Solutions - Areas of Squares - March 4-8, 1996
Highlighted Solutions || Next Problem || Previous Problem || Contents || Search POWs
Correct solutions were submitted by:
Cygnet, Tasmania, Australia Amy Forster, Grade 7, home schooled, Fairfield High School, Fairfield, Connecticut Steve Krubiner, Bilal Seyal, Grade 9 Father Judge HS, Philadelphia, Pennsylvania Brian Edwards, Bill Lamon, Grade 10 Granada High School, Livermore, California Collin Beighley, Grade 11 Debbie Zimmerman & Laura Furrey, Ethan Castor, Zac Crawford, Mike Sue, Daniel Ferguson, Grade 10 Andrew Potter, Grade 9 Gail Watling & Carolyn Watling, Grades 6 & 9, Vineyard EHS & Granada High School, Livermore, Calif. Hanover High School, Hanover, New Hampshire Josh Fischel, Grade 12 Hine Jr. High School, Washington, DC Wallace Lewis and Terrence Mitchell, Grade 8 Holy Heart of Mary High School, St John's, Newfoundland, CA Seth Davenport and Karen Kelly, Grade 10 Marion High School, Marion, Indiana Mike Rutan, Grade 10 Martin County High School, Stuart, Florida Danielle P. Cegelis, Grade 12 Christine Francescani, Grade 10 Julia Schumm and Edna Evans, Grades 10 & 9 Morton Middle School, Vandalia, Ohio Michelle Cook, Grade 8 Murray Junior High, Ridgecrest, California Cassie Gorish, Grade 8 Thomas S. Kuo, Grade 7 Pamlico County High School, Bayboro, North Carolina Angela Davenport & Billy Cauley, Grades 11 & 12 C.Gibbs, B.Piner, & M. Ottinger, Grade 9 Pearl River High School, Pearl River, New York Matt Schneider, Grade 9 Pensacola Catholic High School, Pensacola, Florida Jeremy Krauss, Tyler Soderlind, Grade 10 College Park High School, Pleasant Hill, California Nathan Roberts, Grade 11 Jacqulyne Law, Chris Eskildsen, Philip Horner, Kyaw Soe Mon, Grade 9, Pulaski County, Virginia Kevin Conner, Grade 9 Rockland High School, Rockland, Idaho JaNan Farr, Rick Millward, Grade 10, Roselle Park High School, Roselle Park, New Jersey Max Sheyn, Grade 10 Russell Middle School, Milpitas, California Stephanie Balster, Grade 8 Simon's Rock, Great Barrington, Massachusetts Chris Jurdak and Inna Hughes, Grades 12 & 11 Smoky Hill High School, Aurora, Colorado John MacArthur, Christine Jung, Steve Hansen, Sean Mostafavi, Kary Wyvell, Somsnit Vanprapa, Grade 10 Jennifer Sprangers & Yunny Chen, Grades 9 & 10 Scott Bridger, Mark Overton, Bryon Joel, Tina Barber, Carolyn Jones, Grade 9 Walter Williams High, Burlington, North Carolina Dorothy Moorefield, Grade 11 Waluga Jr. High, Lake Oswego, Oregon Adam Wright, Grade 8 Westwood Jr.-Sr. Regional High School, Westwood, New Jersey Josh Benzwie, Grade 8 Brian Gordon, Dartmouth '92, Middletown, CT Jon Tan, De la Salle University, Manila, Philippines E.Y. MurphEY, Teacher, Haverford High School, Haverford, PA
The solution to the problem is the area of the smaller square is 1/2 the area of the larger. Using the Pythagorean Theorem we find that the side of the small square is the square root of 2 divided by 2 times the length of the side of the large square. By squaring both sides you get the respective areas, r^2 and r times the square root of 2 divided by 2 squared, thus the ratio 2 to 1. JaNan Farr and classmates, Rockland High School, 10th grade. ************************************************ From: Christine Jung Grade: 10 School: Smoky Hill High School The area of the small square compared to the big one is 1/2. I started off in Geometry sketchpad, a little reluctant because I have a cold, and I hadn't learned to use Geometry Sketch pad since the beginning of second semester. Going to school somewhere else, using the computer for geometry was very new to me, not to add the only thing I ever used the computer for was writing and computer graphics. So anyhow, I went into Geometry Sketchpad, and constructed a square using the square by edge script. It worked out fine, and the computer didn't do anything funny so I figured it was all good so far. Then I constructed a midpoint on one of the sides because I wanted a circle with one side of the square as the diameter. So I selected the midpoint and the corner point and went up to construct and chose construct circle by center point. It worked as I planned which made me happy as the bunny on Easter Day. Then I opened another script and used the inscribed square script and it worked well too, I swear I almost cried with joy! Anyhow, I then went to measure, and measured the large square, and then the small inscri- bed one. Then using the calculator, which I might add I've fallen in love with, I then subtracted the first area with the second. In the end I found out that the area of the first square subtrac- ted with the second is equal to the second. Which actually means that the second square is half the area of the first. ************************************************ From: Sean Mostafavi Grade: 10 School: Smoky Hill High School Answer: The ratio of the large square over the small square is 2. Large Triangle -------------- = 2.00 Small Triangle I used the Geometers sketchpad. I constructed a square inscribing a circle. Then I drew a chord (or line) from one of the corners of the square to the other (diagnal). Then I used this diagnal to construct another square using a sample script in the sketch pad. The larger square is 2 times bigger than the small one. The ratio of "pi" will always remain the same, so in any circle that you inscribe a square in, the diagnal of that square will make another square exactly two times bigger than the first one. The definition of a square states that all four sides must be congruent, so you make four sides using that diagnal (or chord of circle) to make the 2nd square. ************************************************ From: Jennifer Sprangers & Yunny Chen Grade: 9,10 School: Smoky Hill High School Answer: We first started with a square. Then we used geometry scetchpad to construct a square. Then using the midpoint of one side we constructed a circle that had the same diameter as the square. Since the other square is inscribed into the circle, we used the inscribed square sketch to construct an inscribed square. Then we used the calculate tool to find the relationship of the areas of the squares. We calculated that the area of the small square is 1/2 the area of the big square. [after being urged to be more general] First we drew our diagram. We started by drawing a circle. Then we drew a square with one of its sides as the diameter of the circle. This is square #1. Our square #2 was drawn inside of the circle, (or inscribed). For both squares we drew the diagonals. The diagonals bisected each other and the bisected segments are congruent. As a result the eight triangles were equilateral. To find the ratio of square #2/#1, we drew the radii of the cirle which was one of the bisected segments of the diagonal in square #2. Since the bisected segments of the square are congruent to their sides, then the sides of the triangles formed by square #2 are the radii of the circle. The radii compared to the diameter is 1/2. After we multiply the lengnth times the height the ratio is still 1/2!!!!!! ************************************************ From: Rick Millward, Rockland High School, Rockland Idaho 1) Assume the raidus of the large circle to be root2. Then the side of the large square would also be root2. 2) Therefore the area of the large square is 2. 3) The diagonals of the small square bisect the 90 degree angles of the corners of the small square. Forming two identical 45-45-90 triangles with hypotenuse of length root 2 4) Therefore, by the 45-45-90 triangle theorom, the side of the small square is 1, and so is the area. 5) And the ratio of the areas is 2:1 ************************************************ From: Debbie Zimmerman & Laura Furrey Grade: 10 School: Granada High School Answer: USING THE GIVEN PICTURE - The big square has the radius of the larger circle as a side so we say: Area of the big square = radius squared The diameter of the smaller circle is an edge of the smaller square so we say: Area of little square = diameter squared This would make the Area of the big square half the area of the smaller square. ************************************************ From: Andrew Potter Grade: 9th School: Granada High School, Livermore, California Answer: The smaller square will be half of the size of the larger square. The circle, if it stays within the big sqaure, but be as big as it can, will be touching the midpoints of each segment of the large square. Then the smaller squares edges will be at hte midpoints of the larger square's segments too, because for the smaller square to be the biggest square it could be, thats where it would be :-) ************************************************ From: Collin Beighley Grade: 11 School: Granada High School Answer: The area of the big square is twice the size of the area of the small square The big square is twice the size of the small square because the big square is made out of the diameter of the circle. Since the smaller square is made inside the circle , the big circle has to be twice as big ************************************************ From: Scott Bridger, Mark Overton, Bryon Joel Grade: 9 School: Smoky Hill High School Answer: The ratio of the large square to the small square is 1:2. By drawing diameters as diagnols of the small square you create four congruent 45-45-90 triangles with a hypotenus that is a radius of the circle. The same triangles when formed from the large square: the hypotenus is the diameter of the circle, making the triangle similar in a ratio of 1:2 ************************************************ From: Tina Barber Grade: 9 School: Smoky Hill High School Answer: To solve this problem I drew a scetch using Sketchpad on the computer. After drawing the problem, I discovered that two sides of the small square and one side of the big square form a 45-45-90 triangle. The proportions of a 45-45-90 triangle are as follows: The base and the altitude are equal to each other and represented by X. The hypotenuse is equal to X times the square root of 2. After recognizing the 45-45-90 triangle you can then find the area of the squares. The area of the small square: x squared The area of the big sqaure:(x times the square root of 2) squared. When you set these proportions up the X squareds cancle out. Leaving the proportion. 1/2 So the area of the small square to the area of the large square is 1/2. :-) ************************************************ From: Kary Wyvell, Somsnit Vanprapa Grade: 10 School: smoky Hill Answer: For the problem of the week we first used the geometry sketchpad. With the geometry skechtpad we used the scripts, square by edge, to form a square. With the square we made a midpoint using the bottiom line of the square. After forming the midpoint we selected the midpoint and the point that was to the right of it. With those two points selected we drew a cirle using the two points. Next we used the same points and drew a square, using another script, with those points. Next we found the are of the two sqaures individually. Next we took the area of one of the squares and subtracted to from the area of the other square. After doing that we learned that the area of the small square is half of the area of the large square. ************************************************ From: Adam Wright Grade: 8 School: Waluga Jr. High Answer: First, I drew a small square inside of the larger square, so it was tilted at a fourty-five degree angle. The made all of the corners of the middle square the midpoints of the sides of the big square. It divided the big square up into the little square, and four right triangles. I made the sides of the big square 2X, so that each side of the right triangles was X. Because of the Pythagorean theorem, I knew that the hypotenuse of each of the right traingles was x times the square root of 2. That means that each of the sides of the small square is x times the square root of 2, and the volume of the square is 2x^2. The volume of the outer square is 4x^2. The ratio is 4x^2/2x^2. We can simplify out the x^2, so we know that the ratio is 4/2 or 2:1. That means that the area of the small square is one half the size of the large square. Thanks, Adam ************************************************ From: C.Gibbs, B.Piner, and M. Ottinger Grade: 9th School: Pamlico County High School Answer: First we drew our picture of the 2 squares and the circle. We made the edge of the square the diameter of the circle.Next we drew the 2nd square inside of our circle. We then drew the digonal of the square inside of the cicle diagonally from left top to right bottom. We knew that in order to find the area of a square, you need to know the diagonal. The formula for finding the area of a square is 1/2diagonal-squared. Then we found the area of the larger square by using the foumula length x width. So our comparison looked like this A=1/2diameter-squared and A=diameter-squared. The area of the small square is half of the big square. ************************************************ From: Dorothy Moorefield Grade: 11 School: Walter Williams High (in North Carolina) Answer: To find the area of a square, square one side of the square. The given squares are set up in the circle such that the diameter is equal the diagonal of the inscribed square and the length of the other square. Lets call the length of the uniscibed square x. The diagonal of the inscribed square = x. Lets call the lenth of the inscibed square z. By the pythgrean theorem, we can find that 2z^2=x^2. So x=z2^(1/2) z = x/2^(1/2). The area of the inscibed square = x^2/2 The area of the other square = x^2 So the relationship between the two squares is The area of the inscibed square is half of the area of the other square. ************************************************ Hi! I have an answer for the geometry forum problem of the week 4-8 th of March 1996 Answer: The area of the small square is half the area of the large square. Solution: The diameter of the small square = the length of one side of the large square. You can cut the small square in half along the diagonal,and place both halves into the large square with the longest sides of the triangles along adjacent edges of the large square. The pieces of the small square cover exactly half of the large square. I am 11 years old and I live at c/-Post office,Cygnet,Tasmania,Australia,7112.I'm in grade 7,and I do home schooling. From Amy Forster. Wilkins/Forster family Crooked Tree Point.Cygnet. Tasmania.Australia ************************************************ From: Angela Davenport & Billy Cauley Grade: Eleven & Twelve School: Pamlico County High Answer: The area of the small square is half that of the area of the big one because there is two formulas to figure out the area of a square (Area= side squared and Area =1/2 diagonal squared). The side of the big square is the diameter of the circle, and being that the small square is inscribed in the same circle, the diagonal is the diameter of the circle and also the same length as the side of the other square. ************************************************ From: Cassie Gorish Grade: 8 School: Murray Junior High Answer: The ratio of the area of the small square to the area of the large square is 1 to 2. I found this out by drawing a picture of the edge of the large square forming the diameter of the circle, and drawing the small square so that its diagonal is the diameter of the circle. By labeling the small square's sides 1 unit, I figured out the sides of the squares and the areas. ************************************************ From: Matt Schneider Grade: 9 School: Pearl River High School Answer: The smaller square is 1/2 the area of the larger because: Let edge of larger square=x which is also the diameter Area of larger square=x^2 Square inscribed a crcle, has diagonal equal to diameter which is x. So the side of the square would be x*(sqr(2)/2) So (x*sqr(2)/2) squared(to find the area) is (x^2)/2 The ratio of the smaller square to the larger is (x^2)/2:x^2 1:2 So the smaller square is one half of the larger ************************************************ From: Brian Edwards Grade: 10th School: Father Judge High School Answer: The diagonal of the concentric sqare is x. The side of the other sqaure is x because the problem states that one square has the diameter of the circle as an edge. The diagonal of the square is the diameter of the circle. The diagonal of the square forms a 45-45-90 triangle because 1) the diagonal bisects two of the 90 degree angles forming two 45 degree angles and 2)the other angle is 90 degrees. Then you have to get one of the sides of the square. In a 45-45-90 triangle the hypotenuse equals the side multiplied by the square root of 2. Because I have the hypotenuse to get the side I have to divide the hypotenuse by the square root of 2. Then I square that. That leaves me with X^2/2 as the area of the smaller sqaure and the area of the larger square is x^2 ************************************************ From: Bill Lamon Grade: 10 School: Father Judge High School Answer: First you inscribe a square inside a circle.Then you draw a diameter in the circle.If the small square had a side of 5,then the area is 25 and the diagnol is 5 times the square root of 2. You then take the small square's diagnol and make it the side of th large square.If you square the large square's side, you get 50. That means that the area of the large square is twice the area of the smaall square. ************************************************ From: Thomas S. Kuo School: Murray Junior High School, Ridgecrest, California Grade: 7th The ratio of the area of the small square to the area of the big one is 1:2. Let r be the radius of the circle and l be the length of the edge of the small square. Since the big square has the diameter of the circle, 2r, as an edge, its area is (2r)^2 = 4r^2 A B * * * * * * * * * * * * * * * * * C * * * * * * * * * * * * * * * * Let the above square be the small square inscribed in the circle. Point C is also the center of the circle. Point A and B are on the cicle. Angle ABC is 90 degrees. I also know that AC = BC = r. Then I can solve the length of the edge of the small square, l, by using Pythagorean Theorem. l^2 = AB^2 = AC^2 + BC^2 = r^2 + r^2 = 2 r^2 Then the area of the small square is l^2 = 2 r^2. Put them into ratio form and simplify it: area of small square / area of big square = 2 r^2 / 4 r^2 = 1/2 ************************************************ Well, it took me a minute of trying to remember back to ninth grade when I was good at these geometry problems, but I think I did manage to get this question right. Okay, pretend that the diameter of the circle is 4 meters. That means that the area of the square whose edge is that diameter is 16. If you inscribe the smaller sqaure inside the circle, from one corner to another is 4 meters. To find the area of that sqaure, you divide it diagonally from corner to corner, and find the area of the two, equal resulting triangles. Each triangle's area is 4 square meters, and if you add the two together, it's 8. Therefore, the area of the inscribed square is half that of the larger sqaure. :0) Josh Fischel Grade 12 Hanover High School Hanover, New Hampshire ************************************************ first of all you let d=the diameter of the circle. Since the circle is as wide as the square the middle of the square is d, then the perimeter of the square is d to the 2nd power. Next you make a diagonal down the inner square and cut it into two 45-45-90 triangles. The hypotenuse of that triangle is d and using the 45-45-90 therom the legs are d/the square root of 2. Since you have to multiply length times width d/the square root of 2 *d/the square root of 2. The area of the inner square is d to the 2nd over 2. So the area of the smaller square is half the area of the larger one. JEREMY KRAUSS 10th GRADE CATHOLIC HIGH PENSACOLA FL. ************************************************ From: max sheyn Grade: 10 School: rossel park Answer: the smaller square is 2 times smaller compared to the large. because I got the area of the liitle one by length times hight.becouse there were no numbers I named them by letters in square. so for the small one I got Y square for the big one I did the same thing and i got X square. then I got the diameter of the big one and it was 2 X square then I divided y square in to 2 x square and I got that the small square is 2 times smaller. ************************************************ large square:36cm.^2 small square:18cm.^2 found area of small square dividing into two triangles altitude in each triangle was 3;base in each triangle was 6;added both areas of the triangles for the small square's area SMALL SQUARE IS HALF AREA OF THE LARGER SQUARE WALLACE LEWIS TERRENCE MITCHELL ************************************************ From: Carolyn Jones Grade: 9 School: SHHS Answer: The area of the smaller square compared to the larger is that the smaller is half of the larger. I found this out by drawing a picture. I made the diameter 4, so the area was 16. In the inscribed square the diagonal was 4 because it was also the diameter of the circle. The new formed 45-45-90 triangle let me know that the formula I could use was to let the hypotenuse equal x times the square root of 2. The legs would equal x. I then found x which was 2 times the square root of 2. Then x times x equaled 8. 8 is half of 16, so there's my answer. Have a nice weekend! ************************************************ From: Chris Jurdak and Inna Hughes Grade: 12 and 11 School: Simon's Rock Answer: Call the inscribed square A, and the square with (side length) = (diameter of circumscribing circle) be called B. Let the (side length of A) = x. Connect opposite vertices of the inscribed square with a line segment D. Then D is a diagonal of the inscribed square, and has length x*(sqrt 2). The vertices of an inscribed polygon lie on the circumference of the circle. Since, by symmetry, each side of the square is the same distance from the center of the circle, each side (chords of the circle) cuts of equal arcs. As D passes through opposite vertices, and the square is symmetrical about D, the arcs that the circle is divided into by D are equal. Hence D is a diameter of the circle, and therefore equal in length to a side of square B. Hence (side length of A) = x, and (side length of B)= x*(sqrt 2). The area of A is equal to X^2, and the area of B is equal to 2*X^2. The ratio (area of A):(area of B) = 1:2 ************************************************ P.O.W.,March 4-8, 1996 Martin County High School, Stuart Fl. Julia Schumm,10, and Edna Evans,9 If you have two squares and a circle, and one square has the diameter of the circle as an edge with the other square inscribed in the circle, the small square's area would be 1/2 the area of the larger square. Firstly, you draw a diameter of the circle and use the line of that diameter to make one side of the larger square. Next you draw a square inside the circle using the line of the diameter to intersect opposite angles of the smaller square. If the diameter of the circle is 4 radical 2, then the area of the larger square can be found by multiplying 4 radical 2 by 4 radical 2, which is 32. If the diameter of the circle is 4 radical 2 and this is used as an angle bisector for the smaller square (opposite angles cut in half) then if you use the diagonal as a hypotenuse cutting the smaller square in half (into a right triangle), you find using the Pythagorean Theorem that one side of the square is 4. Therefore the area of the smaller square is 16. This shows that the ratio of the smaller square to the larger square is 1 to 2 (16/32). ************************************************ POW, March 4 - 8, 1996 Danielle P. Cegelis, grade 12 Martin County High School Stuart, Florida ANSWER: The area of the small square as compared to the area of the larger one is 2:1. EXPLANATION: This is due to the fact that the diagonal of the smaller square is in fact equal to one side of the larger square. Because it was given that the larger square's side is equal to the diameter of the circle, and when you draw the picture of the smaller square within the circle you can see that the diagonal is the diameter. If you were to use the value of 4 for the diameter of the circle then the larger square's area would equal 16 (4*4=16). This would then tell us that the diagonal of the smaller square is 4 thus making the side of the square the value 2* the square root o ************************************************ From: Gail Watling / Carolyn Watling Grade: 6th / 9th School: Vineyard E.H.S./ Granada High School If a square has the diameter of a circle as the length of its edge, and another square is inscribed within that circle, then the smaller square that is within the circle has its four corners at the midpoints of the sides of the larger square. This creates four right triangles which are on the outside of the smaller square,whose hypotenuses are the sides of the smaller square. The area of one of these triangles is, given side 'S', (1/2 S) squared divided by 2 The area of all four right triangles on the outside of the small square is 4 times that amount, or 4(1/2 S)squared divided by 2 When that is reduced it is equal to 1/2 S squared. If the large square ,side length S, has an area of S squared, and the four triangles being cut off of the large square have a combined area of 1/2 S squared, then the remaining area which is the smaller inscribed square, must have an area of 1/2 S squared. The area of the small square compared to the larger is one-half. This is very easy to see using tangrams as you can just flip the small right triangles over to cover the small square. Drawing it on graph paper made it easy to count the area squares too. ************************************************ Here is a solution provided by 2 of my students. The top part of the circle will be positioned exactly in the center of the large square, with the diameter being one side of the large square. Two sides of the smaller square extend to the two corners of the larger square, which is the two ends of the diameter of the circle. Therfore, half of the smaller square is equal to one quarter of the larger square. This means that the large square has twice the area of the smaller square. Seth Davenport, Karen Kelly Grade 10 Holy Heart of Mary High School St. John's, Newfoundland Canada ************************************************ Christine Francescani Martin County High School Grade 10 moroff88h@aol.com First I drew a picture and labeled each side of the large square x. Half of a side of the large square equals x over 2. Each triangle in the corners of the large square is a 45, 45, 90 degree triangle. The hypotenuse of a 45, 45, 90 degree triangle is equal to one side of the triangle multiplied by the square root of 2. The two sides that are not the hypotenuse are equal to x over 2. Therefore, x over 2 times the square root of 2 is equal to one side of the triangle. The area of the large square is length times width which equals x squared. The area of the small square (x times the square root of 2 over two times itself) is x squared over two. Therefore, the area of the small square is equal to half of the area of the large square. ************************************************ From: Michelle Cook Grade: 8 School: Morton Middle, Vandalia, OH Answer: Drawing a picture shows that a diagonal of the small square is a diameter of the circle and a side of the large square. Dividing the small square into two right triangles allows you to use the Pythagorean Theorem to find the the lengths of the legs of the small triangles. Using variables and numbers reveals that the area of the small square is half the area of the larger square. h=diagonal of b (small square) and edge of a (large square) s= side of small square h^2=s^2+s^2 h^2=2s^2 h^2/2=s^2 s=Sq.root of (h^2/2) area of b=s^2 area of a=h^2 The area of b=h^2/2 and the area of a is h^2. The area of the small square compared to the big one is the big square's area divided by 2 to get the little square's area. ************************************************ From: Daniel Ferguson Grade: 10 School: Granada high school Answer: The answer for the problem is that the squares are related in a ratio of one to two, or one half. I found this by drawing a square with diagonals then inscribing it in a circle after which I drew the diameter of the circle and drew a square from that. If the big square's area is 1 then half the side of the big is .5 along with the diagonal of the small square. Using right triangles, you are able to find the hypotenuse, or side of the small square. Which in this case would be .71 meaning the area of the ************************************************ Steve Krubiner, Fairfield HS, grade 9 Draw the circle so that the square is inscribed in the circle. Draw the diameter of the circle where one of the diagonals of the square is and draw the second, larger square, using the diameter as one of its sides. Label the smaller square's segments x and the larger square's segments y. If you use the diagonal of the smaller square and the two sides of the smaller square, you get a right triangle. Using the Pythagorean Theorem, x^2 + x^2 = y^2 or 2x^2 = y^2. Since x^2 is the area of the smaller square, and y^2 is the area of the larger square, 2x^2 = y^2 means that 2 times the area of the smaller square equals the area of the larger square, or the larger square's area is double the smaller square's area. ************************************************ Bilal Seyal, Fairfield HS, grade 9 After I inscribed the square into a circle, I drew the other square, which is supposed to be as long as the diameter or the diagonal of the square inscribed in the circle. I called one of the sides of the smaller square x. So its area is x^2. The diagonal of a square bisects its angles, so the square is divided into two 45-45-90 triangles. There is a theorem that says that in a 45-45-90 triangle, the hypotenuse is sqrt(2) times a leg. So the diagonal, which is also a side of the larger square, is x*sqrt(2). Therefore, the area of the larger square is (x*sqrt(2))^2 or 2x^2. The area of the smaller square is x^2 so the ratio of the area of the smaller square to the area of the larger one is 1:2. ************************************************ From Mike Rutan Grade 10 Marion H.S. Marion, In. Solution: Area of the small square is one-half the size of the large square. Method: Draw circle of diameter 14 on graph paper. Draw the large square with 14 on a side using the diameter of the circle. Draw the small square with the diagonal equal to the diameter of the circle. Find the area of the large square 14 by 14 (196 sq units). The area of the small square with diagonal 14 is 7 times the square root of 2 squared. (45-45-90 triangle) Small square has an area of 98. That is one-half the area of the large square. The Crump's Suzanne and Michael Crump mcrump@iquest.net ************************************************ College Park High School, Pleasant Hill, California Jacqulyne Law The area of the small square is 2r squared, and the big square is 4r squared. First I drew a picture how it would look like, and it ends up that the big square inscribed the circle. According to the definition of circle, any point on the small square is diagonal is r. Then I drew out the 2 diagonal lines and it formed 4 triangles that they are all congruent and they are all right triangle with the base of r and height is r. First I thought if I use the hypotenuse formuls then I will be able to find the sides of the square. So I wrote r squared plus r squared= s squared, but that all I could do, because it cant be solve this way. And I sat and think, but nothing in my mind, so I decided to call my friend and discuss with her. Finally, we spent about 15 minutes together and come up with this solution. As you know I drew the diagonal lines and they formed 4 right triangles, so I put them into a right triangle with the base 2r and height also 2r. The area of a triangle is base times height divided by two, so the big square is 2r times 2r so is 4r squared. 4r squared divided by 2r squared is 2, it means that the big square is 2 times larger than the small one. ************************************************ Chris Eskildsen The area of the small square is 1 half times smaller than the area of the large square. I found this out by drawing a picture. I found the area of the large square because one side is equal to the diameter of the circle. The area of the small square was harder to find using the pythagorean theorem, I created a right triangle usin g a side of the small square and 2 lines from the center of the circle to 2 consectutive vertices of that side, knowing that these 2 lines equal halfd in lenght I used simple math to solve it. ************************************************ Philip Horner The small square is equal to half the area of the larger square. This was found by using pythagorean's theorem to prove that one forth of the small square is equal to one eight of the larger square. To do this, I circumscribed the circle with the small square in it into the larger. I then divided the squares to show one forth of the smaller square is equal to one eight of the larger square. One half the diameter or the radius squared times 2 , equals the side of the smaller square squared or the area of the smaller square. Therefore telling you the areas and then being able to determine the ratios. ************************************************ Kyaw Soe Mon Since the small square is inscribed in the circle, the vertices are equal distince from the center of the circle. Therefore, a diagonal of the square is 2r. I used the pythagorean theorem by using what I know. From all this I got that 2 times the small sides squared is the same as 2 times the area of small square. I also now that diagonal squared is equivalent to the diameter squared, which is same as the side of the larger squared, and that is the area of the big square. So the answer is that the area of small square is half that of the big square. ************************************************ Nathan Roberts: 11th grade College Park High School, Pleasant Hill, Calif. Well, it took me about 1 minute to draw a diagram and see for myself what the correct answer was, 2 minutes to algebraically prove it to myself, and now for the hard part: describing it in words. Well, the diagram I drew was a circle with D as the diameter, and a square with D as the length of the sides. So the area of the larger square is D^2. (D^2 is supposed to read ³D Squared².) The circle was inside the square, so the circle touched the larger square at exactly four points - the midpoints of the sides. Then I drew the smaller square inside the circle. I could "rotate" the smaller square inside the circle to wherever I wanted and the square would be congruent, so I made the 4 points of the smaller square touch the 4 points I mentioned above - the midpoints of the larger square. Now, I already know that if the four points of a square touch the midpoints of a larger square, then the smaller square is exactly half the size, but here is how I proved it algebraically: The diagonal of the smaller square goes from the top midpoint of the larger square to the bottom, so the diagonal of this smaller square is also D (The diameter of the circle and the length of the larger square's sides.) Now I want to find the length of the sides of the smaller square (To find the area with.) Now I have a right triangle. The legs are two sides of the smaller square (Which I'll call "X"), and the hypotenuse is the diagonal of the smaller square. Now I use the Pythagorean theorem: X^2 + X^2 = D^2 or 2X^2 = D^2 Now I solve for X. ˆ2X = D (Square root both sides... The "ˆ2" is supposed to read "The square root of 2".) X = D/ˆ2 (Divide both sides by ˆ2, now I have the length of the sides of the smaller square) Now I know the length of the sides of both squares. D for the larger, D/ˆ2 for the smaller. Now I square them both to get the areas: D^2 Larger square (D/ˆ2)^2 = D^2/2 Smaller square Now you can see, through substitution, that the area of the smaller square is exactly half of the larger. Wow, explaining that turned out to be a lot easier than I thought. ************************************************ From: Brian Gordon Grade: 1992 School: dartmouth Answer: Hi Annie. This one IS easier....The big square is twice the area of the small square. The area of a square is either s^2, or .5 * d^2 (since a square is a rhombus and the diagonals are equal) Well, the diameter of the circle is s for the big square and it's d for the small square. Officialy, if we let x=the diameter of the circle: Big square area: x^2 Small squarea: .5x^2 So it's a 2-to-1 ratio. --bri ************************************************ let s = side of the bigger square. s2 = side of the smaller square. s2 = s/ sqrt(2) or s = s2 * sqrt(2) area of big square is sqr(s) = 2 * sqr(s2) area of small square is sqr(s2) therefore the ratio is 1:2 ! ************************************************ let the diameter of the circle be D this diameter is also the diagonal of the inscribed square. using 45°/45°/90 right triangle stuff then the side of the inscibed circle is (Dˆ2)÷ 2 then the area of this inscribed circle is half of D^2 but the square with edge of D has an arae of D^2 so the inscribed square in a circle of diameter D has half the area of the square whose edge is the diameter D! qed MEF E.Y. MurphEY
The Math Forum is a research and educational enterprise of the Drexel School of Education.