A Math Forum Project

Geometry Forum - Problem of the Week Solutions - Equal Area and Perimeter of a Square and a Circle? - November 7-11, 1994 Annie says: Another good week of responses. Most figured out that the perimeter and area of the square are not equal to those of the circle. Some also commented that this is an understandable and reasonable method for approximating the perimeter, especially for mathematicians who had no concept of irrational numbers such as pi. None noted that the area approximation will get worse for bigger circles since the difference will increase with the square of the dimensions involved. For the same reason it would be a mistake, as contained in one submission, to think of the area of the middle square as the average of the areas of the inner and outer squares. Michele Gibney made the nice connection between her results by noticing that the circle encloses greater area with a perimeter smaller than the square. Annie Chan realized that since the perimeter and area of the circle must involve pi whereas the square would not have to, then the perimeters and areas could not be equated. Michele Gibney A0 = area of outer square P0 = perimeter of outer square AI = area of inner square PI = perimeter of inner square AM = area of middle square PM = perimeter of middle square AC = area of circle PC = perimeter (circumference) of circle D = diameter of the circle PO = 4d AC = 9(D/2)^2 AO = d^2 PC = 9d PI = 2.83d AI = d^2/2 So the PM = PO + PI/ 2 = 4d + 2.83d/2 = 6.83d/2 = 3.415d which does not equal - 3.14d - (PC). Neither the perimeter or the area of the circle is exactly equal to the middle square. They are however extremely close, so if you wanted to have an approximation they would be equal, but technically there is no connection between the two. This problem did show me that the circle had the biggest area and so it is the most economic shape when you want to have the most room inside. Jeff Pieper Area : The first step I took was to find the length of the inner square. From the center of the circle to the corner of the inner square was r (radius). If you draw a line, containing the perpendicular bisector of the 1 side of the inner square, which should go to the center of the circle, you have a right triangle. Next, I used the Pythagorian Theorem to find half of the length of the side of the inner square. 2 sides of the triangle were equal, because the bisector of the side went to the middle of the circle, which is half way to the other side of the square. This means that two a square 2r squared, which equals r dived by the square rect of 2, when reduced. Next, I found the length of 1 side of the outermost square which was simply 2r. Knowing this, I subtracted a. (the distance from the center of the circle to the midpoint of any , side of the innermost square,) from the length of the center of the circle to the midpoint of any side of the outermost square, to get the difference in length between the outer and inner squares. Then I divided this by 2 to get the distance from the edge of the inner square to the edge of the the middle square. I added this to a, to get the distance from the center at the circle to the midpoint of any line in the middle square. This, multiplied by 2 is the length of the middle square. After being reduced and simplified. The length of the side of the middle square was r plus the quantity r over the square root of 2. Next, I squared this to get the area of the riddle square, the set that to 9r^2. After reducing and simpilifying, the answer came out to be 2.914 equals 3.1415, which is not true. The numbers are close, but they are not exact. Perimeters: The perimeter was easy, knowing the length of one side of the middle square. When I found the primeter of the middle square, I set it to 2r9, and found the answer to be 3.4142 equals 3.1415, which also is not true. Once again the numbers are close, but not exact. My dad helped me, so I shouldn't get all the credit! Chad Williams I used graph paper and drew a circle with a radius of 6. The inscribed square had a side of length 8; the circumscribed square had a side of length 12. The square in the middle had a side of length 10. The circumference of the circle is 2 x pi x radius, so you get 2 x pi x 6 and the answer is 37.6. The perimeter of a square is 4 x one side, so you get 4 x 10 = 40. The circumference of the circle is not the same as the perimeter of the square. The circle is 2.4 lower than the square. The area of the circle = pi x radius squared, so you get pi x 6^2 = 113.09. Area of square is side squared, which is 10 ^ 2 = 100, so you see that the area of the circle is larger than the area of the square by 13.09. Michelle Myers The statement is false. I arrived at this conclusion by drawing the three squares and circle on graph. I made the following calculations: Middle Square: 8.5 x 8.5 Middle Square: 34 = perimeter 72.25 = area Small Square: 7 x 7 Large Square: 10 x 10 Circle: 7 pi Circle: 43.982 = circumference 153.93804 = area Sandra Duruisseau Diameter of circle = 6.5cm Circumference = 20.420352 Perimeter of large square = d * 4 = 6.5 * 4 = 26 cm Perimeter of small square = sqrt (r * r + r * r) * 4 = 18.384776 Perimeter of medium square= 26 + 18.384776/2 = 22.192388 Therefore, the perimeter of the middle circle is not equal to the circumference of the circle. Annie Chan Here is what I think the answer for the problem of the week. I don't think the perimeter of the circle and the square would be the same. When I first look at the diagram, I have decided that they wouldn't be the same because the perimeter of the circle must involve pi and the perimeter would be decimal which the perimeter of the square would not be like that. Previous page || Next problem || Previous problem || Table of Contents || Forum Home Page