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Question

In each of the figures given below, AB || CD. Find the value of x in each case.


Solution

Draw EFABCD


Now, ABEFandBEisthetransversal. 
Then,
ABE=BEFAlternateInteriorAnglesBEF=350
Again,   EFCDandDEisthetransversal.
Then,
\angleDEF=\angleFED⇒\angleFED=65°

x0=BEF+FED=350+650
=1000orx=1000

(ii)
Draw  EOABCD
Then, ( \angleEOB+\angleEOD=x ^0 \)
Now, EOABandBOisthetransversal.
EOB+ABO=1800ConsecutiveInteriorAngles
EOB+550=1800

EOB=1250
Again,EOCDandDOisthetransversal
EOD+CDO=1800ConsecutiveInteriorAngles
(\rightarrow \angle EOD+25^0=180^0\)
( \rightarrow \angleEOD=1556^0 \)
x0=EOB+EOD=1250+1550=2800

or, x=2800 

(iii)Draw   EFABCD
Then, AEF+CEF=x0
Now, EFABandAEisthetransversal
AEF+BAE=1800ConsecutiveInteriorAngles
(\rightarrow \angle AEF+116^0 =180^0\)
AEF=640
Again,EFCDandCEisthetransversal
CEF+ECD=1800.ConsecutiveInteriorAngles
CEF+1240=1800
CEF=56°
,x0=AEF+CEF=640+56°0=120°or,x=1200
 

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