9-Pythagorean Theorem One of the proofs of the Pythagorean Theorem
typically uses squares constructed on each side of a right triangle (see figure
below). The area of the square constructed on the hypotenuse (green square) is
equal to the sum of the areas of the squares constructed on each of the legs of
the triangle (blue squares).
Part 1:Instead of using squares, can we use equilateral triangles?(25 Points)
Plan: I needed to draw a right triangle, with equilateral triangles on each side.
Once I did that, I plan to construct them triangles, find the areas and add them
to see if the Pythagorean relationship holds.
Attempted Solution: I constructed a right triangle, then used the midpoints of
the sides, perpendicular lines and a circle to construct equilateral triangles on
each side. I measured each side to verify that they were equilateral.
m = 4.80 cm
k = 5.91 cm
p = 3.44 cm
n = 4.80 cm
j = 5.91 cm
q = 3.44 cm
l = 4.80 cm
r = 5.91 cm
o = 3.44 cm
I then constructed each outside triangle’s interior and calculated the area. I was
expecting to have to use some Algebra, but when I saw the measurements, I
was surprised to see that the area of the triangles on the legs added up to the
area of the triangle on the hypotenuse.
j
D
A
m
Area
ABC = 8.24 cm 2
Area
EBF = 4.22 cm 2
Area
ADE = 12.46 cm 2
k
l
C
r
n
Area
ABC+Area
p
B
EBF = 12.46 cm 2
E
q
o
F
I proceeded to move points and the relationship was maintained.
m = 2.41 cm
k = 2.98 cm
p = 1.75 cm
n = 2.41 cm
j = 2.98 cm
q = 1.75 cm
l = 2.41 cm
r = 2.98 cm
o = 1.75 cm
E
Area
ABC = 2.52 cm 2
Area
EBF = 1.33 cm 2
Area
ADE = 3.85 cm 2
p
r
D
k
j
Area
ABC+Area
EBF = 3.85
q
cm 2
A
l
F
o
B
n
m
C
Closing Analysis:
From the learning window, this exercise is helpful in refining constructions,
understanding the attributes of equilateral triangles, and using the powerful GSP
formula feature. I could see it having the most value after students had some
familiarity with constructions and triangle theorems; I’m not sure students could
proceed without them. It might be hard to draw an equilateral triangle on GSP
without a little experience with a compass (or previous GSP lessons to construct
specific triangles).
From a teaching window, I would probably want to integrate algebra at this point
and show from that standpoint why it works. In my head I see all the numbers
adding up, but I want the proof algebra provides, and would want my students to
see the connections between geometry and algebra. I just couldn’t resist:
In an equilateral triangle:
3
3
height =
(base), so A= ½(base)(height), or
(base)2.
2
4
In the problem, adding the three areas:
p2 3 l2 3 k 2 3


simplifies to p 2  l 2  k 2 when the common factor is
4
4
4
removed.
Reflection: This was a cool problem. For one thing, I felt more comfortable and
secure with GSP than in my first set of problems, and secondly, it was an
interesting extension of a familiar problem. I did find myself still wanting to draw
on paper, and still wanting to do the algebra to back it up. Maybe GSP still seems
“too good to be true” since I’m still a novice. I also wonder how beginning
geometry students approach these exercises—can they do them, and if they can,
is it based on postulates and theorems they are able to understand, or is it based
on being able to manipulate buttons and commands as they go, without
necessarily connecting them to actual geometric assumptions and rules. I find
that as I think about the problem, and how I approach it on GSP, I still go back to
the basic geometry I learned 30 years ago and apply it to GSP. I guess in this
age of technology, my question is, what comes first, geometry or GSP? It might
come down to whether it is an inductive or deductive approach.
How about regular hexagons? (25 Points)
Plan: I planned to approach it just like the previous one, constructing hexagons
on each side of the triangle and comparing sums of areas.
Attempted Solution: I have to admit, I used pencil and paper to draw a
hexagon. As soon as I did, I saw what I needed to do and how GSP was going to
make it really quick. I marked one side of a hexagon as a mirror and reflected a
point around it, repeating it until the hexagon was complete. I even created a
hexagon tool with reflecting to try and speed it up. Once the hexagons were
complete, I constructed their interiors, found their areas and compared the sums
of the hexagons on the legs to the hexagon on the hypotenuse. Once again, it
supported the Pythagorean theorem.
Area AB''A''B'A'B' = 12.79 cm 2
Area BC''B'C'B''C'' = 45.67 cm 2
Area AB''A''B'A'B'+Area BC''B'C'B''C'' = 58.46 cm 2
Area CA''C'A'C''A = 58.46 cm 2
A''
C''
B''
B'
F
A'
A'
A
C''
B B'
E
C
C'
C''
B''
D
A''
B'
C'
Closing Analysis:
From the learning window this was another valuable extension of a problem. For
students, it demands that they understand what a hexagon is and how to
construct it. Once that it done, it doesn’t require much else.
From the teaching window, I would again try to integrate the problem with
algebra and reinforce the attributes of hexagons and 30-60-90 triangles. It can
summarize these other applications quite well. Once they see this works, I would
challenge them with the next question, “Would these figures or any others give
the same results? “
Reflection: I was pleasantly surprised to see it all work again, and felt
comfortable in the construction. The challenge was constructing true regular
hexagons using GSP, but it didn’t turn out to be too much of a challenge, once I
recognized it as a reflection. As I completed the problem, my brain actually “ran
ahead” to the next question, wondering if it would be the same for other figures.
Would these figures or any others give the same results?
I decided that any regular polygon would be fairly boring (other than the
challenge of the actual construction of a heptagon…) since all the polygons are
multiples of the original equilateral triangle. I wanted to think outside the box just
a little, so I tried circles:
Area
Area
EA = 3.87 cm 2
Area
EA
FB = 13.80 cm 2
2
Area FB 
2
Area
Area
DA = 17.67 cm 2
DA
2
Area
2
FB 
+
Area
EA
2
Area
DA
2
= 1.93 cm 2
A
E
= 6.90 cm 2
B
= 8.84
cm 2
D
F
= 8.84 cm 2
= 8.84 cm 2
C
Sure enough, the area of the semi circles on the legs added to the area of the
semicircle on the hypotenuse.