EQ || MT, SMTQE - ?

Okay, let's solve this geometry problem step by step. First, let's analyze the given information. We have a quadrilateral $$MTQE$$ inscribed in a circle, and $$EQ$$ is parallel to $$MT$$. Also, $$OQ = 8$$ and $$\angle EQO = 30^{\circ}$$. We need to find the area of quadrilateral $$MTQE$$. Since $$EQ \parallel MT$$, the quadrilateral $$MTQE$$ is an isosceles trapezoid. This is because if a trapezoid is inscribed in a circle, it must be an isosceles trapezoid. In an isosceles trapezoid, the angles at the base are equal. Therefore, $$\angle MET = \angle QTE$$ and $$\angle MTE = \angle EQM$$. Since $$O$$ is the center of the circle, $$OQ$$ is the radius, so $$OQ = 8$$. In triangle $$EQO$$, since it's an isosceles triangle ($$OE = OQ = 8$$), we have $$\angle OEQ = \angle OQE = 30^{\circ}$$. Therefore, $$\angle EOQ = 180^{\circ} - 30^{\circ} - 30^{\circ} = 120^{\circ}$$. Now, we can find the length of $$EQ$$ using the law of cosines in triangle $$EOQ$$: $$EQ^2 = OE^2 + OQ^2 - 2 cdot OE cdot OQ \cdot \cos(\angle EOQ)$$ $$EQ^2 = 8^2 + 8^2 - 2 cdot 8 cdot 8 \cdot \cos(120^{\circ})$$ $$EQ^2 = 64 + 64 - 128 \cdot (-\frac{1}{2})$$ $$EQ^2 = 128 + 64 = 192$$ $$EQ = \sqrt{192} = 8\sqrt{3}$$ Now, let's find the central angle $$\angle MOQ$$. Since $$EQ \parallel MT$$, the arcs $$EM$$ and $$QT$$ are equal. Therefore, $$\angle EOM = \angle QOT$$. Let $$\angle EOM = x$$. Then $$\angle MOQ = 360^{\circ} - \angle EOQ - \angle EOM - \angle QOT = 360^{\circ} - 120^{\circ} - 2x$$. We know that $$\angle EMT = \frac{1}{2} \angle EOQ = \frac{1}{2} cdot 120^{\circ} = 60^{\circ}$$ because $$\angle EMT$$ is an inscribed angle that subtends arc $$EQ$$. Also, $$\angle EQM = \angle EMT$$ because they are alternate interior angles as $$EQ \parallel MT$$. Therefore, $$\angle EQM = 60^{\circ}$$. Now, $$\angle EOT = 2 \cdot \angle EMT = 2 \cdot 60^{\circ} = 120^{\circ}$$. Because $$\angle EOT$$ is central angle that subtends $$MT$$ arc. Then $$MT = EQ = 8\sqrt{3}$$. Now, find the height of the trapezoid. Consider the perpendicular from $$E$$ to $$MT$$, call the foot $$F$$. Then $$\triangle EMF$$ is right angled. $$\angle EMF = 60^{\circ}$$, then $$EF = EM \sin(60^{\circ})$$. We know that $$EM$$ subtends a central angle of $$60^{\circ}$$ (since $$\angle EQM=60^{\circ}$$), thus the length of the chord $$EM = R = 8$$. Then $$EF = 8 \cdot \sin(60^{\circ}) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}$$. The area of trapezoid $$MTQE$$ is: $$S_{MTQE} = \frac{MT + EQ}{2} cdot EF = \frac{8\sqrt{3} + 8\sqrt{3}}{2} cdot 4\sqrt{3} = 8\sqrt{3} cdot 4\sqrt{3} = 32 cdot 3 = 96$$ Answer: 96