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The Return of the Parallelogram I
February 23-27, 1998

This problem came to me as I read the solutions to Splitting a Parallelogram [http://mathforum.org/geopow/solutions/19980213.geopow.html] from two weeks ago - the parallelogram split by the line. Read this problem carefully, because it's a little different.

You are given parallelogram ABCD. AB is 10. E lies on AB and F lies on CD. If EF is 4, what is the area of the parallelogram?

Here's a hint: the answer doesn't have to be a single number, but tell me as much as you can about the area of the parallelogram. What can it be? I want to know "the answer," which means everything, not just "an answer."

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Solutions

Annie says:

I think some of you learned a little something from the first parallelogram problem a couple of weeks ago. Good work! 76 people got it right this week and only 58 got it wrong.

The important point to notice was that the parallelogram can have just about any area EXCEPT that the height can't be greater than four - if it is, then EF won't be able to reach the top and bottom. Two good explanations of that came from Lance Rashford of Carlisle High School and Tanya Colburn of Mount Saint Joseph Academy. Both solutions are below, and Tanya included some nice pictures.

For EF to be the height of the parallelogram, it doesn't have to be a rectangle. EF doesn't have to be parallel to AD and BC. If you want to play around with the various configurations, I've included Jeff Martin's solution below. Jeff goes to Highland Park Senior High School. He did a nice sketch of the problem, and I converted the sketch to Java. The picture is also there for those of you who can't read Java.

Another way to figure out the area of the parallelogram is to use trig - this lets you figure out the height relative to EF. The angle of EF is really what drives the height of the parallelogram, which is what we need to find the area. If EF is 90 degrees, then we're multiplying by the sine of 90, which is one. If it's less than that, we're multiplying by less than 1. Andrew Cooledge of Odle Middle School provided a good example of this which you can read below.

Many people could not decide how to designate the minimum height of the triangle, so they used 0.1. Multiplied by 10, this gives you one as the minimum area. Way too big. There are a whole lot of numbers between 0.1 and 0, so it's good enough to say "almost zero." In fact, in Lance's solution he said that the area could be zero because the height could be zero. I might argue that it can't quite be, but it can be really, really close.

If you take a look at the Java Sketchpad example and like it, let me know. I would like to try to include more of them in the future. If you want to know how to put them on your web page, you can go to the Key Curriculum Press Java Sketchpad Web site [http://www.keypress.com/sketchpad/java_gsp/].

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:   Lance Rashford
        
Grade:  10
School: Carlisle High School, Carlisle, Pennsylvania

The area could be anywhere from 0cm squared to 40cm squared.  Because the closer 
AB and DC get together the less the area is, even though EF is still 4 cm.  And 
the farther apart AB and DC get from each other, the greater the area would be.  
But the reason you stop at 40cm squared is EF eventually becomes perpendicular, 
becoming the height and not allowing AB and DC to get and further apart.  And 
the reason  you get 0cm squared for the area, is that you could move AB and DC 
until they touched each other.  And if AB, the whole line segment toouched all 
of DC, there would be no area.



From:   Tanya Colburn
        
Grade:  10
School: Mount Saint Joseph Academy, Flourtown, Pennsylvania

POW: The Return of the Parallelogram
The area of the parallelogram is less than or equal to fourty.  
For example:

A  E     B 
**********
*    *   * 
**********
D     F  C

AD and BC could = 3.
AB and DC would = 10.
EF would be a diagonal line  (= 4) drawn from AB to DC.
The area would be 30.
If AD and BC = 2, then EF would still be 4, but it would be a steeper line.  
(The same would be the case if AD and BC were any number less than 4)

*** OR ***

A   E   B
*********
*   *   *
*   *   *
*********
D   F   C

AD and BC could = 4.
AB and DC would still = 10.
EF would be a diagonal line (= 4) drawn from AB to DC.
The area would be 40.

*** HOWEVER ***

AD and BC could not be greater than 4 because then EF wouldn't reach from AB to 
DC!!!

A  E     B
**********
*   *    *
*    *   *
*     F  *
*        *
*        *
**********
D        C



From:   Jeff Martin
        
Grade:  10
School: Highland Park Senior High School, St. Paul, Minnesota

Subject: Feb 27 POW

Jeff Martin, Grade 10, Geometry
Highland Park Senior High School, (612)-293-8940
www.stpaul.k12.mn.us/hphs/highland.html
February 23-27 POW

We know that AB = CD = 10. In other words, the base = 10. We know that EF = 4. 
In the parallelogram POW two weeks ago we figured out that the height changes as 
the tilt of EF changes. So the height can be a max of 4 or any number down to 0. 
(The height can't be 0 because then there wouldn't be a parallelogram.)

The area formula for a parallelogram is A = bh. Since b = 10, and h is between 0 
and 4, the area is between 0 and 40.

My teacher and I tried to construct a parallelogram with Geometer's Sketchpad 
that has all of the properties you gave in the problem. F can move on CD so that 
0 < DF < 10 (the problem from two weeks ago). Also, E can move, but EF has to be 
4. As E moves, the tilt of EF changes and the height of the parallelogram 
changes.

0 < h <= 4
0 < A <= 40

Move E and F and see what happens.



[We've even got a Java version - be patient while it loads!]

Sorry, this page requires a Java-compatible web browser.



From:   Andrew Cooledge
        
Grade:  7
School: Odle Middle School, Bellevue, Washington

Subject: "The Return of the Parallelogram"

Andrew Cooledge
7th Grade
Odle Middle School
Art Mabbott

If EF is perpendicular to AB, then the area of ABCD is 40. But you didn't say
that EF and AB are perpendicular, so it isn't necessarily, but you did say to
tell you everything about the area I can.

If it isn't perpendicular:

Let theta be the angle between segments EF and AB. Construct a perpendicular
from F intersecting segment AB at point G. FG is 4sin(theta), so the area is
40sin(theta).

-Andrew Cooledge




The following students submitted correct solutions this week:

Dazza Baker, Grade 8, R.S.C Rochester, Victoria, Australia
Travis Guy, Grade , Redmond High School, Redmond, Oregon
Jennifer Champine, Grade 9, Southeast Raleigh High School, Raleigh, North Carolina
Jeffrey Burlin, Grade 9, Southeast Raleigh High School, Raleigh, North Carolina
Natasha, Grade 9, Bexley High School, Bexley, Ohio
David Choo, Grade 8, Leonia Middle School, Leonia, New Jersey
Misty Tucker, Grade 11, Columbiana County Career Center, Lisbon, Ohio
Travis Pederson, Grade , Redmond High School, Redmond, Oregon
Katie Anthony, Grade 9, Casady School, Oklahoma City, Oklahoma
Lance Rashford, Grade 10, Carlisle High School, Carlisle, Pennsylvania
Gordon Bockus Jr., Grade Freshman, Eastern Oklahoma State College, Wilburton, Oklahoma
Aaron Mertz, Grade 8, Plum Grove Junior High School, Rolling Meadows, Illinois
Tiffanie Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California
Megan Ross, Grade 9, East Mecklenburg High School, Charlotte, North Carolina
Noam Abrams, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania
Melissa Hackel, Grade 9 , Granada High School, Livermore, California
Dan Chambers, Grade 9, Granada High School, Livermore, California
Angie Holland, Grade 9, Skyview High School, Vancouver, Washington
Jonathan Emmons and Karen Altice, Grade 9, Franklin County High School, Rocky Mount, Virginia
Stephen Nolen, Grade 9, Smoky Hill High School, Aurora, Colorado
Steve Platt, Grade , Redmond High School, Redmond, Oregon
Lyrica Hubbard, Grade , Redmond High School, Redmond, Oregon
Robert Revay, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Alessandro Binotti, Grade 9, American Embassy School, New Delhi, India
Carol Vivaldelli, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania
Tanya Colburn, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania
Chris Lauber, Grade 9, Smoky Hill High School, Aurora, Colorado
Abby Jones, Grade 10, Smoky Hill High School, Aurora, Colorado
Nishant Mathur, Grade 9, American Embassy School, New Delhi, India
Ashley Jaqua and Heather Westendorf, Grade , Redmond High School, Redmond, Oregon
Megan Morgan, Grade , Redmond High School, Redmond, Washington
Jen Erhart, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Jeff Martin, Grade 10, Highland Park Senior High School, St. Paul, Minnesota
Anh Nguyen, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Thao Vuong, Grade 9, Highland Park Senior High School, St. Paul, Minnesota
Yaroslav Bulatov, Grade 12, Corvallis High School, Corvallis, Oregon
David Grant, Grade , Livermore High School, Livermore, California
Laura Roos, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania
Alex Chernyavsky, Grade , Akiba Hebrew Academy, Merion, Pennsylvania
Matt Smawfield, Grade 9, American Embassy School, New Delhi, India
Matt Niederst, Grade , Shaler Area High School, Pittsburgh, Pennsylvania
Jordie Kvidera, Grade , Odle Middle School, Bellevue, Washington
Jason Chiu, Grade 9, Laramie Junior High School, Laramie, Wyoming
Cristin Kenney, Grade 8, homeschooled, Turbot, Pennsylvania
Chester Chan, Grade 6, Odle Middle School, Bellevue, Washington
Alex Chen, Grade 7, Odle Middle School, Bellevue, Washington
Alaina Hollis, Grade 10, Smoky Hill High School, Aurora, Colorado
Jennifer Liang, Grade 8, Odle Middle School, Bellevue, Washington
Larissa Cuccurullo, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania
Ashley Monroe, Grade 9, Casady School, Oklahoma City, Oklahoma
Justin Guy, Grade 11, Redmond High School, Redmond, Oregon
Catalina Anghel, Grade 11, Mackenzie High, Deep River, Ontario, Canada
Jenny Kaplan, Grade 7, Castilleja Middle School, Palo Alto, California
Michelle Wagner, Grade , Mount Saint Joseph Academy, Flourtown, Pennsylvania
Doug Yoder, Grade 12, Highland Park Senior High School, St. Paul, Minnesota
Alison Miller, Grade , homeschooled, Niskayuna, New York
Andrew Cooledge, Grade 7, Odle Middle School, Bellevue, Washington
Kristy Dalrymple, Grade , Granada High School, Livermore, California
Chaim Bloom, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania
Jessica Barclay-Strobel, Grade , Oak Park and River Forest High School, Oak Park, Illinois
Joel Jorgensen, Grade , Redmond High School, Redmond, Oregon
Raun Atkinson, Grade , Redmond High School, Redmond, Oregon
Theo Talbot, Grade , American Embassy School, New Delhi, India
Mark Goodman, Grade , Redmond High School, Redmond, Oregon
Trent Ludwig, Grade , Redmond High School, Redmond, Oregon
Darrin Koski, Grade , Redmond High School, Redmond, Oregon
Nate Hodnefield and Hatim Rockssi, Grade , Redmond High School, Redmond, Oregon
Samantha Schliep, Grade , Redmond High School, Redmond, Oregon
Nathan Walker, Grade , Redmond High School, Redmond, Oregon
Rob Johnson, Grade , Redmond High School, Redmond, Oregon
Salina Wilde, Grade , Redmond High School, Redmond, Oregon
Dallas Witty, Grade , Redmond High School, Redmond, Oregon
David Donlan, Grade , Redmond High School, Redmond, Oregon
Rebecca O'Connell and Sarah Lawrence, Grade , Redmond High School, Redmond, Oregon
Mark Kaye, Grade 10, Smoky Hill High School, Aurora, Colorado
Zimran Douglas, Grade 12, Wingate High School, Brooklyn, New York

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