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IMHO, the triangle OCX, should have the highest area since all the triangles have the same base as that of triangle OCX, but all of them have height (?not sure about S-V agreement) less than that of triangle OCX (which I assumed to be perpendicular to OX)
aox has least area . You are right Jai Hind. Eventhough I knew area = 1/ base in alt, forgot that one can extend the base to drop a perpendicular as in the case of AOX. Digrammatically it has the least height.
Please reject the previous answers to this response as being spurious. The triangle AOX does have the least area of the group, but not due to its smallest height. As you can see, when the height decreases the base increases. The two variables are inversely proportional in the context of this problem. One could have a triangle, resembling EOX, with a smaller area than triangle AOX if the line EX (the base) was short enough. Even though its height would be greater, the area would still be less because of the small base value. Remember, as the height increases, the base decreases (for this problem). The real evidence in the problem is that point A lies "the lowest" on the circle. To explain this without going into detail, if triangle EOX and AOX had supplementary angles at the origin, their areas would be equal, and the points A and E would appear at the same height on the circle. When one decreases the height, the area decreases. In the context of this problem, the triangle with "the lowest point" (A) also has the smallest area. Likewise, the triangle with the highest point (c) has the greatest area.
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