<s>[INST] 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 $2\sqrt2(\sqrt{5}+1)$
 [/INST] you are correct. The answer should be $3\sqrt{3}$
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