Challenging Problem on Area of Trapezium [closed]

Let $ABCD$ be a inscribed trapezium such that sides $AB$ and $CD$ are parallel. If Angle $\angle{AOD}=60^{\circ}$ degrees and altitude of trapezium is $10$. Find area of trapezium.

I am working on this question but unable to get a lead. Any help would be appreciated.

Solution 1:

It is challenging, until you can prove that leg CD is the diameter of the circle. You can do this yourself quite easily, but I'll leave my steps here (my apologies for not being able to get the figure in, please draw it yourself):

Given that ABCD is an cyclic quadrilateral: $$\angle BAD + \angle BCD = 180^{\circ}\ (1)$$ One of the properties of a trapezium is that the adjacent angles are supplementary: $$\angle BAD + \angle ADC = 180^{\circ}\ (2)$$ From (1) and (2), we can deduce that: $$\angle BCD = \angle ADC$$ $\therefore$ ABCD is an isosceles trapezoid (since the adjacent bases are equal). Looking at $\triangle AOD$: $$\angle AOD = 60^{\circ}$$ $$OA = OB$$ $\therefore \triangle$ OAD is an equilateral triangle: $$\angle AOD = \angle ADO = \angle OAD = 60^{\circ}$$ $$\angle BAO = \angle BAD - \angle OAD = 120^{\circ} - 60^{\circ} = 60^{\circ}$$ Similarly, for $\triangle$ BCO: $$\angle BCO = \angle ADO = 60^{\circ}$$ $$OB = OC$$ $\therefore \triangle$ BOC is an equilateral triangle: $$\angle BOC = \angle CBO = \angle BCO = 60^{\circ}$$ $\triangle$ OBA: $$\angle AOB = 180^{\circ} - \angle OAB - \angle OBA = 60^{\circ}$$ $$OA = OB$$ $\therefore \triangle$OBA is an equilateral triangle. Lastly: $$\angle DOC = \angle DOA + \angle AOB + \angle BOC$$ $$\angle DOC = 60^{\circ} + 60^{\circ} + 60^{\circ} = 180^{\circ}$$ $\therefore$ D, O, and C form a straight line with OC = OD $\therefore$ AD is the diameter of the circle (double the radius).

Now, let H $\in$ AB, s.t. OH $\perp$ AB: $$\angle HOA = \angle HOD - \angle AOD = 90^{\circ} - 60^{\circ} = 30^{\circ}$$ $$\frac{OH}{OA} = \cos{\angle HOA}$$ $$OA = \frac{10}{\cos{(30^{\circ})}}=11.547$$ $$AB = OA = 11.547$$ $$DC = OA + OB = 2OA = 23.094$$ Area of ABCD: $$S = \frac{OH(AB+CD)}{2}$$ $$S = \frac{10(11.547 + 23.094)}{2} = 173.205 $$