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

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MCQ(1M)

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28 questions · auto-graded multiple-choice test.

MCQ 11 Mark

If one of the angles of a triangle is $130^\circ$ then the angle between the bisectors of the other two angles can be:

A
$50^\circ$
B
$65^\circ$
C
$90^\circ$
✓
$155^\circ$

Answer

Correct option: D.

$155^\circ$

Let $\angle\text{A}=130^\circ$
In $\triangle\text{ABC},$ by angle sum property,
$\angle\text{B}+\angle\text{C}+\angle\text{A}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}+130^\circ=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=50^\circ$
$\Rightarrow\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=25^\circ$
Now, in $\triangle\text{BOC},$
$\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}+\angle\text{BOC}=180^\circ$
$\Rightarrow25^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=155^\circ$

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MCQ 21 Mark

In the adjoining figure, what is the value of $y$?

A
$36^\circ$
✓
$54^\circ$
C
$63^\circ$
D
$72^\circ$

Answer

Correct option: B.

$54^\circ$

$\text{AOB}$ is a straight line.
$\therefore x^\circ +\ y^\circ\ 90^\circ = 180^\circ$
$\Rightarrow x + y = 90 .....(i)$
Since the angles around a point sum up to $360^\circ ,$
$\Rightarrow x^\circ + 90^\circ + y^\circ + 72^\circ + 3x^\circ = 360^\circ$
$\Rightarrow 4x + y = 198 .....(ii)$
Subtracting $(i)$ from $(ii),$ we get
$3x = 108 \Rightarrow x = 36^\circ$
Substituting in $(i),$ we get
$y = 54^\circ$

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MCQ 31 Mark

The measure of an angle is five times its comlement. The angle measure.

A
$25^\circ$
B
$35^\circ$
C
$65^\circ$
✓
$75^\circ$

Answer

Correct option: D.

$75^\circ$

Let the measure of the angle be $x^\circ ,$
So, its complement $= (90 - x)^\circ$
According to the given condition,
$x = 5(90 - x)$
$\Rightarrow x = 450 - 5x$
$\Rightarrow 6x = 450$
$\Rightarrow x = 75^\circ$
So, the angle measures $75^\circ$ .

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MCQ 41 Mark

In the given figure, straight lines $AB$ and $CD$ intersect at $O$. If $\angle\text{AOC}+\angle\text{BOD}=130^\circ$ then $\angle\text{AOD}=?$

A
$65^\circ$
✓
$115^\circ$
C
$110^\circ$
D
$125^\circ$

Answer

Correct option: B.

$115^\circ$

$\angle\text{AOC}+\angle\text{BOD}=1306^\circ \ ($given$)$
But $\angle\text{AOC}=\angle\text{BOD} \ ($Vartically Opposite angles$)$
$\Rightarrow2\angle\text{AOC}=130^\circ$
$\Rightarrow\angle\text{AOC}=65^\circ$
Since $\text{COD}$ is a straight line,
$\angle\text{AOC}+\angle\text{AOD}=180^\circ$
$\Rightarrow65^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\angle\text{AOD}=115^\circ$

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MCQ 51 Mark

An angle is one fifth of its supplement. The measure of the angle is:

A
$15^\circ$
✓
$30^\circ$
C
$75^\circ$
D
$150^\circ$

Answer

Correct option: B.

$30^\circ$

Let the measure of the angle be $x^\circ .$
So, its supplement $= (180^\circ - x)$
According to the given condition,
$\text{x}=\frac{1}{5}(180^\circ-\text{x)}$
$\Rightarrow5\text{x}=180-\text{x}$
$\Rightarrow6\text{x}=180$
$\Rightarrow\text{x}=30^\circ$

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MCQ 61 Mark

The angles of a triangle are in the ratio $\{2 : 3 : 4\}$. The largest angle of the triangle is:

A
$120^\circ$
B
$100^\circ$
✓
$80^\circ$
D
$60^\circ$

Answer

Correct option: C.

$80^\circ$

By angle sum property,
$2x + 3x + 4x = 180^\circ$
$\Rightarrow 9x = 180^\circ$
$\Rightarrow x = 20^\circ$
Hence, largest angle $= 4x $
$= 4(20^\circ ) = 80^\circ$

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MCQ 71 Mark

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:

A
An isosceles triangle.
B
An obtuse triangle.
C
An equilateral triangle.
✓
A right triangle.

Answer

Correct option: D.

A right triangle.

In a right triangle, one angle is $90^\circ$ and the sum of acute angles of a right triangle is $90^\circ$ .

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MCQ 81 Mark

In the given figure $,\text{AOB}$ is a straight line. If $\angle\text{AOC}=(3\text{x} +10)^\circ$ and $\angle\text{BOC}=(4\text{x}-26)^\circ,$ then $\angle\text{BOC}=?$

A
$96^\circ$
✓
$86^\circ$
C
$76^\circ$
D
$106^\circ$

Answer

Correct option: B.

$86^\circ$

$86^\circ$

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MCQ 91 Mark

In the given figure, $AB \| CD$. If $\angle\text{BAO}=60^\circ$ and $\angle\text{OCD}=110^\circ$ then $\angle\text{AOC}=?$

A
$70^\circ$
B
$60^\circ$
✓
$50^\circ$
D
$40^\circ$

Answer

Correct option: C.

$50^\circ$

Let $\angle\text{AOC}=\text{x}^\circ$
Draw $\text{YOZ} \| CD \| AB$.

Now, $YO \| AB$ and $OA$ is the transversal.
$\Rightarrow\angle\text{YOA}=\angle\text{OAB}=60^\circ\ ($alternate angles$)$
Again, $OZ \| CD$ and $OC$ is the transversal.
$\Rightarrow\angle\text{COZ}+\angle\text{OCD}=180^\circ \ ($interior angles$)$
$\Rightarrow\angle\text{COZ}+110^\circ=180^\circ$
$\Rightarrow\angle\text{COZ}=70^\circ$
Now, $\angle\text{YOZ}=180^\circ\ ($straight angle$)$
$\Rightarrow\angle\text{YOA}+\angle\text{AOC}+\angle\text{COZ}=180^\circ$
$\Rightarrow60^\circ+\text{x}+70^\circ=1806^\circ$
$\Rightarrow\text{x}=50^\circ$
$\Rightarrow\angle\text{AOC}=50^\circ$

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MCQ 101 Mark

An exterior angle of a triangle is $110^\circ$ and its two interior opposite angles are equal. Each of these equal angles is:

A
$70^\circ$
✓
$55^\circ$
C
$35^\circ$
D
$27\frac{1^\circ}{2}$

Answer

Correct option: B.

$55^\circ$

Let each interior opposite angle be $x$.
Then $, x + x = 110^\circ \ ($Exterior angle property of a triangle$)$
$\Rightarrow 2x = 110^\circ$
$\Rightarrow x = 55^\circ$

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MCQ 111 Mark

In the given figure $AB \| CD$ and $CD \| EF$. If $y : z = 3 : 7$ then $x = $?

A
$108^\circ$
✓
$126^\circ$
C
$162^\circ$
D
$63^\circ$

Answer

Correct option: B.

$126^\circ$

$AB \| CD$
$x + y = 180^\circ$ and $y = p\ ($Vertically opposite angles$)$
Also $,CD \| EF$ and t is the transversal.
$\therefore p + z = 180^\circ$
$\Rightarrow y + z = 180^\circ \ (\because\text{p}=\text{y})$
$\therefore x + y = y + z$
$\Rightarrow x = z$
But $y : z = 3 : 7$
$\therefore\text{y}=\Big(180\times\frac{3}{10}\Big)=54^\circ$ and $\text{z}=\Big(180\times\frac{7}{10}\Big)=126^\circ$
$x = 126^\circ\ (\because\text{x}=\text{z})$

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MCQ 121 Mark

The angles of a triangle are in the ration $\{3 : 5 : 7\}$. The triangle is:

✓
Acute $-$ angled.
B
Obtuse $-$ angled.
C
Right $-$ angled.
D
An isosceles triangle.

Answer

Correct option: A.

Acute $-$ angled.

Let the angles measure $(3x)^\circ , (5x)^\circ$ and $(7x)^\circ $.
Then,
$3x + 5x + 7x = 180^\circ$
$\Rightarrow 15x = 180^\circ$
$\Rightarrow x = 12^\circ$
Therefore, the angles are $3(12)^\circ = 36^\circ , 5(12)^\circ = 60^\circ$ and $7(12)^\circ = 84^\circ$ .
Hence, the triangle is acute $-$ angled.

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MCQ 131 Mark

If two angles are complements of each other, then each angle is:

✓
An acute angle.
B
An obtuse angle.
C
A right angle.
D
A reflex angle.

Answer

Correct option: A.

An acute angle.

If two angles are complements of each other, that is, the sum of their measures is $90^\circ ,$ then each angle is an acute angle.

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MCQ 141 Mark

In the adjoining figure, $\text{AOB}$ is a straight line. If $\{x : y : z\} = \{4 : 5 : 6\},$ then $y =$?

✓
$60^\circ$
B
$80^\circ$
C
$48^\circ$
D
$72^\circ$

Answer

Correct option: A.

$60^\circ$

The ratio of the angles is given to be $\{4 : 5 : 6\}$.
So, let the measure of the angles be $4m, 5m$ and $6m.$
Since $\text{AOB}$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow4\text{m}+5\text{m}+6\text{m}=180^\circ$
$\Rightarrow15\text{m}=180$
$\Rightarrow\text{m}=12$
So, $y = 5m = 5(12) = 60^\circ .$

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MCQ 151 Mark

In the given figure $,\text{ AOB}$ is a straight line. The value of $x$ is:

A
$12^\circ$
✓
$15^\circ$
C
$20^\circ$
D
$25^\circ$

Answer

Correct option: B.

$15^\circ$

$\text{AOB}$ is a straight line.
$\Rightarrow\angle\text{AOB}=180^\circ$
$\Rightarrow 60^\circ + 5x^\circ + 3x^\circ = 180^\circ$
$\Rightarrow 60^\circ + 8x^\circ = 180^\circ$
$\Rightarrow 8x^\circ = 120^\circ$
$\Rightarrow x = 15^\circ$

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MCQ 161 Mark

In the given figure, $\angle\text{OAB}=110^\circ$ and $\angle\text{BCD}=130^\circ$ then $\angle\text{ABC}$ is equal to:

A
$40^\circ$
B
$50^\circ$
✓
$60^\circ$
D
$70^\circ$

Answer

Correct option: C.

$60^\circ$

Through $B$ draw $\text{YBZ}\ \| OA \| CD$.

Now, $OA \| YB$ and $AB$ is the transversal.
$\Rightarrow\angle\text{OAB}+\angle\text{YBA}=180^\circ \ ($interior angles are supplementary$)$
$\Rightarrow\angle\text{YBA}=70^\circ$
Also, $CD \| BZ$ and $BC$ is the transversal.
$\Rightarrow\angle\text{DCB}+\angle\text{CBZ}=180^\circ\ ($interior angles are supplementary$)$

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MCQ 171 Mark

In the given figure, $AB \| CD$. If $\angle\text{CAB}=180^\circ$ and $\angle\text{EFC}=25^\circ$ then $\angle\text{CEF}=?$

A
$65^\circ$
✓
$55^\circ$
C
$45^\circ$
D
$75^\circ$

Answer

Correct option: B.

$55^\circ$

Since $AB\| CD,$
$\Rightarrow\angle\text{ACE}=\angle\text{BAC}=80^\circ$
In $\triangle\text{CEF},$
$\angle\text{ACE}=\angle\text{CEF}+\angle\text{CFE}\ ($Exterior angle is equal to sum of the remote interior angles$)$
$\Rightarrow80^\circ=\angle\text{CEF}+25^\circ$
$\Rightarrow\angle\text{CEF}=55^\circ$

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MCQ 181 Mark

In the given figure, straight lines $AB$ and $CD$ interect at $O$.If $\angle\text{AOC}=\phi,\angle\text{BOC}=\theta$ and $\theta=3\phi,$ then $\phi=?$

A
$30^\circ$
B
$40^\circ$
✓
$45^\circ$
D
$60^\circ$

Answer

Correct option: C.

$45^\circ$

$\angle\text{AOD}=\angle\text{COB}=\theta$
$\angle\text{AOC}=\angle\text{BOD}=\phi$
Since the sum of the measures of the angles around a point is $360^\circ ,$
$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$
$\Rightarrow\theta+\theta+\phi+\phi=360$
$\Rightarrow2(\theta+\phi)=360$
$\Rightarrow\theta+\phi=180$
Given that $\theta=3\phi.$
So, $3\phi+\phi=180$
$\Rightarrow4\phi=180$
$\Rightarrow\phi=45$

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MCQ 191 Mark

Two complementary angles are such that twich the measure of the one is equal to three times the measure of the other. The larger of the two measure.

A
$72^\circ$
✓
$54^\circ$
C
$63^\circ$
D
$36^\circ$

Answer

Correct option: B.

$54^\circ$

Let the measure of each angle be $x^\circ$ and $(90 - x)^\circ$ .
According to the given condition,
$2x = 3(90 - x)$
$\Rightarrow 2x = 270 - 3x$
$\Rightarrow 5x = 270$
$\Rightarrow x = 54^\circ$
So $, (90 - x)^\circ = (90 - 54)^\circ = 36^\circ$
So, the larger of the two angles is $54^\circ $.

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MCQ 201 Mark

In the given figure, $\text{AOB}$ is a straight line. If $\angle\text{AOC}=(3\text{x}-10)^\circ,\angle\text{COD}=50^\circ$ and $\angle\text{BOD}=(\text{x}+20)^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$50^\circ$

Answer

Correct option: C.

$80^\circ$

Since $\text{AOB}$ is a straight line,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow(3\text{x}-10)+50+(\text{x}+20)=180^\circ$
$\Rightarrow4\text{x}+60=180$
$\Rightarrow4\text{x}=120$
$\Rightarrow\text{x}=30$
So, $\angle\text{AOC}=3\text{x}-10$
$=3(30)-10=80^\circ$

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MCQ 211 Mark

In the figure $,AB \| CD$. If $\angle\text{APQ}=70 ^\circ$ and $\angle\text{PRD}=120^\circ$ then $\angle\text{QPR}=?$

✓
$50^\circ$
B
$60^\circ$
C
$40^\circ$
D
$35^\circ$

Answer

Correct option: A.

$50^\circ$

Since $AB \| CD,$
$\angle\text{APQ}=\angle\text{PQR}=70^\circ \ ($Alternate angles$)$
$\Rightarrow\angle\text{PRD}=\angle\text{PQR}+\angle\text{QPR}\ ($Exterior angle is equal to sum of the remote interior angles$)$
$\Rightarrow120^\circ=70^\circ+\angle\text{QPR}$
$\Rightarrow\angle\text{QPR}=50^\circ$

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MCQ 221 Mark

In the given figure, $AB$ is a mirror, $PQ$ is the incident ray and $QR$ is the reflected ray. If $\angle\text{PQR}=108^\circ$ then $\angle\text{AQP}=?$

A
$72^\circ$
B
$18^\circ$
✓
$36^\circ$
D
$54^\circ$

Answer

Correct option: C.

$36^\circ$

We know that, angle of incidance $=$ angle reflection.
that is, $\angle\text{AQP}=\angle\text{BQR}$
Since $\text{AOB}$ is a straight line,
$\angle\text{AQP}+\angle\text{BQR}+\angle\text{PQR}=180^\circ$
$\Rightarrow\angle\text{AQP}+\angle\text{AQP}+\angle\text{PQR}=180^\circ$
$\Rightarrow2\angle\text{AQP}=72$
$\Rightarrow \angle\text{AQP}=36^\circ$

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MCQ 231 Mark

An angle which measures more than $180^\circ$ but less than $360^\circ$, is called.

A
An acute angle.
B
An obuse angle.
C
A straight angle.
✓
A reflex angle.

Answer

Correct option: D.

A reflex angle.

An angle which measures more than $180^{\circ}$ but less than $360^{\circ}$ is called a reflex angle.

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MCQ 241 Mark

In the given figure, $\angle\text{OAB}=75^\circ, \angle\text{OBA}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ then, $\angle\text{ODC}=?$

A
$20^\circ$
B
$25^\circ$
✓
$30^\circ$
D
$35^\circ$

Answer

Correct option: C.

$30^\circ$

In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ \ ($Angle sum property$)$
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ \ ($Vertivcally opposite angles$)$
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ \ ($Angle sum property$)$
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$

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MCQ 251 Mark

In the given figure $, \text{AOB}$ is a straight line. If $\angle\text{AOC}=4\text{x}^\circ$ and $\angle\text{BOC}=5\text{x}^\circ$ then $\angle\text{AOC}=?$

A
$40^\circ$
B
$60^\circ$
✓
$80^\circ$
D
$100^\circ$

Answer

Correct option: C.

$80^\circ$

Since $\text{AOB}$ is a straight line,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$
$\Rightarrow4\text{x}+5\text{x}=180^\circ$
$\Rightarrow9\text{x}=180^\circ$
$\Rightarrow\text{x}=206^\circ$
So, $\angle\text{AOC}=4\text{x}=4(20)=80^\circ$

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MCQ 261 Mark

Which of the following statements is false?

✓
Through a given point, only one straight line can be drawn.
B
Through two given points, it is possible to draw one and only one straight line.
C
Two straight lines can intersect only at one point.
D
A line segment can be produced to any desired length.

Answer

Correct option: A.

Through a given point, only one straight line can be drawn.

Option $(a)$ is false, since through a given point we can draw an infinite number of straight lines.

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MCQ 271 Mark

In the given figure $, AB \| CD$. If $\angle\text{EAB}=50^\circ$ and $\angle\text{ECD}=60^\circ$ then $\angle\text{AEB}=?$

A
$50^\circ$
B
$60^\circ$
✓
$70^\circ$
D
$55^\circ$

Answer

Correct option: C.

$70^\circ$

Let $\angle\text{AEB}=\text{x}^\circ$
Now $, AB \| CD$ and $BC$ is the transversal
$\therefore\angle\text{ABE}=\angle\text{BCD}=60^\circ\ ($Alterante angles$)$
In $\triangle\text{ABE},$
$\angle\text{BAE}+\angle\text{AEB}+\angle\text{ABE}=180^\circ\ ($Angle sum property$)$
$\Rightarrow50^\circ+\text{x}^\circ+60^\circ=180^\circ$
$\Rightarrow\text{x}=70^\circ$
$\therefore\angle\text{AEB}=70^\circ$

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MCQ 281 Mark

In the given figure $, AB \| CD$. If $\angle\text{AOC}=30^\circ$ and $\angle\text{OAB}=100^\circ$ then $\angle\text{OCD}=?$

✓
$130^\circ$
B
$150^\circ$
C
$80^\circ$
D
$100^\circ$

Answer

Correct option: A.

$130^\circ$

Construction: Through $O,$ draw $OE \| AB \| CD$
$\Rightarrow\angle\text{BAO}+\angle\text{EOA}=180^\circ$
$\Rightarrow100^\circ+\angle\text{EOA}=180^\circ$
$\Rightarrow\angle\text{EOA}=80^\circ$
So, $\angle\text{EOC}=\angle\text{EAO}-\angle\text{COA}$
$=80^\circ-30^\circ=50^\circ$
Since $CD \| EO$
$\angle\text{OCD}+\angle\text{EOC}=180^\circ$
$\Rightarrow\angle\text{OCD}+50^\circ=180^\circ$
$\Rightarrow\angle\text{OCD}=130^\circ$

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