InterMath | Workshop Support | Write Up Template
Title
Triangle Formula
Problem Statement
Teachers customarily teach a formula to solve for the area of a triangle. What would be a
more concrete way to help students develop the formula for themselves in an
understandable, yet simple, concept?
Problem setup
The problem is asking for a way to explain why the formula A=1/2ba is used to solve for the area
of a triangle…by looking at way the formula could be simply solved by a student geometrically.
What are the simplistic underpinnings of this formula as it relates to 2-D properties and
relationships?
Plans to Solve/Investigate the Problem
Using the Geometry Sketch Pad (GSP), I plan to construct a triangle and play with it.
Why would we use a formula without knowing the basis for use? How can I explain this concept
to my students without resorting to a “formula fix?” I have a rough idea that triangles are tied to
quadrilaterals in some way. Maybe we can start by inscribing a triangle into a rectangle and
seeing how that relationship can possible affect how I view triangles.
Investigation/Exploration of the Problem
Let’s construct a right triangle:
m AB = 4.47 cm
C
m CB = 2.25 cm
m CA = 5.01 cm
Area
A
ABC = 5.03 cm2
B
Here we see a right triangle (created in GSP by constructing a perpendicular line segment (BC)
from line segment AB. I also allowed GSP to calculate the area, since we are not allowed to use
the formula for solving. Next, I want to construct a quadrilateral which inscribes the triangle:
Area
ADC = 5.03 cm2
D
C
m AB = 4.47 cm
m CB = 2.25 cm
m CA = 5.01 cm
B
A
Area
ABC = 5.03 cm2
As you can see, the area for both triangles are the same, but now we have an added trait. The two
similar triangles have constructed a rectangle. NOTICE THIS: If we were to employ the
triangular formula for area, the base and height for that formula are the length and width of the
rectangle. See how much sense the formula makes? If I multiply the length and width of the
rectangle (4.47*2.25) I get TWICE the area of my original triangle! We can synthesize this
information with the first write up (by all rights, this should be an extension of the first write-up,
but I was one Van Heile level too low to make the connection). As long as the base of the
triangle is inscribed in the rectangle, no matter where I move the point along its opposite parallel
line, I should keep the same area. This ALSO means, that all other areas inside quadrilateral
ABCD should be exactly one half of ABCD’s total area…or the same area as the inscribed
triangle. Let’s try a few!
D
E
C
Area
Area
Area
m AB = 4.47 cm
m EB = 2.25 cm
DAC = 2.53 cm2
CBE = 2.50 cm2
DAC+Area
Area
B
A
CBE = 5.03 cm2
ABC = 5.03 cm2
D
E
C
Area
Area
Area
m AB = 4.47 cm
m EB = 2.25 cm
DAC = 1.10 cm2
CBE = 3.93 cm2
DAC+Area
Area
B
A
ABC = 5.03 cm2
E
C
Area
Area
m AB = 4.47 cm
m EB = 2.25 cm
DAC = 4.61 cm2
CBE = 0.42 cm2
DAC+Area
Area
m EA = 5.01 cm
CBE = 5.03 cm2
D
Area
m EA = 5.01 cm
A
CBE = 5.03 cm2
ABC = 5.03 cm2
B
m EA = 5.01 cm
See?! Regardless of where I pull the point, the area remains unchanged in both the original
triangle or the added combination of triangles created within Quadrilateral ABED. Does this
work for all triangles (not just right ones)? Let’s create a random triangle and see!
C
Area
ABC = 4.13 cm2
B
A
In this triangle ABC, I will choose each line segment as a base, and construct a quadrilateral
from that base to inscribe the triangle. The question is…will the areas resulting from each
constructed quadrilateral be exactly twice that of the triangle? Let’s see:
C
I
J
Area
ABC = 4.13 cm2
Area
ABC = 4.13 cm2
Area
ACJ = 6.15 cm2
Area
BCI = 2.02 cm2
B
A
We need to look no further. I was really wrong on this guess! But just by looking at it, I can tell
where I made a huge mistake. I am thinking too far on the side of RIGHT. Notice I made a
rectangle…with all right angles. Instead of being so RIGHT…I think I need to REFLECT
awhile. Let’s do that and draw some new conclusions!
C
Area
ABC = 4.13 cm2
B
A
Okay, I reflected. My thoughts aren’t happy. We have symmetry, but this doesn’t get my
message across. I want a more PARALLEL thought, instead:
Area
ABC = 4.13 cm2
Area
AKC = 4.13 cm2
K
B
A
Oh, yessssss! That’s what I was looking for! The key to this relationship is the understanding of
a parallelogram! Regardless of the triangle’s base, if the two shortest triangle legs create parallel
lines to the opposite point, I can create a parallelogram that is twice the size of the original
triangle! Let’s try a totally different triangle to see if this assumption is correct:
E
D
F
Area
DEF = 2.95 cm2
Area
DFG = 2.95 cm2
G
Okay, I see the pattern emerging. Of course we can determine the area of triangle by exploring
its quadrilateral relationships! The main condition is the triangle must be one half of a
parallelogram! In order to form the parallelogram, we can make parallel lines from the two
shorter side’s line segments. In the above diagram, parallel line segment DG was created from
side EF. Line segment DE needed a parallel line through point F (creating FG).
To summarize, we can help explain the area formula of triangle by using the parallelograms they
are inscribed within. Whether through the simplistic right rectangles or squares or other types of
parallelograms!
Extensions of the Problem
How can we relate pyramids inscribed within a cube to the creation of its formula?
Author & Contact
Lindell Dillon
lindelldillon@yahoo.com
Link(s) to resources, references, lesson plans, and/or other materials
Link 1
Link 2