4 Marks Questions - Maths STD 9 Questions - Vidyadip

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21 questions · timed · auto-graded

Question 14 Marks

Prove that the bisectors of a pair of vertically opposite angles are in the same straight line.

Answer

Given: Lines $AOB $and $COD $intersect at point $O $. Such that $\angle AOC =\angle BOD$. Also, $O E$ is the bisector $\angle AOC$ and $OF $is the bisector $\angle BOD$. To prove: $E O F$ is a straight line. $\angle AOD =\angle BOC =5 x \ldots$ (1) ( [vertically opposite angle] Also, $\angle AOD +\angle BOC$ [vertically opposite angle] $\Rightarrow 2 \angle AOE =2 \angle DOF \ldots$ (2) Now, $\angle AOD +\angle AOC +\angle BOC +\angle BOD =360^{\circ}$ [Sum of all angles around a point is $360^{\wedge} \backslash circ$ ] $\Rightarrow 2 \angle AOD +2 \angle AOE +2 \angle DOF =360^{\circ}$ $\Rightarrow \angle AOD +\angle AOE +\angle DOF =180^{\circ}$ From this we conclude that $EOF $is a straight line.

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Question 24 Marks

Define complementary angles.

Answer

Complementary Angles: Two angles, the sum of whose measures is $90^{\circ}$, are called complementary angles. Thus, angles $\angle BAX$ and $\angle XAC$ are complementary angles. If $x + y =90^{\circ}$

Example 1: Angles of measure $50^{\circ}$ and $40^{\circ}$ are complementary angles, because $50^{\circ}+40^{\circ}=90^{\circ}$ Example 2: Angles of measure $60^{\circ}$ and $30^{\circ}$ are complementary angles, because $60^{\circ}+30^{\circ}=90^{\circ}$

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Question 34 Marks

In figure, Line $l_1$ and $l_2$ intersect at $O$, forming angles as shown in the figure. If $x = 45$, find the value of $y, z$ and $u.$

Answer

Given $x = 45^\circ$
$\therefore z = x = 45^\circ$ [Vertically opposite angle]
Now, $z + u = 180^\circ$ [Linear pair]
$\Rightarrow 45^\circ + u = 180^\circ$
$\Rightarrow u = 135^\circ$
Also, $x + y = 180^\circ$ [Linear pair]
$\Rightarrow y = 180^\circ - x = 180^\circ - 45^\circ = 135^\circ$
$\therefore y = 135^\circ$

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Question 44 Marks

In figure, rays $OA, OB, OC, OD$ and $OE $have the common end point $O$. Show that $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}+\angle\text{EOA}=360^\circ.$

Answer

Given that $OA, OB, OD$ and $OE$ have the common end point $O$. A ray opposite to $OA $is drawn. Since $\angle\text{AOB},\angle\text{BOF}$ are linear pairs, $\angle\text{AOB}+\angle\text{BOF}+180^\circ$ $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COF}=180^\circ\dots(\text{i})$ Also, $\angle\text{AOE}$ and $\angle\text{EOF}$ are linear pairs $\angle\text{AOE}+\angle\text{EOF}=180^\circ$ $\angle\text{AOE}+\angle\text{DOF}+\angle\text{DOE}=180^\circ\dots(\text{ii})$ By adding (i) and (ii) equations we get: $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COF}+\angle\text{AOE}+\angle\text{DOF}+\angle\text{DOE}=180^\circ$ $\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}+\angle\text{EOA}=180^\circ$ Hence proved.

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Question 54 Marks

In figure, lines $PQ $and $RS $intersect each other at point $O$. If $\angle\text{POR}:\angle\text{ROQ}=5:7,$ find all the angles.

Answer

Given
$\angle\text{POR}$ and $\angle\text{ROP}$ is linear pair
$\angle\text{POR}+\angle\text{ROP}=180^\circ$
Given that
$\angle\text{POR}:\angle\text{ROP}=5:7$
Hence,
$\text{POR}=\Big(\frac{5}{12}\Big)\times180=75$
Similarly
$\text{ROQ}=\Big(\frac{7}{7}+5\Big)\times180=105$
Now $POS = ROQ = 105^\circ $ [Vertically opposite angles]
Also,
$SOQ = POR = 75^\circ $ [Vertically opposite angles]

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Question 64 Marks

In figure, find $x.$ Further find $\angle\text{BOC},\angle\text{COD}$ and $\angle\text{AOD}.$

Answer

Since $\angle\text{AOD}$ and $\angle\text{BOD}$ form a line pair,
$\angle\text{AOD}+\angle\text{BOD}=180^\circ$
$\angle\text{AOD}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
Given that, $\angle\text{AOD}=(\text{x}+10)^\circ,\angle\text{COD}=\text{x}^\circ,\angle\text{BOC}=(\text{x}+20)^\circ$
$(\text{x}+10)+\text{x}+(\text{x}+20)=180$
$3\text{x}+30=180$
$3\text{x}=180-30$
$3\text{x}=\frac{150}{3}$
$\text{x}=50$ Therefore, $\angle\text{AOD}=(\text{x}+10)$
$=50+10=60$
$\angle\text{COD}=\text{x}=50^\circ$
$\angle\text{COD}=(\text{x}+20)$
$=50+20=70$
$\angle\text{AOD}=60^\circ,\angle\text{COD}=50^\circ,\angle\text{BOC}=70^\circ$

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Question 74 Marks

If each of the two lines is perpendicular to the same line, what kind of lines are they to each other?

Answer

Let $A B$ and $C D$ be perpendicular to $M N A B D=90 \ldots$$(i)$
$[A B$ perpendicular to $M N] C O N=90 \ldots$
$(ii)$ [CO perpendicular to MN] Now, $A B D=C D N=90$
(From $(i)$ and $(ii)$)
$A B$ parallel to $C D$, since corresponding angles are equal.

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Question 84 Marks

In figure, $\angle\text{AOC}$ and $\angle\text{BOC}$ from a linear pair. If $a - 2b = 30°$, find $a$ and $b.$

Answer

Given that, $\angle AOC$ and $\angle BOC$ form a linear pair If $a - b =30 \angle AOC = a ^{\circ}, \angle BOC = b ^{\circ}$ Therefore, $a + b =180 \ldots$
$(i)$ Given $a -2 b=30$
..$(ii)$ By subtracting
$(i)$ and
$(ii)$
$a+b-a+2 b=180-303 b=150 b=\frac{150}{3} b=50$ Since $a-2 b=$
$30 a-2$(50)$=30 a=30+100 a=130$
Hence, the values of $a$ and $b$ are $130^{\circ}$ and $50^{\circ}$ respectively.

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Question 94 Marks

In figure, arms $BA$ and $BC$ of $\angle\text{ABC}$ are respectively parallel to arms $ED$ and $EF$ of $\angle\text{DEF}$ Prove that$\angle\text{ABC}+\angle\text{DEF}=180^\circ.$

Answer

Given:
$AB \| DE, BC \| EF$
To prove:
$\angle\text{ABC}+\angle\text{DEF}=180^\circ$
Construction: Produce $BC$ to intersect $DE$ at $M$

Proof:
Since $AB \| EM$ and $BL$ is the transversal
$\angle\text{ABC}=\angle\text{EML}\dots(\text{i})$ [Corresponding angle]
Also,
$EF \| ML$ and $EM $is the transversal
By the property of co-interior angles are supplementary
$\angle\text{DEF}+\angle\text{EML}=180^\circ\dots(\text{ii})$
From $(i) $and $(ii)$ we have
Therefore
$\angle\text{DEF}+\angle\text{ABC}=180^\circ$

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Question 104 Marks

If one of the four angles formed by two intersecting lines is a right angle, then show that each of the four angles is a right a angle.

Answer

Given: $AB $and $CD$ are two lines intersecting at $O$
such that $\angle\text{BOC}=90^\circ.$ $R.T.P:$
$\angle\text{AOC}=90^\circ,\angle\text{AOD}=90^\circ$ and $\angle\text{BOD}=90^\circ$
Proof: We have, $\angle\text{BOC}=90^\circ$ [given]
Also, $\angle\text{BOC}=\angle\text{AOD}=90^\circ$ [vertically opposite angles]
$\angle\text{AOC}+\angle\text{BOC}=180^\circ$ [linear pair]
$\Rightarrow\ \angle\text{AOC}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{AOC}=90^\circ$
Now, $\angle\text{AOC}=\angle\text{BOD}=90^\circ$ [vertically opposite angles]
Hence, $\angle\text{AOC}=\angle\text{BOC}=\angle\text{BOD}=\angle\text{AOD}=90^\circ$

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Question 114 Marks

In figure, $\angle\text{AOF}$ and $\angle\text{FOG}$ from a linear pair. If $\angle\text{EOB}=\angle\text{FOC}=90^\circ$ and $\angle\text{DOC}=\angle\text{FOG}=\angle\text{AOB}=30^\circ.$

$i.$ Find the measure of $\angle\text{FOE},\angle\text{COB}$ and $\angle\text{DOE}.$
$ii.$ Name all the right angles.
$iii.$ Name three pairs of adjacent complementary angles.
$iv.$ Name three pairs of adjacent supplementary angles.
$v.$ Name three pairs of adjacent angles

Answer

$i. \angle\text{FOE}=\text{x},\angle\text{DOE}=\text{y}$ and $\angle\text{BOC}=\text{z}$
Since $\angle\text{AOF},\angle\text{FOG}$ is a linear pair
$\angle\text{AOF}+30=180$
$\angle\text{AOF}=180-30$
$\angle\text{AOF}=150$
$\angle\text{AOB}+\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}+\angle\text{EOF}=150$
$30 + z + 30 + y + x = 150$
$x + y + z =150 - 30 - 30$
$x + y + z = 90 ...(1)$
$\angle\text{FOC}=90^\circ$
$\angle\text{FOE}+\angle\text{EOD}+\angle\text{DOC}=90^\circ$
$x + y + 30 = 90$
$x + y = 90 - 30$
$x + y = 60 ...(2)$
Substituting $(2)$ in $(1)$
$x + y + z = 90$
$60 + z = 90$
$z = 90 - 60 = 30$
Given $BOE = 90$
$\angle\text{BOC}+\angle\text{COD}+\angle\text{DOE}=90^\circ$
$30 + 30 + DOE = 90$
$DOE = 90 - 60 = 30$
$DOE = x = 30$
We also know that,
$x + y = 60$
$y = 60 - x$
$y = 60 - 30$
$y = 30$
Thus we have $\angle\text{FOE}=30,\angle\text{COB}=30$ and $\angle\text{DOE}=30$
$ii.$ Right angles are:
$\angle\text{DOG},\angle\text{COF},\angle\text{BOF},\angle\text{AOD}.$
$iii.$ Adjacent complementary angles are:
$(\angle\text{AOB},\angle\text{BOD});(\angle\text{AOC},\angle\text{COD});(\angle\text{BOC},\angle\text{COE}).$
$iv.$ Adjacent supplementary angles are:
$(\angle\text{AOB},\angle\text{BOG});(\angle\text{AOC},\angle\text{COG});(\angle\text{AOD},\angle\text{DOG}).$
$v.$ Adjacent angles are:
$(\angle\text{BOC},\angle\text{COD});(\angle\text{COD},\angle\text{DOE});(\angle\text{DOE},\angle\text{EOF}).$

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Question 124 Marks

Prove that if the two arms of an angle are perpendicular to the two arms of another angle, then the angles are either equal or supplementary.

Answer

Consider be angles $AOB$ and $ACB.$
Given $OA$ perpendicular to $AO$, also $OB$ perpendicular to $BO$.
To Prove:
$\angle\text{AOB}+\angle\text{ACB}=180^\circ$
Proof:
In a quadrilateral $=\angle\text{A}+\angle\text{O}+\angle\text{B}+\angle\text{C}=360^\circ$[Sum of angles of quadrilateral is $360]$
$\Rightarrow 180 + O + B + C = 360$
$\Rightarrow O + C = 360 - 180$
Hence,
$AOB + ACB = 180 ...(1)$
Also,
$B + ACB = 180$
$\Rightarrow ACB = 180 - 90 = ACES = 90^\circ ...(2)$
From $(1)$ and $(2), ACB = A0B = 90$
Hence, the angles are equal as well as supplementary.

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Question 134 Marks

Define adjacent angles.

Answer

Adjacent angles: Two angles are called adjacent angles, if:
$i.$ They have the same vertex.
$ii.$ They have a common arm, and
$iii.$ Uncommon arms are on either side of the common arm.

In the figure above, $\angle\text{AOC}$ and $\angle\text{BOC}$ have a common vertex $O$.
Also, they have a common arm $OC$ and the distinct arms $AO$ and $BO$, lies on the opposite sides of the line $OC$.
Therefore, $\angle\text{AOC}$ and $\angle\text{BOC}$ are adjacent angles.

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Question 144 Marks

In Figure, if a is greater than b by one third of a right angle. Find the value of $a$ and $b.$

Answer

Since $a$ and $b$ are linear $a+b=180 a=180-b \ldots(1)$
From given data, $a$ is greater than $b$ by one third of a right angle $a=b+\frac{90}{3} a=b+30 a-b=30 \ldots (2)$
Equating (1) and (2) $180-b=b+30180-30=2 b$
$b=\frac{150}{2} b=75 \text { From (1) }$
$a=180-b a=180-75 a=105$
Hence the values of $a$ and $b$ are $105^{\circ}$ and $75^{\circ}$ respectively.

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Question 154 Marks

If two straight lines are perpendicular to the same line, prove that they are parallel to each other.

Answer

Given m perpendicular $t$ and $l$ perpendicular to t. $\angle{1}=\angle{2}=90^\circ$
Since, $I$ and $m$ are two lines and it is transversal and the corresponding angles are equal $L \| M$
Hence proved.

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Question 164 Marks

The opposite sides of a quadrilateral are parallel. If one angle of the quadrilateral is $60^\circ $. Find the other angles.

Answer

Given $AB \| CD AD \| BC$ Since $AB \| CD$ and $AD$ is the transversal
Therefore, $A + D = 180$ [Co-interior angles are supplementary] $60 + D = 180 D = 180 - 60 D = 120$
Now, $AD \| BC$ and $AB$ is the transversal $A + B = 180$ [Co-interior angles are supplementary]
$60 + B = 180 B = 180 - 60 = 120$
Hence, $\angle\text{A}=\angle\text{C}=60^\circ$ and $\angle\text{B}=\angle\text{D}=120^\circ$

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Question 174 Marks

If two straight lines intersect each other, prove that the ray opposite to the bisector of one of the angles thus formed bisects the vertically opposite angle.

Answer

Given: $AB$ and $CD$ intersect each other at $O$.
$OE$ bisects $\angle\text{COB}$
To prove: $\angle\text{AOF}=\angle\text{DOF}$
Proof: Let $\angle\text{COE}=\angle\text{EOB}=\text{x}$ $[\because$ OE bisects $\angle\text{COB}]$ $\angle\text{COE}=\angle\text{DOF}=\text{x}\dots(1)$ [vertically opposite angles] $\angle\text{BOE}=\angle\text{AOF}=\text{x}\dots(2)$ [vertically opposite angles]
From $(1)$ and $(2)$
$\angle\text{AOF}=\angle\text{DOF}=\text{x}$

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Question 184 Marks

Two unequal angles of a parallelogram are in the ratio $2 : 3$. Find all its angles in degrees.

Answer

Let $A = 2x$ and $B = 3x$
Now, $A + B = 180$ [Co-interior angles are supplementary] $2x + 3x - 180$ [$AD \| BC$ and $AB$ is the transversal]
$\Rightarrow 5x = 180 \text{x}=\frac{180}{5} x = 36$
Therefore, $A = 2 \times 36 = 72 B = 3 \times 36 = 108$
Now, $A = C = 72$ [Opposite side angles of a parallelogram are equal] $B = D = 108$

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Question 194 Marks

In figure, $OP, OQ, OR$ and $OS$ are four rays. Prove that: $\angle\text{POQ}+\angle\text{QOR}+\angle\text{SOR}+\angle\text{POS}=360^\circ$

Answer

Given that $OP, OQ, OR$ and $OS$ are four ray.
You need to produce any of the ray $OP, OQ, OR$ and $OS$ backwards to a point in the figure.
Let us produce ray $OQ$ backwards to a point $T.$ So that $TOQ$ is a line.
Ray $OP$ stands on the $TOQ$ Since $\angle\text{TOP},\angle\text{POQ}$ is a linear pair $\angle\text{TOP}+\angle\text{POQ}=180^\circ\dots(1)$ Similarly, Ray $OS$ stands on the line $TOQ$ $\angle\text{TOS}+\angle\text{SOQ}=180^\circ\dots(2)$ But $\angle\text{SOQ}=\angle\text{SOR}+\angle\text{QOR}\dots{(3)}$
So, eq. $(2)$ becomes $\angle\text{TOS}+\angle\text{SOR}+\angle\text{OQR}=180^\circ$
Now, adding $(1)$ and $(3)$ you get: $\angle\text{TOP}+\angle\text{POQ}+\angle\text{TOS}+\angle\text{SOR}+\angle\text{QOR}=360^\circ$
$\angle\text{TOP}+\angle\text{TOS}=\angle\text{POS}$ Equation $(4)$ becomes $\angle\text{POQ}+\angle\text{QOR}+\angle\text{SOR}+\angle\text{POS}=360^\circ$

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Question 204 Marks

In figure, ray $OS$ stand on a line $POQ$. Ray $OR$ and ray $OT$ are angle bisectors of $\angle\text{POS}$ and $\angle\text{SOQ}$ respectively. If $\angle\text{POS}=\text{x},$ find $\angle\text{ROT}.$

Answer

Given, Ray $OS$ stand on a line $POQ$ Ray $OR$ and Ray $OT$ are angle bisectors of $\angle\text{POS}$ and $\angle\text{SOQ}$ respectively. $\angle\text{POS}=\text{x}$
$\angle\text{POS}$ and $\angle\text{SOQ}$ is linear pair $\angle\text{POS}+\angle\text{QOS}=180^\circ$
$x + QOS = 180 QOS = 180 - x$
Now, ray or bisector $POS$
$\angle\text{ROS}=\frac{1}{2}\angle\text{POS}$
$=\frac{\text{x}}{2}$
$\text{ROS}=\frac{\text{x}}{2}$ [Since $POS = x$] Similarly ray $OT$ bisector $QOS$
$\angle\text{TOS}=\frac{1}{2}\angle\text{QOS}$
$=\frac{(180-\text{x})}{2}$ [$QOS = 180 - x$] $=90-\frac{\text{x}}{2}$
Hence, $\angle\text{ROT}=\angle\text{ROS}+\angle\text{ROT}$
$=\frac{\text{x}}{2}+90-\frac{\text{x}}{2}$
$=90$
$\angle\text{ROT}=180^\circ$

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Question 214 Marks

In figure $AB \| CD$ and $\angle\text{1}$ and $\angle\text{2}$ are in the ratio $3 : 2$. Determine all angles from $1$ to $8.$

Answer

Let: $\angle\text{1}=3\text{x}$ and $\angle\text{2}=2\text{x}$
$\angle\text{1}$ and $\angle\text{2}$ are linear pair of angle.
Now, $\angle\text{1}$ and $\angle\text{2}$
$\Rightarrow\ 3\text{x}+2\text{x}=180$
$\Rightarrow\ 5\text{x}=180$
$\Rightarrow\ \text{x}=\frac{180}{5}$
$\Rightarrow\ \text{x}=36$
$\angle\text{1}=3\text{x}=108^\circ,\angle\text{2}=2\text{x}=72^\circ$
We know, Vertically opposite angles are equal $\angle\text{1}=\angle\text{3}=108^\circ$
$\angle\text{2}=\angle\text{4}=72^\circ$
$\angle\text{6}=\angle\text{7}=108^\circ$
$\angle\text{5}=\angle\text{8}=72^\circ$
We also know, corresponding angles are equal $\angle\text{1}=\angle\text{5}=108^\circ$
$\angle\text{2}=\angle\text{6}=72^\circ$

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4 Marks Questions - Maths STD 9 Questions - Vidyadip