Quantitative Aptitude Ques 1221
Question: In the given figure, $ AB||CD $ and they cut PQ and QR at E, F and G, H, respectively. $ \angle PEB=70{}^\circ $ and $ \angle QHD=140{}^\circ $ and $ \angle PQR=x. $ Find the value of x.
Options:
A) $ 20{}^\circ $
B) $ 30{}^\circ $
C) $ 24{}^\circ $
D) $ 32{}^\circ $
Show Answer
Answer:
Correct Answer: B
Solution:
- Since, $ AB||CD $ and PQ is the transversal.
$ \therefore $ $ \angle PEF=\angle EGH $ [corresponding angles] $ \angle EGH=70{}^\circ $ $ [\because \angle PEF=70{}^\circ ] $ $ \angle EGH+\angle HGQ=180{}^\circ $ [linear pair]
$ \Rightarrow $ $ \angle HGQ=180{}^\circ -70{}^\circ =110{}^\circ $ $ \angle DHQ+\angle GHQ=180{}^\circ $ [linear pair]
$ \Rightarrow $ $ \angle GHQ=180{}^\circ -140{}^\circ =40{}^\circ $ In $ \Delta GQH, $ $ \angle GQH+\angle GHQ+\angle HGQ=180{}^\circ $
$ \Rightarrow $ $ x+40{}^\circ +110{}^\circ =180{}^\circ $
$ \Rightarrow $ $ x+150{}^\circ =180{}^\circ $
$ \Rightarrow $ $ x=30{}^\circ $
BETA
