Question

In the given figure, ABCD is a square, BCF is an equilateral triangle and AEDF is a rhombus. Find $\angle$EAF.

• A. ${30}^o$
• B. ${120}^o$
• C. ${150}^o$
• D. None of these

Answer 1

The correct option is D None of these

Given that,

$\Delta BCF$ is an equilateral triangle

$\Rightarrow \angle BFC={60}^{\circ }$

$\angle AFB=\angle DFC={90}^{\circ }$ given in the figure

Consider point $F$

sum of all angles around a Point $={360}^{\circ }$

$\Rightarrow \angle AFD+\angle AFB+\angle DFC+\angle BFC={360}^{\circ }$

$\Rightarrow \angle AFD+{90}^{\circ }+{90}^{\circ }+{60}^{\circ }={360}^{\circ }$

$\Rightarrow \angle AFD={120}^{\circ }$

Since $AEDF$ is a Rhombus

$EA$ $||$ $DF$ rhombus is also a parallelogram

$\Rightarrow \angle AFD+\angle EAF={180}^{\circ }$ Co-interoir angles of parallel lines .

$\Rightarrow {120}^{\circ }+\angle EAF={180}^{\circ }$

$\Rightarrow \angle EAF={60}^{\circ }$

Answer is None of these