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# 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 $Y=$ 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, $A = \sqrt{s(s - a)(s - b)(s - c)}$, and $Y1 = \sqrt{x(x - 3)(x - 4)(x - 5)}$ students might get confused with the parenthesis. Make sure all students change their window to the dimensions to the right before graphing.

WINDOW

$Xmin = -1$

$Xmax = 8$

$Xscl = 1$

$Ymin = -1$

$Ymax = 10$

$Yscl = 1$

$Xres = 1$

• 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 $(\infty, 00]$, [3, 4], and $[5, \infty)$ and the range is all real numbers greater than zero. If you have students zoom in further, they will see that there are no $x-$intercepts and one $y$ intercept at (0, 0).

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

## Problem 2

For this triangle, the area is, $A = \frac{1}{2}(15)\left(\frac{4 \sqrt{17}}{3} \right) = 5 \cdot 2 \sqrt{17} = 10 \sqrt{17} \approx 41$

Here, students would graph $y = \sqrt{x(x - 7)(x - 12)(x - 15)}$ and $y = 10 \sqrt{17}$ to determine what $x$, or $s$, is. The graph of the two functions looks like:

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

Feb 23, 2012

Nov 04, 2014