Question

In the figure below, AB = BC = CD = DE = EF = FG = GA. Then, $\angle DAE$ is approximately

• A. ${15}^{\circ }$
• B. ${30}^{\circ }$
• C. ${20}^{\circ }$
• D. ${25}^{\circ }$

Answer 1

The correct option is D ${25}^{\circ }$

Let $\angle EAD=\alpha$, then, $\angle AFG=\alpha$ and also $\angle ACB=\alpha$.
Hence $\angle CBD=2\alpha$ (exterior angle to $\Delta ABC)$
Since CB = CD, hence $\angle CDB=2\alpha$

$\angle FGC=2\alpha$ (exterior angle to $\Delta AFG)$.
Since GF = EF, $\angle FEG=2\alpha$
Now, $\angle DCE=\angle DEC=\beta$ (say)
Then, $\angle DEF=\beta -2\alpha$
Since, $\angle DCB={180}^{\circ }-(\alpha +\beta )$
Therefore, on $\Delta DCB$,
${180}^{\circ }-(\alpha +\beta )+2\alpha +2\alpha ={180}^{\circ }or\beta =3\alpha$
Further $\angle EFD=\angle EDF=\gamma$ (say)
Then, $\angle EDC=\gamma -2\alpha$
If CD and EF meet at P,
then $\angle FPD={180}^{\circ }-5\alpha$
$[\because \beta =3\alpha ]$
Now in $\Delta PED,{180}^{\circ }-5\alpha +\gamma +2\alpha ={180}^{\circ }or\gamma =3\alpha$
Therefore, in $\Delta EFD$,
$\alpha +2\gamma ={180}^{\circ }or\alpha +6\alpha ={180}^{\circ }or\alpha ={26}^{\circ }$
or approximately ${25}^{\circ }$