What's the area of the isosceles trapezoide below?

Question

For reference: Calculate the area of ​​an isosceles trapeze, if its height measures $9$ and its side not parallel if observe from the center of the circumscribed circumference, under an angle of $74^o$.(Answer: $108$)

My progress: $\triangle OBC(isosceles):\angle BOC = \angle COB = 53^o\\ OH \perp BC \implies \triangle OHB(3k, 4k, 5k) \therefore OB=5k\\ \angle BAC = \frac{74^o}{2} = 37^o $

....I can't find the other relationship

Answer 1

If $AG$ and $CH$ are perp to $CD$ extend and $AB$ respectively, please notice that $\triangle ADG \cong \triangle CBH$.

So, $S_{ABCD} = S_{AHCG}$

As $\angle CAB = \frac{1}{2} \angle COB = 37^\circ, AH \approx \frac{4}{3} CH = 12$

That leads you to the answer

Answer 2

Draw the diagonal AC as $\angle CAB = 37$. And let the intersection of the height and AC be E. Now you can use the 3-4-5 right triangle rule for the $\triangle AEM_{AB}, \triangle ECM_{CD}$ as shown in the diagram

Now you can see,

$3x+3y=9$ and $x+y=3$

Also, you can find the lengths of CD and AB with x and y

Now we can write the following equation to find the area of the triangle

$\frac{1}{2}.9.(8x+8y)$

as x+y = 3

$\boxed{\frac{1}{2}*9*24 = 108}$