Btw, I’m not actually sure what Anne’s intended solution is.
I figured out some other ways to solve it using Anne’s suggestion, but those relied on some of the same assumptions & things I had figured out in the course of my own solution. So her suggestion didn’t really make anything simpler for me, it was just another way to look at it. Which is fine - it’s good to have different ways to look at things. But it didn’t like point anything new out to me. I’m not sure if she intended something I’m not seeing, or if she was just relying on the same base existing knowledge.
If you’re stuck on the problem right now, you could try writing down the different things that you know about the image, and how you know them, all in one place. Try to write out your thinking about it, what you have used to figure out what, what things you are assuming, what things you were told in the problem itself, etc.
I’ve labeled points where the new lines intersect the hypotenuses of the “big” triangles we just calculated. These are all at the midpoints of the lines. Given that the length of the entire hypotenuse is 10\sqrt{5}, then the length of half the hypotenuse, such as represented by \overline{OC} and other similar lines in the image, is 5\sqrt{5}. We know that \overline{FC} has a length of 10. So we can get the height of \overline{OF} using the pythagorean theorem:
10^{2} + (\overline{OF})^{2} = (5\sqrt{5})^2
100 + (\overline{OF})^{2} = 125
(\overline{OF})^{2} = 25
\overline{OF} = 5
Since we know that the \overline{FC} and \overline{DF} are 10, we can make some calculations. \triangle OFC has a height of 5 and a base of 10, so the area of that triangle is 25. Since we already know that the area of a triangle like \triangle GDC is 100, that means that the area of the trapezoid GDFO is 75.
\triangle FOL is a right triangle with a hypotenuse of 5. So are triangles \triangle GKP, \triangle JHM and \triangle IEN.
The area of a triangle like \triangle OQG is \frac{10*5}{2} = 25. That’s overinclusive though compared to what we want, since it includes \triangle PKG. What’s the area of that triangle?
I tried this a bit but I think I need to organize stuff more and be more methodical or something. I’m having some kind of trouble doing it that’s hard to describe. I’ll post another update after taking another stab at it.
\triangle OFC is a triangle composed of two triangles, \triangle OFL and \triangle LFC respectively. The height of \triangle OFC is the same as one of the sides of \triangle OFL , namely \overline{FL}. We know a base of \triangle OFC , which is half the hypotenuse or 5\sqrt{5}. We know that its area is 25.
So 5\sqrt{5} * h = 25
and then h = \sqrt{5}.
So \overline{FL} is \sqrt{5}. Now we can figure out \overline{OL}. By the pythagorean theorem
So the area of \triangle OFL is \frac{2\sqrt{5} * \sqrt{5}}{2} or 5.
Now we can figure out the difference between a triangle like \triangle QFN and a triangle like \triangle OFL in order to find the area of the shaded square. We know that the area of \triangle QFN is 25. We now also know that the area of \triangle OFL is 5. So the difference is 20. There are four such sets of triangles in the image, so we multiply 20 by 4 and get 80.
The square has sides of 20. So \overline{HF} is 20.
Q is the midpoint of \overline{HF}. So \overline{QF} is 10.
Now I kind of want to continue and say that O is the midpoint of \overline{QF}, so \overline{OF} is 5. But then I wasn’t sure about it. With Q, well, the square has to have a center, so we can just say Q is at that center point. But can we definitively say that the midpoint between Q and F is on the boundary between the shaded square and \triangle GDC? I think I had an intuition that the answer was “no” which is why I went and used pythagorean.
I tried adding another line to my image in order to sub-divide the bottom half the square further into two halves:
What have you read from earlier in the math book? What problems have you done? What have you skipped? When did you do stuff (any significant time gaps)? Do you have any notes, trees, outlines, records, etc., for what you already did?
Have you taken a problem and tried to understand it in depth and solve it many ways before? What you’re trying to do right now … have you done that successfully with any simpler or easier problems? Do you have any practice at it?
I went through the entire pre-algebra book a few years ago. I took a couple of diagnostic tests from the people who make the book out of curiosity. The tests indicated that I had retained stuff from the first half of the book but not so much from the second half (which is more geometry focused). So I figured I’d try working on those. I have made some flash cards, mostly focused on 1) problems where I either made a conceptual error or where their explanation was significantly different than mine, or 2) geometry terminology. I haven’t really done additional notes besides those.
I have not. Have often tried to solve something two ways, often in order to check that the answer I got is correct before checking the answer in the book. But haven’t tried to solve something from a bunch of different angles. In terms of high effort math stuff, this is probably the closest thing I’ve done to what I’ve been attempting in the thread, but still it wasn’t quite the same cuz it was about gradually working my way up to understanding one solution
You don’t know the contents of the book well enough and are skipping too much. You also don’t have practice analyzing problems or good notes to refresh your memory.
