2 Marks Questions

Question 12 Marks

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

Answer

Draw a line RU parallel to ST through point R.

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

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Question 22 Marks

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

Answer

$\angle$AGE = $\angle$GED = 126^o . . . [Alternate interior angles]
$\angle$GED = 126^o
^{$\therefore$ $\angle$}GEF + $\angle$FED = 126^o
^{$\therefore$ $\angle$}GEF + 90^o = 126^o . . . [As EF ⊥ CD $\therefore$ $\angle$FED = 90^o]
^{^$\therefore$ $\angle$}GEF = 126^o – 90^o = 36^o
$\angle$GEC + $\angle$GEF + $\angle$FED = 180^o
$\angle$GEC + 36^o + 90^o = 180^o
$\angle$GEC + 126^o = 180^o
$\angle$GEC = 180^o – 126^o = 54^o
$\angle$FGE = $\angle$GEC = 54^o . . . . [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^o . . . (2)
z + y = 180^o . . . [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^o

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Question 42 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^o, 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^o.
$\angle $XYZ + $\angle $ZYP = 180^o
But $\angle $XYZ = 64^o
$\Rightarrow$ 64^o + $\angle $ZYP = 180^o
$\Rightarrow$ $\angle $ZYP = 116^o
Ray YQ bisects $\angle $ZYP,or
$\angle $QYZ = $\angle $QYP = $\frac{116^o}{2}$ = 58^o
$\angle $XYQ = $\angle $QYZ + $\angle $XYZ
= 58^o + 64⁰ = 122^o.
Reflex $\angle $QYP = ${360^\circ }$ - $\angle $QYP
= 360⁰ - 58⁰
= 302⁰.
Therefore, we can conclude that $\angle $XYQ = 122^oand Reflex $\angle $QYP = 302^o

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Question 52 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 62 Marks

Fig., AB || CD and CD || EF. Also, EA $\perp$ AB. If $\angle$BEF = 55°, 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^o = 180^o
$\Rightarrow$ y = 180^o - 55^o = 125^o
So, x = y = 125^o
Since AB || CD and CD || EF.
$\therefore$ AB || EF
$\Rightarrow$ $\angle$EAB + $\angle$FEA = 180^o [$\because$ Interior angles on the same side of the transversal EA are supplementary]
$\Rightarrow$ 90^o + z + 55^o = 180^o
$\Rightarrow$ z = 35^o

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Question 72 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 82 Marks

In Fig.,if PQ || RS, $\angle$MXQ = 135° and $\angle$MYR = 40°, 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^o
$\Rightarrow$ 135^o + $\angle$XMB = 180^o
$\Rightarrow$ $\angle$XMB = 180^o - 135^o = 45^o
Now, AB || RS and $\angle$BMY and $\angle$MYR are alternate angles.
$\therefore$ $\angle$BMY = $\angle$MYR
$\Rightarrow$ $\angle$BMY = 40^o
Hence, $\angle$XMY = $\angle$XMB + $\angle$BMY = 45^o + 40^o = 85^o

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Question 92 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 102 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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