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Maths 3 Mark Question

Properties of Parallel Lines · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 9 questions.

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3 Mark Question

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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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