Find the area of the square

Oh no! There goes a bunch of time trying to figure that out.?ÿ That site frequently leaves out critical information that you have to assume.?ÿ Here, I'm going to assume all intersections are right angles, and the square they are asking about is the total figure.

@bill93

correct assumptions. ?????ÿ

The first conclusion I reach is that what looks like little squares are not actually square, because that leads to contradictions.

Area=144

This is one of those problems where it is much easier to ignore the given dimension and give some other line an assumed dimension. Work that problem and scale the result to the given info. I assumed a dimension for the height of the little almost-square.

Edit: use a letter for the width of the almost-square, and you can easily solve for it after making some of the figures the same area.

One of those problems that's not terribly hard but will suck up an hour that could be spent watching TV.?ÿ ????ÿ

Some guy, junior hs teacher or maybe hs, on YouTube had a polygon shaped sort of like a C with overlapping dimensions and some other piece of information and I said screw that, tell the architect to supply the missing dimensions because Iƒ??m not taking responsibility for it.

an hour that could be spent watching TV.

Watching somebody be nasty to someone else? Or otherwise people with problems I can't solve for them. That's most of what seems to be on tv and I'd rather work on something g I can solve.

@bill93

I suppose it depends on what you decide to watch.

Edit: use a letter for the width of the almost-square, and you can easily solve for it after making some of the figures the same area.

Each of the colored figures within the square are equal in area. ?????ÿ

144 is the correct answer, which means each color is 28.8 in area. 28.8 divided by 3 is 9.6 for orange.?ÿ That leaves 2.4 by 12 or 28.8 for blue.?ÿ Then yellow must be 28.8 divided by 9 wide or 3.2.?ÿ Pink and green must be equal so 9.6 minus 3.2 leaves a width of 6.4?ÿ and a height of 9 divided by 2 or 4.5 for each.

@flga-2-2?ÿ

I said "some" because you don't need to do anything with blue except as a check.

Call the height and width of pink h and w. All have area hw and the whole 5hw.

Green shares w so must also have the same h.

Then yellow must be 2h by w/2.

So orange width is 3w/2.?ÿ Its area hw=3*3w/2 so h=9/2.

Yellow height then is 9 and the whole figure 12, so area is 12 squared =144.

You can go ahead and check by putting numbers on all the Individual figures.

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The catch is that we like to assume all dimensions are integers.

I used simultaneous equations: 3 equations, 3 unknowns

I called the area of each polygon "y". The long orange side "x", the short blue "a" and the long blue "b".

3x=y

ab=y

(x+a)b=5y

The only integer that solves is b=12 and since it is a square the area=144

y=12a and x=4a?ÿ

@antcrook

Slick.?ÿ Nothing given says it has to be integer, but your equations work.

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Or.................you recognize that the three colors below the top one have the same total width as the top one so must have three times the height of the top one.?ÿ Thus 3 plus 9 is 12.?ÿ The answer to the question is 12 times 12 is 144.?ÿ?ÿ

if you like, you can calculate each one out to prove your answer is correct.

Had their been six different colors below the top one, then their would be 3 times 6 = 18 beside them?ÿ Then add the three above for a total of 21 on that side?ÿ Then 21 times 21 is 441.

@holy-cow I like this reasoning and solution very much.