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6.1: Analyzing Heron’s Formula

Difficulty Level: At Grade Created by: CK-12
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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=s(sa)(sb)(sc), and Y1=x(x3)(x4)(x5) 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 (,00], [3, 4], and [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 xintercepts 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 12(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 than one manner, this will always work as a way to solve for s.

Problem 2

For this triangle, the area is, A=12(15)(4173)=5217=101741

Here, students would graph y=x(x7)(x12)(x15) and y=1017 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 (17,1017).

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Date Created:
Aug 19, 2014
Last Modified:
Nov 04, 2014
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