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This is the second of our many quizzes on Transversals. Click on this to access more such competitive quizzes on Geometry.

Transversals Quiz 2

mathematics

 $l$ and $m$ are two parallel lines and $h$ is a transversal, find the value of the unknown variable $x$. <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/ec003c2b-1432-4a96-b9ce-dae4dfdc04575079355641812880868.png"/>

 Equate the corresponding angles to get the required value.

  $\text{Here, } 23x-5=21x+5 \\ \text{ (Corresponding angles are equal) } \\ \Rightarrow 23x=21x+10\Rightarrow 2x=10 \\ \therefore x=5$

 If $MN$ is a transversal cutting two pairs of lines, then find the pair of the parallel lines for the figure:<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a86ee83b-f0e6-4613-b0ec-ef36aebf254c7605768731429952514.png"/>

 By equating the corresponding angles, we get $\angle MOX=\angle MPZ$, then conclude which lines are parallel.

  $\text{Here, } \angle MOW\ne \angle MPY \\ \text{So, OW and PY are not parallel} \\ \angle MPZ=52{}^\circ +28{}^\circ =80{}^\circ \\ \text{So},\, \angle MOX=\angle MPZ \\ \text{Since corresponding angles are equal, so } OX||PZ$ 

 Three parallel lines are such that when they are cut by two transversals, $\angle BEF={{45}^\circ}$ as shown: $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/b93bfcde-752a-4d49-a9eb-2c598afc4b802902280616604992113.png"/>$ \\ $ Find the value of the three unknown variables $x, y$ and $z$.

 By considering co-interior angles we get the value of $y$, then by considering corresponding angles we get $x$, finally by complementary angles, we get $z$.

$\text{Here, } \angle FED=\angle EDC \text{ as they are alternate angles.} \\ \text{Hence},\,  \angle y=180{}^\circ -45{}^\circ =135{}^\circ \text{ (Co-interior angles) } \\ \angle y=\angle x  \text{ (Corresponding angles) } \\ \text{Therefore, } \angle x=135{}^\circ \\ \text{Also, } 90{}^\circ +\angle z+45{}^\circ =180{}^\circ \\ \text{Therefore},\, 135{}^\circ +\angle z=180{}^\circ \\ \text{Therefore, } \angle z=180{}^\circ -135{}^\circ =45{}^\circ $ 

 A transversal $l$ cuts three parallel lines as shown below, if $\dfrac{y}{z}=\dfrac{4}{5}$ find the value of the three unknown variables.<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f0f5fb20-d3f4-4b62-aa26-4fe8803d1c533594543384221892852.png"/>

 By alternate angles, $\angle z=\angle m$, then by considering corresponding angles $\angle m=\angle x$, we get an expression of $\angle x\text{ and }\angle y$ then we get all three values by putting the ratio values.

 $\text{Let the common ratio be a}\\ \text{Then } y=4a\text{ and }z=5a \\ \text{Also},\,  \angle z=\angle m \text{ (Alternate interior angles) } \\ \text{Since},\, z=5a, \text{ so } \angle m=5a [RS\parallel XY \text{ cut by transversal } t]\\ \text{Now},\, \angle m=\angle x \text{ (Corresponding angles) }\\ \text{So},\, \angle x=5a \\ [PQ\parallel RS \text{ cut by transversal } l]\\ \angle x+\angle y=180{}^\circ  \text{ (Co-interior angles) } \\ 9a=180{}^\circ \Rightarrow a=20 \\ \text{Since},\, y=4a,y=4\times 20=80{}^\circ \\  z=5a,z=5\times 20=100{}^\circ \\ x=5a,x=5\times 20=100{}^\circ \\ \text{Therefore},\, \angle x=100{}^\circ ,\angle y=80{}^\circ ,\angle z=100{}^\circ $ 

 In the below given diagram, $MN\parallel OP,OP\parallel QR,OQ\parallel TP$ . Also,  $\angle NTP=60{}^\circ ,\angle QOP=55{}^\circ $, then find the three unknown variables $\angle MNT,\angle TPO\text{ and }\angle OQR$. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/996ec083-c513-43c3-9cb4-7cd8e11fb1611001993655198984065.png"/>

 Take the transversal $TP$ first, then find the value of $\angle 2$, then find the value of $\angle 1$ using complementary angles, and finally take the transversal $OQ$ to get $\angle 4$.

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/23a5c636-8cfe-4339-b100-97ff013b49263005968466190046326.png"/>$\text{Since }, OQ||PT \text{ cut by transversal } PO \\ \text{ Therefore},\, \angle QOP=\angle OPT=\angle 5=55{}^\circ \\ \text{Therefore},\, \angle OPT=\angle x=55{}^\circ \\ \text{Also},\, \angle x+\angle 1+\angle y=180{}^\circ \\ 55{}^\circ +60{}^\circ +\angle y=180{}^\circ ,115{}^\circ +\angle y=180{}^\circ   \\ \angle y=180{}^\circ -115{}^\circ =65{}^\circ   \\ \text{Now},\, \angle y+\angle 2=180{}^\circ \text{ (Co-interior angles) } \\ 65{}^\circ +\angle 2=180{}^\circ ,\angle 2=180{}^\circ -65{}^\circ =115{}^\circ \\ \text{Since},\, OP\parallel QP \text{ cut by transversal } OQ \\ \text{Therefore},\, \angle 3+\angle 4=180{}^\circ ,55{}^\circ +\angle 4=180{}^\circ \Rightarrow \angle 4=125{}^\circ $ 