Question

In the given figure (not drawn to scale), $ABCD$ is a parallelogram, $ADF$ is an isosceles triangle with $AD=AF$, $FAB$ and $EDC$ are straight lines. Find $y$.

• A. $112$
• B. $102$
• C. $132$
• D. $110$

Answer 1

The correct option is A $112$

FAB and EDC are straight lines, and Df is crossing these lines,
So, $\angle EDF=\angle DFA=34$
& ADF is an isosceles triangle with AD=AF
$\angle ADF=\angle DFA=34$
In triangle AFD,
$\angle FAD+\angle ADF+\angle DFA=180$
$34+34+\angle FAD=180$
$\angle FAD=112$
FAB is straight line and $AD\parallel BC$
So, $\angle FAD=\angle y=112$
Answer (A) $112$