQ 51 Mark

The angle in a semicircle measures:

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

Answer

Correct option: C.

$90^\circ$

The angle in a semicircle measures $90^\circ .$

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

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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, $AOB$ 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 $AOB$ 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, $BOC$ 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 $BOC$ 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 $OD.$

$O B=O D$
$\Rightarrow O B=O E+E B$
$\Rightarrow O B=O E+4$
$\Rightarrow O E=O B-4 \ldots \text { (i) }$
In right $\triangle \mathrm{OED}$,
$O D^2=O E^2+D E^2$
$\Rightarrow O D^2=(O B-4)^2+D E^2[\text { From (i) }]$
$\Rightarrow O D^2=O B^2-80 A+16+8^2$
$\Rightarrow 8 O D=16+64$
$\Rightarrow 8 O D=80$
$\Rightarrow O D=\frac{80}{8}=10 \mathrm{~cm}$
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 $AOD$ 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 $ABCD$ 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 $ABCD$ 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, $AOB$ is a diameter and $ABCD$ 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 $BOC$ 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, $ABCD$ 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. $ABCD$ 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 $ABCD$ 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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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, $ABCD$ and $ABEF$ 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 $ABCD$ 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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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 $\angle\text{AOC}=360^\circ-130^\circ$
$⇒$ 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, $ABCD$ 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 $ABCD$ 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, $BOC$ 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]
$\Rightarrow\ \triangle\text{BEO}\cong\triangle\text{CFO} [ASA$ 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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MCQ 311 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 $OD = 6\ cm,$ 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 $BOC$ 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$]$
$\Rightarrow\ \frac{6}{\text{CA}}=\frac{1}{2}$
$⇒ CA = 12\ cm$ 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 $BOA$ 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,
$A P=5 \mathrm{~cm}, B P=3 \mathrm{~cm}$ and $A B=4 \mathrm{~cm}$
$\mathrm{AP}^2=5^2=25$
$\mathrm{BP}^2+\mathrm{AB}^2=3^2+4^2=25$
In $\triangle \mathrm{ABP}$
Since $A P^2=B P^2+A B^2$
$\triangle \mathrm{ABP}$, is a right-angled triangle and $\mathrm{PQ}=2 \mathrm{BP}=2(3)=6 \mathrm{~cm}$.

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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.

$8.5\ cm$

Join $AC.$
$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$
$⇒ AE × BE = DE × CE ...(i)$
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
$14 × 3 = 3.5 × (CD + 3.5)$
$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$
$⇒ 12 = CD + 3.5$
$⇒ CD = 8.5\ cm$

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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°
B
80°
✓
110°
D
40°

Answer

Correct option: C.

110°

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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MCQ 391 Mark

In the given figure $ABCD$ 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 $\angle\text{AOB}=360^\circ-130^\circ$
$⇒$ 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 $OABC$ 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 $AECB$ forms a cyclic quadrilateral.
$\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$

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M.C.Q - Maths STD 9 Questions - Vidyadip