A Math Forum Project

Geometry Forum - Problem of the Week

    Solutions - Squaring the Square, Feb. 6-10, 1995

    Annie says:

    This week's problem was indeed inspired by the cover of Scientific American, circa November 1958. It's entitled "'Perfect' Rectangle" (and the Scientific American cost 50 cents!). The solution to 'squaring' the rectangle that we used was the first published solution to the problem, as it is as close to a square as you can get (as you can see in the solutions).

    I won't tell more about the problem, in case some of you want to try your hand at other solutions. A number of students did state that you could generate other rectangles split into squares by doubling the edgelengths of the squares in the original problem. Good answer, but I think we'll look for alternative answers before awarding the t-shirts!

    Dan Asimov adds:

    There do exist "squared squares" -- contrary to the apparent intent of the above quote -- i.e., squares composed of square tiles all of different side lengths, all oriented orthogonally. An article about these appeared in, I think, the first collection of Martin Gardner columns, the Scientific American Book of mathematical Puzzles and Diversions. (Or maybe it was the second.) --Dan Asimov

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    Bill Becht, Bill Glisson, Mike Donovan, and Tim Pappas

    Here is our answer to the problem of the week. We are an eighth grade class from Drexel Hill School of the Holy Child. Our names are Bill Becht, Bill Glisson, Mike Donovan and Tim Pappas.

    We had your drawing on the board and we filled in the measurements as we figured them out one square at a time. You told us that the area of C is 64, so that means that each side of C is 8. And you told us that the area of D is 81, so that means that each side of D is 9. We used this information to figure out the rest. Then we figured out I because we had some of the information we needed.We said that 8 is one less than 9 so I has to have an area of 1 because 1 x 1 = 1. Next we figured out H . We knew that H is one less than C so H must have sides equal to 7 and an area of 49. The sum of D and I =E so 9 + 1 = 10. So E must have sides of 10 and an area of 100.Next on to square G. We know that H - I = 7-1=6 and 6 + ? = 10 so the side of G must be 4 and the area of G , 16. Now we figured out F . This was easy.Just add E = G (10+4)to get 14 on the sides and an area of196. Then we were able to get A, which is the same as F + G (14+4) = 18 on each side and an area of 324.Last but not least we figured out B. Since the side of B= H+C( 7+8) = 15 . B has an area of 225 and sides of 15. Now that we figured out all the separate squares we just added the sides to check and see if this is a square or a rectangle. (We guessed a square). On the right hand side we added 9 + 8 + 15 = 32 and since this is either a square or a rectangle the opposite side is also 32. But, we checked by adding 14 + 18 = 32. Now to decide if it's a square or a rectangle we have to check the top and bottom measurements. We added the top first 18 + 15 = 33 Then we added the bottom 14 + 10 + 9 = 33. To be a square all sides must have the same measure. And 33 does not equal 32. We just proved that this figure is NOT a square. It is a RECTANGLE.

    Mike Veltri and friends

    Hey there it's me Mike Veltri and friends here to tell you that we figured out the answer to the problem of the week for Feb. 6-10 we got 33 for the horizontal height and 32 for the the base we got this answer by finding out the squares of all the squares and adding them together, so the answer is NO it is not a perfect square, it is a rectangle.

    Eric Chen

    Hi, I am Eric Chen. I am a student in Burnaby South Secondary School. I am in grade 11. This is my first time to send the answer to week problem. I get the answer for the first one. The squares consist of the diagram is not the square. The length of squareA, B, C, D, E, F, G, H, and I are 18, 15, 8, 9, 10, 14, 4, 7, and 1 unit beacuse I figure out from the imformation(the area of C is 64 and the area of D is 81) of the quation. Then the lengths of the diagram are (18+14=)32, 32, (15+18=)33, and 33. From the steps I get the diagram is a rectangle so the diagram consisted of these 9 square is not a square. I wish you can understand what I have written. Hope you have a good weekend.

    Imran Hayat, Nancy Chen, Anita Lau, and Pawel Gobis

    Since the area of C is 64 and the area of D is 81, we can calculate the length of each (equal) side to be 8 and 9, respectively. The length of I can be found by subtracting D's length by C's length, and therefore getting 9-8=1. H's length will hence be 7 as it will be C's length minus I's length, which is 8-1=7. E's length can be found by adding D's length and I's length, resulting in 9+1=10. B's length can now be calculated by adding C and H's length, therefore 8+7=15. G's length can be found by first calculating the length of I subtracted from H, this being 7-1=6. This 6 is then subtracted from 10 (E's length) to give 10-6=4, therefore 4 is the length of G.

    F's length will be G's length added to E's length, hence 10+4=14. A's length can be found by adding F and G's lengths together, resulting in 14+4=18. There, we have now found all of the sides of each square!! :)

    For a recap, the (equal) sides of each square are:

    A=18, B=15, C=8, D=9, E=10, F=14, G=4, H=7, and I=1.

    We can now find the length's of the sides of the BIG rectangle (or square) by adding A and B to give 33 and by adding B, C, and D to give 32. Therefore, the enclosing shape is not square but is a rectangle, 33 units x 32 units in area.

    -=-=-=-=-=-=-=-=-=-

    For the second part of the problem, we tried and tried but could not come up with another square or rectangle with each square's length diffent inside it. I suppose you could double (or triple, etc..) each square's length to give a similar square doubled or tripled in size, but I suppose, if this is correct, it will be cheating. (I suppose)

    This problem done by Imran Hayat, Nancy Chen, Anita Lau, and Pawel Gobis of the Experimental Math class at Burnaby South Secondary School.

    David Love

     __________18__________ _______15___ ___
|                      |                |
|                      |                |
|                      |                |
|18                    |                |
|                      |                |
|           A          |        B       |15
|                      |                |
|                      |                |
|                      |                |
|                      |                |
|                      |___7____ __8____|
|_____14_________ ___4_|       |        |
|                |     |   H   |7       |8
|                4| G  |       |    C   |
|                |_____|_______|        |
14               |        I->|1|________|
|        F       |           |          |
|               10     E     |9         |
|                |           |    D     |9
|                |           |          |
|______14__ _____|____10_____|____9_____|
    The area of the whole rectangle is 1056 (32 times 33). The area of A is 324. The area of B is 225. The area of C is 64. The area of D is 81. The area of E is 100. The area of F is 196. The area of G is 16. The area of H is 49. The area of I is 1. The dimensions of the rectangle are 32 by 33. Other squares and rectangles can be split into squares with different areas by simply doubling, tripling, quadrupling etc the values of all sides. An example rectangle would be 66 by 64 where all lengths are doubled. The area would be 4224.

    David Love
    Akiba Hebrew Academy grade 10
    Merion PA

    Aron Freidenreich, Ariel Collins, Daniel A. Gabriel, and Lili Grunwald

     ______________________ ________________
|       18             |        15      |
|                      |                |
|                      |                |
|                      | 15             |
|18                    |                |
|           A       18 |        B     15|
|                      |                |
|                      |                |
|                      |                |
|                      |         15     |
|        18            |_______ ________|
|________________ _4___|    7  |    8   |
|      14        |     |7  H  7|        |
|                |4 G 4|       |8    C  |
|                |__4__|_7_____|        |
| 14             |     10 I->|1|____8___|
|        F     14|           |    9     |
|                |10   E  10 |          |
|                |           |9    D   9|
|                |           |          |
|_________14_____|______10___|_____9____|
    Large shape is not a square. It is 32 X 33 and is a rectangle. To find another rectangle composed of squares,each, with diferent areas all you have to do is use multiples of the areas of the squares that make up the rectangle given.

    Aron Freidenreich, grade 10
    Ariel Collis, 10th Grade
    Daniel A. Gabriel, grade 10
    Lili Grunwald, grade 10
    Akiba Hebrew Academy

    Eric Wahl

    Annie,

    Here is the response from Eric Wahl Grade 11 Masterman Philadelphia

    The shape is a rectangle 33 x 32.

    Sides of figures are as follows: D: 9 E:10 F: 14 side = 33
    F:14 A:18 side = 32
    A:14 B:15 side = 33
    B:15 C:8 D: 9 side = 32

    Betsy McCormick

    Betsy McCormick, Fairfield HS, grade 9

    I conclude that is is not a square as the left and right side each are 32 units in length and the top and bottom sides are each 33 units. You are given that the area of square C is 64. Because the area of a square is determined by the formula A = s*s and sqrt(64) is 8, each side must be 8 units. Next you are given that the area of square D is 81 and you can conclude that each side is 9. From this point you just add or subtract units for each side of the similar squares until there is a length for each section of the four sides. Then all the segments of a side are added together for the total. As all four sides are not equal in length, the figure is not a square under the definition of a square.

    Eric Robert

    Eric Robert, Fairfield HS, grade 10

    To solve the problem of the week, I first found the square roots of 64 and 81 to find the length of each segment in squares C and D. I then found the length of the remaining boxes by subtracting and adding the different lengths of segments. I then added up the outer edges of the squares to find the perimeter of the large box. The measurements were 33, 33, 32, and 32. Therefore, the large box is not a square because the sides are not all equal.

    Brad Warner

    Brad Warner, Fairfield HS, grade 9

    Since the area of C is 64, each side is 8 and since the area of D is 81, each side is 9. I and C share common sides with D. The side of D is 9 and the side of C is 8, leaving 1 for the side of I, and all sides of I. This means that one side of E is 10, so every side of E is 10. This also means that one side of H is 7 (8 - 1), so every side of H is 7. This gives you 15 for each side of B. E shares part of a common side with H. However, since part of H is shared with I and not with E, H only shares a measure of 6 with E, leaving 4 for the part of E shared with G. Each side of F is therefore 14 and each side of A is 18. The measures of the sides of the whole figure are 33, 32, 33, and 32 making it a rectangle but not a square.

    Samantha Brenner

    Samantha Brenner, Fairfield HS, grade 9

    The large quadrilateral is not a square, it is a rectangle. The length of the sides are 32 and 33 and the angles are right angles therefore it's a rectangle. I went from square to square to figure out the side measures. The sides of C are 8 and D are 9. You can figure out I by subtracting 8 from 9 leaving a measure of 1. I + C = D. Then I added 9 + 1 or D + I to get E's measure of 10. I then subtracted 1 from 8 and got 7 which was the side length of H. I added 8 + 7 to get B's measure of 15. I added I + E to get 11 which was equal to G + H and H was 7 so G was 4. G + E was 14 which was equal to F. F + G was 14 + 4 or 18 which is the measure of side A. Now I had all the side measures of all the squares so I added B + C + D to get 32 which was one side of the quadrilateral. I added A + B to get 33. Since the sides were not all equal, the figure is not a square.

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2 July 1995