Question

In the figure, $AB=BC=CD=DE=EF=FG=GA$, then find $\angle DAE$ ( approximately).

• A. $24{}^{\circ }$
• B. $25{}^{\circ }$
• C. $26{}^{\circ }$
• D. None of the above

Answer 1

Answer: (C)

$26{}^{\circ }$

Let us assume, $\angle DAE=x$

$△ABC$ is isosceles as $AB=BC$ $⟹\angle BCA=\angle CAB=x$

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

$△BCD$ is isosceles as $BC=CD$ $⟹\angle CBD=\angle CDB=2x$

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

$△CDE$ is isosceles as $CD=DE$ $⟹\angle DCE=\angle DEC=\angle AED=3x$

Similarly,

$\angle ADE=\angle EFD=\angle AEF+\angle DAE=\angle EGF+\angle DAE$

$=(\angle DAE+\angle GFA)+\angle DAE=\angle DAE+\angle DAE+\angle DAE=3x$

Hence, in $△ADE$,

$\angle ADE+\angle DAE+\angle AED=3x+x+3x=7x$

Hence, $7x={180}^{\circ }$ $⟹x=\frac{180}{7}={25.7}^{\circ }\approx {26}^{\circ }$.

Hence, the answer is ${26}^{\circ }$.