Question

In the following figure, $FD\parallel BC\parallel AE$ and $AC\parallel ED$. Find the value of x.

[3 marks]

Answer 1

In $\Delta ABC$,
$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$ [Angle sum property of triangle]
$\Rightarrow {64}^{\circ }+\angle BCA+{52}^{\circ }={180}^{\circ }$
$\Rightarrow \angle BCA={180}^{\circ }-{64}^{\circ }-{52}^{\circ }={64}^{\circ }$
[1 mark]

$\angle FAE=\angle BCA$ [$\because$ Alternate angles; $AE\parallel BC$, AC is the transversal]
$\Rightarrow \angle FAE={64}^{\circ }$
[1 mark]

Now, $\angle FAE+x={180}^{\circ }$ [Adjacent angles in a parallelogram are supplementary]
$\Rightarrow x={180}^{\circ }-{64}^{\circ }\therefore x={116}^{\circ }$
[1 mark]