Given that DE is the diameter and that AB and DE are perpendicular, some test-takers will intuitively perceive the relationships shown in the figure above.
That said, here's a proof:
The diagonals of ADBE form 4 right angles.
A quadrilateral whose diagonals form 4 right angles is a kite, a rhombus, or a square.
An inscribed angle that intercepts the diameter is a right angle.
Thus, ∠DAE = 90 and ∠DBE = 90.
A rhombus with two opposite right angles is a square.
Implication:
ADBE is either a kite or a square.
In a square, the diagonals form 4 congruent triangles.
In a kite, the diagonals form 2 congruent smaller triangles and 2 congruent larger triangles.
Implication:
Whether ADBE is a square or a kite, ∆AEF and ∆BEF must be CONGRUENT.
Statement 1: arc AB = 120º
The degree measurement of an inscribed angle is 1/2 the degree measurement of the arc intercepted by the inscribed angle.
Since inscribed ∠AEB intercepts arc AB, ∠AEB = 60º.
The result is the following figure:
The figure shows that ∆ABE is EQUILATERAL, but there is no way to determine the perimeter of ∆ABE.
INSUFFICIENT.
Statement 2: AB = 2
The result is the following figure:
No way to determine the perimeter of ∆ABE.
INSUFFICIENT.
Statements combined:
Thus, the perimeter of ∆ABE = 2+2+2 = 6.
SUFFICIENT.
The correct answer is
C.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at
[email protected].
Student Review #1
Student Review #2
Student Review #3
