In the figure, each side of square ABCD has length 1, the length of line Segment CE is 1, and the length of line segment BE is equal to the length of line segment DE. What is the area of the triangular region BCE?

∆ECF:
Since AB || CF, ∠ECF = 45 degrees.
Thus, is a ∆ECF is a 45-45-90 triangle.
In a 45-45-90 triangle, the sides are in the following ratio:
x : x : x√2.
Since it is given the CE=1, EF = 1/√2 = (1*√2)/(√2*√2) = √2/2.

∆BCE:
If we consider BC the base, then EF is the corresponding height.
Thus, the area of ∆BCE = (1/2)(b)(h) = (1/2)(1)(√2/2) = √2/4.

The correct answer is B.

Hi GMATGuruNY,

∆ECF:
Since AB || CF, ∠ECF = 45 degrees.
Thus, is a ∆ECF is a 45-45-90 triangle.

How can we assume that CE will overlap with the diagonal of the square extended ? an you please explain how we can take that angle 45 ?
Can the figure not be a 3D figure also?

prachi18oct wrote:Hi GMATGuruNY,

∆ECF:
Since AB || CF, ∠ECF = 45 degrees.
Thus, is a ∆ECF is a 45-45-90 triangle.

How can we assume that CE will overlap with the diagonal of the square extended ? an you please explain how we can take that angle 45 ?
Can the figure not be a 3D figure also?

On page 20 of the OG13, the instructions for the Problem Solving section read as follows:
Figures are drawn as accurately as possible.
All figures lie in a plane unless otherwise indicated.

In my solution above, I was able to discern from the figure given in the question prompt that CE could be extended to vertex A, yielding diagonal AC.
That said, here is a proof:



Comparing triangles BCE and ECD, we get:
BE = DE (as stated in the problem).
BC = CD = 1.
Both triangles share side CE.
Since all corresponding sides are equal, ∆BCE and ∆ECD are CONGRUENT.

In congruent triangles, corresponding angles are EQUAL.
Thus, as shown in the figure above, ∠BCE = ∠ECD = 135.

Since ∠BCE = 135, we get the figure shown in my initial post:

Alternate approach:



In square ABCD, since s=1, BD = √2 and CF = √2/2.
It is given that CE=1.
Thus, EF = √2/2 + 1.

∆BED = (1/2)(BD)(EF) = (1/2)(√2)(√2/2 + 1) = (1/2)(1 + √2) = 1/2 + √2/2.

∆BCD = (1/2)(square ABCD) = 1/2.

Quadrilateral BEDC = ∆BED - ∆BCD = (1/2 + √2/2) - 1/2 = √2/2.

∆BCE = (1/2)(quadrilateral BEDC) = (1/2)(√2/2) = √2/4.

The correct answer is B.