Rectangle geometry ??
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Dr.Peterson
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I don't think there's one special property to use; just do what you can! That's commonly how problem solving works.
I'd start by drawing a proper figure, showing a rectangle and including DE. Maybe define AD=x, AE=y, EB=z and express the given areas in terms of the variables, then try to express the area of the rectangle similarly and eliminate variables. (You'll have two equations in three unknowns, so you can't solve fully, but will be able to solve for the area.)
Make an effort and show your attempt, so we can guide you further.
Dr.Peterson
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That's a nice approach; but it clearly is not enough. If I wanted to proceed in that direction rather than do what I suggested before, I might try doing something related to the ratio EF:FC. Did you notice that EBF and CDF are similar?
You start by systematically naming each relevant unknown with a distinct letter.
[MATH]x = \text {length of lines AD and BC;}\\ y = \text {length of lines AB and CD;} \\ z = \text {area of rectangle ABCD;}\\ p = \text {length of line AE;}\\ q =\text {length of line BE; and}\\ r = \text {height of triangle BEF.}[/MATH]There may be other lengths or areas that will become relevant, but we have a start. That is six relevant unknowns so far. Thus, we need six independent equations Involving those unknowns.
Some are obvious.
[MATH]xy = z,\\ p + q = y,\\ \dfrac{1}{2} * px = 50, \text { and}\\ \dfrac{1}{2} * fq = 40.[/MATH]That means we must find at least two more equations (two if we discover no other relevant unknowns).
This has been relatively mechanical, naming things and looking for obvious relationships, but now we must begin to think. I can see immediately two ideas that may or may not be helpful.
I see two similar triangles, one with a known area. I don’t see an immediately obvious way that helps, but I haven't really thought about it . I know I can decompose this figure into five triangles, two with known areas. Again, the only way to know whether that helps or not is to think about it. How do we think about such amorphous things? We try them and see where they lead.
[MATH]x = \text {length of lines AD and BC;}\\ y = \text {length of lines AB and CD;} \\ z = \text {area of rectangle ABCD;}\\ p = \text {length of line AE;}\\ q =\text {length of line BE; and}\\ r = \text {height of triangle BEF.}[/MATH]There may be other lengths or areas that will become relevant, but we have a start. That is six relevant unknowns so far. Thus, we need six independent equations Involving those unknowns.
Some are obvious.
[MATH]xy = z,\\ p + q = y,\\ \dfrac{1}{2} * px = 50, \text { and}\\ \dfrac{1}{2} * fq = 40.[/MATH]That means we must find at least two more equations (two if we discover no other relevant unknowns).
This has been relatively mechanical, naming things and looking for obvious relationships, but now we must begin to think. I can see immediately two ideas that may or may not be helpful.
I see two similar triangles, one with a known area. I don’t see an immediately obvious way that helps, but I haven't really thought about it . I know I can decompose this figure into five triangles, two with known areas. Again, the only way to know whether that helps or not is to think about it. How do we think about such amorphous things? We try them and see where they lead.
Dr.Peterson
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What is the ratio of areas EBF:CDF? What does that tell you about the ratio of sides? What does that tell you about other ratios of areas?
Saying you can't do something isn't helpful, either to you or to us. You need to do something, look at what you've done, let us look at what you've done, and think some more. This is how problem solving works.
JeffM's approach, and my original suggestion, are routine approaches you can at least fall back on if you see nothing else to do. You have a creative approach, perhaps based on some examples you've seen, that has taken you almost to the end, and I have encouraged you to continue. So get moving!
Dr.Peterson
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Yes, that is correct. (And the method is very nice: EF:EC = 2:3, so x = 60, then just add everything up.)