Final Project Part A

By Amber Candela

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The Problem

Produce a GSP sketch to show a square with a quarter circle arc from each corner producing a region bordered by four smaller arcs

Determine the area of the shaded region in terms of the square on side lengths "s". Prove the conclusion of the area.

Given square ABCD, construct four circles. Circle with center D, radius CD; circle with center C, radius CB; circle with center B, radius BA; circle with center A, radius AD

Construct radii DE, DH, JH and EI. Since the sides of the square are radii, it gives that DE = DH = JH = EI = S where S = side of square ABCD

Look at triangle DEH and find the measure of angle DEH.

First find the measure of angle HDI. Using the trigonometric functions, we know that DI = s/2 since it is half of the side and DE = s so we know that cos θ=(S/2)/2 = 1/2, this gives angle EDH to be 60 degrees, similarly measure of angle JDE is cosθ=(S/2)/2 = 1/2 which gives it to be 60 degrees. Since angle CDA is a right angle and angle JDE = 60 and angle HDI = 60, it gives angle EDH = 30 degrees.

Using the area of a sector formula, which is Area = (θ/360) πr². This gives the area of sector DEH = (30/360)πs² = 1/12πs.

In order to find the area of the section between the triangle and arc EH we must find the area of triangle EDH.

Another way to find the area of a triangle is to use the formula A = 1/2(side1)(side2)(sin (angle between 1 and 2)). This gives the area of triangle EDH to be A = 1/2 (S)(S)(Sin30) = (1/2)(1/2) S² =1/4(S² )

To find the area between arc EH and triangle DEH we take the difference between the area of the sector and the area of the triangle.

[1/12πS] -[1/4(S² )] = 1/12πS - 1/4S²

Since there are four of these sections, we must multiply by 4 to get :

Area of four little arcs = (1/3πS - S² )

We know need to find the area of square FEHI so it can be added to the four little spaces.

First find one side, say EH.

To find EH we must first find EG. To find EG we must use our trig functions again and find EI. We know that sin(60) = EI/(s) which gives √3/2= EI / (s) which gives EI = S√3/2. To find GI take S - S√3/2. This is GI so EG = EI - GI which is S√3/2 - (S - S√3/2) = S√3-S = EG.

To find EH use trig functions again. This gives sin45 = x/(S√3-S) so x = (s√6-s√2)/2 = EH

This gives the area of square EFHG to be ((s√6-s√2)/2)²

Thus the area of the blue shaded region is given as

((s√6-s√2)/2)²+ (1/3πS - S²)

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Final Project Part B

Tree Data

From Assignment 12

By Amber Candela

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The Problem:

The following is tree data from a lumber company on age of trees in a forest and average number of board feet of lumber per tree that will be collected.

Find a function to fit the data

I first plotted the points and fit a quadratic formula to it. This gave me an equation of y = 0.011x²-.0682x+13.31. I then tried a power function and got an equation of y = .0006x^(2.4926). Both lines seemed to fit the data. Below is the graph with both functions.

Which one is better?

In order to figure out which would be a better fit, I put both functions into excel and used the x-values to see which would give me values closes to those found by the lumber company. Below is the excel spreadsheet of my findings.

While most may be comfortable using a quadratic function to fit the graph, from the values the power function gives, I conclude that it is a better fit. Both graphs could be used and if there was more data it might be found that one is better than the other, but with what is given I believe the power function fits the data better.

To look at the entire excel sheet, click below

TreeData

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