Intermath Workshop Support s2

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:

C m AB = 4.47 cm m CB = 2.25 cm m CA = 5.01 cm Area ABC = 5.03 cm2 A 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

A B 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 m AB = 4.47 cm m EB = 2.25 cm Area DAC = 2.53 cm2 2 m EA = 5.01 cm Area CBE = 2.50 cm A B

Area DAC+Area CBE = 5.03 cm2 Area ABC = 5.03 cm2

Area DAC+Area CBE = 5.03 cm2 Area ABC = 5.03 cm2 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!

Area ABC = 4.13 cm2

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 Area ABC = 4.13 cm2 Area ABC = 4.13 cm2 Area ACJ = 6.15 cm2 I 2 J Area BCI = 2.02 cm B

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

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

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:

2 D Area DEF = 2.95 cm F Area DFG = 2.95 cm2

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 [email protected]

Link(s) to resources, references, lesson plans, and/or other materials Link 1 Link 2