3D Geometry Question

[student22]

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?

A. 1/3
B. root(2)/4
C. 1/2
D. root(2)/2
E. 3/4

OA: B

I attached the figure in question:



I'm having a really hard time visualizing this shape, let alone figuring out how to solve it. Is it some sort of pyramid? If it is, then basically it's asking me to solve for one of its faces. If that's the case, then it's telling you that a triangle that cuts through it in the middle is equilateral.

I'm completely lost with this problem and the only solutions I found were here:

https://www.urch.com/forums/gmat-problem ... n-bce.html

However, I couldn't really understand the steps involved. Can anyone help explain what this shape is, as well as the logic you would use to solve this problem. Thanks!

[student22]

I actually just found this topic on this post and the explanation makes alot of sense. For some reason I must've missed it when I googled the question the first time.

https://www.beatthegmat.com/area-of-tria ... html#62848

The problem was, that I was imagining a 3-Dimensional shape, when this was actually drawn on a 2-Dimensional plane. Crazy how these questions trick you.

[kvcpk]

This is not so tricky a question if we know that the area of triangle is
(1/2)*a*b*sinC

Here the line CE is the angular bisector of the angle BCD, because triangles BCE and CDE are equivalent triangles.
So angle BCE = 90/2 = 45

now in triangle BCE, we know that BC =1, CE=1, angle BCE=45.
Therefore area=(1/2)*1*1*(sin45) = (1/2)*(1/root(2))=1/2(root(2))

[student22]

My problem with this question was visualizing what the shape was. I thought it was a pyramid when I first saw it, and I wasn't even sure if it had 3 faces or 4.

Once I saw from that explanation that the shape was actually 2-dimensional and that you could extend CE to the center of the square, it all became clear.

[ceilidh.erickson]

Here are images of the 2 possible solutions: