Pythagorean Theorem Part Two

PYTHAGOREAN THEOREM – PART TWO

INTRODUCTION

The objective for this lesson on the Pythagorean Theorem – Part Two is, the student will apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

The skills students should have in order to help them in this lesson include, the Pythagorean Theorem, Squares and Square Roots.

We will have three essential questions that will be guiding our lesson. Number one, how does knowing the vertical and horizontal length of a right triangle plotted on the coordinate plane help you to determine the hypotenuse? Justify your thinking. Number two, how can you determine the length of the hypotenuse of a right triangle when it is plotted on a coordinate plane? Explain your thinking. Number three, describe the process you would use to find the hypotenuse of a right triangle, given the coordinates of the three vertices.

Begin by completing the warm-up on finding the hypotenuse of a triangle to prepare for Pythagorean Theorem – Part Two.

SOLVE PROBLEM – INTRODUCTION

The SOLVE problem for this lesson is, Terrence walked to the store from his house. On the map he started at the point three, negative four. He walked six blocks west and eight blocks north to get to the store. If each block is exactly the same distance in length, how far would he have to walk if he could walk a straight line directly from the store back to his house?

In Step S, we Study the Problem. First we need to identify where the question is located within the problem and underline the question. The question for this problem is, how far would he have to walk if he could walk a straight line directly from the store back to his house?

Now that we have identified the question, we need to put this question in our own words in the form of a statement. This problem is asking me to find the distance from the store to his house in one straight line.

REVIEW OF VERTICAL AND HORIZONTAL DISTANCES ON THE COORDINATE PLANE

What are the coordinates of Point A? two, five

What are the coordinates of Point B? five, two

What are the coordinates of Point C? two, two

What are the coordinates of Point B and Point C? five, two and two, two

Which are the coordinates are the same for these two points? The y-coordinates are the same so we know that this is a horizontal line.

Explain how we can find the distance between the two points by looking at the graph? We can count from two to five.

Is there another method we can use? Explain your thinking. We can subtract the values of the x-coordinates five minus two equals three.

What is the horizontal distance between Point B and C? Three units

What points can we use to determine the vertical distance of the leg that forms the triangle? Point A and Point C

What are the coordinates of Point A and Point C? A is two, five and C is two, two.

What do you notice about the coordinates of the two points that we are using to determine the measure of the vertical leg? The x-coordinates are the same.

How can we find the distance between the two points by looking at the graph? We can count from two to five.

Is there another method we can use to determine the distance? Explain another method we can use. Subtract the values of the y-coordinates five minus two is equal to three.

What is the vertical distance between Points A and C? Three units

DISCOVERY ACTIVITY – PYTHAGOREAN THEOREM

Plot Points A, B and C on the coordinate plane and label each. Then connect to form a triangle. Point A at three, four; Point B at three, eight; and Point C at six, four.

What is the length of line segment AB? Four units. Defend your thinking. We counted the units or we subtracted the y-coordinates.

What is the length of AC? Three units. Defend your thinking. We counted the units or subtracted the x-coordinates.

What is the length of line segment BC? We cannot tell by looking at the graph.

Let’s make a ruler by cutting the bottom row of the squares off of the grid.

Place the ruler next to the diagonal line to determine its length.

What is the length of the hypotenuse of our triangle? Five units

To the left of line segment AB let’s draw a square so that line segment AB is one side of the square.

How long is each side of the square? Four units

What is the area of the square? Sixteen square units. Justify your answer. I used the formula for the area of a square and multiplied the side measures. I can also count the number of unit squares inside the larger square.

Below line segment AC let’s draw a square so that line segment AC is one side of the square.

How long is each side? Three units

What is the area of the square? Nine square units. Justify your answer. I used the formula for the area of a square and multiplied the side measures. I can also count the number of unit squares inside the larger square.

To the right of line segment BC let’s draw a square so that line segment BC is one side of the square.

How long is each side? Five units

What is the area of the square? Twenty five units squared or twenty five square units. Justify your answer. I used the formula for the area of a square and multiplied the side measures. I can also count the number of unit squares inside the larger square.

Identify the area of the three squares. Nine, sixteen and twenty five square units

Do you see a relationship between the first two and the third numbers? Yes. Defend your answer. The sum of nine and sixteen is twenty five.

Explain what this means in relationship to the sides of the triangles. The measure of line segment AB squared, which is four squared plus the measure of line segment AC squared, which is three squared is equal to the measure of line segment BC squared, which is five squared. Or three squared plus four squared is equal to five squared. Nine plus sixteen is equal to twenty five.

APPLYING THE PYTHAGOREAN THEOREM ON THE COORDINATE PLANE

Using the same graph, plot the three points given on the coordinate system, label them, and connect them to form a triangle.

What is the length of line segment DE? Defend your thinking. Line segment DE is five units in length. I counted the units or I subtracted the y-coordinates.

What is the length of line segment DF? Defend your thinking. The length of line segment DF is twelve units. I counted the units or I subtracted the x-coordinates.

What is the length of line segment EF? Defend your thinking. I can’t tell by looking at the graph. Let’s use the strip cut from the bottom of the page as a ruler to find the length of line segment EF. What is the length? The length of line segment EF is thirteen units.

We will not have room to draw all the squares next to each side of the triangle, but we can still use the same steps as in the first example to figure out the area of each square.

What would the area of a square be that had line segment DE as one of its sides? Explain your thinking. The area would be twenty five units squared. I would multiply side times side to determine the area, five times five is equal to twenty five.

What would the area of a square be that had line segment DF as one of its sides? Explain your thinking. The area would be one hundred forty four units squared. I would multiply side by side to determine the area. Twelve times twelve is equal to one hundred forty four.

What would the area be of a square be that had line segment EF as one of it sides? Explain your thinking. The area would be one hundred sixty nine units squared. I would multiply side times side to determine the area. Thirteen times thirteen is equal to one hundred sixty nine.

Identify the areas of the three squares that would be formed by using the measure of each side of the triangle. Twenty five units squared, one hundred forty four units squared and one hundred sixty nine units squared.

Explain the relationship between the three numbers. The sum of the area of the first two squares equals the area of the third square.

What does this mean in relationship to the sides of the triangles? The measure of DE squared, which is five squared plus the measure of DF squared, which is twelve squared is equal to EF squared, which is thirteen squared. Or five squared plus twelve squared is equal to thirteen squared. Twenty five plus one hundred forty four is equal to one hundred sixty nine.

How did we get the numbers nine, sixteen and twenty five in Question nine on the previous page and the numbers twenty five, one hundred forty four and one hundred sixty nine in Question nine on this page? They were the areas of the square we made from the sides of the triangles.

Explain how to find the area of a square. Multiply a side by a side, or side squared.

If the sides are identified as a and b, and the hypotenuse, which is always the longest side of the triangle and across from the right angle, was side c, what equation could you write to relate the areas of the three squares? A squared plus b squared is equal to c squared or the area of the square with side a, plus the area of the square with side b, equals the area of the square with side c.

FINDING THE HYPOTENUSE ON THE COORDINATE PLANE USING EDGE LENGTHS OR COORDINATES

We will be using the triangle and substituting the horizontal and vertical lengths of the triangle from the coordinate grid into the Pythagorean Theorem to determine the hypotenuse.

What is the Pythagorean Theorem? a squared plus b squared is equal to c squared.

What is the length of side a? six units

What is the length of side b? eight units

What is the value of six squared? Thirty six

What is the value of eight squared? Sixty four

What is the value of c squared? One hundred

Explain how we can determine the length of Side C if we know the value of c squared. We can find the square root of c squared.

What is the value of the square root of one hundred? The value of the square root of one hundred is ten.

Triangle ABC has the following coordinates: A is at five, four; Point B is at five, seven; and Point C is at ten, four.

How is Problem four different from Problems one through three? Problems one through three have a triangle on a coordinate grid and Problems four through six give the coordinates of the triangle, but do not show a picture.

How can we determine the side lengths of the triangle that is created given the coordinates? We can find the vertical distance by finding the difference between the y-coordinates of two points and we can find the horizontal distance by finding the difference between the x-coordinates.

What is the distance from Point A to Point B? Three units

What is the measure of line segment AB? Three units

What is the distance from Point A to Point C? Five units

What is the measure of line segment AC? Five units

Explain how we will find the distance from Point B to Point C. Use the Pythagorean Theorem. Three squared plus five squared is equals c squared. Nine plus twenty five is equal to c squared. C squared is equal to thirty four. C then is equal to the square root of thirty four or approximately five point eight three.

SOLVE PROBLEM – COMPLETION

We are going to go back to the SOLVE problem from the beginning of the lesson. The question was, Terrence walked to the store from his house. On the map he started at the point three, negative four. He walked six blocks west and eight blocks north to get to the store. If each block is exactly the same distance in length, how far would he have to walk if he could walk a straight line directly from the store back to his house?

In the S Step, we Study the Problem. We underline the question and we complete this statement. This problem is asking me to find the distance from the store to his house in one straight line.

In the O Step, we Organize the Facts. We first begin by identifying the facts. Let’s go back and read our SOLVE problem again. Each time we read a fact we’ll strike the fact with a mark. Terrance walked to the store from his house./ On the map he started at the point three, negative four./ He walked six blocks west and eight blocks north to get to the store./ If each block is exactly the same distance in length,/ how far would he have to walk if he could walk a straight line directly from the store to his house?

After we identify the facts, we eliminate any unnecessary facts. When we go back to our SOLVE problem we identify one fact as unnecessary. And that was that Terrence walked to the store from his house. We can strike through that fact. All the rest of the facts are necessary to solve the problem.

After we eliminate the unnecessary facts we list the necessary facts. He walked six blocks west; he walked eight blocks north; the blocks are all equal distance; and he started at the point three, negative four on the map.

Our next Step is L, Line Up a Plan. Write in words what your plan of action will be. Draw a picture of the blocks he walked, then make a diagonal line to make a triangle. Use the Pythagorean Theorem to find the hypotenuse.

The next part of L is to choose an operation or operations. We will be using three operations in this problem. Multiplication, addition and determining square roots

In the V Step we Verify Your Plan with Action. We begin with an estimate. We estimate your answer. The estimate we have here is the number of blocks will be larger than nine.

Then we carry out your plan. We use the Pythagorean Theorem, a squared plus b squared is equal to c squared. We substitute in our given values from our O Step, six and eight. So that six squared plus eight squared is equal to c squared. Thirty six plus sixty four is equal to c squared. One hundred is equal to c squared. In order to identify the value of C we take the square root of c and the square root of one hundred. So that c is equal to ten.

In the E Step, we Examine Your Results.

Does your answer make sense? We compare it to the question. Yes, because I found the distance from the store to his house.

Is your answer reasonable? Compare your answer to the estimate. Yes, I thought the distance would be greater than nine blocks, and it is.

Is your answer accurate? This is where you would check your work. Yes.

Write your answer in a complete sentence. The distance from the store to Terrence’s house is ten blocks.

CLOSURE

Now let’s go back and review the essential questions from this lesson.

Our first question was, how does knowing the vertical and horizontal length of a right triangle, plotted on the coordinate plane, help you to determine the hypotenuse? Justify your thinking. You can substitute the vertical and horizontal side lengths into the Pythagorean Theorem as the values for a and b in order to find the value of c which is the hypotenuse.

Question two, how can you determine the length of the hypotenuse of a right triangle when it is plotted on a coordinate plane? Explain your thinking. You can count the units to determine the measure of the two legs of the triangle and then substitute those values into the Pythagorean Theorem to find the value of the hypotenuse.

And question three, describe the process you would use to find the hypotenuse of a right triangle, given the coordinates of the three vertices. Find the two vertices that have the same x-coordinates and subtract the y-values to determine the length of the vertical leg. Use the two vertices that have the same y-coordinates and subtract the x-values to determine the length of the horizontal leg. Substitute in those two values for a and b in the Pythagorean Theorem and solve for c to find the hypotenuse.