4 Marks Questions

Question 14 Marks

In Fig. $\text{BA}||\text{ED}$ and $\text{BC}||\text{EF}$ . Show that $\angle\text{ABC}=\angle\text{DEF}$
[Hint: Produce DE to intersect BC at P (say)].

Answer

Produce DE to intersect BC at P(say). $\text{EF}||\text{BC}$ and DP is the transversal,

$\therefore\angle\text{DEF}=\angle\text{DPC}.....(1)$ $[\text{Corres}.\angle\text{S}]$
Now, $\text{AB}||\text{DP}$ and BC is the transversal, $\therefore\angle\text{DPC}=\angle\text{ABC}..(2)$ $[\text{Corres}.\angle\text{S}]$ From (1) and (2), we get
$\angle\text{ABC}=\angle\text{DEF}$ Hence, Proved.

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Question 24 Marks

In Fig. OD is the bisector of $\angle\text{AOC},$ OE is the bisector of $\angle\text{BOC}$ and $\text{OD}\perp\text{OE}.$ Show that the points A, O and B are collinear.

Answer

Given: In figure, $\text{OD}\perp\text{OE}$ $(\text{i.e.}\angle\text{DOE}=90^\circ),$ OD and OE are the bisector of $\angle\text{AOC}$ and $\angle\text{BOC}.$
To prove: points A, O and B are collinear i.e., AOB is a straight line.
Proof: Since, OD and OE bisect angles $\angle\text{AOC}$ and $\angle\text{BOC}$ respectively.
$\therefore\ \angle\text{AOC}=2\angle\text{DOC}...(1)$
And $\angle\text{COB}=2\angle\text{COE}...(2)$
On adding euations (1) and (2), we get
$\angle\text{AOC}+\angle\text{COB}=2\angle\text{DOC}+\angle\text{COE}$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=2\big(\angle\text{DOC}+\angle\text{COE}\big)$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=2\angle\text{DOE}$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=2\times90^\circ$ $\big[\because\text{OD}\perp\text{OE}\big]$
$\Rightarrow\angle\text{AOC}+\angle\text{COB}=180^\circ$
$\therefore\ \angle\text{AOB}=180^\circ$
So, $\angle\text{AOC}+\angle\text{COB}$ are forming linear pair or AOB is a straight line. Hence, points A, O and B are collinear.

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Question 34 Marks

In Fig.$\text{BA}||\text{ED}$ and $\text{BC}||\text{EF}$ Show that $\angle\text{ABC}+\angle\text{DEF}=180^\circ$

Answer

Produce ED to meet BC at P(say)

Now, $\text{EF}||\text{BC}$ and EP is the transversal. $\angle\text{DEF}+\angle\text{EPC}=180^\circ....(1)$ Again, $\text{EP}||\text{AB}$ and BC is the transversal. $\therefore\angle\text{EPC}=\angle\text{ABC}....(2)$ $[\text{corresponding}\angle\text{S}]$ From (1) and (2), we get $\angle\text{DEF}+\angle\text{ABC}=180^\circ$ $\Rightarrow\angle\text{ABC}+\angle\text{DEF}=180^\circ$ Hence, proved.

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Question 44 Marks

Prove that through a given point, we can draw only one perpendicular to a given line.
[Hint: Use proof by contradiction].

Answer

From the point P, a perpendicular PM is drawn to the given line AB. $\therefore\ \angle\text{PMB}=90^\circ$ Let if possible, we can draw another perpendicular PN to the line AB. then, $\angle\text{PMB}=90^\circ$ $\angle\text{PMB}=\angle\text{PNB},$ which is possible only when PM and PN coincide with each other.

Hence, through a given point, we can draw only one perpendicular to a given line.

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Question 54 Marks

In Fig. $\text{DE}||\text{QR}$ and AP and BP are bisectors of $\angle\text{EAB}$ and respectively. Find $\angle\text{APB}$

Answer

Given, $\text{DE}||\text{QR}$ and AP and PB are the bisectors of $\angle\text{EAB}$ and $\angle\text{RBA},$ respectively. We know that, the interior angles on the same side of transversal are supplementary.
$\therefore\ \angle\text{EAB}+\angle\text{RBA}=180^\circ$
$\Rightarrow\frac{1}{2}\angle\text{EAB}+\frac{1}{2}\angle\text{RBA}=\frac{180^\circ}{2}$ [dividing both sides by 2]
$\frac{1}{2}\angle\text{EAB}+\frac{1}{2}\angle\text{RBA}=90^\circ....(\text{i})$
Since, AP and BP are the bisector of $\angle\text{EAB}$ and $\angle\text{RBA},$ respectively.
$\therefore\ \angle\text{BAP}=\frac{1}{2}\angle\text{EAB}....(\text{ii})$
and $\angle\text{ABP}=\frac{1}{2}\angle\text{RBA}....(\text{iii})$
On adding Eqs. (ii) and (iii), we get
$\angle\text{BAP}+\angle\text{ABP}=\frac{1}{2}\angle\text{EAB}+\frac{1}{2}\angle\text{RBA}$
From Eq. (i)
$\Rightarrow\angle\text{BAP}+\angle\text{ABP}=90^\circ..(\text{iv})$
In $\Delta\text{APB},$ $\angle\text{BAP}+\angle\text{ABP}+\angle\text{APB}=180^\circ$ [sum of all angles of a triangle is 180°]
$\Rightarrow90^\circ+\angle\text{APB}=180^\circ$[from Eq.(iv)]
$\Rightarrow\angle\text{APB}=180^\circ-90^\circ=90^\circ$

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Question 64 Marks

AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (Fig.). Show that $\text{AP}||\text{BQ}.$

Answer

Given In the figure $\text{l}||\text{m},$ AP and BQ are the bisectors of $\angle\text{EAB}$ and $\angle\text{ABH},$ respectively. To prove $\text{AP}||\text{BQ}$ Proof Since, $\text{l}||\text{m},$ and t is transversal. Therefore, $\angle\text{EAB}=\angle\text{ABH}$ [alternate interior angles]

$\frac{1}{2}\angle\text{EAB}=\frac{1}{2}\text{ABH}$ [dividing both sides by 2] $\angle\text{PAB}=\angle\text{ABQ}$ $\big[$AP and BQ are the bisectors of $\angle\text{EAB}$ and $\angle\text{ABH}\big]$ Since, $\angle\text{PAB}$ and $\angle\text{ABQ}$ are alternate interior angles with two lines AP and BQ and transvetrsal AB. Hence, $\text{AP}||\text{BQ}.$

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Question 74 Marks

If in Fig. bisectors AP and BQ of the alternate interior angles are parallel, then show that $\text{l}||\text{m}.$

Answer

Given, In the figure $\text{AP}||\text{BQ},$ AP and BQ are the bisectors of alternate interior angles $\angle\text{CAB}$ and $\angle\text{ABF}.$ To show $\text{l}||\text{m}$ Proof Since,$\text{AP}||\text{BQ}$ and t is transversal, therefore $\angle\text{PAB}=\angle\text{ABQ}$ [alternate interior angles] $=>2\angle\text{PAB}=2\angle\text{ABQ}$ [multiplying both sides by 2]

So, alternate interior angles are equal.
We know that, if two alternate interior angles are equal, then lines are parallel Hence, $\text{l}||\text{m}.$

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Question 84 Marks

Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other.

Answer

Two lines p and n are respectively perpendicular to two parallel line l and m, i.e., $\text{p}\perp\text{l}$ and $\text{n}\perp\text{m}$
We have to show that p is parallel to n.
As $\text{n}\perp\text{m},$ So $\angle1=90^\circ....(1)$
Again, $\text{p}\perp\text{l},$ So $\angle2=90^\circ.$
But, l is parallel to m, so
$\angle1=\angle3$ $[\text{corres}.\angle\text{s}]$
$\therefore\angle2=\angle90^\circ...(2)$ $[\because\angle2=90^\circ]$
From (1) and (2), we get
$\Rightarrow\angle1=\angle3$ $[\text{Each}=90^\circ]$
angles.
Hence, p||n.
But, these are corresponding.

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