M.C.Q

MCQ 11 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. The value of $x$ is:

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

Answer

Correct option: D.

$60^\circ$

$\angle\text{FEC}+\angle\text{ECD}=180^\circ$ (Sum of $2$ supplementary angles is $180^\circ )$
$\angle\text{ECD}=\angle180^\circ-150^\circ=30^\circ$
$\angle\text{x}=\angle\text{BCE}=\angle\text{ECD}$
$\angle\text{x}=30^\circ+30^\circ=60^\circ.$

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MCQ 21 Mark

Side $BC$ of $\triangle\text{ABC}$ has been produced to Don left-hand side and to Eon right-hand side such that $\angle\text{ABD}=125^\circ$ and$\angle\text{ACE}=130^\circ$ then $\angle\text{A}=?$

A
$55^\circ $
B
$50^\circ$
✓
$75^\circ$
D
$65^\circ$

Answer

Correct option: C.

$75^\circ$

$\angle\text{ABD}+\angle\text{ABC}=180^\circ$(Linear Pair)
$\angle\text{ABC}=180^\circ-125^\circ=55^\circ$
$\angle\text{ACE}+\angle\text{ACB}=180^\circ$(Linear Pair)
$\angle\text{ACB}=180^\circ-130^\circ=50^\circ$
In $\triangle\text{ABC}$
$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$(Angle sum property)
$\angle\text{BAC}=180^\circ-50^\circ-55^\circ$
$\angle\text{BAC}=75^\circ$

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MCQ 31 Mark

In figure, $\text{PQ}||\text{RS},\angle\text{QPR}=70^\circ,\angle\text{ROT}=20^\circ$ find the value of $x.$

A
$20^\circ$
B
$70^\circ$
✓
$50^\circ$
D
$110^\circ$

Answer

Correct option: C.

$50^\circ$

$\text{PQ}||\text{RS}$
$\angle\text{QPR}=\angle\text{SRO}=70^\circ$ (Corresponding, Angle)
$\text{NOW IN}\triangle\text{RTO}$
$\text{x}+20^\circ=70^\circ$ (exterior angle)
$\text{x}=70^\circ-20^\circ$
$\text{x}=50^\circ$

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MCQ 41 Mark

Each angle of an equilateral triangle is:

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

Answer

Correct option: C.

$60^\circ$

Let the angle of an equilateral triangle be $xox + x + x = 180^\circ $ (Angle sum property)
$3x = 180^\circ$
$x = 60^\circ .$

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MCQ 51 Mark

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

A
$60^\circ$
B
$100^\circ $
C
$12^\circ$
✓
$80^\circ$

Answer

Correct option: D.

$80^\circ$

Suppose $\triangle\text{ABC}$ such that $\angle\text{A}:\angle\text{B}:\angle\text{C} = 2 : 3 : 4$
Let $\angle\text{A}=2\text{k},\angle\text{B}=3\text{k}$ and $\angle\text{C} = 4\text{k}$ where k is some constant
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
$\Rightarrow 2k + 3k + 4k = 180^\circ$
$\Rightarrow 9k = 180^\circ$
$\Rightarrow k = 20^\circ $.

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MCQ 61 Mark

Which of the following pairs of angles are complementary?

✓
$25^\circ , 65^\circ$
B
$32.1^\circ , 47.9^\circ$
C
$70^\circ , 110^\circ$
D
$30^\circ , 70^\circ$

Answer

Correct option: A.

$25^\circ , 65^\circ$

Complementary angles always add up to $90^\circ$
$25^\circ + 65^\circ = 90^\circ$
Therefore this is the right option (by substitution).

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MCQ 71 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 81 Mark

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

✓
$85^{\circ}$
B
$90^{\circ}$
C
$70^{\circ}$
D
$75^{\circ}$

Answer

Correct option: A.

$85^{\circ}$

Given that,
$l1 \| l2$
Let transversal $P$ and $Q$ cuts them
$\angle1=37^\circ$
$\angle4=58^\circ$
$\angle5=\text{x}^\circ$
$\angle1=\angle2=37^\circ$(Corresponding angles) (i)
$\angle2=\angle3$ (Vertically opposite angle)
$\angle3=37^\circ$
$\angle3+\angle4+\angle5=180^\circ$(Linear pair)
$37^{\circ}+58^{\circ}+x=180^{\circ}$
$x=85^{\circ}$.

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MCQ 91 Mark

In the given figure, the measure of $\angle1$ is:

A
$48^\circ$
✓
$42^\circ$
C
$138^\circ$
D
$158^\circ$

Answer

Correct option: B.

$42^\circ$

The two given angles are,"Vertically Opposite angles" which are known to be equal
Therefore, $\angle1=42^\circ.$

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MCQ 101 Mark

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

A
$36$
✓
$54$
C
$63$
D
$72$

Answer

Correct option: B.

$54$

$AOB$ 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 111 Mark

In the adjoining figure, $\angle\text{a}$ and $\angle\text{g}$ are called:

A
Co-interior angles.
✓
Alternate exterior angles.
C
Corresponding angles.
D
Alternate interior angle.

Answer

Correct option: B.

Alternate exterior angles.

$\angle\text{a}$ and $\angle\text{g}$ are on alternate side and are exterior.

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MCQ 121 Mark

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

A
$36^\circ $
✓
$108^\circ$
C
$72^\circ$
D
$54^\circ$

Answer

Correct option: B.

$108^\circ$

Let a and b are two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio $2 : 3$ and we know that If a transversal intersects two parallel lines, then each pair of interior angles on the same side of the transversal is supplementary.Let common ratio is $x,$
$a = 2x$ and $b = 3x$
$a + b = 180^\circ$
$2x + 3x = 180^\circ$
$5x = 180^\circ$
$\text{x} = \frac{180^\circ}{5} = 36^\circ$
$x = 36^\circ$
$3x = 3 \times 36^\circ = 108^\circ .$

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MCQ 131 Mark

The number of lines that can pass through a given point is:

A
Only one.
✓
Infinity.
C
One.
D
Two.

Answer

Correct option: B.

Infinity.

As seen from the above image, any number of lines can be drawn through a given point.
Hence the answer may be given as $"$Infinity$"$.

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MCQ 141 Mark

In figure, which of the following statement must be true?
$(i) a + b = d + c$
$(ii) a + c + e = 180^\circ$
$(iii) b + f = c + e$

A
$(i)$ only
B
$(ii)$ only
C
$(iii)$ only
✓
$(ii)$ and $(iii)$ only

Answer

Correct option: D.

$(ii)$ and $(iii)$ only

From figure, we can see that
$\angle\text{a}^\circ+\angle\text{b}^\circ+\angle\text{c}^\circ=\angle\text{FOC}=180^\circ$
Also,
$\angle\text{b}^\circ=\angle\text{e}^\circ$ [Opposite angles]
So,
$\angle\text{a}^\circ+\angle\text{e}^\circ+\angle\text{c}^\circ=180^\circ$
$\Rightarrow (ii)$ is correct
Now,
$\angle\text{FOB}\neq\angle\text{DOB}$
$\Rightarrow\ \angle\text{a}^\circ+\angle\text{b}^\circ\neq\angle\text{d}^\circ+\angle\text{c}^\circ$
$\Rightarrow (i)$ is correct
Now,
$\angle\text{b}^\circ=\angle\text{e}^\circ$ and $\angle\text{f}^\circ=\angle\text{c}^\circ$ [Opposite angles are equal]
Thus,
$\angle\text{b}^\circ=\angle\text{f}^\circ=\angle\text{e}^\circ+\angle\text{c}^\circ$
$\Rightarrow (iii)$ is correct.

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MCQ 151 Mark

Write the correct answer in the following: In Fig. if $\text{AB}||\text{CD}||\text{EF},\text{PQ}||\text{RS},$ $\angle\text{RQD}=25^\circ$ and $\angle\text{CQP}=60^\circ,$ then $\angle\text{QRS}$ is equal to.

A
$85^\circ$
B
$135^\circ$
✓
$145^\circ$
D
$110^\circ$

Answer

Correct option: C.

$145^\circ$

Given, $\text{PQ}||\text{RS}$
$\angle\text{PQC}=\angle\text{BRC}=60^\circ$$\big[$alternate exterior angles and $\angle\text{PQC}=60^\circ$ (given)$\big]$ and $\angle\text{DQR}$
$=\angle\text{QRA}=25^\circ$ [alternate interior angles]
$$$\big[\angle\text{DQR}=25^\circ,\text{(given)}\big]$
$\angle\text{QRS}=\angle\text{QRA}+\angle\text{ARS}$
$=\angle\text{QRA}+\big(180^\circ-\angle\text{BRS}\big)$ [linear pair axiom]
$=25^\circ+180^\circ-60^\circ=205^\circ-60^\circ=145^\circ$

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MCQ 161 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 171 Mark

In the adjoining figure, the three lines $AB, CD$ and $EF$ all pass through the point O. If $\angle\text{EOB}=90^\circ$ and $x : y = 2 : 1$ then $\angle\text{BOD}$ and $\angle\text{COE}:$

✓
$30^\circ , 60^\circ$
B
$80^\circ , 20^\circ$
C
$45^\circ , 45^\circ$
D
$60^\circ , 60^\circ$

Answer

Correct option: A.

$30^\circ , 60^\circ$

$x + y + 90^\circ = 180^\circ $ (Linear Pair)
$2a + a + 90^\circ = 180^\circ $ (Since, $x : y = 2 : 1)$
$a = 30^\circ$
$\text{x}=\text{2a}=\angle\text{COE}=60^\circ$ (Vertically opposite angles)
$\text{y}=\angle\text{BOD}=30^\circ $ (Vertically opposite angles).

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MCQ 181 Mark

Measurement of reflex angle is:

✓
Between $180^\circ $ and $360^\circ $
B
$90^\circ $
C
Between $0^\circ $ and $90^\circ $
D
Between $90^\circ $ and $180^\circ$

Answer

Correct option: A.

Between $180^\circ $ and $360^\circ $

Let $x$ be the angle then its reflex angle is $360^\circ - x$ and in any triangle, the angle lies between $0$ to $180^\circ .$

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MCQ 191 Mark

In the given figure $x = 30^\circ $, the value of $Y$ is:

A
$36^\circ $
B
$10^\circ$
✓
$40^\circ$
D
$45^\circ$

Answer

Correct option: C.

$40^\circ$

In the given figure we have
$3Y + 2X = 180^\circ $ (Linear - Pair)
$X = 30^\circ$
$3Y + 2 \times 30^\circ = 180^\circ$
$3Y + 60^\circ = 180^\circ$
$3|Y = 180^\circ - 60^\circ$
$3Y = 120^\circ$
$\text{Y}=\frac{120^\circ}{3}$
$Y = 40^\circ .$

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MCQ 201 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 $COD$ 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 211 Mark

The measure of an angle is five times its complement. The angle measures.

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

Answer

Correct option: B.

$75^\circ$

Let the measure of the required angle be $x^\circ$
Then, the measure of its complement will be $(90 − x)^\circ$
Therefore, $x = 5 (90 - x)$
$\Rightarrow x = 450 - 5x$
$\Rightarrow 6x = 450$
$\Rightarrow x = 75^\circ $.

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MCQ 221 Mark

$A, B, C$ are the three angles of a triangle. If $A - B = 15^\circ $ and $B - C = 30^\circ $, then angles$ A, B, C$ are respectively:

✓
$80^\circ , 65^\circ , 35^\circ$
B
$65^\circ , 80^\circ , 35^\circ$
C
$80^\circ , 35^\circ , 65^\circ$
D
$35^\circ , 65^\circ , 80^\circ$

Answer

Correct option: A.

$80^\circ , 65^\circ , 35^\circ$

Since $ABC$ is a triangle
$A + B + C = 180^\circ$ (Angle sum property) $(i)$
$A - B = 15^\circ$
$A = 15^\circ + B (ii)$
$B - C = 30^\circ$
$C = B - 30^\circ (iii)$
From $(i)$ equation
$15^\circ + B + B + B - 30^\circ = 180^\circ$
$B = 65^\circ$
From equation $(ii)$ and $(iii)$
$A = 15^\circ + B = 15^\circ + 65^\circ = 80^\circ$
$C = B - 30^\circ = 65^\circ - 30^\circ = 35^\circ .$

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MCQ 231 Mark

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

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

Answer

Correct option: D.

$50^\circ$

$\angle\text{APQ}=\angle\text{PQR}=70^\circ$ (Alternate interior angles)
$\angle\text{PRQ}+\angle\text{PRD}=180^\circ$ (Linear Pair)
$\angle\text{PRQ}=180^\circ-120^\circ=60^\circ$
In $\triangle\text{PQR}$
$\angle\text{PQR}+\angle\text{PRQ}+\angle\text{QPR}=180^\circ$ (Angle sum property)
$\angle\text{QPR}=180^\circ-70^\circ-60^\circ=50^\circ.$

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MCQ 241 Mark

The angles of a triangle in ascending order are $x, y, z$ and $y - x = z - y = 10^{\circ}$. The smallest angles is:

A
$60^{\circ}$
✓
$50^{\circ}$
C
$70^{\circ}$
D
$40^{\circ}$

Answer

Correct option: B.

$50^{\circ}$

$x+y+z=180^{\circ} \text { (i) (Angle sum property) }$
$y-x=10^{\circ}$
$y-10^{\circ}=x \text { (ii) }$
$z-y=10^{\circ}$
$z=10^{\circ}+y \text { (iii) }$
On putting the value of $x$ and $z$ in equation (i)
$y-10^{\circ}+y+10^{\circ}+y=180^{\circ}$
$y=60^{\circ}$
$x=50^{\circ}(\text { From equation ii })$
$z=70^{\circ}(\text { From equation iii) }$
Smallest angle is $x=50^{\circ}$.

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MCQ 251 Mark

If 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
$60^\circ$
✓
$80^\circ$
D
$100^\circ$

Answer

Correct option: C.

$80^\circ$

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

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MCQ 261 Mark

In Fig. $AB \| CD \| EF$ and $GH \| KL$. The measure of $\angle\text{HKL}$ is:

A
$85^\circ$
B
$135^\circ$
C
$215^\circ$
✓
$145^\circ$

Answer

Correct option: D.

$145^\circ$

Given, $AB \| CD \| EF$ and $GH \|$ KLProduce $HG$ to $M$ and KL to $N$
$\angle\text{MHD}$ and $\angle\text{CHG}=60^\circ$ (Vertically opposite angle)
Since, $MG \| NL$ and transversal cuts them
So, $\angle\text{MHD}+\angle1=180^\circ$ (Interior angles)
$60^\circ+\angle1=180^\circ$
$\angle1=120^\circ$
$\angle3=\angle\text{HKD}=25^\circ$ (Alternate angles) $(i)$
$\angle1=\angle\text{MKL}=120^\circ$ (Corresponding angles) $(ii)$
Now, $\angle\text{HKL}=\angle3+\angle\text{MKL}$
$= 25^\circ + 120^\circ$
$= 145^\circ .$

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MCQ 271 Mark

In the adjoining figure, if $QP \| RT$, then $x$ is equal to:

A
$55^\circ$
✓
$75^\circ$
C
$65^\circ$
D
$70^\circ$

Answer

Correct option: B.

$75^\circ$

$\angle\text{QPR}=\angle\text{PRT}=40^\circ$ (Alternate interior angles)
In $\triangle\text{QPR}$
$\angle\text{PQR}+\angle\text{QPR}+\angle\text{PRQ}=180\circ $ (Angle sum property)
$65^\circ+40^\circ+\text{x}^\circ=180^\circ$
$\text{x}^{\circ}=180^\circ-40^\circ-65^\circ$
$\text{x}^{\circ}=75$

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MCQ 281 Mark

In the given $x =?$

A
$\alpha-\beta-\gamma$
✓
$\alpha+\beta+\gamma$
C
$\alpha+\beta-\gamma$
D
$\alpha+\gamma-\beta$

Answer

Correct option: B.

$\alpha+\beta+\gamma$

$OBCA$ is a quadrilateral$\angle\text{OAC}+\angle\text{BOA}+\angle\text{ACB}+\angle\text{CBO}=360^\circ$
$\gamma+\beta+\angle\text{ACB}+\alpha=360^\circ$
$\angle\text{ACB}=360^\circ-\gamma-\beta-\alpha$
$\text{x}=360^\circ-\angle\text{ACB}$
$\text{x}=\alpha+\beta+\gamma.$

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MCQ 291 Mark

Write the correct answer in the following: The angles of a triangle are in the ratio $5 : 3 : 7$ The triangle is.

✓
An acute angled triangle.
B
An obtuse angled triangle.
C
A right triangle.
D
An isosceles triangle.

Answer

Correct option: A.

An acute angled triangle.

Let the angles of the triangle be $5x, 3x$ and $7x$.As the sum of the angles of a triangle is $180^\circ $ then
$5x + 3x + 7x = 180^\circ$
$\Rightarrow 15x = 180^\circ$
$\Rightarrow x = 180^\circ ÷ 15 = 12^\circ$
Therefore, the angle of the triangle are:
$5 \times 12^\circ , 3 \times 12^\circ $ and $7 \times 12^\circ , i.e., 60^\circ , 36^\circ $ and $84^\circ$
As the measure of each angle of the triangle is less than $90^\circ $, so the angles of triangle are acute angles.
Therefore, the triangle is an acute angled triangle.
Hence, $(a)$ is the correct answer.

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MCQ 301 Mark

In the given figure, $\angle\text{OEB} = 75^\circ,\angle\text{OBE}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ Then $\angle\text{ODC}=?$

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

Answer

Correct option: A.

$30^\circ$

In $\triangle\text{OEB}$
$\angle\text{OEB}+\angle\text{EBO}+\angle\text{BOE}=180^\circ$ (Angle sum property)
$75^\circ+55\circ+\angle\text{BOE}=180^\circ$
$\angle\text{BOE}=50^\circ$
$\angle\text{BOE}=\angle\text{COD}=50^\circ$ (vertically opposite angle)
In $\triangle\text{ODC}$
$\angle\text{ODC}+\angle\text{DOC}+\angle\text{DCO}=180^\circ$
$\angle\text{ODC}=180^\circ-100^\circ-50^\circ$
$\angle\text{ODC}=30^\circ$

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MCQ 311 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 321 Mark

In Fig., the value of $x$ is:

A
$15^\circ$
B
$12^\circ$
C
$8^\circ$
✓
$20^\circ$

Answer

Correct option: D.

$20^\circ$

Let,$AB, CD$ and $EF$ intersect at $O$
$\angle\text{COB}=\angle\text{AOD}$ (Vertically opposite angle)
$\angle\text{AOD}=3\text{x}+10\text{ (i)}$
$\angle\text{AOE}+\angle\text{AOD}+\angle\text{DOF}=180^\circ$ (Linear pair)
$x + 3x + 10^\circ + 90^\circ = 180^\circ$
$4x + 100^\circ = 180^\circ$
$4x = 80^\circ$
$x = 20^\circ .$

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MCQ 331 Mark

Two straight lines $AB$ and $CD$ cut each other at $O$. If $\angle\text{BOD}=63^\circ,$ then $\angle\text{BOC}=$

A
$63^\circ$
✓
$117^\circ$
C
$17^\circ$
D
$153^\circ$

Answer

Correct option: B.

$117^\circ$

$\angle\text{BOD}$ and $\angle\text{BOC}$ from a linear pair.
$\therefore\ \angle\text{BOD}+\angle\text{BOC}=180^\circ$
$\Rightarrow\ 63^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\ \angle\text{BOC}=117^\circ$

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MCQ 341 Mark

In a given figure, if $AB \| CD \| EF, PQ \| RS$, $\angle\text{RQD}=25^\circ$ and $\angle\text{CQP}= 60^\circ,$ then $\angle\text{QRS}$ is equal to:

✓
$145^\circ $
B
$110^\circ$
C
$85^\circ$
D
$135^\circ$

Answer

Correct option: A.

$145^\circ $

Given, $PQ \| RS$
$\angle\text{PQC}=\angle\text{BRS}=60^\circ$ [alternate exterior angles and $\text{PQC}=60^\circ$ (given)] and $\angle\text{DQR}=\angle\text{QRA}=25^\circ$ [alternate interior angles]
$[\angle\text{DQR}=25^\circ, \text{ given}]$
$\angle\text{QRS}=\angle\text{QRA}+\angle\text{ARS}$
$=\angle\text{QRA}+(180^\circ–\angle\text{BRS})$ [linear pair axiom]
$=25^\circ+180^\circ-60^\circ=205^\circ- 60^\circ=145^\circ.$

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MCQ 351 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 361 Mark

In the given figure, the measure of $\angle\text{a}$ is:

A
$150^\circ$
✓
$30^\circ$
C
$15^\circ$
D
$50^\circ$

Answer

Correct option: B.

$30^\circ$

In the given figure
$150^\circ\angle\text{a}=180^\circ$ (linear - pair)
$\angle\text{a}=180^\circ=150^\circ$
Therefore,
$\angle\text{a}=30^\circ$

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MCQ 371 Mark

The angle which is equal to 8 times its complement is:

A
$88^\circ$
B
$72^\circ$
✓
$80^\circ$
D
$90^\circ$

Answer

Correct option: C.

$80^\circ$

We know that two angles, whose sum is $90^\circ $, are called the complementary angle.
Let one angle be x then its complementary angle be 8x,
$x + 8x = 90^\circ$
$9x = 90^\circ$
$x = 10^\circ$
Its complementary angle is $8 \times 10 = 80^\circ .$

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MCQ 381 Mark

Given $\angle\text{POR}=3\text{x}$ and $\angle\text{QOR}=2\text{x}+10^\circ.$ If $\angle\text{POQ}$ is a straight line, then the value of $x$ is:

A
$30^\circ $
B
$36^\circ $
✓
$34^\circ $
D
None of these

Answer

Correct option: C.

$34^\circ $

Given,
$POQ$ is a straight line
$\angle\text{POR}+\angle\text{QOR}=180^\circ$ (Linear pair)
$3x + 2x + 10^\circ = 180^\circ$
$5x = 170^\circ$
$x = 34^\circ .$

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MCQ 391 Mark

In the adjoining figure, $AB \| CD$ and $AB \| EF$. If $EA$ $\bot$ $BA$ and $\angle\text{BEF}$ then the values of $x, y$ and $z:-$

✓
$125^\circ , 125^\circ , 35^\circ$
B
$60^\circ , 60^\circ , 60^\circ$
C
$120^\circ , 130^\circ , 25^\circ$
D
$35^\circ , 125^\circ , 120^\circ$

Answer

Correct option: A.

$125^\circ , 125^\circ , 35^\circ$

$x + 55 = 180^\circ$ (Sum of supplementary angles or co-interior angles)$x = 125^\circ $
$x = y = 125^\circ $ (Corresponding angles)
$\text{z}+\angle\text{EAB}$ (Exterior angle property)
$\text{z}=125^\circ-90^\circ=35^\circ$

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MCQ 401 Mark

In the given figure, $AB \| CD$ and $O$ is a point joined with $B$ and $D$, as shown in the figure such that $\text{ABO}=35^\circ$ and $\angle\text{CDO}=40^\circ$ Reflex $\angle\text{BOD}=?$

A
$265^\circ$
✓
$285^\circ$
C
$275^\circ$
D
$255^\circ$

Answer

Correct option: B.

$285^\circ$

$\angle\text{ABO}+\angle\text{BOE}=180^\circ$ (Sum of supplementary angles)
$\angle\text{BOE}=180^\circ-35^\circ=145^\circ$
$\angle\text{CDO}+\angle\text{DOE}=180^\circ$ (Sum of supplementary angles)
Reflex of $\angle\text{BOD}=\angle\text{BOE}+\angle\text{DOE}=145^\circ+140^\circ$
Reflex of $\angle\text{BOD}=285^\circ.$

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MCQ 411 Mark

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

✓
$15$
B
$20$
C
$25$
D
$12$

Answer

Correct option: A.

$15$

It is given that, $A O B$ is a straight line.
$\therefore$ $60^\circ + (5x^\circ + 3x^\circ ) = 180^\circ $ (Linear pair)
$\Rightarrow 8x^\circ = 180^\circ - 60^\circ = 120^\circ$
$\Rightarrow x^\circ = 15^\circ$
Thus, the value of $x$ is $15.$

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MCQ 421 Mark

For what value of $x$ shall we have $l \| m?$

A
$x = 60^\circ$
✓
$x = 50^\circ$
C
$x = 70^\circ$
D
$x = 45^\circ$

Answer

Correct option: B.

$x = 50^\circ$

$(2x - 30)^\circ = (x + 20)^\circ$ (corresponding angle)
$2x -30^\circ = x + 20^\circ$
$2x - x = 30^\circ + 20^\circ$
$x = 50^\circ .$

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MCQ 431 Mark

In the given figure, $AOB$ 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
$100^\circ$
B
$40^\circ$
✓
$80^\circ$
D
$60^\circ$

Answer

Correct option: C.

$80^\circ$

We have,
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [Since $AOB$ is a straight line]
$\Rightarrow 4x + 5x = 180^\circ$
$\Rightarrow 9x = 180^\circ$
$\Rightarrow x = 20^\circ$
$\therefore\angle\text{AOC}=4\times20^\circ=80^\circ.$

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MCQ 441 Mark

In the given figure, $AB \| DC$, $\angle\text{BAD}=90^\circ,\angle\text{CBD}=28^\circ$ and $\angle\text{BCE}=65^\circ.$ Then $\angle\text{ABD}=?$

A
$32^\circ$
✓
$37^\circ$
C
$43^\circ$
D
$53^\circ$

Answer

Correct option: B.

$37^\circ$

In $\triangle\text{DBC}$
$\angle\text{BCE}=\angle\text{DBC}+\angle\text{BDC}$ (Exterior angle property)
$65^\circ=28^\circ+\angle\text{BDC}$
$\text{BDC}=37^\circ$
As, $AB$ is parallel to $CD$
$\angle\text{ABD}=\angle\text{BDC}=37^\circ$ (Alternate interior angle).

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MCQ 451 Mark

In the given below figure, the measure of $\angle\text{AED}$ is:

A
$140^\circ$
✓
$130^\circ$
C
$120^\circ$
D
$110^\circ$

Answer

Correct option: B.

$130^\circ$

in $\triangle\text{ABC}$$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ (Angle sum property)
$\angle\text{ACB}=180^\circ-25^\circ-45^\circ$
$\angle\text{ACB}=110^\circ$
$\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair)
$\angle\text{ACD}=180^\circ-110^\circ=70^\circ$
In $\triangle\text{CED}$
$\angle\text{AED}+\angle\text{EDC}+\angle\text{EDC}$ (Exterior angle is equal to sum of its two interior opposite angles)
$\angle\text{AED}=60^\circ+75^\circ=130^\circ.$

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MCQ 461 Mark

In the adjoining figure, $y =?$

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

Answer

Correct option: D.

$54^\circ$

We have,$3x + 72 = 180^\circ $[$\because$ AOB is a straight line]
$\Rightarrow 3x = 108$
$\Rightarrow x = 36$
Also,
$\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$ [$\because$ $AOB$ is a straight line]
$\Rightarrow 36^\circ + 90^\circ + y = 180^\circ$
$\Rightarrow y = 54^\circ .$

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MCQ 471 Mark

In figure, if line segment AB is parallel to the line segment $CD$, what is the value of $y?$

A
$12$
B
$15$
C
$18$
✓
$20$

Answer

Correct option: D.

$20$

From figure,
$\angle\text{ABD}+\angle\text{EBD}=180^\circ$
$\Rightarrow\ \angle\text{EBD}=180^\circ-\angle\text{ABD}\dots(1)$
Now,
$\angle\text{ABD}=\text{y}^\circ+2\text{y}^\circ+\text{y}^\circ$
$\Rightarrow\ \angle\text{ABD}=4\text{y}^\circ\dots(2)$
Substituting (2) in (1), we have
$\angle\text{EBD}=180^\circ-4\text{y}^\circ$
Now,
$\angle\text{EBD}=\angle\text{BDC}$ [Alternate angles]
$\Rightarrow\ 180^\circ-4\text{y}^\circ=5\text{y}^\circ$
$\Rightarrow\ 180^\circ=9\text{y}^\circ$
$\Rightarrow\ \text{y}^\circ=20^\circ$

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MCQ 481 Mark

In Fig., which of the following statements must be true?
$i. a + b = d + c$
$ii. a + c + e = 180^\circ$
$iii. b + f = c + e$

A
$(iii)$ only
B
$(i)$ only
C
$(ii)$ only
✓
$(ii)$ and $(iii)$ only

Answer

Correct option: D.

$(ii)$ and $(iii)$ only

Let $AB, CD$ and $EF$ intersect at $O$
$\angle\text{AOD}=\angle\text{COB} ($Vertically opposite angle$)$
$b = e (i)$
$\angle\text{EOC}=\angle\text{DOF} ($Vertically opposite angle$)$
$f = c (ii)$
Adding $(i)$ and $(ii),$ we get
$b + f = c + e (iii)$
Now,
$\angle\text{AOD}+\angle\text{EOC}+\angle\text{COB}=180^\circ$
$a + f + e = 180^\circ$
$a + c + e = 180^\circ $ [From $(ii)].$

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MCQ 491 Mark

The number of angles formed by a transversal with a pair of parallel lines are:

✓
$8$
B
$6$
C
$3$
D
$4$

Answer

Correct option: A.

$8$

As we can see there are $4$ angles formed at every point of intersection thus giving a total of $8$ angles.

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MCQ 501 Mark

If $\angle\text{A}=4\angle\text{B} = 6\angle\text{C},$ then $A : B : C?$

A
$6 : 4 : 3$
✓
$12 : 3 : 2$
C
$2 : 3 : 4$
D
$3 : 4 : 6$

Answer

Correct option: B.

$12 : 3 : 2$

Let A be x$\text{B}=\frac{1}{4}\text{x}$
$\text{C}=\frac{1}{6}\text{x}$
$A : B : C$
$\text{x}=\frac{1}{4}\text{x}:\frac{1}{6}\text{x}$
$LCM$ of $4$ and $6$ is $12$
$12 : 3 : 2.$

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