01 Area enclosed by line rays inside a square

From the right triangle ADE

$\tan \alpha = 2/4$

$\alpha = 26.565^\circ$

 

From triangle ABC

$\beta = 180^\circ - 45^\circ - \alpha$

$\beta = 180^\circ - 45^\circ - 26.565^\circ$

$\beta = 108.435^\circ$
 

By Sine law
$\dfrac{x}{\sin \alpha} = \dfrac{2}{\sin \beta}$

$\dfrac{x}{\sin 26.565^\circ} = \dfrac{2}{\sin 108.435^\circ}$

$x = 0.9428 \, \text{ in.}$

 

Area of triangle ABC
$A_{ABC} = \frac{1}{2}(2x) \sin 45^\circ$

$A_{ABC} = \frac{1}{2}(2)(0.9428) \sin 45^\circ$

$A_{ABC} = 0.6667 \, \text{ in.}^2$
 

Required area
$A = 8A_{ABC} = 8(0.6667)$

$A = 5.33 \, \text{ in.}^2$           answer

From right triangle BAD

$\tan \alpha = 4/2$

$\alpha = 63.435^\circ$

 

From triangle ABC

$\beta = 180^\circ - 45^\circ - \alpha$

$\beta = 180^\circ - 45^\circ - 63.435^\circ$

$\beta = 71.565^\circ$

By Sine law
$\dfrac{x}{\sin \alpha} = \dfrac{2}{\sin \beta}$

$\dfrac{x}{\sin 63.435^\circ} = \dfrac{2}{\sin 71.565^\circ}$

$x = 1.8856 \, \text{ in.}$

 

Area of triangle ABC
$A_{ABC} = \frac{1}{2}(2x) \sin 45^\circ$

$A_{ABC} = \frac{1}{2}(2)(1.8856) \sin 45^\circ$

$A_{ABC} = 1.3333 \, \text{ in.}^2$
 

Required area
$A = 4^2 - 8A_{ABC} = 16 - 8(1.3333)$

$A = 5.33 \, \text{ in.}^2$           answer