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. ${28}^{\circ }$

Answer 1

Answer: (C)

${26}^{\circ }$

Let $\angle DAE=x$

$△ABC$ is an isosceles triangles since, $AB=BC$

$\therefore$ $\angle BCA=\angle CAB=x$

$\Rightarrow$ $\angle CBD=\angle CAB+\angle BCA$ [ External angle of $△ABC$ ]

$\Rightarrow$ $\angle CBD=x+x$

$\therefore$ $\angle CBD=2x$

$△BCD$ is an isosceles triangle as $BC=CD$

$\therefore$ $\angle CBD=\angle CDB=2x$

$\Rightarrow$ $\angle DCE=\angle DAE+\angle CDA$ [ External angle of $△ACD$ ]

$\Rightarrow$ $\angle DCE=x+2x$

$\therefore$ $\angle DCE=3x$

$△CDE$ is an isosceles triangle as $CD=DE$

$\therefore$ $\angle DCE=\angle DEC=\angle AED=3x$

Similarly,

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

$=\angle EGF+\angle DAE$

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

$=\angle DAE+\angle DAE+\angle DAE$

$=3x$

In $△ADE,$

$\Rightarrow$ $\angle ADE+\angle DAE+\angle AED={180}^o$

$\Rightarrow$ $3x+x+3x={180}^o$

$\Rightarrow$ $7x={180}^o$

$\Rightarrow$ $x=\frac{{180}^o}{7}$

$\Rightarrow$ $x={25.7}^o\approx {26}^o$