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Given the following image, image Determine the area of FEC given that the total area is 1 area unit. The correct answer should be 1/12 a.u. but I cannot get all the way to that conclusion. Note, one's not allowed to use sin or cos, which would make for an ugly solution any how..

The obvious parts here are that FEC and ABF are mirror images. Resulting in proportional lengths for each. Meaning:

FE/FB = FC/FA = EC/AB

Area of Δ = base*height/2

The base being EC=AB/2, and the height being..?

I'm certain that I've missed some vital part of the puzzle. Any hint the the right direction would be much appreciated.

Best regards

Raoul
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  • Is E the mid-point of CD? Is ABCD a square or can it also be a rectangle? I think you are missing some information here. – user625 Jun 14 '18 at 21:45
  • Yes, ED = AB/2, so half way. And ABCD is a square specifically. Though a general solution would be interesting too. – Raoul Jun 14 '18 at 21:50
  • I added solution for square below. Same logic (and answer) applies to a rectangle also. Just assume the adjacent sides are l and w (instead of 1 each) and follow along. – user625 Jun 14 '18 at 22:15

2 Answers2

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I assume ABCD is a square and E is mid-point of CD. As you found out FEC and FBA are similar triangles. Now EC = AB/2, thus the height of FEC = half of height of FBA. Also, height of FEC + height of FBA = 1 = length of side of the square. Thus, height of FEC = 1/3. We already know EC = $\frac{1}{2}$. So, area of FEC = $\frac{1}{2} \frac{1}{3} \frac{1}{2}$ = $\frac{1}{12}$

user625
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If we draw the line DF and then create four triangles, AFD, DFC, CFB, and BFA, and pair together the opposite triangles (AFD with BFC; CFA with DFC), we can easily realize that these pairs both compromise half of the total area of the square because each base is one and if you sum the heights they sum to one, meaning the combined area is 1*1*1/2 for each of these pairs. Now, using mass geometry and assuming each corner point has mass one, point E has mass 2, meaning the ratio of BF to FE is 1:2. Now note that AFB is similar to CFE due to AAA (because opposite sides of squares are parallel and then AC and BE are transversals). The ratio of similarity is then 1:2, so the ratio of areas between AFB and DFC is 1:4. Note that since EC = DE, that the area of EFC is 1/2 of the area of DFC.

Now, we know from the beginning that AFB + CFD is 1/2 and DFE has the same area as EFC, so this can be rewritten as AFB + 2*EFC, and because the ratio of the areas EFC to AFB is 1:4, then EFC can be rewritten as 1/4*AFB yielding AFB + 1/2AFB = 1/2. Simplifying this equation yields AFB = 1/3.

Now we're almost done, because AFB + 2*EFC = 1/2 and AFB = 1/3, then 2*EFC = 1/6 so finally EFC = 1/12

This may not be the most time-efficient solution however as far as I'm aware it's fully rigorous

Michael
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