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I was at the building supply store the other day and these two people were
having an argument. I think they were going to build an enclosure for
their dog (I hope it's a small dog!).
The guy was leaning against three sections of fencing that were 6, 8, and 12 feet long. The woman was telling him to get rid of the 12-foot section and buy a 10-footer - they would save money, and would be able to fence in a bigger area. He didn't seem to believe her. What do you think? Go with the 10-footer or the 12-footer? Make sure you tell me why - I want to be able to explain it to them if I ever see them again! SolutionsThis proved to be a fun problem, and generated a very wide range of methods, which I always like! 124 people got it right, and 65 got it wrong. I don't remember ever learning Heron's Formula for figuring out the area of a triangle, but a lot of you seem to know it. I should write to my high school geometry teacher and ask him if we learned it - I'll let you all know what I find out :-) The nifty part of this problem is that it doesn't seem right that the shorter sides would give you the bigger triangle. Tony Caddington of Nashoba Tech points out the illogic of this idea, saying that it's obvious that the triangle with longer legs has greater area... or does it? You can read his solution below. Tony uses Heron's Formula to find the areas, which was a very popular method. Another way to find the areas is to use trigonometry. Use the Law of Cosines to find the value of an angle, then use the formula 1/2*absinc to find the areas. Glenn Coats of Concordia Lutheran High School did just this, and his solution is included below. Many of you know that I am an advocate of doing as little work as possible if you can still adequately explain your answer. In this case, that would involve looking at the triangles in a particular way - if they are sitting on a base of 8, one has a height of 6 (since it's a right triangle), and the other has a height of less than six - the obtuse triangle makes the height shorter. Done! You don't even have to figure out the areas of either one. Alex Chen of Odle Middle School did this, and I've included his solution below. It's an excellent explanation of this idea. (Rick Peterson included some super pictures of this idea in his solution, but you'll have to delve into the full solutions to read it as I didn't include it here). If you really want to find the area of the obtuse triangle, but can't remember Heron's Formula, you can use the Pythagorean theorem to find the height of that triangle. A couple of people did that, and I've included the solution of Frannie Laks of Akiba Hebrew Academy. This is how I would have done it if I felt the need to find the area, since I would never think of Heron's Formula! A number of students made comments like, "To solve this problem, you need to figure out the areas of the two triangles." You should always be careful when you say that something MUST be done to solve a problem. Usually there will be more than one way to solve a problem, and you have just managed to think of one of them. Better to say something like, "One way to solve this problem would be...." This isn't a big deal, but it makes you sound as if you know there are probably other ways :-) Other people said things like, "the area of a right triangle will always be greater than that of an acute or obtuse triangle." Well, um, not really. You need to add something here about how this is true ONLY WHEN they have equal legs. Emily Demich of Shaler Area High School did just this, and you should read her solution below. Still others commented on the fact that the woman was right, but I won't add anything to that idea... 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: Glenn Coats
Grade: 10
School: Concordia Lutheran High School, Fort Wayne, Indiana
The 10 footer would be better. Here's why:
First you must find one angle measure using the law of cosines.
For 6,8,12:
8^2=6^2+12^2-2*6*12*cos A
64=36+124-144*cos A
144*cos A=180-64
144*cos A=116
/144 /144
cos A=116/144
Angle A=cos-1*116/144
A=about 36 degrees
Then to find the area of that triangle, you must use K=0.5ab sinC
K=0.5*12*6*sin 36
K=36*sin36
K= about 21.160 feet squared.
Then I did the same for the 6,8,10:
8^2=6^2+10^2-2*6*10*cos A
64=136-120*cos A
120*cos A=72
cos A=72/120
A=cos-1 *72/120
Angle A= 53 degrees.
K=0.5*6*10*sin 53
K=30*sin 53
K=23.959 feet squared
Therefore, the area of the 6,8,10 Triangle is larger than the area of the 6,8,12
triangle.
From: Tony Caddigan
Grade: 11
School: Nashoba Tech, Westford, Massachusetts
Because there are only 3 sections of fencing, there is only one
possible area (3 lines in a triangle can only be arranged one
way.) That area would obviously be lessened if a 10 foot piece
of fence was used instead of the 12. Or would it?
According to Heron's formula, if the 10 foot piece was
used, the area would be larger:
K=Sq. rt. of s(s-a)(s-b)(s-c)
K=area
S=a+b+c/2
First, lets find the area using the 12 footer:
S=a+b+c/2
S=6+8+12/2
S=13
K=Sq. rt. of 13(13-6)(13-8)(13-12)
K=Sq. rt. of 13(7)(5)
K=SQ. rt. of 455
Now, lets find the area using the 10 footer:
S=6+8+10/2
S=12
K=Sq. rt. of 12(12-6)(12-8)(12-10)
K=Sq. rt. of 12(6)(4)(2)
K=Sq. rt. of 576
As illogical as it seems, using less fencing (in this situation)
would allow for more area.
From: Alex Chen
Grade: 7
School: Odle Middle School, Bellevue, Washington
In this problem, I will use the side of length eight for both
triangles as the base. The 6, 8, 10 triangle is a right triangle
(multiple of 3, 4, 5) so the height is six. The area of this
triangle would therefore be 24, but that is not important.
So the 6, 8, 10 triangle has a base of eight, a height of six,
and a hypotinuse of ten. If the side with the length of eight
is used as the base of the 6, 8, 12 triangle, then we see that
the height is less than six. We could explain this by saying
that to get the 6, 8, 12 triangle from a 6, 8, 10 triangle, we
just have to "extend" the hypotinuse. To do this, we would have
to tilt the side with length six, to keep it the same length.
Therefore, the right angle becomes an obtuse angle, and the
height is no longer six. The height must now be drawn from the
vertex created by the sides with lenghts 6 and 12 and drawn to
an extended part of the base. This length must be less
than six because it forms a right triangle with the side of
length six the hypotinuse. So since the height of the 6, 8, 12
triangle is shorter than the height of the 6, 8, 10 triangle,
and their bases are the same length, the 6, 8, 10 right triangle
will have a bigger area.
From: Emily Demich
Grade:
School: Shaler Area High School, Pittsburgh, Pennsylvania
Subject: pow
Hi, this is Emily Demich from Pittsburgh, Pa. I'm Mr. Lishack's Honors
Geometry class. Here is my POW.
In this Pow they asked which fenced in area would have a larger
area-one with sides 6,8, and 10 ft long or 6,8, and 12 ft long. These
would form triangles. The 6-8-12 triangle has an area of 21.33 ft. I
found this using Heron's Formula, the square root of s(s-a)(s-b)(s-c)
where s is the semiperimeter and a,b,c are the triangle's sides. The
6-8-10 trianlge is a right triangle if the legs are 6 and 8 and the
hypotenuse is 10. Its area is 1/2(6)(8)=24 ft. Therefore they should
buy the 10 footer because they can fence in a bigger area for less
money. This problem shows that right triangles have a larger area than
obtuse triangles (with the same length legs). This is also true for
right triangles compared to acute triangles that have the same length
legs-right triangles have the larger area.
The following students submitted correct solutions this week:Arden McAllister, Grade 10, Livermore High School, Livermore, CaliforniaNathan Betke, Grade 11, Columbiana County J.V.S. Gina Christenson, Grade 9, Southeast Raleigh High School, Raleigh, North Carolina Jennifer Nelson, Grade 9, Smoky Hill High School, Aurora, Colorado Noam Abrams, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Beth Bowman, Grade 9, Franklin County High School, Rocky Mount, Virginia Gordon Bockus Jr., Grade Freshman, Eastern Oklahoma State College, Wilburton, Oklahoma Kurt Sigmon, Grade 12, Bunker Hill High School, Claremont, North Carolina Chris Brown, Grade , Penn High School, Mishawaka, Indiana Julia Tarlo, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Lee Rachakonda, Grade 11, Phillips Academy Glenn Coats, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Anna Hanke, Grade 9, Concordia High School, Fort Wayne, Indiana Adam Esslinger, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Ben Miller, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania June Lin, Grade 5, Sahuaro Elementary School, Tuscon, Arizona Stephen Nolen , Grade 9, Smoky Hill High School, Aurora, Colorado Emily Ly, Grade 9, Granada High School, Livermore, California Karen Altice and Megan Via, Grade 9, Franklin County High School, Rocky Mount, Virginia Jonathan Emmons and Amey Adkins, Grade 9, Franklin County High School, Rocky Mount, Virginia Tammy Davis, Grade 11, Shelby County High School, Columbiana, Alabama Chloe Plumb, Grade 9, Summit School, Salem, North Carolina Arielle Cohen, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Kathy Krug, Grade 10, Lakeside School, Seattle, Washington Rachel Bethlahmy and Bethany Cooper, Grade 10, Lakeside School, Seattle, Washington Tiffanie Lam, Grade 8, Sequoia Middle School, Pleasant Hill, California Mallory Vander Kooy, Grade 9, Franklin County High School, Rocky Mount, Virginia Meghan Hanson, Grade 9, Concordia Lutheran High School, Fort Wayne, Indiana Julie Hoover, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Renee Cohen, Grade 9, Bexley High School, Bexley, Ohio Ben Barnes, Grade 10, Marple Newtown Senior High School, Newtown Square, Pennsylvania Aaron Mertz, Grade 8, Plum Grove Junior High School, Rolling Meadows, Illinois Jennifer Mamer, Grade 9, Stoughton Central High School, Stoughton, Saskatchewan, Canada Anne Howard, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Kendra McKaig, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Tony Caddigan, Grade 11, Nashoba Tech, Westford, Massachusetts Rick Peterson, Grade Class of 1970, Lyman Hall High School, Wallingford, Connecticut Linda Hoffman, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Catalina Anghel, Grade 11, Mackenzie High School, Deep River, Ontario, Canada Tessa Goodhew, Grade 10, Lakeside School, Seattle, Washington Mark McIntyre, Grade 10, Lakeside School, Seattle, Washington Cornelius, Grade 7, Ephraim Curtis Middle School, Sudbury, Massachusetts John Kremer, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Dan McCoy, Grade 9, Marple Newtown Senior High School, Newtown Square, Pennsylvania Greg, Grade 9, Sparta High School, Sparta, New Jersey Zimran Douglas, Grade 12, George Wingate High School, Brooklyn, New York Adam Hendricks , Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Betsy Feldman, Grade 9, Marple Newtown Senior High School, Newtown Square, Pennsylvania Bill Braund, Grade , Dave Schoenefeld, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana John Allen, Grade 7, Louis Pasteur Middle School, Orangevale, California Alicia Newton, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Bradley Hamilton, Grade 10, Washington County High School, Springfield, Kentucky Matthew Hallien, Grade 9, Concordia Lutheran High School, Fort Wayne, Indiana Joey Whirrett, Grade 9, Concordia Lutheran High School, Fort Wayne, Indiana Avrum Tilman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Laura Meyer, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Nathan Countryman, Grade 8, Challenge School, Denver, Colorado Sara Clagg, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Amber Josephson, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Crystal Cannaday, Grade 9, Franklin County High School, Rocky Mount, Virginia Jon Amt, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Allison Bultmeier, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Lily Tian and Rachel Meyer, Grade 8, Challenge School, Denver, Colorado Kevin Bryan, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Molly Parker, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Margaret Mackenzie, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Kyle Smith, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Joe Rodenbeck, Grade 9, Concordia Lutheran High School, Fort Wayne, Indiana Nicole Sweet, Grade 10, Concordia Lutheran High School, Fort Wayne, Indiana Jon Gantman, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania J. Dee Itri, Grade , Marple Newtown Senior High School, Newtown Square, Pennsylvania Hannah Margoles, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Clayton Smith, Grade 12, Lambton-Kent Composite School, Windsor, Ontario, Canada Norman Fine, Grade , retired old-timer Mary Kenney and Mary Diamond, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Matthew Roitstein, Grade 8, Placerita Junior High School, Newhall, California Dan Chambers, Grade 9, Granada High School, Livermore, California Alex Chen, Grade 7, Odle Middle School, Bellevue, Washington Frannie Laks, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Sonia Teas, Grade 10, Oak Park and River Forest High School, Oak Park, Illinois Allison Shoemaker, Grade , Grafton High School, Yorktown, Virginia Tom Geary, Grade , Grafton High School, Yorktown, Virginia Rachael Pinsker, Grade 10, Southeast Raleigh High School, Raleigh, North Carolina Rocky Reed, Grade 10, Monahans High School, Monahans, Texas Vicki Scott, Grade 9, Franklin County High School, Rocky Mount, Virginia Aimee Johnson, Grade 11, Granada High School, Livermore, California Katie Schouten, Grade 10, Oak Park and River Forest High School, Oak Park, Illinois Maureen Brady, Grade 10, Mount Saint Joseph Academy, Flourtown, Pennsylvania Brian Porch, Grade , Redmond High School, Redmond, Oregon Peter Hoover, Grade , Redmond High School, Redmond, Oregon Raun Atkinson, Grade , Redmond High School, Redmond, Oregon Lyrica Hubbard, Grade , Redmond High School, Redmond, Oregon Lindsey Fangman, Grade , Redmond High School, Redmond, Oregon Alison Miller, Grade 6, homeschooled, Niskayuna, New York Erik Van de Vreugde, Grade , Livermore High School, Livermore, California Paul Didomenico, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Bridget Timony, Grade , Redmond High School, Redmond, Oregon Alec Rzoski, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Clayton Dillaway, Grade 8, Odle Middle School, Bellevue, Washington Kelly Flis, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Lauren Moser, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Erin Roll, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Bridget Timony and Rebecca O'Connell and Sarah Lawrence, Grade , Redmond High School, Redmond, Oregon Michelle P, Grade , Redmond High School, Redmond, Oregon Sarah Fladwood, Grade , Redmond High School, Redmond, Oregon Billy and Dallas and Peter, Grade , Redmond High School, Redmond, Oregon Clint Soose, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Colleen Perrotte, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Samantha Schliep, Grade , Redmond High School, Redmond, Oregon Crystal Bussiere, Grade 10, Roselle Park High School, Roselle Park, New Jersey Emily Demich, Grade , Shaler Area High School, Pittsburgh, Pennsylvania Ana Villalobos, Grade , Redmond High School, Redmond, Oregon Laura Roos, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Melanie Hudak, Grade 10, Shaler Area High School, Pittsburgh, Pennsylvania Alex Chernyavsky, Grade , Akiba Hebrew Academy, Merion, Pennsylvania Shameica Edwards, Grade 12, Wingate High School, Brooklyn, New York Katie McDonald, Grade , Redmond High School, Redmond, Oregon Walter Cannon, Grade , Chaim Bloom, Grade 10, Akiba Hebrew Academy, Merion, Pennsylvania Nathan Reynolds, Grade , Franklin County High School, Rocky Mount, Virginia Jenny Kaplan, Grade 7, Castilleja Middle School, Palo Alto, California Jason Chiu, Grade 9, Laramie Junior High School, Laramie, Wyoming Bethany Kim, Grade , Redmond High School, Redmond, Oregon |
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