• Evaluate every proportion with respect to the proper assignment of corresponding parts before assessing the accuracy of the computation. Equivalencies between ratios are valid only with appropriate relationships between numerator and denominator.
• In class discussion throughout the lesson and during the review of the lesson, note the accuracy of students’ representation of each ratio.
• Examine the arithmetic of addition and subtraction of the values of ordered pairs on a coordinate grid. Note the distinction between direction (left to right for positive, and right to left for negative) and distance. For example, the absolute value of the distance between (0, 1) and (−10, 1) is 10.

[IS.9 - Struggling Learners]

Part 1: Distance and Perimeter

Tell students, “When discussing the topic of distance, we can discuss the length of a segment, the distance from Point A to Point B, the total distance around an object (perimeter), the distance between coordinates on a coordinate plane, distance as a function of the product of rate and time, or the height of an object.” [IS.3 - All Students]

“We can use distance to find a missing length, a missing height, or the perimeter of a shape or object. We can use a variety of methods to find such distances.”

“When we discuss distance, do we think in terms of positive and negative distance? If so, give an example.”

“Say, for instance, that Sarah faced a stop sign and walked 3 steps backwards. How far did she walk? How would we represent that distance? In order to denote the negative direction, we would represent the steps as −3. However, she only walked 3 steps. She didn’t walk negative steps. In a common sense context, we know that we can’t go a negative distance. Therefore the actual distance is 3, which can be described as |-3| = 3. We are touching on the concept of absolute value here. To interpret the equation in words, we are saying the absolute value of −3 is 3.” [IS.4 - All Students]

“Now, suppose she takes 5 steps forward. How far did she walk? How would we represent that distance? In order to represent the positive direction, we would represent the steps as +5. She walked 5 steps. Thus, using the absolute value description as shown above, the distance can be described as |5| = 5. The absolute value of 5 is 5.”

“If going forward 5 steps from the point of 3 steps behind the stop sign, where does this place her? In other words, what is ? The answer is 2. Now, how many steps did she walk in all? We can determine the answer to that question by using the absolute value of the two numbers, e.g., |-3| + |5| = 3 + 5 , which equals 8. So, she walked a total of 8 steps.”

Use a picture to visually represent this concept while stating the above statements to the class. [IS.5 - Struggling Learners] Draw a number line where the stop sign is at zero and show Sarah walking 3 steps behind the stop sign (−3) to 3 steps past the stop sign (+3). This will show students where the 3 − (−3) = 6 comes from visually. Remind students that when subtracting, make sure to absolute value the answer since distance is always positive, as discussed above.

“Note the + and − distances are comparable to right and left on an x-axis, respectively, or the number line. Now that you have an understanding of direction of a distance and the absolute value, thereof, let’s move on to measurement of a distance, using a ruler!”

“We can use a ruler to draw a line segment of a given length. We can find the length of a line segment, using a ruler. Let’s do both.”

“First, suppose we wish to draw a line segment of length 2.375 inches. We must determine the number of intervals we have between each successive inch. We find that we have 8 intervals. Thus, the ruler is marked in eighths, with each interval representing  inch.”

“Now, what must we decide? We must convert 2.375 to a fraction. What is 2.375 as a fraction?” ()

“Thus, we will draw a line from 0 inches to 2 inches and then continue on until we reach the third mark after the 2.”

Instruct students to draw this on their paper while following along with you. “Second, let’s measure the given line segment.”

Hand out to students the Measurement activity sheet (M-8-5-1_Measurement.docx) with the line segment pre-drawn at  cm.

“Notice that both lines on the activity sheet are larger than the actual measurement scale in order to make them easy to see and read.”

Instruction with this activity is intended to emphasize the partitions of the two measurement scales rather than the absolute distances they represent. In the inch scale, the partitions are in sixteenths and in the centimeter/millimeter scale, the partitions are in tenths.

Use the following instructions to verbally guide students through the process of measuring with centimeters: [IS.6 - All Students] “Let’s place the straightedge directly under the line segment. We must line up the left- side of the line segment with 0 cm on the straightedge. We can tell that the line segment is more than 4 cm. Each interval between successive centimeters is . Our segment goes seven one-tenths past 4 cm, or past 4 cm. Thus, the length of the segment is  cm. Note that this measurement is also 4.7 cm; it is also 4 centimeters and 7 millimeters.”

Point out to students that each centimeter is divided into 10 millimeters.

“Notice that decimal measures for feet and inches are quite common, too, but they can cause problems and must be reported correctly to eliminate any confusion. For example, when writing 3.5 ft, you might think of this measurement as 3 feet 5 inches. However, it is actually 3 feet 6 inches, since 0.5 or one-half of one foot is 6 inches.”

“Likewise, 3.6 feet is not 3 feet 6 inches! It is actually 3 ft 7.2 inches because 0.6 of a foot is more than half a foot, so we know it is more than 6 inches. To determine exactly how many inches we have, multiply .6 by 12 (the number of inches in a foot). Examples like this illustrate why it’s much more common to see metric measurements in decimal form and customary measurements expressed with fractions.”

“Now suppose we are not only interested in the length of a line segment. Instead, we wish to find the total distance around a shape, i.e., the perimeter. Given the following shape, let’s find the perimeter.”

“The perimeter is the total distance around the triangle. Thus, the perimeter is  7 + 4 + 9, which equals 20.”

“What if we looked at perimeter in a real-world context? Suppose Michael must build a border around a rectangular garden. The width can only be 4 feet, with a total maximum perimeter of 26 feet. What length of border must Michael use?” (9 ft)

The steps for solving this problem:

Activity—Perimeter

Part 1: Draw a rectangle, square, and pentagon. Label each side length with values of your choosing. Show the process for finding the perimeter and record that perimeter.

Part 2: Write a word problem, describing the real-world need for finding perimeter. Show the process for finding the perimeter or distance, and record the perimeter/distance.

Perimeter of Compound Figures

Tell students, “Given the following compound figure, find the total distance around the figure.” (42 cm)

“Notice that we did not count the length of the bottom of the triangle, since it is adjoined to the rectangle.”

“What if we had a circle? Suppose I stake a peg into a football field, with a long rope attached. I then stretch out the rope and use a paint gun to mark a spot. I continue to stretch the rope tight, mark spots, and move around in a circular motion. When I am finished, I will have a bunch of markers that form a circle, with my center at the staked peg. Now, what if I wish to find the distance around the circle, or the distance that I just walked? How could I find that distance? What is another term for the distance around a circle?”

Give students time to ponder various measurement possibilities. (Some might say you could stretch a measuring tape around the circle. Others might think of creating a square around the circle, finding that perimeter, and estimating the distance around the circle.)

“The distance around a circle, or the perimeter of a circle, is known as the circumference. Circumference can be found using the following formula:

 or C = 2πr where π is estimated to be approximately 3.14; d = the diameter of the circle, and r is the radius of the circle.”

“Suppose our circle had a diameter of 24 feet. Substituting into our formula we have:”

Note: Remind students that, by the commutative property of multiplication, 24p and p24 are one and the same.

“Therefore, we could say that the distance around the circle was 75.36 feet.”

Divide students into groups of three to four. Students will use a tape measure to find how close they can get to pi. Measure the circumference (distance around the circle) and diameter of the three different circles found on the floor (using cm). Use the formula for circumference to find each value of pi. Due to potential creases in the tape measure, students will likely get three approximations of the value of pi. Each group will report on the three values for pi, indicating the value closest to 3.14. Copy or print the chart below for student groups to record their results and calculations:

Tell students, “Often, we are asked to determine the percent of a distance, or calculate the distance for a given percent. Let’s explore this topic within the discussion of circumference of a circle. We will explore the topic twofold.”

“First, we will determine the percent of a distance. Suppose we are walking along a circular track. We have walked 189 meters of a 400 meter circular track. What percentage of the perimeter/circumference of the circle have we walked? We would set up the following question:

189 meters is what percent of 400 meters?

We can convert this question to an equation by writing:

Thus, we have walked approximately 47% (0.47 * 100 converts the decimal to the percent) of the distance around the circular track. This percentage seems reasonable, since 189 is less than, but close to 200.” Note: Point out to students that most circular running tracks that surround playing fields are not strictly circular, but composite figures. They typically consist of two semicircles at either end whose radii are about 36.4 meters. They are joined by an 85-meter straight line on each side.

“Second, let’s determine the distance for a given percent. Suppose the circumference of our circle is 16π, or approximately 50.27. What is 60% of the circumference, or distance around the circle?” (approximately 30.16)

“Therefore, if we trace 60% of the perimeter of the circle, we will have covered 30.16 units. Keep in mind that when we use percents, we have to use the decimal version of the percent. This is found by moving the decimal place two spaces to the left or dividing 100, thus 60% is 0.60 as a decimal.”

“A group of university students is given the following space for their fundraising booth.” Draw the approximate locations of the following four ordered pairs on the board: “The coordinates of the four vertices of the rectangle are: A (−10, 20), B (10, 20), C (10, −20), and D (−10, −20).” (Distance from A to B = 20; Distance from B to C = 40; Distance C to D = 20; Distance D to A = 40).

“If the units are feet, what is the perimeter of the space they are given?” Give students time to discuss.

“We need to find the horizontal distance from either Point A to Point B or Point D to Point C, and the vertical distance from either Point A to Point D or Point B to Point C.”

“Let’s find the horizontal distance from Point A to Point B. As we discussed earlier, distance is positive, as evidenced by the absolute value. We will take the absolute value of the difference between the x-coordinates of each point. Thus, the length of their space can be indicated by:

“Now, let’s find the vertical distance from Point A to Point D.”

“In order to find the width of the space, we will find the absolute value of the difference between the y-coordinates of each point. Thus, the width of their space can be indicated by:

The total distance around the space they are given, or the perimeter, can be found by adding: 20 + 20 + 40 + 40, which equals 120. The perimeter of the space is 120 feet.”

Tell students, “Suppose you are given the following choices and accompanying diagram:”

The parking lot at the city mall is being reconstructed. For the front lot of the mall, the construction manager must use a width of 428 feet. If the diagram illustrates two different widths (with the lot starting at different distances from the drive), which width should s/he choose?

Students should realize that Width Choice A is the desired width, since the distance can be found by:

Comparing the points on the graph with the calculated values shows that they are approximations.

Create a real-world problem, in which an individual must choose the correct path, based upon a desirable/preferred distance. Include an appropriate diagram, drawn on a coordinate grid, to illustrate the lengths/distances. Show the process for solving the problem, state your answer, and provide supporting reasons.

“Sometimes we are given only two sides of a right triangle and must find the length of the third side. How would we go about finding a missing side length of a right triangle?”

“We could use the Pythagorean Theorem. The Pythagorean Theorem states that the sum of the squares of two legs equals the square of the hypotenuse. Written in equation form, we have:

where a and b are legs, and c is the hypotenuse.”

“Let’s look at an example. Suppose you are given:

In this example, we are looking for the length of the hypotenuse, thus, we will solve for c.

Therefore, c is approximately 3.61 cm.”

“Now, let’s look at another example. This is actually a very common triangle. In this example, we will be given the lengths of the hypotenuse and one leg. Note that we must identify the right angle to determine which side is the hypotenuse and which sides are the two legs.”

“If we want to find the length of the missing leg, we simply substitute the lengths that we do know into the Pythagorean Theorem formula:

Therefore, the length of the missing leg is 4 inches.”

“Now, let’s take a look at a real-world problem:” Read the problem aloud and then diagram it on the board: [IS.7 - All Students]

Monique is apartment hunting. She has chosen two apartment complexes to view. To get from Apartment 1 to Apartment 2, she can either walk 0.8 miles north and then 0.6 miles east, or she can cut across a park. Find the distance she would travel if she cuts across the park. Will that path be shorter or longer than the route walking directly north and then directly east? If so, how much shorter or longer?

“The indirect path is represented by the hypotenuse. Therefore, we will substitute the values of our legs into the formula for the Pythagorean Theorem.”

“The path across the park is 1 mile in distance. The distance of the path is the sum of 0.8 and 0.6, which equals 1.4 miles. Therefore, the path across the park is 0.4 miles shorter than the path, going directly north and then directly east.”

“Now, suppose we are to find distances on a coordinate grid. In the example, the distances form a right triangle, therefore enabling us to use either the Pythagorean Theorem to find the missing distance, or use the actual distance equation. We will look at both. It should be noted that the distance equation can be used to find the distance between any two coordinate pairs, regardless of whether or not there is more than one line segment. We are simply extending our discussion of the Pythagorean Theorem.”

“Consider a triangle with vertices at (20, 56), (20, 32), and (38, 32). Do you recall how we can find a horizontal distance and vertical distance?” (Subtract the coordinates.)

“In order to find the horizontal distance between (20, 32) and (38, 32), we compute 38 − 20, which equals 18. In order to find the vertical distance between (20, 32) and (20, 56), we compute 56 − 32, which equals 24.”

“We can now use those two lengths and the Pythagorean Theorem to find the length of the hypotenuse:

Therefore, the length of the hypotenuse is 30.”

“Now, let’s use the distance formula to find the distance between coordinates (20, 56) and (38, 32). Does anyone have an idea of what this formula will look like?” (the Pythagorean Theorem)

“In order to find the distance between two points, use the following formula:

Substituting in our values, we have:

Simplifying gives:

Again, we find the distance of the hypotenuse to be 30.”

Tell students, “Using what you have learned about distances, the Pythagorean Theorem, and the formula used to find the distance between any two points, create a short, animated PowerPoint to illustrate a sample real-world problem. The PowerPoint must include at least one animated illustration. For example, when looking at triangles included in the architectural plan for a bridge, you might illustrate how the appropriate lengths are calculated. In this case, it would be most appropriate to use the Pythagorean Theorem.” Note: You can refer to the article at http://blogs.techrepublic.com.com/five-tips/?p=316 to find helpful information on creating animations within the 2010 version of PowerPoint.

Part 2: Other Applications and Reasoning

Distance via Rate (d = rt)

“We can also speak of distance when discussing the idea of distance traveled. Suppose we have the following problem:

Hannah drives for 4 hours at 65 miles per hour. How far has she driven?”

Ask,

• “How would we solve this problem? (multiply 4 by 65)
• Do you agree that we are indeed looking for distance traveled?”

Give students time to ponder ideas. Some may go straight for the distance formula, while others may realize the answer is to multiply the hours by the rate. “We must multiply the number of hours she drives by the rate at which she drives. In this case, our rate is a unit rate, since we are discussing number of miles driven in one hour.”

“We can divide by hours as a common denominator t, and are thus left with:”

Assign this task for students to create: “Write a science-based word problem, involving the concepts of distance, time, and rate. [IS.8 - All Students] Show your work, solve, and be prepared to share the problem with the class.”

Give students time to research a distance/rate idea that is scientific in nature.

Proportion

Tell students, “Proportion can be used to find missing lengths or heights. Recall that a proportion is an equation that sets two ratios equal to one another.”

“Foresters use proportional reasoning throughout their careers, as they must determine the heights of trees.” Ask,

• “How would you use a proportion to find the height of a tree?”

(Use the length of its shadow to solve a proportion.)

• “What would you use?” (tape measure)
• “What measurements would be included in the proportion?” (the length of an object of known height and the length of its shadow)

“Suppose you go out into the field, with only yourself and a measuring tape.” Give students time to discuss and determine that the individual’s height, length of his/her shadow, and the length of the shadow of the tree are all required in the setup.

Divide students into groups of three or four. If possible, outside in an area with trees, have students actually perform an experiment to determine the height of a tree. If not possible, adapt the activity to involve finding the height of another object, such as a building, pole, tower, etc.

Once students finish the activity, have each group present the findings from their experiment. Discuss questions, problems encountered, ideas, etc. Provide the following sample problem to go into their notes:

Suppose Amanda is 5.5 feet tall. She casts a 7.5 foot shadow. She must determine the height of a tree that casts an 18 foot shadow. How would she go about solving this problem?

Tell students, “Amanda would set up a proportion in either two ways. First, she could compare the ratios of the heights to the shadows. Second, she could compare the ratios of the heights to heights and shadows to shadows. Looking at the equations will help to clarify.”

“If she compares the ratios of heights to shadows, she will set up a proportion as:”

“Solving for x, she will get:”

“Now, let’s try it the other way. We will set up a proportion, comparing her height to the tree’s height, and her shadow length to the tree’s shadow length. The proportion is set up as:”

“Solving for x, she will get ”

“Either way, she can say that the height of the tree is approximately 13 feet. After discussing the example problem, allow students time to use the similar procedures to solve their problem in the activity.”

“Under the general ideas of proportional reasoning and missing lengths, we can look at the topic of determining reasonableness of a length of a figure, given the measurements of another similar figure, and the measurement of at least one side of the figure in question.”

“Let’s examine two similar triangles:”

• The side lengths of Triangle 1 are a = 9.2, b = 4.4, and c = 6.1units.
• The side lengths of Triangle 2 are d and e not known, and f = 12.2 units.
• Side a corresponds to side d; side b corresponds to side e; and side c corresponds to side f.

“Would an estimate of 18 units for side d be a reasonable estimate? Why or why not?”

Students should be encouraged to set up proportions mentally and visualize the relationship between the similar triangles. It should be discerned that the second triangle appears to have side lengths double that of the other triangle. Therefore, if side a in a triangle corresponds to side d in another triangle, it is reasonable that d would be approximately 18, since 9.2(2) = 18.4.

For review, hold a class discussion related to aspects of the activities that proved most beneficial to students. What connections did they make in this lesson? Possible questions to get the discussions going are:

1. Is there a connection between the distance formula (coordinates) and the Pythagorean Theorem?
2. Where else in real life could you use any of these types of distances?
3. Which of the topics do you see yourself using most often?

Students should complete the Lesson 1 Exit Ticket (M-8-5-1_Lesson 1 Exit Ticket and KEY.docx).

Extension:

• Invite students to dig deeper into one of the activities/explorations. Specifically, have them work with circumference of a circle and pi. For example, since all circles are similar to each other in the same way that all squares are similar to each other, what is the relationship between the following two circles:
  • Circle X has a radius of 5 units.
  • Circle Y has a radius of 20 units.
  • How much longer is the circumference of Circle Y than the circumference of Circle X? (4 times, because 5 × 4 = 20)
• Provide students problems to practice and apply the concepts in Lesson 1 independently and/or as homework.