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14 Feb 2018, 03:44
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
37% (01:00) correct
63%
(01:01)
wrong
based on 367
sessions
History
Date
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Not Attempted Yet
14 Feb 2018, 10:21
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Bunuel
There is a smart way to prove that ABCD may or may not be a square.
Case 1: consider ABCD to be a square inscribed in a circle.
A = C = 90º, AB = AD and ABCD is a square.
Case 2: consider upper right semicircle BCD, with diameter BD. If the diameter of a circle is also the triangle’s side, then that triangle is a right triangle (the reverse is also true: a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle.) According to this property ANY point C on this semicircle will create a right angle BCD. For example, angle BCD will also be a right angle.
A = C = 90º, AB = AD and ABCD is NOT a square.
Answer: E.
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General Discussion
14 Feb 2018, 03:47
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