6.5 - Applications of Matrices and Determinants

6.5 - Applications of Matrices and Determinants -

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
<?xml version="1.0" encoding="iso-8859-1"?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>6.5 - Applications of Matrices and Determinants</title> <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" /> <link href=". ./m116.css" rel="stylesheet" type="text/css" /> <link href="matrix.css" rel="stylesheet" type="text/css" /> </head> <body> <h1>6.5 - Applications of Matrices and Determinants</h1> <h2><img src="triangle.gif" alt="Area of a Triangle" width="232" height="214" class="imgrt" />Area of a Triangle</h2> <p>Consider a triangle with vertices at (x<sub>1</sub>,y<sub>1</sub>), (x<sub>2</sub>,y<sub>2</sub>), and (x<sub>3</sub>,y<sub>3</sub>). If the triangle was a right triangle, it would be pretty easy to compute the area of the triangle by finding one-half the product of the base and the height. </p> <p>However, when the triangle is not a right triangle, there are a couple of other ways that the area can be found.</p> <h3>Heron's Formula</h3> <p><img src="herons.gif" alt="Triangle with vertices at (-2,2), (1,5), and (6,-1)" width="368" height="273" class="imgrt" />If you know the lengths of the three sides of the triangle, you can use Heron's Formula to find the area of the triangle. </p> <p>In Heron's formula, s is the semi-perimeter (one-half the perimeter of the triangle). </p> <p>s = 1/2 ( a + b + c )<br /> Area = sqrt ( s ( s-a) ( s-b) ( s-c) )</p> <p>Consider the triangle with vertices at (-2,2), (1,5), and (6,1). </p> <p>Using the distance formulas, we can find that the lengths of the sides (arbitrarily assigning a, b, and c) are a = 3 sqrt(2), b = sqrt(61), and c = sqrt(73).</p> <p>Using those values gives us . ..</p> <p>s = 1/2 ( 3 sqrt(2) + sqrt(61) + sqrt(73) )<br /> s - a = 1/2 ( - 3 sqrt(2) + sqrt(61) + sqrt(73) )<br /> s - b = 1/2 ( 3 sqrt(2) - sqrt(61) + sqrt(73) )<br /> s - c = 1/2 ( 3 sqrt(2) + sqrt(61) - sqrt(73) )</p> <p>s ( s - a ) ( s - b ) ( s - c ) = 1089 / 4</p> <p>When you take the square root of that, you get 33/2, so the area of that triangle is 16.5.</p> <h4><strong>Problems with Heron's Formula include</strong> </h4> <ul> <li>Must know the lengths of the sides of the triangle. If you don't then you have to use the distance formula to find the lengths of the sides of the triangle.</li>
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
<li>You have to compute the semi-perimeter, so chances are you will have fractions to work with.</li> <li>Lots of square roots are involved. For the lengths of the sides of the triangle and for the area of the triangle.</li> <li>It's not the easiest thing in the world to work with.</li> </ul> <h3>Geometric Technique</h3> <p><img src="geoarea.gif" alt="Triangle with vertices at (-2,2), (1,5), and (6,-1) bounded by rectangle" width="368" height="278" class="imgrt" />The triangle can be enclosed in a rectangle. The vertices of the triangle will intersect the rectangle in three places, forming three right triangles. These triangles are denoted A, B, and C in the
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

6.5 - Applications of Matrices and Determinants -

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online