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I am putting a new garden in part of my yard. I already have a rectangular
garden, a square garden, and a semi-circular garden, so I figure I need
a trapezoidal garden. I am trying to figure out the dimensions of one
part of it, so let me give you the problem. Make sure you read carefully
and draw a good picture!
Call the garden plot trapezoid ABCD with sides AB and CD parallel. Point E is on AD and F is on BC, with EF parallel to AB. The distance from A to E is 3/4 of the distance from E to D. If segment BC is 14 feet long, how long is segment FC? Bonus: How does the area of DCFE compare to the area of EFBA? SolutionsThings went okay this week, but way too many of you assumed it was an isosceles triangle. 73 people got it right, and 47 got it wrong. Of those who got it wrong, at least 20 assumed it was isosceles, and probably another 15 didn't give an adequate explanation. A few others read the problem wrong - getting those relations right took some careful reading. The most interesting part, to me, was the bonus part. Only five folks got it right, and I read about 47 solutions before the first correct one came in. What made it interesting at first is that for some reason, I couldn't figure it out! It's not really that complicated once you figure it out, but I just couldn't get it to come out. I knew I wasn't right because each thing I tried I would do in Sketchpad, and it would tell me that yes, that was fine, as long as you didn't change the length of the top, or make the top longer than the bottom, or other things. I've highlighted three solutions this week. The first two folks got the first part right as well at the bonus. Tiffanie Lam of Sequoia Middle School got the bonus wrong the first time, making at least one of the same mistakes I had made, and then sent me a new version. In it, she points out that you don't know what kind of trapezoid it is. It could even be a rectangle, and you have to allow for the fact that we don't know for sure which base is longer and which is shorter, or how they're related, or anything. Maja White of Highland Park Senior High School included a nice picture with her solution, and I've made a Java version of it for y'all to play with. One of the things that a number of people said was that the "bottom" will always be bigger than the "top." Now, this doesn't tell us what's top and what's bottom - I started my trapezoid with A and B on the bottom, but that doesn't seem to be the most popular method. So let's say that the part with the greater height (DCFE) is always bigger than the other part (ABFE). But even this isn't true. Make AB long enough (or DC short enough), and that half will be bigger. Try dragging the sketch around to see if you can do that. I had a number of guesses for the bonus, including 12:9, 1.33x, 1.25x, top is bigger, 2x, 3/4 the size, and 16/9. The 16/9 is a good guess because it shows that you're squaring some factor you know goes with the areas, but it doesn't account for the fact that the bases can be any length. Using the "scale factor" for that would work for triangles, but not trapezoids. I've also included the solution of Alex Chen of Odle Middle School. He didn't use the theorem that was very popular with most folks, that a line parallel to the bases of a trapezoid cuts the legs proportionally, but he proved that instead. Take a look. Also, look at Alex's bonus. It's not quite right, so see if you can find where he made his mistake. One last note. Be very careful when you label quadrilaterals (or anything with more than three sides). I got a number of references to trapezoid ABEF, which simply won't work. 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: Tiffanie Lam
Grade: 8
School: Sequoia Middle School, Pleasant Hill, California
Given BC = 14.
Let ED = x. Then AE = (3/4)*x and AD = x + (3/4)*x = (7/4)*x.
Using proportion, ED/AD = FC/BC or x/[(7/4)*x] = FC/14
or FC = 14*(4/7) = 8.
[Tiffanie missed the bonus first try, but then sent this]
My error is obviously made when I assumed that AB=(3/4)*EF and EF=(3/4)*DC.
They may not have the same proportion, as matter of fact, you did not indicate
what kind of trapezoid ABCD is. ABCD could even be a rectangle with BC=14. In
that case, AB=EF=DC.
Anyway, my calculation should be:
Area of Trapezoid DCEF (DC+EF)*h1/2 (DC+EF)*(3/4)*h2/2 3*(DC+EF)
---------------------- = ------------ = ------------------ = ---------
Area of Trapezoid EFBA (EF+BA)*h2/2 (EF+BA)*h2/2 4*(EF+BA)
Tiffanie Lam
Sequoia Middle School
From: Maja White
Grade: 9
School: Highland Park Senior High School, St. Paul, Minnesota
Subject: Apr 10 POW
Maja White, Grade 9, Geometry IB
Highland Park Senior High School, (612) 293-8940
Apr 6 - Apr 10 POW
[And a Java version for those of you who are sophisticated:]
AB, EF, and DC are parallel. Therefore the ratio BF/FC is the same as the ratio
AE/ED, which is 3/4. So BF/FC = 3/4, and BC = 14. Since 6/8 = 3/4 and 6 + 8 =
14, BF = 6 and FC = 8.
The area formula for a trapezoid is A = (1/2)(height)(base1 + base2).
A(DCFE) = (1/2)(4h)(b3 + b2).
A(EFBA) = (1/2)(3h)(b1 + b3).
The bottom trapezoid is bigger than the top trapezoid, and the ratio A(DCFE)/
A(EFBA) = (1/2)(4h)(b3 + b2)/(1/2)(3h)(b1 + b3).
The 1/2's and the h's cancel. That leaves us with
A(DCFE)/A(EFBA) = 4(b3 + b2)/3(b1 + b3) = (4/3)(b2 + b3)/(b1 + b3).
I wonder we can replace b3 in terms of b1 and b2, but I don't know how.
From: Alex Chen
Grade: 7
School: Odle Middle School, Bellevue, Washington
A***************************B
* + *
* + *
E***************************G*******F
* + *
* + *
* + *
D***************************H*****************C
Draw a line BH so that BH is parallel to AD. BH intersect EF
at point G. Since EF is parallel to AB and DC, it follows that
Angle BGF = Angle BHC, Angle BFG = Angle BCH
Triangle BGF and Triangle BHC share a common vertex with
Angle GBF. Therefore these two triangles are similar.
In addition, AEGB is a parallelogram, therefore BG=AE,
similarly GH=ED. We therefore obtain
BF/FC = BG/GH = AE/ED = 3/4
BC = 1.75 FC = 14
Therefore FC = 8
EXTRA:
The previously discussed relations between the two silimar
triangles hold true if line BFC is vertical to the 3 parallel
lines. Therefore,
the distance between AB & EF = 3/4 the distance between EF & CD
For simplicity we assume the distance between AB & EF = 3 and
the distance between EF & CD = 4
the area of DCFE = 3*(AB + 0.5*GF)
the area of EFBA = 4*[AB + 0.5*(GF+HC)] = 4*[AB + 0.5*(10/3)GF]
the area of DCFE/the area of EFBA
=(3/4)*[1+0.5*(GF/AB)]/[1+(5/3)*GF/AB)]
For positive GF, [1+(5/3)*GF/AB)] > [1+0.5*(GF/AB)]
Therefore [1+0.5*(GF/AB)]/[1+(5/3)*GF/AB)] < 1
We have (the area of DCFE)/(the area of EFBA) < 3/4
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