enniguy wrote:Dana and everyone,
I still feel that the answer should be C. Can you please help me out here? This is how I solved it.
(1) Consider a line DB=7 and a circle around it. Draw the line BE such that BCA=60 degrees. So, without knowing that DE || CA, I can draw a line in many ways. So option 1 is not sufficient alone.
(2) Consider X to be 75 degrees or 80 degrees. You can still find a point A on the circle. Using this point A, draw your new BE line. (Extend your BE line forever). Now, draw a line DE such that DE || CA. This DE will satisfy the figure but with a different length and not 7.
Together, we can fix both the angle X - 60 degrees - (through which we will fix the point A) and length of DE (because DE || CA).
Is there anything wrong in my thinking?
That is what I thought as well initially and got it wrong.
It's true that, under condition (2), you could draw point A anywhere on the circle and that would result in a different DE. However, regardless of where you drew point A (in other words regardless of angle x), the length of DE would not change.
This is illustrated if you imagined angle x to be 1 degree. This would make point A just above point B, line BE almost completely vertical, and DE almost horizontal, but the same length as the diameter.
Alternatively, you could draw out different versions of the diagram with different angles x, and then measure the length of DE. It will be equal to the diameter regardless of the angle (0 & 180 being exceptions as those angles would cause DE cease to exist).
This problem exploits the common assumption that if a figure can be constructed differently, the dimensions of the different constructions would be different as well. Because the second condition makes the triangles similar, the different constructions would still result in the same dimensions.
Definitely a tricky one.
-
Justas
