Square Puzzle

You can't do it with simple whole squares, but if it is allowed to assemble the 7 smaller squares from triangles, then it can be done. It might be an interesting challenge to do this with the minimum number of pieces, or with the fewest number of different shapes.

If you search on "geometric dissections" you might find some good ideas.

y^2=7*x^2

y^2/x^2=7

y/x= srqt(7)=2.6457513....

So, since y/x is not a whole number, it is not possible.

Not all 7 equal to each other, but 7 squares:

Nothing against usbport, but who voted this as a GA? The question is very clearly stated. It's asking for "equal squares". This solution doesn't solve the problem in any way whatsoever.

cute

I was thinking of seven different sizes, but the three of one size and four of another looks more elegant. Now, he says he wants seven equal size squares. Even though they wouldn't fit together to make a square, you could take the total area, divide by seven, and then make a square using that 1/7 area. I think we need more clarification.

A bit late to get into this Usbport

I kept counting but always found 8 squares.

I once made a 7 cornered frame for my boss when I was working in picture framing. I had young Charlie as an apprentice. He said it was impossible but I showed him how to do it. He was the co-creator so it was an Xmas present from us both. The boss couldn't believe it either but he had two left hands at the best of times.

A few circles here and there, a straight edge and jigs and bobs your uncle. Same goes for glass or cutting in marble floors. My back hurts just thinking about it. Stuff all the numbers, if the customer asks for it its gotta be done by some one.

Try it, its fun when it works out, Ky.

Basis: Area of square = 100in²

100in²/7 = 14.28571in²

The square root of 14.28571 = 3.780.

Your seven squares are each 37.8% on a side compared with the original.

NO

I do not think that it is possible to divide a square into a prime number of equal squares.

I was thinking "odd number" but that too.

It seems to me the 'logic' is; the divisor must be even* or the square of a whole number.

But I could be wrong - just something about the 'square concept' seems to whisper that to me

_{*fixed in edit}

You are right. It has to be a number that is a square, odd or even. Sudoku has 81 squares, consisting of 9 squares each with 9 small sqaures.

I do not think even number per se work. For example, one cannot divide into two equal squares or eight.

Cancelled. (My question was already answered.)

4², 6 ..... _{^{Wot? canceled - rats}}

We don't know yet if the puzzle allows the smaller squares to be assembled from yet smaller pieces.

As has been noted, if you divide a square into smaller equal squares whose total area equals that of the original square, then the number of smaller squares must be a a perfect square (4, 9, 16,...). But that is too trivial to be of interest.

In the present puzzle, you can get at most 4 smaller squares of the right size; the remaining 3 squares must be assembled from yet smaller rectangles or triangles. I haven't yet found an elegant solution for this, such as the smallest number of pieces or the smallest number of different shapes.

Well I got insecure and 'edited in' "even number" thinking of 'even sides' but as "8" pointed out that flaw, I'll just stay with 'square of a number'.

The cutting to triangles is interesting. Essentially a diagonal division. I guess the target is 28 x ¼ squares, or 56 x ^_1/8 squares - but all I get is multiples of 4. LCM seems a way lot of cutting

Of course if we are cutting, we could assume a scrap allowance, so divide into 8 and just chuck the scrap one?

But on thinking further - it might be done if you divide the area by 7 then draw 6 squares of that 1/7th area, then divide the remaining 'border areas' (side and top) into squares the size of the remnant in the top corner - which should then assemble into the 7th square.

Good point! (Although I agree with somebody else, I suspect Usbport's answer is the desired answer, with some corruption in the question.)

Taking your approach, maybe only 4 squares can be drawn directly (calculation below), then 3 squares assembled from the left over bits.

Just using a 10x10 square, area = 100, 1/7 of that is 14.285..., so to make a square of that area, take the square root (= 3.779...), so in a 10x10 square you can draw 4 "whole" squares 3.779... x 3.779 (in a 2 x 2 configuration).

Could you do something with the remaining area to make 3 more squares--something like creating triangles and (short of grinding it up in a blender and "painting" 3 more squares).

From some quick sketches it doesn't look easy / straightforward.

Well I had some time so;

Gets a bit "mouse milking' in the last annoying bit.

Total 4900 mm² - 4 x 700 mm² + 4 x 350mm² half squares = 2, so total 6 equal squares, at first pass.

And, as said, 'building the remnant on the corner' - now demonstrated = 1 = total 7

But the way I did it: it's just down to getting out the very sharp scissors and microscope, because to get the tiny 45⁰ pointy bit I had to cut one resultant 90⁰ end in a half mm wide bit, at 45⁰, at the mid point, and glue it back on.

Perhaps the new competition can be doing 'more elegantly'

Very difficult to have a square as remnant.

The nearest I got was 1764 / 7 = 252 which is the square of a fractional number 15.87450786638750...?.

A square number is X = a²

(X/7 - Int(X/7)) = b such that 6 * b = Int(X/7) and Int(X/7) = c²

Therefore, 6*c² + 6b = 6*6*b + 6b = 6b(6+1) = 7*6*b = X or = a²

Then 6*b = a²/7 and b = a²/42

Take b=42

Then 6*b = 252

Then a² = 7*6*b = 1764 this is the square

Divide this 1764 by 7 = 252

And 252 =(15.8745078663875000??)²

Sajal,

Bravo88 and Mikerho have shown why it's impossible (Good Answers).....

Do you need any more explanation?

Tornado and USBport have also had good ideas. (I bet USBport's is the answer and the question has somehow been corrupted).

Forget 7, can you divide it in two equal squares? answer is no.

you can only divided it in equal pieces of 4, 9, 16, 25, 36, ---------

If the square is a market square. It can be divided in to equal area, equal number of shops, or maybe equal in the fact they all have one fountain. Each one named as a separate market square.

Ok play on words But it can be done!

Easy, divide it into 8 squares & throw away the spare bit. The OP says the 7 squares have to be equal but not that the total has to equal the original square.

(OK, I know I'm clutching at straws here).

Think about it. If you are dividing by cutting the square, one edge of the smaller square is the same as the edge of the original square. So the length of the edge of the original square is a multiple of the length of the edge of the smaller square.

Let's call the multiple, n.

And, as we are cutting squares, the multiple along one of the adjacent edges is also n.

So we are cutting the original square into n*n smaller pieces, or n-squared.

(As has been said elsewhere in the thread, you can only cut it into a square number of equal sided squares)

So unless it is a carpet-remnant problem where it is allowed to stitch together separate pieces...

Divide it into 9 equal squares, and hide two of them.

or since, I prefer dividing by 2's than 3's, might as well divide it into 16, keep your seven, and I'll take the other 9.

ok, since my answer of "NO" wasn't considered very good, go to a search engine of your choice (google, bing, yahoo, etc) and look for a square root calculator and enter 7 then find, which would be the needed length of one edge of a square. If you don't understand that answer, make sure you find a calculator that has about 100 decimal points. Dang, I forgot to ask, is this a metric square? Maybe they have different rules for them.

I wouldn't rule out your answer of "NO" as not being a good answer. It may indeed be the only correct answer.

"is it possible to divide (a) square into 7 equal squares?"

It turns out the answer is simple, no. But you can cut it into smaller squares that can then be assembled into 7 equally sized squares. It takes a minimum of four squares to assemble into a larger square, or think of it as 2 squared. The next multiple would be nine, or 3 squared. All we have to find is the lowest multiple of 4, or 9, or 16, or 25, etc., and our key value of 7 that is the square of a whole number. Oviously, 7 times 49 works, but is there a smaller one?

Okay, time to earn my name. Tell me if this satisfies all the requirements:

1. You take a 7" x 7" square piece of paper.

2. You draw 6 parallel lines on it, 1" apart.

3. You fold it on each line in a zig-zag fold.

4. This gives you a strip that is 1" x 7".

5. You draw 6 parallel lines on this, perpendicular to the folds, 1" apart.

6. You crease on each line, in the same direction, and form a heptagon.

7. You now have 7 equal squares, forming that heptagon.

Nope, when you divide something, you should always be able to add it back together to be whole (not me , Einstein said it). You trying to sell that 7 one inch squares would be equal to 49 square inches that the original square was.

I like it origami!

"Satisfy"? except for the part where you make it 3D - so now have 14

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hello everyone,

after reading all of your answer it is look like it is not possible

or possible in the squares of numbers or in whole or if it is unequal squares.

now if question is changing little bit let now we have to divide the square in 7 parts in which some of them are equal squares and some of them are equal rectangles and only two values of area is possible one for square and another one for rectangle.

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Haha, bravo88 answered this question perfectly some 30 posts ago... GA

In Euclidean 2-space, it can't be done, as others have ably shown previously, however...

In mathematics, a square could easily have sides named ''s'' such that the area

of such a square would be such that A = s^2.

And, one seventh of such an area could be expressed mathematically as:

A/7 = (s^2)/7

Then, the side of such a square would mathematically be of a length;

l = _/\/(A/7) = _/\/((s^2)/7) = s / ( _/\/(7) )= ( s _/\/7 ) / 7

which has one mathematically real root, and one imaginary root.

This is a mathmatically sufficient solution, but not a physically sufficient solution...

So, yes it is possible in math, but not in two dimensional real space, really...

Hmmm imaginary root... is that like the imaginary collateral that lead to the banking collapse not so many months ago? Let just imagine we gave you a score.

This is a 2-dimensional problem that doesn't involve high-dimension hypersquares, isn't it?

....or put another way, can 7 equal squares be arranged into 1 big square.......

hello everyone,

everyone said that division of square into 7 equal squares is not possible but now i am chnaged the problem in comment no. 35

But no one comment on it what happened it is also not possible with some number of squares or rectangles????????/

if anyone know the logic please help me.

Like one of these?

thank you randall

for your answers

you are very quick