M.C.Q - Page 3 - Maths STD 9 Questions - Vidyadip

Questions · Page 3 of 5

M.C.Q

MCQ 1011 Mark

In two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $5 : 4$, then the smaller of the two angles is:

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

Answer

Correct option: D.

$80^\circ$

We know that sum of two interior angles on the same side of atransversal intersecting two parallel lines is $180^\circ$
let the common ratio is $x$ so the angles are $5x ,4x$
$So 5x + 4x = 180^\circ$
$9x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}$
$\text{x}=20^\circ$
So the angles are $5x = 100^\circ$
$4x = 80^\circ$
So smallest angle is $80^\circ$

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MCQ 1021 Mark

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

A
$75^\circ$
B
$15^\circ$
✓
$30^\circ$
D
$15^\circ$

Answer

Correct option: C.

$30^\circ$

Let one angle be$ x^\circ $
Its supplementary angle will be $180^\circ - x^\circ $
According to question
$\text{x}=\frac{1}{5}(180^\circ-\text{x})$
$5x + x = 180^\circ $
$6x = 180^\circ $
$\text{x}=\frac{180}{6}$
$x = 30^\circ .$

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MCQ 1031 Mark

The measure of the Complementary angle of $63^{\circ}$ is:

✓
$27^{\circ}$
B
$117^{\circ}$
C
$30^{\circ}$
D
$36^{\circ}$

Answer

Correct option: A.

$27^{\circ}$

Sum of complementary angles is 90$^{\circ}$.If one angle is 63$^{\circ}$
Then the other angle will be 90$^{\circ}$ - 63$^{\circ}$= 27$^{\circ}$.

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MCQ 1041 Mark

In figure, if lines l and m are parallel lines, then $x =$

A
$70^\circ$
B
$100^\circ$
✓
$40^\circ$
D
$30^\circ $

Answer

Correct option: C.

$40^\circ$

From figure,
$\angle\text{ABC}=\angle\text{DCE}\dots(1)$ [Corresponding angles]
$\angle\text{ECF}=180^\circ-\angle\text{DCE}$ [Supplementary]
$=180^\circ-\angle\text{ABC}$ [From (1)]
$=180^\circ-70^\circ$
$\Rightarrow\ \angle\text{ECF}=110^\circ$
Now, in $\triangle\text{CEF}$
$\angle\text{ECF}+\angle\text{CFE}+\angle\text{FEC}=180^\circ$
$\Rightarrow\ 110^\circ+\text{x}+30^\circ=180^\circ$
$\Rightarrow\ \text{x}=40^\circ$

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MCQ 1051 Mark

Write the correct answer in the following: Angles of a triangle are in the ratio $2 : 4 : 3$. The smallest angle of the triangle is,

A
$60^\circ$
✓
$40^\circ$
C
$80^\circ$
D
$20^\circ$

Answer

Correct option: B.

$40^\circ$

Given that: The Ratio of angles of a triangle is $2 : 4 : 3$
Let the angles of the triangle be $\angle\text{A},\angle\text{B},$ and $\angle\text{C},$
$\therefore\ \angle\text{A}=2\text{x},\angle\text{B}=4\text{x}$ and $\angle\text{C}=3\text{x}$
In $\angle\text{ABC},$ $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
[$\because\ $Sum of angles of a triangle is $180^\circ ]$
$\Rightarrow2\text{x}+4\text{x}+3\text{x}=180^\circ\Rightarrow9\text{x}=180^\circ\Rightarrow\text{x}=\frac{180^\circ}{9}=20^\circ$
$\therefore\ \angle\text{A}=2\text{x}=2\times20^\circ=40^\circ$
$\angle\text{B}=4\text{x}=4\times20^\circ=80^\circ$
And $\angle\text{C}=3\text{x}=3\times20^\circ=60^\circ$
Hence, the smallest angles of a triangle is $40^\circ $ and option $(b)$ is correct answer.

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MCQ 1061 Mark

Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is:

A
$45^\circ$
✓
$36^\circ$
C
$30^\circ$
D
None of these

Answer

Correct option: B.

$36^\circ$

Let x and $(90^\circ - x)$ be two complimentary angles
According to question,
$2x = 3 (90^\circ - x)$
$2x = 270^\circ - 3x$
$x = 54^\circ $
The angles are:
$54^\circ $ and $90^\circ - 54^\circ = 36^\circ$
Thus, smallest angle is $36^\circ .$

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MCQ 1071 Mark

An exterior angle of a triangle is $80^\circ $ and two interior opposite angles are equal. Measure of each of these angles is:

A
$100^\circ$
B
$120^\circ$
✓
$40^\circ$
D
$60^\circ$

Answer

Correct option: C.

$40^\circ$

We know that the exterior angle so formed is equal to the sum of the two interior opposite angles.Let the two interior opposite angles be $x.$
So, $x + x = 80^\circ$
$2x = 80^\circ$
$x = 40^\circ .$

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MCQ 1081 Mark

In figure, if $CP \| BQ$, then the measure of $x$ is:

✓
$130^\circ$
B
$105^\circ$
C
$175^\circ$
D
$125^\circ$

Answer

Correct option: A.

$130^\circ$

From figure,
$\angle\text{QBA}=\angle\text{CEA}$ [Correspondence angles]
$\Rightarrow\ \angle\text{CEA}=105^\circ\dots(1)$
In $\triangle\text{ACE},$
$\angle\text{CEA}+\angle\text{EAC}+\angle\text{ACE}=180^\circ$
$\Rightarrow\ 105^\circ+25^\circ+\angle\text{ACE}=180^\circ$ [From (1)]
$\Rightarrow\ 130^\circ+\angle\text{ACE}=180^\circ$
$\Rightarrow\ \angle\text{ACE}=50^\circ$
Now,
$\text{x}=\angle\text{ACP}=180^\circ-\angle\text{ACE}$
$=180^\circ-50^\circ=130^\circ$

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MCQ 1091 Mark

$AB$ and $CD$ are two parallel lines. $PQ$ cuts $AB$ and $CD$ at $E$ and $F$ respectively. $EL$ is the bisector of $\angle\text{FEB}.$ If $\angle\text{LEB}=35^\circ,$ then $\angle\text{CFQ}$ will be:

✓
$55^\circ$
B
$70^\circ$
C
$110^\circ$
D
$130^\circ$

Answer

Correct option: A.

$55^\circ$

From figure,
$\angle\text{LEB}=\angle\text{FEL}$ [EL is bisector of $\angle\text{FEB}$]
Now,
$\angle\text{FEB}=2\angle\text{LEB}=2\times35^\circ=70^\circ$
Also,
$\angle\text{FEB}=\angle\text{CFE}$ [Alternate interior angles]
$\Rightarrow\ \angle\text{CFE}=70^\circ$
Now,
$\angle\text{CFE}+\angle\text{CFQ}=180^\circ$
$\Rightarrow\ 70^\circ+\angle\text{CFQ}=180^\circ$
$\Rightarrow\ \angle\text{CFQ}=180^\circ-70^\circ=110^\circ$

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MCQ 1101 Mark

In Fig. if $l1 \| l2$, what is the value of $y?$

A
$120$
✓
$135$
C
$100$
D
$150$

Answer

Correct option: B.

$135$

Given that,$l1 \| l2$ and $l3$ is transversal
$\angle1=3\text{x}$ (Vertically opposite angle)
$\text{y}=\angle1$ (Corresponding angle)
$y=3 x \text { (i) }$
$y+x=180^{\circ} \text { (Linear pair) }$
$3 x+x=180^{\circ}[\text { From (i) }]$
$4 x=180^{\circ}$
$x=45^{\circ}$
Therefore,
$y=3 x=3 \times 45^{\circ}$
$=135^{\circ}$.

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MCQ 1111 Mark

If two supplementary angles are in the ratio $2 : 7$, then the angles are:

✓
$40^\circ , 140^\circ$
B
$50^\circ , 130^\circ$
C
$70^\circ , 110^\circ$
D
$35^\circ , 145^\circ$

Answer

Correct option: A.

$40^\circ , 140^\circ$

We know that supplementary angles are those angles whose sum is $180^\circ $
The two given supplementary angles are in the ratio $2 : 7$
Let the commom ratio be $x,$
So angles are $2x$ and $7x$ respectively $2x + 7x = 180^\circ$
$\text{x}=\frac{180^\circ}{9}=20^\circ$
$2x = 2 \times 40 = 40^\circ$
$7x = 7 \times 20^\circ = 140^\circ .$

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MCQ 1121 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 1131 Mark

In the given figure, $AB \| CD$, If $\angle\text{ABO}=45^\circ$ and $\angle\text{COD}=100^\circ$ then $\angle\text{CDO}=?$

A
$45^\circ$
✓
$35^\circ$
C
$25^\circ$
D
$30^\circ$

Answer

Correct option: B.

$35^\circ$

$\angle\text{ABC}=\angle\text{BCD}=45^\circ$ (Alternate interior angles)
In $\triangle\text{COD}$
$\angle\text{COD}+\angle\text{CDO}+\angle\text{DCO}=180^\circ$ (Angle sum property)
$\angle\text{CDO}=180^\circ-100^\circ-45^\circ$
$\angle\text{CDO}=35^\circ.$

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MCQ 1141 Mark

In Fig. if 1$_1 \| 1_2$, what is $x + y $ in terms of w and $z?$

✓
$180 - w + z$
B
$180 + w + z$
C
$180 + w - z$
D
$180 - w - z$

Answer

Correct option: A.

$180 - w + z$

Given that,
$1_1 \| 1_2$
Let m and n be two transversal cutting them
$\angle\text{w}+\angle\text{x}=180^\circ$ (Consecutive interior angle)
$x=180^{\circ}-w(i)$
$z = y (Alternate angles) (ii)$
From (i) and (ii), we get
$x+y=180^{\circ}-w+z$.

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MCQ 1151 Mark

The complement of $(90^{\circ} - a)$ is:

✓
$a^{\circ}$
B
$-a^{\circ}$
C
$90^{\circ}$+ a
D
$90^{\circ}$ - a

Answer

Correct option: A.

$a^{\circ}$

Two angles, whose sum is$ 90^{\circ}$, are called the complementary angle.Let x is a complimentary angle of $(90^{\circ} - a)$
$x+\left(90^{\circ}-a\right)=90^{\circ}$
$x=90^{\circ}-\left(90^{\circ}-a\right)$
$x=90^{\circ}-90^{\circ}+a$
$x=a^{\circ} .$

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MCQ 1161 Mark

In the adjoining figure, $AOB$ 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 $4\ m, 5\ m$ and $6\ m$.
Since $AOB$ 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 1171 Mark

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

A
$12$
✓
$15$
C
$20$
D
$25$

Answer

Correct option: B.

$15$

$AOB$ 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 1181 Mark

Write the correct answer in the following: In Fig. $POQ$ is a line.The value of $x$ is.

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

Answer

Correct option: A.

$20^\circ$

We have $3x + 4x + 40^\circ = 180^\circ $ (Angles on the straight line)
$\Rightarrow 7x + 40^\circ = 180^\circ $
$\Rightarrow 7x = 180^\circ - 40^\circ = 140^\circ $
$\Rightarrow x = 140^\circ ÷ 7 = 20^\circ $
Hence, $(a)$ is the correct answer.

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MCQ 1191 Mark

Write the correct answer in the following: An exterior angle of a triangle is 105° and its two interior opposite angles are equal. Each of these equal angles is,

A
$37\frac{1}{2}^\circ$
✓
$52\frac{1}{2}^\circ$
C
$72\frac{1}{2}^\circ$
D
$75^\circ$

Answer

Correct option: B.

$52\frac{1}{2}^\circ$

An exterior angle of triangle is $150^\circ $
Let each of to two interior opposites angles be x.
We know that exterior angle of a equal to the sum of two interior opposite angles.
$105^\circ=\text{x}+\text{x}\Rightarrow2\text{x}=105^\circ$
$\Rightarrow\text{x}=\frac{1}{2}\times105^\circ=52\frac{1}{2}^\circ$
So, each of equal angle angle is $52\frac{1}{2}^\circ$
Hence, $(b)$ is the correct answer.

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MCQ 1201 Mark

In the adjoining figure, the value of $x$ is:

A
$18^\circ$
✓
$15^\circ$
C
$12^\circ$
D
$10^\circ$

Answer

Correct option: B.

$15^\circ$

$7x + 2x + 3x = 180^\circ$
$12x = 180^\circ$
$x = 15^\circ .$

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MCQ 1211 Mark

The incorrect statement is:

A
Two lines drawn in a plane always intersect at a point.
B
A line segment has definite length.
C
Three lines are concurrent if and only if they have a common point.
✓
One and only one line can be drawn passing through a given point and parallel to a given line.

Answer

Correct option: D.

One and only one line can be drawn passing through a given point and parallel to a given line.

If two lines intersect then they must lie in one plane but its converse is not necessarily true.

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MCQ 1221 Mark

Two straight lines $AB$ and $CD$ intersect one another at the point $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ, $ then $\angle\text{AOD}=$

✓
$86^{\circ}$
B
$137^{\circ}$
C
$90^{\circ}$
D
$94^{\circ}$

Answer

Correct option: A.

$86^{\circ}$

Given,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ\text{ (i)}$
$\angle\text{AOD}+\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=360^\circ$(Angles at a point)
$\angle\text{AOD}+274^\circ=360^\circ$
$\angle\text{AOD}=86^\circ.$

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MCQ 1231 Mark

If two angles of a triangle are $30^\circ $ and $45^\circ $, what is the measure of the third angle?

✓
$105^\circ$
B
$95^\circ$
C
$90^\circ$
D
$60^\circ$

Answer

Correct option: A.

$105^\circ$

We know that the sum of all angles of a triangle is $180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
Let $\angle\text{A}=30^\circ,\angle\text{B}=45^\circ$
$30^\circ+45^\circ+\angle\text{C}=180^\circ$
$75^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-75^\circ$
$\angle\text{C}=105^\circ.$

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MCQ 1241 Mark

In Fig., if lines $l$ and $m$ are parallel, then the value of $x$ is:

A
$75^{\circ}$
B
$65^{\circ}$
C
$55^{\circ}$
✓
$35^{\circ}$

Answer

Correct option: D.

$35^{\circ}$

Given that,
$l \| m$ and n cuts them
Let, $\angle1=\text{x}$
$\angle2=90^\circ$
$\angle3=125^\circ$
$\angle3+\angle5=180^\circ$(Linear pair)
$125^\circ+\angle5=180^\circ$
$\angle5=55^\circ\text{ (i)}$
$\angle4=99^\circ\text{ (ii)}$
Now,
$\angle1+\angle4+\angle5=180^\circ$ (Angle sum property)
$\text{x}+90^\circ+55^\circ=180^\circ$
$\text{x}=35^\circ.$

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MCQ 1251 Mark

Two angles measure $(70 + 2x)^\circ $ and $(3x - 15)^\circ $. If each angle is the supplement of the other, then the value of $x$ is:

✓
$25$
B
$250^\circ $
C
$30$
D
$20$

Answer

Correct option: A.

$25$

$70 + 2x + 3x - 15 = 180$ (Supplementary angles)
$5x = 180 - 55$
$x = 25.$

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MCQ 1261 Mark

In figure, if lines $l$ and $m$ are parallel, then $x =$

A
$20^\circ$
✓
$45^\circ$
C
$65^\circ$
D
$85^\circ$

Answer

Correct option: B.

$45^\circ$

From figure,
$\angle\text{ABD}=\angle\text{CDF}$ [Correspondence angles]
$\Rightarrow\ \angle\text{CDF}=65^\circ$
Now,
$\angle\text{FDE}=180^\circ-\angle\text{CDF}=180^\circ-65^\circ$
$\Rightarrow\ \angle\text{FDE}=115^\circ$
In $\triangle\text{EDF},$
$\angle\text{FDE}+\angle\text{DEF}+\angle\text{EFD}=180^\circ$
$\Rightarrow\ 115^\circ+\text{x}+20^\circ=180^\circ$ [Sum of all interior angles of a $\triangle$ as 180°]
$\Rightarrow\ \text{x}=180^\circ-20^\circ-115^\circ=45^\circ$

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MCQ 1271 Mark

In figure, if $I_1 \| I_2$, what isb the value of $y?$

A
$100$
B
$120$
✓
$135$
D
$150$

Answer

Correct option: C.

$135$

Let angle supplement of $3 x^{\circ}$ be $Z^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow\ \angle\text{AHF}+\angle\text{FHB}=180^\circ$
$\Rightarrow z^{\circ}+3 x^{\circ}=180^{\circ}$
$\Rightarrow z^{\circ}=180^{\circ}-3 x^{\circ}$
Now,
$x^{\circ}+y^{\circ}=180^{\circ}$
Also,
$x^{\circ}=z^{\circ} \text { [Correspondence angles] }$
$\Rightarrow x^{\circ}=180^{\circ}-3 x^{\circ}$
$\Rightarrow 4 x^{\circ}=180^{\circ}$
$\Rightarrow x^{\circ}=45^{\circ}$
$x^{\circ}+y^{\circ}=180^{\circ}$
$\Rightarrow y^{\circ}=180^{\circ}-x^{\circ}=180^{\circ}-45^{\circ}=135^{\circ}$.

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MCQ 1281 Mark

An exterior angle of a triangle is $80^\circ $ and the interior opposite angles are in the ratio $1 : 3$. Measure of each inte4rior opposite angle is:

A
$30^\circ , 60^\circ$
✓
$20^\circ , 60^\circ$
C
$30^\circ , 90^\circ$
D
$40^\circ , 120^\circ$

Answer

Correct option: B.

$20^\circ , 60^\circ$

let the common ratio is $x$
the ratio of interior angles are $1 : 3$
so angles are $x$ and $3x$
$x + 3x = 80$
$\text{x}=\frac{80}{4}$
$x = 20$
So angles are $20^\circ $ and $60^\circ $

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MCQ 1291 Mark

In Fig. if $AB \| HF$ and $DE \| FG$, then the measure of $\angle\text{FDE}$ is:

✓
$80^\circ$
B
$90^\circ$
C
$108^\circ$
D
$100^\circ$

Answer

Correct option: A.

$80^\circ$

Given that,
$AB \| HF$ and $CD$ cuts them
$\angle\text{HFC}=\angle\text{FDA}$ (Corresponding angle)
$\angle\text{FDA}=28^\circ$
$\angle\text{FDA}+\angle\text{FDE}+\angle\text{EDB}=180^\circ $(Linear pair)
$28^\circ+\angle\text{FDE}+72^\circ=180^\circ$
$\angle\text{FDE}=80^\circ.$

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MCQ 1301 Mark

In the adjoining figure, if $AB \| DE$, then the measure of $\angle\text{ACD}$ is:

A
$70^\circ$
B
$80^\circ$
C
$90^\circ$
✓
$100^\circ$

Answer

Correct option: D.

$100^\circ$

$x + 110^\circ = 180^\circ $(Supplementary angles)
$x = 70^\circ $
$y + 150^\circ = 180^\circ $(Supplementary angles)
$y = 30^\circ $
$\angle\text{ACD}=70^\circ+30^\circ=100^\circ.$

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MCQ 1311 Mark

The sum of all the angles of a quadrilateral is:

A
$400^\circ$
B
$180^\circ$
✓
$360^\circ$
D
$320^\circ$

Answer

Correct option: C.

$360^\circ$

Sum of the angles of a polygon $= (n - 2) \times 180^\circ $
Quadrilateral has $4$ sides,
So sum of interior angles $= (4 - 2) \times 180^\circ = 360^\circ .$

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MCQ 1321 Mark

In the given figure, $AB \| CD$, If $\angle\text{CAB}=80^\circ$ and $\angle\text{EFC}=25^\circ$ then $\angle\text{CEF}=?$

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

Answer

Correct option: A.

$55^\circ $

$\angle\text{BAF}=\angle\text{DCF}=80^\circ$ (Corresponding angle)
In $\triangle\text{CEF}$
$\angle\text{CFE}+\angle\text{CEF}=\angle\text{DCF}$ (exterior angle is equal to the sum of its two opposite interior angles)
$25^\circ+\angle\text{CEF}=80^\circ$
$\angle\text{CEF}=55^\circ$

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MCQ 1331 Mark

In the adjoining figure, $BE$ and $CE$ are bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ respectively. If $\angle\text{BEC}=25^\circ$ then $\angle\text{BAC}$ is equal to:

✓
$50^\circ$
B
$25\frac{1}{2}^\circ$
C
$12\frac{1}{2}^\circ$
D
$65^\circ$

Answer

Correct option: A.

$50^\circ$

$\angle\text{BEC}+\angle\text{EBC}=\angle\text{ECD} ($Exterior angle property$)$
$\angle\text{BEC}=\angle\text{ECD}-\angle\text{EBC}$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BAC}=\angle\text{ACD}$
$\angle\text{ABC}+2\angle\text{EBC}=2\angle\text{ECD}$
$\angle\text{ABC}=2(\angle\text{ECD}-\angle\text{EBC})$
$\angle\text{ABC}=2(\angle\text{BEC})$
$\angle\text{ABC}=50^\circ.$

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MCQ 1341 Mark

In the adjoining figure $\angle\text{QPR}=62^\circ$ and $\angle\text{PRQ}=64^\circ$ If $OQ$ and $OR$ and bisectors of $\angle\text{PQR}$ and $\angle\text{PRQ}$ respectively, then $\angle\text{OQR}$ and $\angle\text{QOR}:$

✓
$27^\circ , 121^\circ$
B
$20^\circ , 80^\circ$
C
$26^\circ , 124^\circ$
D
$121^\circ , 20^\circ$

Answer

Correct option: A.

$27^\circ , 121^\circ$

In $\triangle\text{PQR}$
$\angle\text{QPR}+\angle\text{PQR}+\angle\text{PRQ}=180^\circ$ (Angle sum property)
$\angle\text{PQR}=180^\circ-62^\circ-64^\circ$
$\angle\text{PQR}=54^\circ$
$\angle\text{ORQ}=32^\circ$ (OR is a bisector)
$\angle\text{OQR}=27^\circ$ OQ is a bisector)
In $\triangle\text{OQR}$
$\angle\text{OQR}+\angle\text{ORQ}+\angle\text{QOR}=180^\circ$ (Angle sum property)
$\angle\text{QOR}=180^\circ-32^\circ-27^\circ=121^\circ.$

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MCQ 1351 Mark

Two straight lines $AB$ and $CD$ intersect one another at the point $O$. If $\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ,$ then $\angle\text{AOD}=$

✓
$86^\circ$
B
$90^\circ$
C
$94^\circ$
D
$137^\circ$

Answer

Correct option: A.

$86^\circ$

$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}+\angle\text{AOD}=360^\circ\dots(1)$
Now,
$\angle\text{AOC}+\angle\text{COB}+\angle\text{BOD}=274^\circ\dots(2)$ [Given]
From (1) and (2).
$274^\circ+\angle\text{AOD}=360^\circ$
$\Rightarrow\ \angle\text{AOD}=360^\circ-274^\circ$
$\Rightarrow\ \angle\text{AOD}=86^\circ$

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MCQ 1361 Mark

In the adjoining figure, $m \| n$, if $\angle1=50^\circ$ then $\angle2$ is equal to:

✓
$130^\circ$
B
$120^\circ$
C
$40^\circ$
D
$50^\circ$

Answer

Correct option: A.

$130^\circ$

$\angle2=180^\circ-\angle1$
$\angle2=180^\circ-50^\circ=130^\circ.$

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MCQ 1371 Mark

If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3$, then the measure of the larger angle is:

A
$54^\circ$
B
$120^\circ$
✓
$108^\circ$
D
$136^\circ$

Answer

Correct option: C.

$108^\circ$

Let $AB$ and $CD$ are two parallel lines and $PQ$ is transverce to it.
According to question,
$\frac{\angle\text{BRS}}{\angle\text{DSR}}=\frac{2}{3}$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\angle\text{DSR}\dots(1)$
Now,
$\angle\text{CSR}=\angle\text{BRS}$ [Alternate angles]
$\Rightarrow\ \angle\text{CSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{BRS}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \frac{2}{3}\angle\text{DSR}+\angle\text{DSR}=180^\circ$
$\Rightarrow\ \angle\text{DSR}=\frac{180\times3}{5}=108^\circ$
$\Rightarrow\ \angle\text{BRS}=\frac{2}{3}\times108^\circ=72^\circ$
Thus,
$\angle\text{DSR}=108^\circ$ and $\angle\text{BRS}=72^\circ$
$\Rightarrow$ Larger angle is $\angle\text{DSR}.$

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MCQ 1381 Mark

The sum of all the angles of a quadrilateral is:

A
$180^\circ$
✓
$360^\circ$
C
$400^\circ$
D
$320^\circ$

Answer

Correct option: B.

$360^\circ$

Sum of the angles of a polygon $= (n - 2) \times 180$
Quadrilateral has $4$ sides,
So sum of interior angles $= (4 - 2) \times 180^\circ = 360$

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MCQ 1391 Mark

In figure,$ AOB$ is a straight line. If $\angle\text{AOC}+\angle\text{BOD}=85^\circ,$ then $\angle\text{COD}=$

A
$85^\circ $
B
$90^\circ$
✓
$95^\circ$
D
$100^\circ$

Answer

Correct option: C.

$95^\circ$

From figure, we can see
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
But,
$\angle\text{AOC}+\angle\text{BOD}=85^\circ$ [Given]
$\Rightarrow\ 85^\circ+\angle\text{COD}=180^\circ$
$\Rightarrow\ \angle\text{COD}=95^\circ$

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MCQ 1401 Mark

In figure, if $I_1 \| I_2$ and $I_3 \| I_4$, what is $y$ in terms of $x?$

A
$90+\text{x}$
B
$90+2\text{x}$
✓
$90-\frac{\text{x}}{2}$
D
$90-2\text{x}$

Answer

Correct option: C.

$90-\frac{\text{x}}{2}$

From figure,
$\angle\text{EPR}=\angle\text{PQS}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{PQS}=\text{x}^\circ$
Also,
$\angle\text{PQS}=\angle\text{RSD}$ [Correspondence angles are equal]
$\Rightarrow\ \angle\text{RSD}=\text{x}^\circ$
Now,
$\angle\text{RSD}+\text{y}^\circ+\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{x}^\circ+2\text{y}^\circ=180^\circ$
$\Rightarrow\ \text{y}^\circ=\frac{180^\circ-\text{x}^\circ}{2}$
$\Rightarrow\ \text{y}^\circ=90^\circ-\frac{\text{x}^\circ}{2}$

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MCQ 1411 Mark

The number of triangles that can be drawn having angles as $50^\circ , 60^\circ $ and $70^\circ $ are:

A
None of these
B
Two
C
Only one
✓
Infinite

Answer

Correct option: D.

Infinite

As we know similar triangles can be drawn for any given triangle.
These similar triangles will have the same angles as the original triangle
$(\text{i.e}\angle50^\circ,\angle60^\circ$ and $\angle70^\circ)$ and will be infinite in number.

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MCQ 1421 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 $YBZ \| OA \| CD.$

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 1431 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 1441 Mark

How many triangles can be drawn having angles as $45^\circ , 60^\circ $ and $85^\circ ?$

A
Two.
B
Infinitely many.
✓
None.
D
Only one.

Answer

Correct option: C.

None.

If we add up the three given angles we get
$45^\circ + 60^\circ + 85^\circ = 190^\circ$
But as we know that the angles of any triangle only add-up to $180^\circ .$
Hence the above-given triangle is not possible.

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MCQ 1451 Mark

When two straight lines intersect:
$i$. Adjacent angles are complementary
$ii$. Adjacent angles are supplementary.
$iii$. Opposite angles are equal.
$iv$. Opposite angles are supplementary.
Of these statements:

A
$(ii)$ and $(iv)$ are correct
✓
$(ii)$ and $(iii)$ are correct
C
$(i)$ and $(iii)$ are correct
D
$(i)$ and $(iv)$ are correct

Answer

Correct option: B.

$(ii)$ and $(iii)$ are correct

When two straight lines intersect them, Adjacent angles are supplementary and opposite angles are equal.

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MCQ 1461 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 360º,
$\angle\text{AOD}+\angle\text{COB}+\angle\text{AOC}+\angle\text{BOD}=360^\circ$
$\Rightarrow\theta+\theta+\phi+\phi=360$
$\Rightarrow2(\theta+\phi)=360$
$\Rightarrow\theta+\phi=180$
Given that $\theta=3\phi.$
So, $3\phi+\phi=180$
$\Rightarrow4\phi=180$
$\Rightarrow\phi=45$

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MCQ 1471 Mark

Which of the following statements is false?

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

Answer

Correct option: C.

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

This statement is false because we can draw infinitely many straight lines through a given point.

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MCQ 1481 Mark

In Fig. if $CP \| BQ$, then the measure of $x$ is:

A
$125^\circ$
✓
$130^\circ$
C
$175^\circ$
D
$105^\circ$

Answer

Correct option: B.

$130^\circ$

Given that,

$CP \| BQ$
Produce $CP$ to $E$
So, $PE \| BQ$ and $AB$ cuts them
$\angle\text{QBE}=\angle\text{QBA}=105^\circ$ (Corresponding angles)
In $\triangle\text{ECA}$
$\angle\text{CEA}+\angle\text{ECA}+\angle\text{EAC}=180^\circ$
$105^\circ+\angle\text{ECA}+25^\circ=180^\circ$
$\angle\text{ECA}=50^\circ$
$\angle\text{PCA}+\angle\text{ECA}=180^\circ$ (Linear pair)
$x + 50^\circ = 180^\circ$
$x = 130^\circ .$

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MCQ 1491 Mark

The number of triangles that can be drawn having angles as $50^\circ , 60^\circ $ and $70^\circ $ are:

A
Only one.
B
Two.
✓
Infinite.
D
None of these.

Answer

Correct option: C.

Infinite.

As we know similar triangles can be drawn for any given triangle.
These similar triangles will have the same angles as the original triangle (ie., $\angle50^\circ,\angle60^\circ$and $\angle70^\circ$) and will be infinite in number.

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MCQ 1501 Mark

In figure, if lines $l$ and $m$ are parallel, then the value of $x$, is:

✓
$35^\circ$
B
$55^\circ$
C
$65^\circ$
D
$75^\circ$

Answer

Correct option: A.

$35^\circ$

From figure,
$\angle\text{ACB}=180^\circ-\angle\text{ACD}=180^\circ-125^\circ=55^\circ$
Or
$\angle\text{BCA}=55^\circ$
In right $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{CAB}=180^\circ$
$\Rightarrow\ 90^\circ+55^\circ+\text{x}=180^\circ$
$\Rightarrow\ \text{x}=35^\circ$

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M.C.Q - Page 3 - Maths STD 9 Questions - Vidyadip