6.1: Analyzing Heron’s Formula
This activity is intended to supplement Trigonometry, Chapter 5, Lesson 2.
Time Required: 20 minutes
Activity Overview
In this activity, students will use their graphing calculators to determine the relationship between Heron’s Formula and the basic area formula.
Topics Covered
 Finding the area of a triangle
 Points of intersection
 Interpreting a graph
Teacher Preparation and Notes
 Make sure students have cleared \begin{align*}Y=\end{align*}
Y= menu before starting.  You may need to remind students how to TRACE and find points of intersection.
Associated Materials
 Student Worksheet: Analyzing Heron's Formula http://www.ck12.org/flexr/chapter/9703
Problem 1: The 3, 4, 5 right triangle
 Students should know this is a right triangle, with hypotenuse 5. Make sure that they have that the legs are 3 and 4.
 The area of this triangle is 6.
 With Heron’s Formula, \begin{align*}A = \sqrt{s(s  a)(s  b)(s  c)}\end{align*}
A=s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√ , and \begin{align*}Y1 = \sqrt{x(x  3)(x  4)(x  5)}\end{align*}Y1=x(x−3)(x−4)(x−5)−−−−−−−−−−−−−−−−−−√ students might get confused with the parenthesis. Make sure all students change their window to the dimensions to the right before graphing.
WINDOW
\begin{align*}Xmin = 1\end{align*}
\begin{align*}Xmax = 8\end{align*}
\begin{align*}Xscl = 1\end{align*}
\begin{align*}Ymin = 1\end{align*}
\begin{align*}Ymax = 10\end{align*}
\begin{align*}Yscl = 1\end{align*}
\begin{align*}Xres = 1\end{align*}
 The graph is to the right. Have students analyze the domain and range and why there are blank spaces in the graph. The domain is \begin{align*}(\infty, 00]\end{align*}
(∞,00] , [3, 4], and \begin{align*}[5, \infty)\end{align*}[5,∞) and the range is all real numbers greater than zero. If you have students zoom in further, they will see that there are no \begin{align*}x\end{align*}x− intercepts and one \begin{align*}y\end{align*}y intercept at (0, 0).

\begin{align*}Y2= 6\end{align*}
Y2=6 represents the area of this triangle. The horizontal line crosses the graph at (0.435, 6) and (6, 6). The first point, however is invalid because \begin{align*}x\end{align*}x cannot be negative here. Ask students why. Explain that \begin{align*}x\end{align*}x is actually \begin{align*}s\end{align*}s and that \begin{align*}s\end{align*}s cannot be negative, because by definition it is \begin{align*}\frac{1}{2}(a + b + c)\end{align*}12(a+b+c) .  Specifically, the point (6, 6) represents (\begin{align*}s\end{align*}
s , area). So, from this system of equations, we have determined what \begin{align*}s\end{align*}s is such that we have the correct area, in this case, also 6. If we can find the area in more that one manner, this will always work as a way to solve for \begin{align*}s\end{align*}s .
Problem 2
For this triangle, the area is, \begin{align*}A = \frac{1}{2}(15)\left(\frac{4 \sqrt{17}}{3} \right) = 5 \cdot 2 \sqrt{17} = 10 \sqrt{17} \approx 41\end{align*}
Here, students would graph \begin{align*}y = \sqrt{x(x  7)(x  12)(x  15)}\end{align*}
Again, \begin{align*}s\end{align*}