Calculate the area of trapezoid.

In an isosceles trapezoid $$ABCD$$ the leg $$AB$$ and the smaller base $$BC$$ are 2 cm long, and $$BD$$ is perpendicular to $$AB$$.Calculate the area of trapezoid.

Let $$\angle BAD=\theta,$$
$$BD=h$$,
$$\angle ABD=90^\circ$$
$$\angle CBD=90^\circ-\theta$$
$$CD=2$$because trapezoid is isosceles

Apply cosine law in triangle BDC,

$$\cos(90-\theta)=\frac{h^2+2^2-2^2}{2\times 2\times h}=\frac{h}{4}$$

$$\sin\theta=\frac{h}{4}..(1)$$

In right triangle $$ABD,\sin \theta=\frac{h}{\sqrt{h^2+4}}..(2)$$
From $$(1)$$ and $$(2)$$,$$h=2\sqrt3$$

Area of $$ABCD=\frac{1}{2}\times 2\times h+\frac{1}{2}\times 2\times 2\times \sin2\theta=3\sqrt3$$
But the answer given is $$2\sqrt2(\sqrt{5}+1)$$

I've got $$3\sqrt 3$$ using a slightly different method. $$\angle BAD=\theta$$, $$\angle ABD =90^\circ$$ means $$\angle CBD=\angle BDA=90^\circ-\theta$$. Since the trapezoid is isosceles, $$\angle CDA=\theta$$, and you can get $$\angle CDB=2\theta -90^\circ$$. Since $$BC=AB=CD$$ you get $$\angle CDB=\angle CDB$$ or $$2\theta-90^\circ=90^\circ-\theta$$ so $$\theta=60^\circ$$. Drawing perpendiculars from $$B$$ and $$C$$ to $$AD$$, you can get $$AD=BC+2AB\cos 60^\circ=4$$, and the height $$h=AB\sin60^\circ=\sqrt 3$$. Therefore the area is $$\frac12 (BC+AD)\cdot h=\frac12 6\sqrt3=3\sqrt 3$$
you are correct. The answer should be $$3\sqrt{3}$$