Associated Topics || Dr. Math Home || Search Dr. Math

### Expected Value of Area of Triangle

```
Date: 05/10/2000 at 10:16:17
From: Aled Jones
Subject: Uniform Distribution

AB is the diameter of a circle, with radius r. AP is a chord. X is the
angle between AB and AP. X is uniformly distributed over the interval
0 to pi/2. Find the expected value of the area of triangle APB.
```

```
Date: 05/10/2000 at 14:59:55
From: Doctor Anthony
Subject: Re: Uniform Distribution

The area of triangle

APB = (1/2)2r.2r.cos(x).sin(x)

=  2r^2.cos(x).sin(x)

=  r^2.sin(2x)

The expected value of the area is then

INT(0 to pi/2)[r^2.sin(2x).dx/(pi/2)]

= (2/pi)r^2.(-1/2)cos(2x)   from  0 to pi/2

= (-1/pi)r^2[cos(pi) - cos(0)]

= (-1/pi)r^2[-1 - 1]

= 2.r^2/pi

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability
High School Probability

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/