Question

In the figure AB = BC = CD = DE = EF = FG = GA, then $\angle DAE$ is equal to

• A. ${24}^{\circ }$
• B. ${25}^{\circ }$
• C. ${27}^{\circ }$
• D. $\frac{{180}^{\circ }}{7}$

Answer 1

The correct option is D $\frac{{180}^{\circ }}{7}$

Let us assume, DAE = x

Triangle ABC is isosceles as $AB=BC-->BCA=CAB=x$

Hence, $CBD=CAB+BCA=x+x=2x$.............. [External angle of triangle ABC]

Triangle BCD is isosceles as $BC=CD-->CBD=CDB=2x$

Hence, $DCE=DAE+CDA=x+2x=3x$.............. [External angle of triangle ACD]

Triangle CDE is isosceles as $CD=DE-->DCE=DEC=AED=3x$

Similarly, $ADE=EFD=AEF+DAE=EGF+DAE=(DAE+GFA)+DAE=DAE+DAE+DAE=3x$

Hence, in triangle $ADE,ADE+DAE+AED=3x+x+3x=7x$

Hence, $7x=180--->x=180/7$