3 Marks Question

Question 13 Marks

In each of the given figures, two lines/ and m are cut by a transevrsal t. Find whether $\text{l || m}.$

Answer

$\angle2+\angle3=180^\circ$ (linear pair)
$35^\circ+\angle3=108^\circ$
$\angle3=145^\circ=145^\circ=\angle1$
$\therefore\ \text{l}\neq\text{m}$

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Question 23 Marks

In each of the given figures, two lines/ and m are cut by a transevrsal t. Find whether $\text{l || m}.$

Answer

$\angle1+\angle2=180^\circ$ (linear pair)
$130^\circ+\angle2=180^\circ$
$\angle2=50^\circ\neq40^\circ=\angle3$
$\therefore\ \text{l}\neq\text{m}$

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Question 33 Marks

In the given figure, $AB \| DC$ and $AD \| BC$, and $AC$ is a diagonal. If $\angle\text{BAC}=35^\circ,\ \angle\text{CAD}=40^\circ,\ \angle\text{ACB}=\text{x}^\circ$ and $\angle\text{ACD}=\text{y}^\circ,$ find the values of $x$ and $y.$

Answer

Given: $AB \| DC AD \| BC$
$\angle\text{BAC}=35^\circ$ $\angle\text{CAD}=40^\circ$
$\therefore\ \angle\text{BAC}=\text{y}=35^\circ$ (alternate angles when $AB \| DC)$
$\angle\text{CAD}=\text{x}=40^\circ$ (alternate angles when $AD \| BC)$
$\therefore\ \text{x}=40$ $\text{y}=35$

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Question 43 Marks

In the adjoining figure, it is given that $AB \| CD$, $\angle\text{ABO}=50^\circ$ and $\angle\text{CDO}=40^\circ.$ Find the measure of $\angle\text{BOD}.$

Answer

Given: $AB \| CD$
$\angle\text{ABO}=50^\circ$ $\angle\text{CDO}=40^\circ$
Construction: Through O, draw $EOF \| AB.$
$\angle\text{ABO}=\angle\text{BOF}=50^\circ$
(alternate angles, when $AB \| EF$ and $OB$ is a transversal) $\angle\text{FOD}=\angle\text{ODC}=40^\circ$
(alternate angles, when $CD \| EF$ and $OD$ is a transversal) $\angle\text{BOD}=\angle\text{BOF}+\angle\text{FOD}$
$\angle\text{BOD}=50^\circ+40^\circ=90^\circ$

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Question 53 Marks

In the given figure $\angle\text{B}=65^\circ$ and $\angle\text{C}=45^\circ$ in $\triangle\text{ABC}$ and $\text{DAF}\ ||\ \text{BC}.$ If $\angle\text{DAB}=\text{x}^\circ$ and $\angle\text{EAC}=\text{y}^\circ,$ find the values of $x$ and $y.$

Answer

Given: $\angle\text{B}=65^\circ$
$\angle\text{C}=45^\circ$ $\text{DAE || BC}$
The given lines are parallel. $\therefore\ \text{x}^\circ=\angle\text{B}=65^\circ$
(alternate angles when $AB$ is taken as trasversal) $\text{y}^\circ=\angle\text{C}=45^\circ$
(alternate angles when $AC$ is taken as trasversal)
$\therefore\ \text{x}=65$ $\text{y}=45$

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Question 63 Marks

In each of the given figures, two lines/ and m are cut by a transevrsal t. Find whether $\text{l || m}.$

Answer

$\angle2+\angle3=180^\circ$ (linear pair)
$\angle3=108^\circ-125^\circ=55^\circ$
$\angle3=55^\circ\neq60^\circ=\angle1$
$\therefore\ \text{l}\neq\text{m}$

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Question 73 Marks

In the adjoining figure, it is being that $\text{AO || CD, OB || CE}$ and $\angle\text{AOB}=50^\circ.$ Find the measure of $\angle\text{ECD}.$

Answer

Given: $\text{AO || CD}$
$\text{OB || CE}$ $\angle\text{AOB}=50^\circ$ $\angle\text{AOD}
=\angle\text{CDB}=50^\circ$ (When $AO \| CD$ and $OB$ is the transversal)
$\angle\text{ECD}+\angle\text{CDB}=180^\circ$
(consecutive interior angles are supplementary, $OB \| CE$ and $CD$ is the transversal)
$\angle\text{ECD}=180^\circ-50^\circ=130$

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Question 83 Marks

In the given figure, $AB \| CD$ and a transversal $EF$ cuts them at $G$ and $H$ respectively.
If $GL$ and $HM$ are the bisectors of the alternate angles $\angle\text{AGH}$ and $\angle\text{GHD}$ respectively, prove that $GL \| HM$.

Answer

Given: $AB \| CD$
GL and HM are angle bisectors of $\angle\text{AGH}$ and $\angle\text{GHD},$ respectively.
$\angle\text{AGH}=\angle\text{GHD}$ (alternate angles)
$\frac{1}{2}\angle\text{AGH}=\frac{1}{2}\angle\text{GHD}$
$\angle\text{LGH}=\angle\text{GHM}$ (given)
Therefore, $GL \| HM$ as we know that if the angles of any pair of alternate interior angles are equal, then the lines are parallel.

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Question 93 Marks

In the given figure, $ABCD$ is a quadrilateral in which $AB \| DC$ and $AD \| BC.$
Prove that $\angle\text{ADC}=\angle\text{ABC}.$

Answer

Given: $AB \| CD , AD \| BC$ $\angle1+\angle2=180^\circ$
$(AB \| CD$ and $AD$ is the transversal) $....(i)$
$\angle2+\angle3=180^\circ$ $(AD \| BC$ and $AB$ is the transversal) $....(ii)$ From $(i)$ and $(ii):$
$\angle1+\angle2=180^\circ=\angle2+\angle3$
$\angle1+\angle3$ $\angle\text{ADC}=\angle\text{ABC}$

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