2 Marks Questions

Question 12 Marks

In figure, if $PQ \ || \ ST, \angle PQR = 110^ \circ$ and $\angle RST = 130 ^\circ.$ Find $\angle QRS.$

Answer

Draw a line $RU$ parallel to $ST$ through point $R.$

$\angle RST + \angle SRU = 180^\circ$
$\therefore 130^\circ + \angle SRU = 180^\circ$
$ \therefore \angle SRU = 180^\circ – 130^\circ = 50^\circ . . . (1)$
$\angle QRU = \angle PQR = 110^\circ . . . . [$Alternate interior angles$]$
$\therefore \angle QRS + \angle SRU = 110^\circ$
$\therefore \angle QRS + 50^\circ = 110^\circ . . . [$Using $(1)]$
$\therefore \angle QRS = 110^\circ – 50^\circ = 60^\circ$

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Question 22 Marks

In figure, if $AB \ || \ CD, EF ⊥ CD$ and $\angle GED = 126^\circ.$ Find $\angle AGE, \angle GEF$ and $\angle FGE.$

Answer

$\angle AGE = \angle GED = 126^\circ . . . [$Alternate interior angles$]$
$\angle GED = 126^\circ$
$\therefore \angle GEF + \angle FED = 126^\circ$
$\therefore \angle GEF + 90^\circ = 126^\circ . . . [$As $EF \perp CD \therefore \angle FED = 90^\circ]$
$\therefore \angle GEF = 126^\circ – 90^\circ = 36^\circ$
$\angle GEC + \angle GEF + \angle FED = 180^\circ$
$\angle GEC + 36^\circ + 90^\circ = 180^\circ$
$\angle GEC + 126^\circ = 180^\circ$
$\angle GEC = 180^\circ– 126^\circ = 54^\circ$
$\angle FGE = \angle GEC = 54^\circ . . . . [$Alternate interior angles$]$

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Question 32 Marks

In figure, if $AB \ || \ CD, CD \ || \ EF$ and $y : z = 3 : 7,$ find $x.$

Answer

As $AB \ || \ CD$ and $CD \ || \ EF$
$\therefore AB \ || \ EF$
$\therefore x = z . . . . [$Alternate interior angles$] . . . (1)$
$x + y = 180^\circ . . . (2)$
$z + y = 180^\circ . . . [$From $(1)$ and $(2)]$
$y : z = 3 : 7$
Sum of the ratios $= 3 + 7 = 10$
$\therefore y={3\over10}\times180^0=54^0$ and $ z={7\over10}\times180^0=126^0$
$\therefore$ $x = z = 126^\circ$

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Question 42 Marks

In figure, find the values of $x$ and $y$ and then show that $AB \ || \ CD.$

Answer

$\angle AEG + \angle AEH = 180^\circ . . . [$Linear pair$]$
$\therefore 50^\circ + x = 180^\circ$
$\therefore x = 180^\circ – 50^\circ = 130^\circ . . . (1)$
$y = 130^\circ . . . [$Vertically opposite angles$] . . . (2)$
$x = y . . . . [$From $(1)$ and $(2)]$
But $x$ and $y$ are alternate interior angles and they are equal.
$\therefore AB \ || \ CD$

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Question 52 Marks

If the $\angle XYZ = {64^\circ }$ and $XY$ is produced to point $P.$ Draw a figure from the given information. If ray $YQ$ bisects $\angle ZYP, $ find $\angle XYQ$ and reflex $\angle QYP.$

Answer

We are given that $\angle XYZ = 64^\circ, XY$ is produced to $P$ and $YQ$ bisects $\angle ZYP$ We can conclude the given below figure for the given situation:

We need to find $\angle XYQ$ and reflex $\angle QYP$
From the given figure, we can conclude that $\angle XYZ$ and $\angle ZYP$ form a linear pair.
We know that sum of the angles of a linear pair is $180^\circ.$
$\angle XYZ + \angle ZYP = 180^\circ$
But $\angle XYZ = 64^\circ$
$\Rightarrow 64^\circ + \angle ZYP = 180^\circ$
$\Rightarrow \angle ZYP = 116^\circ$
Ray $YQ$ bisects $ \angle ZYP,or$
$\angle QYZ = \angle QYP = \frac{116^\circ}{2} = 58^\circ$
$\angle XYQ = \angle QYZ + \angle XYZ$
$= 58^\circ + 64^\circ = 122^\circ.$
Reflex $\angle QYP = {360^\circ } - \angle QYP$
$= 360^\circ - 58^0$
$= 302^\circ.$
Therefore, we can conclude that $\angle XYQ = 122^\circ$ and Reflex $\angle QYP = 302^\circ$

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Question 62 Marks

In the given figure, $\angle$PQR = $\angle$PRQ, then prove that $\angle$PQS = $\angle$PRT

Answer

We need to prove that $\angle$PQS = $\angle$PRT
We are given that $\angle$PQR = $\angle$PRQ
From the given figure, we can conclude that $\angle$PQS and $\angle$PQR, and $\angle$PRQ and $\angle$PRT form a linear pair.
We know that sum of the angles of a linear pair is ${180^\circ }$
$\therefore \angle PQS + \angle PQR = {180^\circ },$ and ...(i)
$\angle PRQ + \angle PRT = {180^\circ }.$ ...(ii)
From equation (i) and (ii), we can conclude that
$\angle PQS + \angle PQR = \angle PRQ + \angle PRT.$
But, $\angle$PQR = $\angle$PRQ
$\therefore$ $\angle$PQS = $\angle$PRT
Hence, proved.

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Question 72 Marks

Fig.,$ AB || CD$ and $CD || EF$. Also$, EA \perp AB.$ If $\angle BEF = 55^\circ ,$ find the values of $x, y,$ and $z$.

Answer

Since corresponding angles are equal.
$\therefore x = y ... (i)$
We know that the interior angles on the same side of the transversal are supplementary.
$\therefore y + 55^\circ = 180^\circ$
$\Rightarrow y = 180^\circ - 55^\circ = 125^\circ$
So$, x = y = 125^\circ$
Since $AB || CD$ and $CD || EF.$
$\therefore AB || EF$
$\Rightarrow \angle EAB + \angle FEA = 180^\circ [\because$ Interior angles on the same side of the transversal $EA$ are supplementary$]$
$\Rightarrow90^\circ + z + 55^\circ = 180^\circ$
$\Rightarrow z = 35^\circ$

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Question 82 Marks

If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.

Answer

Given AD is transversal intersect two lines PQ and RS
To prove PQ $\parallel$ RS
Proof: BE bisects ABQ
$\angle 1=\angle A B E=\angle E B Q=\frac{1}{2} \angle A B Q$ ...(i)
Similarity CG bisects $\angle$ BCS
$\therefore \angle 2=\frac{1}{2} \angle B C S$ ...(ii)
But BE $\parallel$ CG and AD is the transversal
$\therefore {\text{ }}\angle {\text{1 = }}\angle {\text{2}}$
$\therefore {\text{ }}\frac{1}{2}\angle ABQ = \frac{1}{2}\angle BCS$ [by (i) and (ii)]
$\Rightarrow {\text{ }}\angle {\text{ABQ = }}\angle {\text{BCS}}$ [$\because$ corresponding angles are equal]
$\therefore$ PQ $\parallel$ RS

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Question 92 Marks

In Fig.,if $PQ || RS,\angle MXQ = 135^\circ$ and $\angle MYR = 40^\circ ,$ find $\angle XMY.$

Answer

Through point $M$ draw a line $AB$ parallel to the line $PQ$ as shown in Fig. Thus, we have

$AB || PQ$ and $PQ || RS$
$\Rightarrow AB || RS$
Now, $AB || PQ$ and $\angle QXM$ and $\angle XMB$ are interior angles on the same side of the transversal $XM.$
$\therefore \angle QXM + \angle XMB = 180^\circ$
$\Rightarrow 135^\circ + \angle XMB = 180^\circ$
$\Rightarrow \angle XMB = 180^\circ - 135^\circ = 45^\circ$
Now, $AB || RS$ and $\angle BMY$and $\angle MYR$ are alternate angles.
$\therefore \angle BMY = \angle MYR$
$\Rightarrow \angle BMY = 40^\circ$
Hence, $\angle XMY = \angle XMB + \angle BMY = 45^\circ + 40^\circ = 85^\circ$

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Question 102 Marks

In figure ray OS stands on a line POQ, ray OR and ray OT are angle bisector of $\angle$POS and $\angle$SOQ respectively. If $\angle$POS = x, find $\angle$ROT.

Answer

Ray OS stands on the line POQ
$ \therefore {\text{ }}\angle {\text{POS + }}\angle {\text{SOQ = 18}}{{\text{0}}^0}$
But $\angle$PQS = X
$\therefore$ x + $\angle$SOQ = 180°
$\angle$SOQ = 180° - x
Now ray OR bisects $\angle$POS,
Therefore $ \angle {\text{ROS = }}\frac{1}{2} \times \angle POS$ $ {\text{ = }}\frac{1}{2} \times x = \frac{x}{2}$
Similarly, $ \angle {\text{ SOT = }}\frac{1}{2} \times \angle SOQ$ $ {\text{ = }}\frac{1}{2} \times ({180^0} - x) = {90^ \circ } - \frac{x}{2}$
$ \angle ROT = \angle ROS + \angle SOT$ $ {\text{ = }}\frac{x}{2} + {90^ \circ } - \frac{x}{2} = {90^0}$

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Question 112 Marks

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

Answer

$\angle POR + \angle ROQ = {180^0}$ [linear pair]
But,$ \angle {\text{POR: }}\angle ROQ$ = 5:7 [Given]
$\therefore {\text{ }}\angle {\text{ POR = }}\frac{5}{{12}} \times {\text{18}}{{\text{0}}^0}{\text{ = 7}}{{\text{5}}^0}{\text{ }}$
Similarly, $\angle {\text{ROQ = }}\frac{7}{{12}} \times {\text{18}}{{\text{0}}^0}{\text{ = 10}}{{\text{5}}^0}$
Now $\angle {\text{POS = }}\angle {\text{ROQ = 10}}{{\text{5}}^0}$ [vertically opposite angle]
And $\angle {\text{SOQ = }}\angle {\text{POR = 7}}{{\text{5}}^0}$ [vertically opposite angle]

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