Area of pentagon

Construct line segments $AC$ and $CE$ .

Let $b = AB = BC$ , $d = CD = DE$ , $f = AC$ and $h = CE$ .

From isosceles triangles $ABC$ and $CDE$ , we get

$\angle BCA = 15$ and $\angle DCE = 75$

From the given information, we get

$f = 2b\cos(15)$

$h = 2d\cos(75) = 2d\sin(15)$

Thus, $fh = (2b)(2d)\cos(15)\sin(15) = 2bd\sin(30) = bd$

Letting $x = \angle ACE$ and applying the Cosine Law to triangle $BCD$ , we get

$4 = 2^2 = b^2 + d^2 - 2bd\cos(x + 90) = b^2 + d^2 + 2bd\sin(x)$ Hence,

Area( $ABCDE$ ) = Area( $ABC$ ) + Area( $CDE$ ) + Area( $ACE$ )

= $\frac{1}{2}b^2\sin(150) + \frac{1}{2}d^2\sin(30) + \frac{1}{2}fh\sin(x)$

= $\frac{1}{4}[b^2 + d^2 + 2bd\sin(x)] = \frac{1}{4}(4) = 1$