# 6.1: Analyzing Heron’s Formula

Difficulty Level: At Grade Created by: CK-12

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*} menu before starting.
• You may need to remind students how to TRACE and find points of intersection.

Associated Materials

## 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*}, and \begin{align*}Y1 = \sqrt{x(x - 3)(x - 4)(x - 5)}\end{align*} 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*}, [3, 4], and \begin{align*}[5, \infty)\end{align*} 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*}intercepts and one \begin{align*}y\end{align*} intercept at (0, 0).

• \begin{align*}Y2= 6\end{align*} 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*} cannot be negative here. Ask students why. Explain that \begin{align*}x\end{align*} is actually \begin{align*}s\end{align*} and that \begin{align*}s\end{align*} cannot be negative, because by definition it is \begin{align*}\frac{1}{2}(a + b + c)\end{align*}.
• Specifically, the point (6, 6) represents (\begin{align*}s\end{align*}, area). So, from this system of equations, we have determined what \begin{align*}s\end{align*} 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*}.

## 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*} and \begin{align*}y = 10 \sqrt{17}\end{align*} to determine what \begin{align*}x\end{align*}, or \begin{align*}s\end{align*}, is. The graph of the two functions looks like:

Again, \begin{align*}s\end{align*} cannot be negative, so we eliminate the negative point of intersection. Therefore the answer is \begin{align*}\left(17, 10 \sqrt{17}\right)\end{align*}.

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