# 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 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, , and students might get confused with the parenthesis. Make sure all students change their window to the dimensions to the right before graphing.

WINDOW

- 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 , [3, 4], and and the range is all real numbers greater than zero. If you have students zoom in further, they will see that there are no intercepts and one intercept at (0, 0).

- 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 cannot be negative here. Ask students why. Explain that is actually and that cannot be negative, because by definition it is .
- Specifically, the point (6, 6) represents (, area). So, from this system of equations, we have determined what 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 .

## Problem 2

For this triangle, the area is,

Here, students would graph and to determine what , or , is. The graph of the two functions looks like:

Again, cannot be negative, so we eliminate the negative point of intersection. Therefore the answer is .

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Feb 23, 2012## Last Modified:

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