This guide does not give you the answers, but rather explains in a rather terse way how to solve it logically, without using algebra or fractions, with simple area formulas (rectangle, triangle), whole number ratios, and logic. It's more like a set of hints than an explanation, and I recommend doing the work as you read these notes. Ideally, work through a few puzzles that you've already solved to understand this approach better.

Pythagoria sometimes uses hatch marks like − and = to indicate sides of equal length.

I don't use pictures because it would have tripled the effort and given away all the puzzles (copyright!). If you want pictures, use the game. You can open the game's data.dat file with 7zip[www.7-zip.org] or similar; the puzzles png images are in the texture\puzzle folder. I lifted the icon for this guide from the same file.

If you see a better, more logical way to solve a puzzle, please leave a comment!

The "missing" rectangle in the lower right corner is 2 cm by 3 cm = 6 cm². It adds up with the 10 cm² rectangle to 16 cm². Together, they' have the same width as the 16 cm² rectangle, and because they're the same size, they must have the same height as well.

The 16 cm² rectangle is 11-3=8 cm wide; thus it must be 16/8=2 cm high.

Compute by division the height of the 15 cm² rectangle, the width of the 6 cm² rectangle, and the height of the 10 cm² rectangle.

Compute the area of the outer rectangle as 3*5=15 cm²; subtract 14 cm² to get the remaining "?" area.

Width ratios: 15:10 = 3:2 = 9:?. (Also works with height ratios.)

Compute the "missing" upper left rectangle to 5*9=45 cm². That makes the upper row 45+35=80 cm². Since that is the same area as the lower row and has the same width, the lower row has the same height as the upper row.

"?" is twice as wide as the 4 cm² rectangle and has the same height, so it must have twice the area.

Comparing the 13 cm² and 26 cm² rectangles, we see that the lower row is half as high as the upper row. Thus, the "missing rectangle" in the lower left has 16 cm². That is twice as much as the 8 cm² rectangle, which has the same height, so that must have half the width.

Because heights and widths are equal, 8+7+6 = 7+6+?.

The part that juts out at the top is 3 cm by 5 cm = 15 cm². Subtracting that from the 20 cm² leaves a 5 cm² rectangle at the bottom. Because it is as high as the 5 cm² rectangle next to it, they have the same width = 3 cm. So the "?" rectangle is also 9-3-3 = 3 cm wide, and has the same area.

The 6 cm wide part has an area of 3 times 7 cm². Since 2 cm is a third of 6 cm, the "?" is a third of the wider area: 6:2 = 21:?

Draw a vertical line at the right end of the 10 cm² rectangle that bisects the "?" rectangle. The right part of the "?" rectangle is 2 cm by 1 cm. The left part of the "?" is as wide, but half as high, as the 10 cm² rectangle and thus has half the area. Add to find the solution.

The 6 cm by 2 cm rectangle at the bottom has an area of 12 cm². That means the small rectangle in the lower right corner is 12-9=3 cm². The "?" rectangle above it is as wide and twice as high, therefore it has twice the area.

Extend the vertical line in the center into the upper rectangle, splitting it into a left rectangle 3 cm by 3 cm = 9 cm² and a remaining 20 cm² on the right. Since it has the same width and area as the 20 cm² below, the heights are also equal, and therefore the left-hand rectangles are equal as well.

Since the two 3 cm² areas have the same width, they also share the same height: both rows of the drawing have the same height. Draw a vertical line at the left end of the 2 cm² area, bisecting the "?" rectangle. The left part is the same as the short upper row, the right part is equal to the 2 cm² rectangle above. Add.

The "missing rectangle" in the upper left corner is 4-2=2 cm wide, so it must have 10 cm² like the rectangle beside it. Summing up the 3 rectangles on the left gives 10+10+8=28 cm². The right rectangle has twice the area (2*28=56) and the same height, so it must have twice the width.

Comparing the column at the right with th column on the left, we see from the first row that its width must be halved. That means there are 10 cm² at the right divided into the 1 cm² marked at the bottom and 9 cm² unmarked above. Since the center column also has 9 cm² in the center row, it must be the same width as the right column, wihch means that their areas correspond 1:1.

Because the 12 cm² rectangle is 4 cm high, it is 3 cm wide ⇒ the 20 cm² rectangle is 5 cm wide ⇒ the 20 cm² rectangle is 4 cm high ⇒ the "?" rectangle is 3 cm high ⇒ compute the area.

The 6 cm² rectangle is 3 cm wide. 15+12 =27 cm² on 9 cm width in the lower row. 9:3 = 27:?

Bisect the 7 cm² rectangle vertically so that it has 2 cm² on the left and 5 cm² on the right. Because the 5 cm² section and the "?" are 6 cm wide (see upper measurement), the width of the 2 cm² section must be 2 cm. That makes its height 1 cm. It follows that the width of the 5 cm² rectangle is 5 cm; subtract from 6 to get the width of the "?" and multiply with the height.

The height of both triangles is the same, thus the area is proportional to the width of the base.
3:6 = 4:?

Since each green triangle pairs up with a white triangle it is symmetric to, the yellow triangles comprise exactly half the area of the outer rectangle, which is 4 by 3 cm.

There are various ways to solve this with ratios, here's one: Calling the area of the bottom unmarked rectangle X, we have 9:3 = 12:X ⇒ X=4. The, going up, 6:3 = X:?.

The white strips have areas of 1*5 and 1*3 cm², respectively, and they overlap by 1*1 cm² ⇒ they cover 5+3-1=7 cm² of the 3*5=15 cm² outer rectangle. Subtract to get the yellow area.

Extend the side of the 5 cm² rectangle to bisect the 12 cm² rectangle. The right part of it is 2 cm by 3 cm = 6 cm² ⇒ the left part is 12-6 = 6 cm² as well ⇒ they have the same width, 2 cm. That makes the "missing rectangle" above the 5 cm² rectangle have 2 cm by 2 cm = 4 cm². Add to the 5 cm² rectangle to make 9 cm², which is half of 18 cm² ⇒ the 18 cm² rectangle must be twice as wide ⇒ 4 cm width. Add the extra 2 cm noted on the upper left to get the answer.

Assume that the red area is square. Draw the diagonals of the outer square (they're also diagonals of the red square). You will now see 16 equal triangles, 4 of which are red.That makes the red area exactly one fourth of the area of the outer square.

Flip the white triangle down into the blue area, dividing it into 3 equal 1 cm² triangles. Iso lateral triangles are fun!

Extend the vertical lines to bisect the 16 cm² and the "?" rectangle. The left part of the 16 cm² rectangle is 3*4=12 cm², so the right part is 16-12=4 cm²; since the rectangle below has half the area of that above (compare 12 and 6), the left part of the "?" rectangle has 2 cm². The right part is the same as the 6 cm² rectangle on the left. Add.

The left stack is 32 cm² and its height is 8 cm, so its width must be 4 cm. That makes the height of the 12 cm² rectangle 3 cm. Divide the 6 cm² rectangle by its 2 cm height to get 3 cm width. Compute "?".

Draw the "missing rectangle" in the upper right corner. Together with the 1 cm² rectangle below it, the area is 3 cm by 1 cm = 3 cm², which means the missing rectangle is 3-1=2 cm². That's the same area as the rectangle in the center; as it has the same height, its width is also 3 cm. Since the 3 cm² rectangle has the same height as the 1 cm² rectangle, it must be thrice as wide, making it 9 cm wide. Subtract the width of the 2 cm² rectangle which we computed before to get the width of the "?" rectangle. Its height is given, compute the area.

Draw the enclosing 6 cm by 8 cm rectangle. You now see 8 equal triangles. Divide the area of the outer rectangle by 8 to get the area of 1 triangle.

Because the two 5 cm² rectangles have the same width, they also have the same height: 2 cm each, the same as the height of the bottom "?" rectangle. Extend the vertical lines to trisect the "?" rectangle. Compute the caps from width and height. The center must be the same 5 cm², because of equal width and height 2 cm. Add.

Complete the 5 cm by 6 cm rectangle. The "missing corner" must have 30 cm² - 20 cm² - 5 cm² = 5 cm² as well. That means the vertical line is centered and bisects the 6 cm in the middle, making the "?" rectangle 3 cm wide. Together with the 5 cm² rectangle, it forms a rectangle 3 cm wide and 4 cm high of 12 cm²; subtract the 5 cm² to get the answer.

The "curve" is supposed to be a semicircle, making the height of the figure 2 cm. Add the two colored segments to form a triangle. Add a vertical center line to see 4 of these triangles, making the are of one exactly a fourth of the rectangle's area.

If you double the width of the 1 cm² rectangle, you get the 2 cm² rectangle shown in the drawing ⇒ they're the same height ⇒ the 18 cm² rectangle and the "?" rectangle are the same height ⇒ the "?" rectangle is half as wide, so it has half the area.

Draw a cross in the center, see 8 equal triangles, 2 of which are colored in. This makes the blue area be exactly one fourth of the rectangle's area.

Compute the base of the 4 cm² triangle to be 4 cm (formula: area = base * height / 2). The 4 cm² and 6 cm² triangles make a 10 cm² triangle with the same height and twice the base as the red triangle ⇒ twice the area.

Extend the right edge of the 3 cm² rectangle so it bisects the 8 cm² rectangle into a 2 cm² cap and a 6 cm² rectangle below the 3 cm² rectangle. Because they share the same width, the 3 cm² rectangle must be half as high = 1 cm. That makes the 2 cm² rectangle 2 cm wide, which adds up to 4 cm width for the "?" rectangle. Its height is given ⇒ compute the area.

The outer 7 cm square is 49 cm², the inner 5 cm square is 25 cm². That means the 4 outer white triangles are 49-25 = 24 cm², and since they're mirror images, the inner triangles have the same area. Subtract to get the "?" square.

Add the "missing" 3 cm by 4 cm rectangle next to the 6 cm² rectangle. Note that its twice as big as the 6 cm² rectangle, so the other side of that is 1.5 cm, making a length of 4.5 cm for one side of the 29 cm² rectangle. Extend the edge of the "?" rectangle to bisect the 29 cm² rectangle. The lower part has 6 * 4.5 = 27 cm², leaving a slice of 2 cm² that is 4.5 cm wide. The "?" rectangle has the same height as that slice, but double the width, so it has double the area.

Draw a vertical line down the center. Assemble the parts into a triangle that is half the size of the outer square.

The small triangle on the right is the 3-4-5 right triangle, the smallest right triangle with integer sides. Use the Pythagorean theorem to compute the vertical line in the center to √(5²-3²) = 4 cm. Take it as the base for the "?" triangle, with its corresponding height of 3 cm.

The big 2cm by 6 cm rectangle on the bottom is 12 cm², leaving 12-9=3 cm² for the small unmarked rectangle. That's the same size as the 3 cm² above, so both rows have the same height, making the "?" rectangle the same as the one below it.

There's a lot of misdirection in this one, just look at the 4 rectangles on the bottom right.
8:12 = 2:3 = 6:?
You need the other rectangles if you want to know its height and width.

The "?" triangle and the 4 cm² triangle have equal angles, which makes them scale versions of each other. The area scales with the square of it defining lengths, so since the base lines are 6:9 = 2:3, the areas are 2²:3² or 4:9.

Extend the diagonals to bisect the two big rectangles and add the missing 4 cm by 5 cm rectangle at the bottom. The "outsides" of the bisected rectangles can be computed, and the insides found by subtraction to be 10 cm² and 18 cm². With the 20 cm² "missing" rectangle, we can do ratios again, such as 20:10 = 2:1 = 18:?

You can mirror (rotate 180°) the drawing about its center point, which proves that the white and red ares are equal, and thus the red area is half the total area of the rectangle. Since all of the small right-leaning diagonals are parallel, the center rectangles must be the same size as the outer rectangles, making the whole are 4 times the area of the indicated square.

Bisect the 10 cm and draw a horizontal line across. This bisects the 26 cm² rectangle into two 13 cm² rectangles that are 5 cm high, same as the one on the bottom left; they must have the same width. Thus, the 38 cm² rectangle has the same width as the 12 cm² rectangle, and we cut the mirror image of the latter off the former when we drew the horizontal, leaving 26 cm² in the top part of the 38 cm² rectangle. Since this is twice as large as the 13 cm² rectangle to its right (the top half of the 26 cm² that we split), it must be twice as wide, which makes the whole drawing 3 times as wide as the right column. Since the left column has the same width, the center column has as well. We can now extend a vertical to split the 12 cm² rectangle, and since we know the width on its left and right are equal, the parts are both an equal 6 cm². The left part and the "?" rectangle are the same size as the 13 cm² next to it (same width), and subtraction gives us the answer.

The green triangle has the same base and twice the height of the 4 cm² triangle, so it must have twice the area.

This one stumped me initially. Measure the 6 cm at the top starting from the right edge of the "?" rectangle going left. That means adding another 10 cm² on the upper left to make up the full 15 cm², and adding another 5 cm² at the bottom left to make up the full 2 cm by 6 cm = 12 cm². From this we can see that the height of the upper row is twice that of the lower row, since both additions have the same width but differ by a factor of 2. This means that the area above the 7 cm² must be 14 cm²; subtract the 5 cm² and there's the answer.

Extend a horizontal line to show the 2 cm by 8 cm 16 cm² rectangle; the nely created rectangle in the lower right must then have 16-12=4 cm². Since that is the same area and width as the 4 cm² rectangle at the top, that one must have a height of 2 cm as well. Because the top and bottom row are each 2 cm high, the center row is also 6-2-2=2 cm high. Since the right column is 2 cm wide and 6 cm high, its area is 12 cm²; add to that the 12 cm² rectangle and the 10 cm² rectangle for a U-shaped area of 34 cm². Since the outer rectangle has 6x8=48 cm², that leaves 48-34=14 cm² for the inside of the U, which is the "?" rectangle and the one above it Since both rows are the same height, these have the same area, which is half of 14 cm².

Because the red rectangle is 3 cm high, its upper edge bisects the diagonal line in its center, which is also the center of the undrawn 6 cm by 6 cm square. Thus, the red area is a quarter of that square.

The big 6 by 6 cm diagonal bisects the horizontal line in the center, making its right side 3 cm long. This is the base of a down-pointing triangle that shares a tip with the yellow triangle and has the same corner angles. Because the ratio of their base lines is 6:3, the smaller triangle is scaled down by a factor of 2; this also applies to its height, which means the smaller triangle has a heightof 1 cm and the yellow triangle has a height of 2 cm, making up 3 cm total. Now the height and base of the yellow triangle are known, and its area can be computed.

Extend the horizontal lines to trisect the 23 cm² rectangle. Since its top and bottom are 3*3=9 cm², the center part is 23-9-9=5 cm². The 17 cm² rectangle and the unmarked one to its right have the same size as the 26 cm² rectangle, which makes the unmarked rectangle 9 cm². Because that's the same as the 9 cm² that's the lower part of the 23 cm² on the left, the "?" rectangle corresponds to the center of that.

Because the "?" triangle is an isosceles right triangle, all of the angles in the drawing must be either 45° or 90°. Symmetry then makes the 2-slash line exactly half as long as the 1-slash line (draw the other diagonal of the square to see it), and that makes the short sides of the "?" triangle 2 cm long.

Flip the green area over to make a green triangle exactly half the size of the square.

Rotate the center square 45° and draw its diagonals to see it is exactly half the size of the outer square.

There are two blue triangles with a common (vertical) base line of 2 cm and heights that add up to 8 cm. That means their area adds up the area of a single triangle with base 2 cm and height 8 cm.

If the 38 cm² rectangle was only 3 cm wide, its area would be 19 cm². Extend the right edge of the "?" rectangle down to make such a rectangle. The height of the 35 cm² rectangle is 5, so the left part of it is 3 cm by 5 cm = 15 cm², and adds up to 19 cm² with the "?" rectangle. Subtract for the answer.

Cut 1 cm off on the left of the "?" rectangle to make a 2 cm² cap, leaving 6 cm of the indicated width, same as on the bottom. Because the rectangle at the bottom right has the same width as the capped part of the "?" rectangle at the top left, but is three times the height, its area is also three times as large, which makes the capped part 21/3=7 cm². Add the cap for the answer.