3D Geometry Question

Expert replies

student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Sun Jun 13, 2010 7:36 pm

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!

Source: Beat The GMAT — Problem Solving | Sun Jun 13, 2010 7:36 pm

student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Sun Jun 13, 2010 7:53 pm

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: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

by kvcpk » Sun Jun 13, 2010 8:49 pm

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: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Mon Jun 14, 2010 11:49 am

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.

GMAT/MBA Expert

ceilidh.erickson: GMAT Instructor; Posts: 2095; Joined: Tue Dec 04, 2012 3:22 pmFollowed by: 247 members; Thanked: 1443 times

by ceilidh.erickson » Wed Jun 10, 2015 5:45 am

Here are images of the 2 possible solutions:

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