Dale Pearson is a friend of mine who happens to be a teacher, and who happens to use the POW in his classes. He suggested this extension to last week's problem. Yes, more tangrams! (Play with them in your spare time. They are lots of fun, trying to make different shapes and things.)A _________________________ D |* *| | * 1 * | | * F * | This is a picture of a tangram, which is a | * * | puzzle that consists of a square cut into | * * | 4 | seven pieces. If you did last week's problem, | * E * | | this oughta look pretty familiar! | 2 * 3 | *|G | * * | * | | H * * | * | | * 5 * | If AB = 1, what is the PERIMETER of each | * * * I 7 | of each of the seven pieces? Yes, perimeter, | * 6 * * | not area. B|*___________*____________|C JTo make life easier for you, I have numbered each of the pieces in the picture, so you can use those numbers in your explanation. (And please do explain it, don't just give me numbers.)BONUS: Do the pieces with equal area also have equal perimeter? Explain this part!
Annie says: Solutions
This problem really was suggested by my friend Dale Pearson - it's a very good extension of last week's problem, and I think it's a bit harder because there is a lot to keep track of. And there is not good way to check yourself as you could with the area - perimeters of figures don't add up like areas do!
74 people got this one right, and 40 got it wrong. 26 people got the bonus.
Making your solution to a problem like this easy to read should be one of your goals. We have two pretty good examples of good presentations. John Martin of Shaler Area High School provides a very nice solution that you all should read. First he says that he went back to his notes from the previous week to see if there was anything useful. Yes, on occasion the POW might refer to past problems, though you won't ever _have_ to have done them. He also stated the assumptions that he made, which are again that a tangram is as a tangram seems. Then he gives a very thorough explanation of the answers, which is well done. Give it a read.
Christy Thornburg of Shelby County High School also gave a very thorough solution. She adds in a nice summary of the perimeters, and then gives a very good explanation of the bonus. No, shapes that have the same area do NOT have to have the same area, though they might. Read her solution below as well.
Another interesting idea that came out of several of the solutions is that people tried to compare perimeters based on the areas of the figures. So if one of the small triangles has half the area of a big one, does that mean its perimeter is also half? Nope, it's a quarter - take a look at piece 1 and piece 6. I smell a possible POW in that concept, too.
A few more things. This was a great exercise in adding radicals! And we could all use the practice, I know. To make my job easier, and just as a matter of habit, you should simplify those radicals. sqrt(1/2) is right sometimes, but sqrt2/2 is easier to read, and tells us more about where the piece came from - in this case, we know that it's half of the diagonal. sqrt(1/2) doesn't tell us that as clearly, if at all.
Use the numbers that you are given for the pieces in your solution. I did that to make it easier for you AND for me. And try not to use special characters in your solution - there are a lot of 1/2's that look like pi's in this problem. Just use the "regular" characters available to you, so that we can be sure that everyone will be able to read your solution on their computer.
A list of all the people who got this problem right follows the highlighted solutions below, and most of the solutions are also available.
From: John Martin Grade: 10 School: Shaler Area High School, Pittsburgh, Pennsylvania POW: December 8-12, 1997 Since this problem is based off of the last POW we were given, I went back to my solution from last week to see if it had any useful information that could be applied to this weeks problem: To find the perimeter of each of the seven pieces of AB=1. Again, several assumptions were made to simplify the problem. They are as follows: 1) Segments BD and AI are diagonals of the square. And since the square is a rhombus, they bisect each other to form right angles. 2) Figure EIJH is a square. 3) Figure FDGI is a parallelogram. 4) AI is three-fourths of the diagonal going from A to C. 5) E is the midpoint of BD; H is the midpoint of BE; F is the midpoint of ED; E would be the midpoint of AC; I would be the midpoint of EC; I is the midpoint of JG; J is the midpoint of BC; G is the midpoint of DC; segment FI is one-half the length of segment DC; a segment drawn from segment F perpendicular to segment DG is one-fourth the length of segment AD Now that these assumptions were again made, I could start in at the problem. Again I went to last week1s solution to find the length of the diagonal. To do this, I used the Pythagorean Right-Triangle Theorem in which C or the hypotenuse is found by using the following equation: a^2 + b^2 = c^2 (in this example AD is a, AB is b, and BD is C). Since AD and AB are both equal to one because they are sides of the square, the equation would appear as follows: 1^2 + 1^2 = c^2. When solved, it is found that c is equal to the square root of 2, or 1.414. Based upon this number and the assumptions, all measurements should be easily found. Figure 1 - This figure has 1 side that has a length of 1 (segment AD a side of the square), and 2 sides with lengths one-half of that of the diagonal (which are segments AE and ED, both with length of .707). I then added up .707 + .707 + 1 and found the perimeter of figure 1 to be 2.414 units. Figure 2 - Since it was proven last week that Figures 1 & 2 are congruent, the perimeter of Figure 2 is also 2.414 units. But for anyone who did not see last weeks solution, you go about finding the perimeter the same way as above, but substitute segment AB for AD, segment BE for ED, and segment AE stays the same (all substituted segments have the same lengths as the segment they are substituted for). Thus, again you get .707 + .707 + 1 = 2.414 units Figure 3 - This figure has two sides that are one-fourth of the diagonal (segments EI and EF) or have a length of .3535 units each. Segment FI is one-half of segment DC, which is a side of the square. Therefore, FI is .5 units in length. When added together .3535 + .3535 + .5 = 1.207. Figure 3 has a perimeter of 1.207 units. Figure 4 - Since this figure is a parallelogram, if 2 adjacent sides are found, you can find the perimeter of the whole figure, because opposite sides are congruent. Segment FD is one-fourth of the diagonal or .3535 units. Segment IG is congruent to FD and has an identical length. Segment FI has a length of .5 as explained above in Figure 31s solution. Segment DG is congruent to segment FI and thus has an identical length. Therefore the perimeter of this figure is .3535 +.3535 + .5 + .5 The perimeter of Figure 4 is 1.707 units. Figure 5 - Figuring the perimeter of this figure was relatively easy. Since it a square, all sides are congruent. Since we know that segment HE is one-fourth of the diagonal, or .3535 units, the equation to figure the perimeter is .3535 * 4. The perimeter of Figure 5 is 1.414 units. Figure 6 - First off, we know that Segment BJ has a length of .5 because it is one-half of segment BC. We also know that segment BH is one-fourth of the diagonal, or .3535. It was figured out in the last problem that all sided of the square are .3535 units in length. Since Figure 6 shares a side of the square(Segment HJ), it has another side with a length of .3535. Thus, the equation to find the perimeter of this figure is .3535 * 2 + .5. The perimeter of Figure 6 is 1.207 units. Figure 7 - In this figure we know that segments JC and GC each have a length of .5 because they are each one-half of a side of the square. Segment JG has a length of one-half of the diagonal, because it is made up of a side of the square(JI) with a length of .3535, and Segment IG is congruent to Segment FD, which we found in Figure 4 equaled .3535. We have established earlier that .3535 is equal to one-fourth of the diagonal, so added together we have one-half of the diagonal or .707. Thus the equation to find the perimeter of Figure 7 is .707 + .5 * 2. The perimeter of Figure 7 is 1.707.
From: Christy Thornburg Grade: 10 School: Shelby County High School, Columbiana, Alabama I split this problem into a 45-45-90 triangle. In this triangle the hypotenuse equals a side times (radical 2). Each side of a square is congruent so they are all 1 because line segment AB is 1. I used 2/2 to be 1 because I wanted the same denominators. You times the side with the hypotnuse and (radical 2) and that helps you find the other sides. So then I added 2/2 + (2 * (radical 2))/2 and that gave me 1 + (radical 2). Since Piece #1 and Piece #2 are congruent then that makes Piece # and Piece #2 equal the perimeter 1 + (radical 2) or if you do it in decimals it would be 1.4 + 1 = 2.4 (rounded to the tenths). Because Pieces #3 and #6 are congruent, I added (radical 2)/4 + 1/2. But I wanted the denominators to be the same so I changed it to be (radical 2)/4 + (radical 2)/4 + 2/4 and this equalled (2* (radical 2) + 2) / 4 = (radical 2)/2 + 1/2 = ((radical 2)+ 1) / 2. And in decimal form it would be 2.4/2 and equal 1.2. For piece #5, it is a square. So I added (radical 2)/4 + (radical 2)/4 + (radical 2)/4 + (radical 2)/4 and it equaled (radical 2). In decimal form it equaled 1.4. For piece #4, I added the 4 sides together. 1/2 + 1/2 +(radical 2)/4 + (radical 2)/4 But I changed it to make the denominator the same. 2/4 +2/4 + (radical 2)/4 + (radical 2)/4 which equalled (4 + 2 * (radical 2))/4 = 1 + (radical 2)/2. In decimal form it would be 1 + 1.4/2 = 1 + 0.7 = 1.7 Finally I tried to find the perimeter of piece #7. I added 1/2 + 1/2 + (radical 2)/2 and it equalled (2 + (radical 2))/2 = 1 = (radical 2)/2 and in decimal form it equaled 1.7. #######Summary of perimeters.######## Piece #1 and Piece #2 is (radial 2)+ 1 or 2.4 units. Piece #3 and Piece #6 is ((radial 2) + 1)/2 or 1.2 units Piece #4 and Piece #7 is 1 + (radical 2)/2 or 1.7 units Piece #5 is (radical 2) or 1.4 units ##################################### BONUS Sometimes if they have equal area they also have equal perimeter. Only if the figures are congruent or composed of congruent figures such as in Piece #1 and Piece #2. Piece # 4 and Piece #7 are made up of 2 congruent triangles; however, Piece #5, Piece #4, and Piece #7 have equal areas but the perimeter of Piece #5 is not the same as Piece #4 and Piece #7. You can have equal area and still not have equal perimeter.
The following students submitted correct solutions this week:
Jason Chiu, Grade 9, Laramie Junior High School, Laramie, Wyoming
John Martin, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Milan Fillmore, Grade 9, Smoky Hill High School, Aurora, Colorado
Josh Sarfaty, Grade 8, Valley View Middle School, Simi Valley, California
Tiffanie Lam, Grade 8, Sequoia Middle School, Pleasant Hill,California
Jeff Knepper, Grade 10, Smoky Hill High School, Aurora, Colorado
Libbie Gies, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Math Mob, Grade 6, Ridge Mills Middle School, Rome, New York
Katie Madden, Grade 9, Mount Saint Joseph Academy, Flourtown, Pennsylvania
James Tong, Grade 11, Klein High School, Klein, Texas
Mark Kaye, Grade 10, Smoky Hill High School, Aurora, Colorado
Kellyan and Laura, Grade 8, Challenge School, Denver, Colorado
Erika Mitchell, Grade 11, Shelby County High School, Columbiana, Alabama
Gordon Bockus Jr., Grade Freshman, Eastern Oklahoma State College, Wilburton, Oklahoma
Teneal Dollar, Grade 10, Shelby County High School, Columbiana, Alabama
Avrum Tilman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania
Jackie Evans, Grade 9, Smoky Hill High School, Aurora, Colorado
Neil Seifried, Grade 10, Pullman High School, Pullman, Washington
Chris Lauber, Grade 9, Smoky Hill High School, Aurora, Colorado
Lily Tian, Grade 8, Challenge School, Denver, Colorado
Nathan Countryman, Grade 8, Challenge School, Denver, Colorado
Mark Van Arsdale, Grade 8, The Challenge School, Cenver, Colorado
Sara Clagg, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana
Darren Kerstein and Rachel Meyer and Jessica Black and Brandi Moore and Rob Eagle and John Yi, Grade 7 & 8, The Challenge School, Denver, Colorado
Catherine Mangasi, Grade 12, Wilburton High School, Wilburton, Oklahoma
Kendra McKaig and Anne Howard, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana
Noam Abrams, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania
Molly Parker, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana
Linda Hoffman, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana
Cherise Lewis, Grade 9, North Pole Middle School, North Pole, Alaska
David Grant, Grade , Livermore High School, Livermore, California
Roger Dieterich, Grade 10, Smoky Hill High School, Aurora, Colorado
Dane Skilbred, Grade 9, North Pole High School, North Pole, Alaska
Saurabh Sarkar, Grade 8, Albright Middle School, Houston, Texas
Brandi Grohman, Grade 9, North Pole High School, North Pole, Alaska
Matt, Grade 9, Skyview High School, Vancouver, Washington
Natko Bajic, Grade 7, Pojisan Primary School, Split, Croatia
Ashley Monroe, Grade 9, Casady School, Oklahoma City, Oklahoma
Kate Jaggard, Grade 10, Delaware County Christian School, Newtown Square, Pennsylvania
Jenny Kaplan, Grade 7, Castilleja Middle School, Palo Alto, California
Sonia Teas, Grade 10, Oak Park and River Forest High School, Oak Park, Illinois
Christy Thornburg, Grade 10, Shelby County High School, Columbiana, Alabama
Sara FitzSimmons and Jane Milton, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania
Matt Niederst, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Rachel Pytlik, Grade ,
Clint Soose, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Adam Hendricks, Grade , Concordia Lutheran High School, Fort Wayne, Indiana
Siobhan Ganster, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Amanda Gelik, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Jen Erhart, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Niki Weber, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Emily Demich, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Chris Casavale, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Sarah Lawrence, Grade , Redmond High School, Redmond, Oregon
Lori Matesic, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Kelsey Long, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Stacie Graham, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Heather Beck, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Emily Buzicky, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Alison Falkenhagen, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Allen Hsu, Grade , Nitschmann Middle School, Bethlehem, Pennsylvania
Laura Roos, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Lydia Wang, Grade 9, Smoky Hill High School, Aurora, Colorado
Alison Miller, Grade 6, homeschooled, Niskayuna, New York
Dan Resnick and Jeff Beer, Grade 9, Germantown Academy, Fort Washington, Pennsylvania
Emily Rozak, Grade 10, Germantown Academy, Fort Washington, Pennsylvania
Mike Buckler, Grade 10, Germantown Academy, Fort Washington, Pennsylvania
Poornima Kumar, Grade 10, Delaware County Christian School, Newtown Square, Pennsylvania
Zach Dillon, Grade 7, Odle Middle School, Bellevue, Washington
Anne Hines and Lauren Rossi, Grade , Germantown Academy, Fort Washington, Pennsylvania
Jennifer Liang, Grade 8, Odle Middle School, Bellevue, Washington
Candace Murray, Grade , Redmond High School, Redmond, Oregon
Kyle Herron, Grade , Redmond High School, Redmond, Oregon
Jessie Nelson, Grade , Redmond High School, Redmond, Oregon
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