What's the height?

Construct $F$ such that $AECF$ is a rectangle:

Then $\angle BAE = \angle BAD - \angle EAD = 90° - \angle EAD = \angle EAF - \angle EAD = \angle DAF$ , and $\angle BEA = 90° = \angle DFA$ , and since $AB = AD$ , $\triangle BEA \cong \triangle DFA$ by AAS congruency.

Since $\triangle BEA \cong \triangle DFA$ , $A_{ABCD} = A_{AECF} = 100$ . Also since $\triangle BEA \cong \triangle DFA$ , $AE = AF$ , which makes $AECF$ a square.

As the side of a square with an area of $100$ , $AE =\sqrt{100} = \boxed{10}$ .

Let $m\angle{B} = \theta$ and $h = \overline{BD} = x\sin(\theta)$ .

$100 = A_{\triangle{ABD}} + A_{\triangle{BDC}} = \dfrac{x^2}{2} + \cos(\theta - 45^{\circ})\sin(\theta - 45^{\circ})x^2 =$

$\dfrac{x^2}{2} + \dfrac{\sin(2\theta - 90^{\circ})}{2}x^2 \implies 200 = (1 - \cos(2\theta))x^2 = 2\sin^2(\theta)x^2 \implies$

$100 = (x\sin(\theta))^2 = h^2 \implies h = \boxed{10}$ .