Simple Math Question: [Solved!]

Underdog

Ok so I'm teaching a supplementary Math class and I have to fill in the answers to the homework. The only problem I've had is trying to figure out this question. It's not that I don't know how to do the math, it's that it seems like it's really poorly labelled and so I can't figure out where to start. Here's an image of it:

http://img835.imageshack.us/img835/9655/photo000t.jpg

I can't for the life of me figure out what A, B, C and D are actually referring to. The sides? The vertices? Is it just me?

Edit: All right, so I overestimated my math abilities. I'm actually not sure how to go about solving this. I think the answer is 1/3 but my process isn't very... mathy. I'm making a lot of assumptions about the lengths of the parallelogram to get to my answer and I have this feeling that's not how you're supposed to do it. I'll put my "solution" in a spoiler just so it doesn't look like I'm having you all do my work for me.

Ok assuming sides of the shape are equal, then splitting the shape in two along the dotted line gives me at least an isosceles triangle to work with. Now because AB = CD, then that must mean the shaded triangle has the same height as the large triangle. The base however is 1/3 the length of AD. Same height, base 1/3, that means the area of the shaded triangle is 1/3 of the large one.

Example
AD = 6, height = 10
Area large triangle = 6x10/2 = 30
Area small triangle = 2x10/2 = 10

This doesn't feel right though.

GoodOmens

Single capital letters almost always refer to the points. AB, BC and CD refer to the lengths of those line segments.

Hevach

The letters aren't well placed, they're too big for the figure. A and D are the outside corners along the center line, B and C are the matching corners of the shaded region. AB, BC, and CD are the segments connecting those points.

Underdog

Ok thanks. I thought that was what it was but B and C were so far from the vertices that I was unsure.

Hirocon

Your solution is correct, and it does not depend on whether or not the figure is actually a parallelogram. If the figure is a non-parallelogram kite (i.e. if the left and right isosceles triangles have different heights) then the shaded triangle on the left is still 1/3 the area of the left isosceles triangle, and the shaded triangle on the right is 1/3 the area of the right isosceles triangle. Combined together, the entire shaded region is 1/3 the area of the entire figure.

Underdog

Ohhh yeah because the height is always going to be the same, parallelogram or not, and the base is 1/3. Then it's just a simple plug in the formula deal. Ok, great!

soxbox

The more maths-y way:

The area of a triangle is 1/2 width * height.

Area of BC triangle = 1/2 * BC * height
Area of AD triangle = 1/2 * 3 * BC * height

Put that as a fraction, eliminate the common terms, and you get you 1/3.

Underdog

Ok great! I was really just unsure if the height would be the same but I think I've figure out that it is so that makes things a lot easier.