Grade 9 – Math – Lines and Angles – Important Questions and Answers

    Question 1: In the figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that,

    \[\angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS})\]


    $\begin{array}{ll} \angle \mathrm{ROS}=\angle \mathrm{ROP}-\angle \mathrm{POS} & \ldots(1) \\ \angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{QOR} & \ldots(2)\end{array}$

    Adding (1) and (2),

    $\angle \mathrm{ROS}+\angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{QOR}+\angle \mathrm{ROP}-\angle \mathrm{POS}$
    $\Rightarrow 2 \angle \mathrm{ROS}=\angle \mathrm{QOS}-\angle \mathrm{POS}\hspace{1cm}(\because \angle \mathrm{QOR}=\angle \mathrm{ROP}=90^{\circ})$
    $\Rightarrow \angle \mathrm{ROS}=\frac{1}{2}(\angle \mathrm{QOS}-\angle \mathrm{POS}) \quad$ Proved.

    Question 2: It is given that $\angle X Y Z=64^{\circ}$ and $X Y$ is produced to a point P. Draw a figure from the given information. If ray $Y Q$ bisects $\angle$ ZYP, find $\angle X Y Q$ and reflex $\angle Q Y P$ .

    Answer. From figure,$\angle \text{XYZ}=64^{\circ} \quad[\text { Given }]$
    Now,ZYP+XYZ=180 [Linear pair axiom] ZYP+64=180ZYP=18064=116

    Question 3. Let $O A, O B, O C$ and $O D$ are rays in the anti-clockwise direction such that $\angle A O B=\angle C O D=100^{\circ}, \angle B O C=82^{\circ}$ and $\angle A O D=78^{\circ}$ Is it true to say that AOC and BOD are lines?

    Answer. $\angle \mathrm{AOB}+\angle \mathrm{BOC}=100^{\circ}+82^{\circ}=182^{\circ} \neq 180^{\circ} $
    Hence, AOC is not a line.
    $\angle \mathrm{BOC}+\angle \mathrm{COD}=82^{\circ}+100^{\circ}=182^{\circ} \neq 180^{\circ}$
    Hence, BOD is not a line.

    Question 4: In the given figure, $\angle{AOC}$ and $\angle{BOC}$ form a linear pair. If $a – b = 80°$, find the values of $a$ and $b$.

    We have, $a – b = 80°$ [Given]
    Also, $a + b = 180°$ [Linear pair]
    $\Rightarrow 2a = 260° $
    $\Rightarrow a = 130° $ and
    $b = 180° – 130° = 50°$.
    So, $a$ and $b$ are 130° and 50° respectively.

    Question 5:. In the given figure, if $\angle C O D=90^{\circ},$ then find$\angle A O C$ and $\angle B O D .$

    Answer:. since $\mathrm{AOB}$ is a line, therefore, $\angle \mathrm{AOB}+\angle \mathrm{COD}+\angle \mathrm{BOD}=180^{\circ}$
    $\Rightarrow 2 x+90^{\circ}+x=180^{\circ} \Rightarrow 3 x=90^{\circ} \Rightarrow x=30^{\circ}$
    $\therefore 2 x=2 \times 30^{\circ}=60^{\circ}$
    $\therefore \angle \mathrm{AOC}=60^{\circ}$ and $\angle \mathrm{BOD}=30^{\circ}$

    Question 5:. In the given figure, ray $O E$ bisects $\angle A O B$ and $O F$ is a ray opposite to OE. Show that $\angle F O B=\angle F O A$

    Since ray OE bisects $\angle \mathrm{AOB}$
    $ \therefore\angle \mathrm{EOB}+\angle \mathrm{EOA}\ldots$(i)
    Now, ray OB stands on the line EF.
    $\therefore \angle \mathrm{EOB}+\angle \mathrm{FOB}=180^{\circ} \ldots$ (ii) $[$ Linear pair $]$
    Again, ray OA stands on the line EF.
    $\therefore \angle \mathrm{EOA}+\angle \mathrm{FOA}=180^{\circ}\ldots$(iii)
    From (ii) and (iii), we get
    $ \angle \mathrm{EOB}+\angle \mathrm{FOB}=\angle \mathrm{EOA}+\angle \mathrm{FOA}$
    $\Rightarrow \angle \mathrm{EOA}+\angle \mathrm{FOB}=\angle \mathrm{EOA}+\angle \mathrm{FOA}\hspace{1cm}[\because \angle \mathrm{EOB}=\angle \mathrm{EOA} \text { from }\text{(ii)}]$
    $\Rightarrow \angle \mathrm{FOB}=\angle \mathrm{FOA}$

    Question 6: If ray $O C$ stands on line $A B$ such that $\angle A O C=\angle C O B,$ then show that $\angle A O C=90^{\circ}$

    Answer. since ray $\mathrm{OC}$ stands on line $\mathrm{AB}$. Therefore, $\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ} \quad[$ Linear pair $]$
    But $\angle \mathrm{AOC}=\angle \mathrm{COB}\hspace{0.5cm}$[Given]
    $\therefore \angle \mathrm{AOC}+\angle \mathrm{AOC}=180^{\circ}$
    $\Rightarrow 2 \angle \mathrm{AOC}=180^{\circ}$
    $\Rightarrow \angle \mathrm{AOC}=90^{\circ}$

    Question 7: In the figure, ray OX bisects $\angle \mathrm{AOC}$ and ray OY bisects $\angle B O C$. If $\angle \mathrm{XOY}=90^{\circ}$, show that A, $O, B$ are collinear.

    Answer. We have, $\angle \mathrm{AOC}=2 \angle \mathrm{XOC}\hspace{1cm}\ldots$(i)
    $[\because$ ray $\mathrm{OX}$ bisects $\angle \mathrm{AOC}]$
    $\mathrm{And}, \angle \mathrm{BOC}=2 \angle \mathrm{YOC}\hspace{1cm}\ldots$(ii)
    $[\because \text { ray OY bisects } \angle \mathrm{BOC}]$
    Adding, (i) and (ii) we get
    \angle \mathrm{AOC}+\angle \mathrm{BOC}=2 \angle \mathrm{XOC}+2 \angle \mathrm{YOC}\\
    =2(\angle \mathrm{XOC}+\angle \mathrm{YOC}) \\
    =2 \angle \mathrm{XOY}\\=2 \times 90^{\circ} &\left(\because \angle \mathrm{XOY}=90^{\circ}\right) \\
    = 180^{\circ}
    Therefore, $\angle \mathrm{AOC}$ and $\angle \mathrm{BOC}$ form a linear pair. Consequently OA and OB are two opposite rays. Hence, $\mathrm{A}, \mathrm{O}, \mathrm{B}$ are collinear.

    Question 8: If two lines intersect, prove that the vertically opposite angles are equal.

    Given : Two lines AB and CD intersect at O
    To prove :
    (i) $\angle \mathrm{AOC}=\angle \mathrm{BOD}$
    (ii) $\angle \mathrm{AOD}=\angle \mathrm{BOC}$

    Proof : since ray OC stands on the line AB, we have
    $\angle \mathrm{AOC}+\angle \mathrm{COB}=180^{\circ} \quad \ldots$ (i) $[$ Linear pair $]$
    since, ray OA stands on the line $\mathrm{CD}$, we have
    $\angle \mathrm{AOC}+\angle \mathrm{AOD}=180^{\circ}\hspace{1cm} \ldots$(ii) [Linear pair]
    From (i) and (ii), we have $\angle \mathrm{AOC}+\angle \mathrm{COB}=\angle \mathrm{AOC}+\angle \mathrm{AOD}$
    $\Rightarrow \angle \mathrm{COB}=\angle \mathrm{AOD}$
    Similarly, $\angle \mathrm{AOC}=\angle \mathrm{BOD}$.

    Question 9: In the given figure, two straight lines PQ and RS intersect each other at O. If $\angle{POT} = 75° $. Find the values of b and c.

    Answer. since $O R$ and $O S$ are in the same line.
    Therefore, $\angle \mathrm{ROP}+\angle \mathrm{POT}+\angle \mathrm{TOS}=180^{\circ}$
    $\Rightarrow 4 b+75^{\circ}+b=180^{\circ}$
    $\Rightarrow 5 b+75^{\circ}=180^{\circ}$
    $\Rightarrow 5 b=105^{\circ}$
    $\Rightarrow b=21$
    since, $P Q$ and $R S$ intersect at $O$. Therefore, $\angle \mathrm{QOS}=\angle \mathrm{POR} \quad$ [Vertically opp. angles]
    $\Rightarrow a=4 a$
    $\Rightarrow a=4 \times 21=84 \quad[\therefore b=21]$
    Now, OR and OS are in the same line. Therefore, $\angle \mathrm{ROQ}+\angle \mathrm{QOS}=180^{\circ}$
    $\Rightarrow 2 c+a=180^{\circ}$
    $\Rightarrow 2 c+84=180^{\circ}$
    $\Rightarrow 2 c=96$
    $[\because a=84]$
    $\Rightarrow c=48$
    Hence, $b=21$ and $c=48$

    Question 10: In the given figure, OP bisects $\angle \mathrm{BOC}$ and $O Q$ bisects $\angle \mathrm{A} O C$. Show that $\angle P O Q=90^{\circ}$

    Answer: since OP bisects $\angle \mathrm{BOC}$
    $\therefore \angle \mathrm{BOC}=2 \angle \mathrm{POC}\hspace{1cm}\ldots$(i)
    Again, $\mathrm{OQ}$ bisects $\angle \mathrm{AOC}$
    $\therefore \angle \mathrm{AOC}=2 \angle \mathrm{QOC}$
    since ray $\mathrm{OC}$ stands on line $\mathrm{AB}$.
    $\therefore \angle \mathrm{AOC}+\angle \mathrm{BOC}=180^{\circ}\hspace{1cm}\ldots$(i)
    $\Rightarrow 2 \angle \mathrm{Q} \mathrm{OC}+2 \angle \mathrm{POC}=180^{\circ}[\mathrm{Using}(\mathrm{i})$ and $(\mathrm{ii})]$
    $\Rightarrow 2(\angle \mathrm{POC}+\angle \mathrm{QOC})=180^{\circ}$
    $\Rightarrow \angle \mathrm{POC}+\angle \mathrm{QOC}=90^{\circ}$
    $\Rightarrow \angle P O Q=90^{\circ}, \quad$ Proved.

    Question 11: In the figure, find the values of x and y and then show that AB || CD.

    In the given figure, a transversal intersects two lines AB and CD such that x + 50° = 180° [Linear pair axiom]
    $\Rightarrow \text{x} = 180° – 50° = 130°$
    y = 130° [Vertically opposite angles]
    Therefore, x = y = 130° [Alternate angles]
    ∴ AB || CD [Converse of alternate angles axiom]

    Question 12:. In the figure, if ${A B}|| {C D}, C D || EF $and $y: z=3: 7,$ find $x .$

    Answer: In the given figure, $\mathrm{AB}||\mathrm{CD}, \mathrm{CD}|| \mathrm{EF}$ and $y: z=3: 7$
    Let $y=3 a$ and $z=7 a$
    $\angle \mathrm{DHI}=y \quad$ [Vertically opposite angles $]$
    $\angle \mathrm{DHI}+\angle \mathrm{FIH}=180^{\circ}[$ Interior angles on the same side of the transversal]
    \Rightarrow \quad y+z=180^{\circ} & \\
    \Rightarrow 3 a+7 a=180^{\circ} \\
    \Rightarrow \quad 10 a=180^{\circ} \Rightarrow a=18^{\circ}
    Also, $\quad x+y=180^{\circ}$
    $\Rightarrow \quad x+54^{\circ}=180^{\circ}$
    $\therefore \quad x=180^{\circ}-54^{\circ}=126^{\circ}$

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