Unit 1 Portfolio: The Orchard Hideout
Cover Letter
Solving the unit problem requires knowledge of both the Pythagorean Theorem and coordinate geometry, both which we’ve learned. There are many interesting and helpful ways to find missing values in shapes or on a coordinate plane. Most of these methods are formulas designed to give clear instructions. There are also many proofs to consider when using these methods, which we’ll go over later.
The first concept to understand when solving this unit problem is the Pythagorean Theorem (1). The Pythagorean Theorem is an equation used to find the length of a right triangle’s longest side. Right triangles have a 90° angle, and their longest side is called the hypotenuse.
The Pythagorean Theorem is “a squared + b squared = c squared.” This can be used to find the length of a side of any right triangle if we’re given the length of the other two sides. Using some coordinate geometry which we’ll cover next, we can discover not only crucial right triangles hidden within the orchard hideout, but the length of its sides.
Using coordinate geometry, we can take our first steps towards solving the unit problem. Coordinate geometry allows us to find the distance between two points as well as finding the point equally distant from two points (midpoint). Let’s dive deeper into each of these concepts.
Finding the distance between two points can be done using this equation::
The formula can be used to find the distance between any two coordinates.
When looking between two growing trees to find the last line of sight, you’re going to look at the spot equidistant from each: the midpoint. Here you will find handy the midpoint formula:
This formula allows us to determine the exact point equally distant from both trees, or in other words, a point that the last line of sight must intersect. Using the coordinates of our midpoint, we can double the coordinate’s values to find the location of the furthest tree bordering our line of sight shown in the image below.
The final concept to uncover is that of circles. There are formulas that allow us to find the value of a circle’s circumference and its area, both which will be vital to solving the unit problem. The equation for a circle’s circumference is C = 2 x Pi x R. The equation for a circle’s area is A = Pi x R squared. This will help us solve the final portion of the unit problem when we start comparing the tree sizes with the information for how much they grow.
Unit Problem
The Orchard Hideout is a problem that requires the appliance of all the geometrical concepts previously discussed. The problem is that there is a grid of trees growing at a particular rate. If you stand in the center of the orchard you will be able to see out of it in every direction, but as the trees grow, your vision of beyond the orchard will decrease. The question is, “By what time will the trees have grown so thick, you can no longer see outside the orchard?”
To help us solve this problem, let’s use a simplified version of the orchard hideout by reducing our trees from 100 on each axis to six. The first step to take is finding where our last line of sight will be? By exploring our options for possible last lines of sight, we’ll notice that the hideout is divisible by 8. This means 1/8 of the hideout is being reflected 8 times to create the whole orchard. Since this is true, we only have to analyze 1/8 of the hideout in order to find a best line of sight because there’s no sense in doing the same math eight times.
Now that our orchard is totally simplified, let’s find the last line of sight. From what we know about equidistant lines, this line will be as distant from every tree as possible, which means it will fall somewhere between the trees (1, 0) and (3, 1). We also know it will intersect the midpoint of those trees because it must be equidistant from them. Now, let’s find the midpoint of (1, 0) and (3, 1) using the midpoint formula.
Finally, we can draw our last line of sight.
Now that we have a last line of sight, let’s take a look at the radius of our tree. The radius of the tree is determined by the distance from its center to its outside. Since the outside of the tree is tangent to our last line of sight during the undetermined time, the radius is the (shortest) distance from the center of the tree to the last line of sight.
At this point, we can create some right triangles. The first triangle consists of the radius of the tree, a segment of the last line of sight, and the distance from (1, 0) to (2, 0). The second, larger triangle consists of the distance from (1, 0) to (4, 0), the distance from (1, 0) to (4, 1) and the distance from (4, 1) to (4, 0). Notice that these triangles are similar, as they share the same angle. Because of this, we can use the proportion of these triangles’ side lengths to determine r -- the leg of our small triangle and the radius of our tree.
Selection of Work
Reflection
This unit gave me lots of new knowledge about algebra and geometry, and how they work together. I’ve always been curious about the relationship between coordinates and equations, and this unit was exactly about that. As a baseline, I learned several formulas for finding the distance of lines on a grid as well as the measurements of a right triangle and a circle. These later helped me find patterns on a graph, allowing me to calculate missing values like the hypotenuse of a triangle and the side-lengths of a similar triangle. It was the complex combination of these principles that caused me to fall off when finalizing the solution, but rest assured, the knowledge was still retained. As someone with an interest in statistics, I’ll be sure to carry everything I learned with me to college.
In addition to the complexity of the unit problem, I was unable to complete it due to my procrastination. Throughout the unit, my stack of uncomplete work rose higher and higher, and gave me great anxiety. Instead of getting the work done, I shunned it and decided to scrape by and do the bare minimum. At some point, I was so far behind that I felt I couldn’t ask the teacher for help without feeling embarrassed. Procrastination has always been a challenge for me, and I believe it occurs when I don’t feel pressured to do work. COVID restrictions did not help pressure me to do my work. Being in a classroom surrounded by peers and teachers is a lot more pressuring than sitting at home with an open pantry and total access to Netflix. Since COVID is also going to be a part of my life next semester, I’m making damn sure to prepare by doing therapy and making an effort to change my habits. Hopefully I’ll see you then with my nose to the grindstone.
Solving the unit problem requires knowledge of both the Pythagorean Theorem and coordinate geometry, both which we’ve learned. There are many interesting and helpful ways to find missing values in shapes or on a coordinate plane. Most of these methods are formulas designed to give clear instructions. There are also many proofs to consider when using these methods, which we’ll go over later.
The first concept to understand when solving this unit problem is the Pythagorean Theorem (1). The Pythagorean Theorem is an equation used to find the length of a right triangle’s longest side. Right triangles have a 90° angle, and their longest side is called the hypotenuse.
The Pythagorean Theorem is “a squared + b squared = c squared.” This can be used to find the length of a side of any right triangle if we’re given the length of the other two sides. Using some coordinate geometry which we’ll cover next, we can discover not only crucial right triangles hidden within the orchard hideout, but the length of its sides.
Using coordinate geometry, we can take our first steps towards solving the unit problem. Coordinate geometry allows us to find the distance between two points as well as finding the point equally distant from two points (midpoint). Let’s dive deeper into each of these concepts.
Finding the distance between two points can be done using this equation::
The formula can be used to find the distance between any two coordinates.
When looking between two growing trees to find the last line of sight, you’re going to look at the spot equidistant from each: the midpoint. Here you will find handy the midpoint formula:
This formula allows us to determine the exact point equally distant from both trees, or in other words, a point that the last line of sight must intersect. Using the coordinates of our midpoint, we can double the coordinate’s values to find the location of the furthest tree bordering our line of sight shown in the image below.
The final concept to uncover is that of circles. There are formulas that allow us to find the value of a circle’s circumference and its area, both which will be vital to solving the unit problem. The equation for a circle’s circumference is C = 2 x Pi x R. The equation for a circle’s area is A = Pi x R squared. This will help us solve the final portion of the unit problem when we start comparing the tree sizes with the information for how much they grow.
Unit Problem
The Orchard Hideout is a problem that requires the appliance of all the geometrical concepts previously discussed. The problem is that there is a grid of trees growing at a particular rate. If you stand in the center of the orchard you will be able to see out of it in every direction, but as the trees grow, your vision of beyond the orchard will decrease. The question is, “By what time will the trees have grown so thick, you can no longer see outside the orchard?”
To help us solve this problem, let’s use a simplified version of the orchard hideout by reducing our trees from 100 on each axis to six. The first step to take is finding where our last line of sight will be? By exploring our options for possible last lines of sight, we’ll notice that the hideout is divisible by 8. This means 1/8 of the hideout is being reflected 8 times to create the whole orchard. Since this is true, we only have to analyze 1/8 of the hideout in order to find a best line of sight because there’s no sense in doing the same math eight times.
Now that our orchard is totally simplified, let’s find the last line of sight. From what we know about equidistant lines, this line will be as distant from every tree as possible, which means it will fall somewhere between the trees (1, 0) and (3, 1). We also know it will intersect the midpoint of those trees because it must be equidistant from them. Now, let’s find the midpoint of (1, 0) and (3, 1) using the midpoint formula.
Finally, we can draw our last line of sight.
Now that we have a last line of sight, let’s take a look at the radius of our tree. The radius of the tree is determined by the distance from its center to its outside. Since the outside of the tree is tangent to our last line of sight during the undetermined time, the radius is the (shortest) distance from the center of the tree to the last line of sight.
At this point, we can create some right triangles. The first triangle consists of the radius of the tree, a segment of the last line of sight, and the distance from (1, 0) to (2, 0). The second, larger triangle consists of the distance from (1, 0) to (4, 0), the distance from (1, 0) to (4, 1) and the distance from (4, 1) to (4, 0). Notice that these triangles are similar, as they share the same angle. Because of this, we can use the proportion of these triangles’ side lengths to determine r -- the leg of our small triangle and the radius of our tree.
Selection of Work
- Circle Formulas https://docs.google.com/document/d/1nR7ZUpUoLfttkb37Ks1u3_yWpnLgHsLMP96NzjLsAGE/edit
- More on Circle Formulas [https://classroom.google.com/u/1/c/MTc5NDE0OTUzMTEx/sa/MTk4NDkwOTA0OTE3/details]
Reflection
This unit gave me lots of new knowledge about algebra and geometry, and how they work together. I’ve always been curious about the relationship between coordinates and equations, and this unit was exactly about that. As a baseline, I learned several formulas for finding the distance of lines on a grid as well as the measurements of a right triangle and a circle. These later helped me find patterns on a graph, allowing me to calculate missing values like the hypotenuse of a triangle and the side-lengths of a similar triangle. It was the complex combination of these principles that caused me to fall off when finalizing the solution, but rest assured, the knowledge was still retained. As someone with an interest in statistics, I’ll be sure to carry everything I learned with me to college.
In addition to the complexity of the unit problem, I was unable to complete it due to my procrastination. Throughout the unit, my stack of uncomplete work rose higher and higher, and gave me great anxiety. Instead of getting the work done, I shunned it and decided to scrape by and do the bare minimum. At some point, I was so far behind that I felt I couldn’t ask the teacher for help without feeling embarrassed. Procrastination has always been a challenge for me, and I believe it occurs when I don’t feel pressured to do work. COVID restrictions did not help pressure me to do my work. Being in a classroom surrounded by peers and teachers is a lot more pressuring than sitting at home with an open pantry and total access to Netflix. Since COVID is also going to be a part of my life next semester, I’m making damn sure to prepare by doing therapy and making an effort to change my habits. Hopefully I’ll see you then with my nose to the grindstone.