GRADE 8 β’ MODULE 2
(Optional) Topic D:
The Pythagorean Theorem
8.G.B.6, 8.G.B.7
Focus Standard: 8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
Instructional Days: 2
Lesson 15: Informal Proof of the Pythagorean Theorem (S)1
Lesson 16: Applications of the Pythagorean Theorem (P)
In Topic D, students are guided through the square within a square proof of the Pythagorean theorem, which requires students to know that congruent figures also have congruent areas. Once proved, students will practice using the Pythagorean theorem and its converse in Lesson 16 to find unknown side lengths in right triangles. Students apply their knowledge of the Pythagorean theorem to real-world problems that involve two-and three-dimensional figures.
1 Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
Topic D: The Pythagorean Theorem
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8β’2 Lesson 15
Lesson 15: Informal Proof of the Pythagorean Theorem
Student Outcomes
Students are introduced to the Pythagorean theorem and will be shown an informal proof of the theorem. Students will use the Pythagorean theorem to find the length of the hypotenuse of a right triangle.
Lesson Notes Since 8.G.B.6 and 8.G.B.7 are post-test standards, this lesson is designated as an extension lesson for this module. However, the content within this lesson is prerequisite knowledge for Module 7. If this lesson is not used with students as part of the work within Module 2, it must be used with students prior to beginning work on Module 7. Please realize that many mathematicians agree that the Pythagorean theorem is the most important theorem in geometry and has immense implications in much of high school mathematics in general (e.g., especially when studying quadratics and trigonometry). It is crucial that students see the teacher explain several proofs of the Pythagorean theorem and practice using it before being expected to produce a proof on their own.
Classwork
Concept Development (5 minutes)
The Pythagorean theorem is a famous theorem that will be used throughout much of high school mathematics. For that reason, you will see several proofs of the theorem throughout the year and have plenty of practice using it. The first thing you need to know about the Pythagorean theorem is what it states.
Pythagorean Theorem: If the lengths of the legs of a right triangle are ππ and ππ, and the length of the hypotenuse is ππ, then ππ2 + ππ2 = ππ2.
Given a right triangle π΄π΄π΄π΄π΄π΄ with π΄π΄ being the vertex of the right angle, then the sides π΄π΄π΄π΄ and π΄π΄π΄π΄ are called the legs of βπ΄π΄π΄π΄π΄π΄, and π΄π΄π΄π΄ is called the hypotenuse of βπ΄π΄π΄π΄π΄π΄.
Take note of the fact that side ππ is opposite the angle π΄π΄, side ππ is opposite the angle π΄π΄, and side ππ is opposite the angle π΄π΄.
Scaffolding: Draw arrows, one at a time, to show that each side is the opposite of the given angle.
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8β’2 Lesson 15
Discussion (15 minutes)
The first proof of the Pythagorean theorem that you will see requires you to know some basic facts about geometry.
1. Congruent triangles have equal areas.
2. All corresponding parts of congruent triangles are congruent.
3. The triangle sum theorem. (β sum of β³)
a. For right triangles, the two angles that are not the right angle have a sum of 90Β°. (β sum of rt. β³)
What we will look at next is what is called a square within a square. The outside square has side lengths (ππ + ππ), and the inside square has side lengths ππ. Our goal is to show that ππ2 + ππ2 = ππ2. To accomplish this goal, we will compare the total area of the outside square with the parts it is comprised of, i.e., the four triangles and the smaller inside square.
Ask students the following questions during the discussion:
Looking at the outside square only, the square with side lengths (ππ + ππ); what is its area?
The area of the outside square is (ππ + ππ)2 = ππ2 + 2ππππ + ππ2. Are the four triangles with sides lengths ππ and ππ congruent? If so, how do you know?
Yes, the triangles are congruent. From the diagram, we can see that each triangle has a right angle, and each triangle has side lengths of ππ and ππ. Our rigid motions will preserve those measures, and we can trace one triangle and use rigid motions to prove that they are congruent.
What is the area of just one triangle?
12ππππ
Does each triangle have the same area? If so, what is the sum of all four of those areas?
Yes, each triangle has the same area because they are congruent. The sum of all four triangles is
4 οΏ½12πππποΏ½ = 2ππππ.
Note to Teacher: Remind students to use the distributive law to determine the area of the outside square. Also remind them to use what they know about exponential notation to simplify the expression.
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8β’2 Lesson 15
We called this entire figure a square within a square, but we want to make sure that the figure in the center is indeed a square. To do so, we need to look at the angles of the triangles. First of all, what do we know about corresponding angles of congruent triangles?
Corresponding angles of congruent triangles are also congruent and equal in measure.
So we know that the angles marked by the red arcs are equal in measure, and the angles marked with the blue
arcs and line are equal in measure. What do we know about the sum of the interior angles of a triangle (β sum of β³)?
The sum of the interior angles of a triangle is 180Β°.
What is the sum of the two interior angles of a right triangle, not including the right angle (β sum of rt. β³)? How do you know? For right triangles, we know that one angle has a measure of 90Β°. Since the sum of all three angles
must be 180Β°, then we know that the other two angles must have a sum of 90Β°.
Now look at just one side of the figure. We have an angle with a red arc and an angle with a blue arc. In between them is another angle that we do not know the measure of. All three angles added together make up the straight side of the outside figure. What must be the measure of the unknown angle (the measure of the angle between the red and blue arcs)? How do you know? Since the angle with the red arc and the angle with the blue arc must have a sum of 90Β°, and all three
angles together must make a straight angle measuring 180Β° (β π π on a line), then the unknown angle must equal 90Β°.
That means that the figure with side lengths ππ must be a square. It is a figure with four equal sides and four right angles. What is the area of this square?
The area of the square must be ππ2.
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8β’2 Lesson 15
Recall our goal: To show that ππ2 + ππ2 = ππ2. To accomplish this goal, we will compare the total area of the
outside square with the parts it is comprised of, i.e., the four triangles and the smaller, inside square. Do we have everything we need to accomplish our goal?
Yes, we know the area of the outside square, (ππ + ππ)2 = ππ2 + 2ππππ + ππ2, the sum of the areas of the
four triangles, 4 οΏ½12πππποΏ½ = 2ππππ, and the area of the inside square, ππ2.
Show students the end of the square within a square proof:
Total area of the outside square = area of four triangles + area of inside square
ππ2 + 2ππππ + ππ2 = 2ππππ + ππ2ππ2 + 2ππππ β 2ππππ + ππ2 = 2ππππ β 2ππππ + ππ2
ππ2 + ππ2 = ππ2.
Thus, we have shown the Pythagorean theorem to be true using a square within a square.
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8β’2 Lesson 15
Example 1 (2 minutes)
Example 1
Now that we know what the Pythagorean theorem is, letβs practice using it to find the length of a hypotenuse of a right triangle.
Determine the length of the hypotenuse of the right triangle.
The Pythagorean theorem states that for right triangles ππππ + ππππ = ππππ, where ππ and ππ are the legs and ππ is the hypotenuse. Then,
ππππ + ππππ = ππππππππ + ππππ = ππππππππ + ππππ = ππππππππππ = ππππ.
Since we know that ππππππ = ππππππ, we can say that the hypotenuse ππ = ππππ.
Example 2 (3 minutes)
Example 2
Determine the length of the hypotenuse of the right triangle.
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8β’2 Lesson 15
Based on our work in the last example, what should we do to find the length of the hypotenuse?
Use the Pythagorean theorem and replace ππ and ππ with 3 and 7. Then,
ππ2 + ππ2 = ππ232 + 72 = ππ2
9 + 49 = ππ258 = ππ2.
Since we do not know what number times itself produces 58, for now we can leave our answer as 58 = ππ2. Later this year, we will learn how to determine the actual value for ππ for problems like this one.
Exercises 1β5 (10 minutes) Exercises 1β5
For each of the exercises, determine the length of the hypotenuse of the right triangle shown. Note: Figures not drawn to scale.
1.
2.
3.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ ππ = ππ
ππππ + ππππ = ππππ ππππ + ππππππ = ππππ ππππ+ ππππππ = ππππ
ππππππ = ππππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
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8β’2 Lesson 15
4.
5.
Closing (5 minutes)
Summarize, or have students summarize, the lesson.
We were shown a proof for the Pythagorean theorem that required us to find the area of four congruent triangles and two squares.
We learned that right triangles have sides ππ and ππ known as legs and a side ππ known as the hypotenuse.
We know that for right triangles, ππ2 + ππ2 = ππ2. We learned how to use the Pythagorean theorem in order to find the length of the hypotenuse of a right
triangle.
Exit Ticket (5 minutes)
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
Lesson Summary
Given a right triangle π¨π¨π¨π¨π¨π¨ with π¨π¨ being the vertex of the right angle, then the sides π¨π¨π¨π¨ and π¨π¨π¨π¨ are called the legs of βπ¨π¨π¨π¨π¨π¨, and π¨π¨π¨π¨ is called the hypotenuse of βπ¨π¨π¨π¨π¨π¨.
Take note of the fact that side ππ is opposite the angle π¨π¨, side ππ is opposite the angle π¨π¨, and side ππ is opposite the angle π¨π¨.
The Pythagorean theorem states that for any right triangle, ππππ + ππππ = ππππ.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
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8β’2 Lesson 15
Name Date
Lesson 15: Informal Proof of the Pythagorean Theorem
Exit Ticket 1. Label the sides of the right triangle with leg, leg, and hypotenuse.
2. Determine the length of ππ in the triangle shown.
3. Determine the length of ππ in the triangle shown.
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8β’2 Lesson 15
Exit Ticket Sample Solutions
1. Label the sides of the right triangle with leg, leg, and hypotenuse.
2. Determine the length of ππ in the triangle shown.
3. Determine the length of ππ in the triangle shown.
leg
leg
hypotenuse
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππππ = ππππ ππππ = ππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
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8β’2 Lesson 15
Problem Set Sample Solutions Students practice using the Pythagorean theorem to find the length of the hypotenuse of a right triangle.
For each of the problems below, determine the length of the hypotenuse of the right triangle shown. Note: Figures not drawn to scale.
1.
2.
3.
4.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
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8β’2 Lesson 15
5.
6.
7.
8.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ ππ = ππ
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππππ + ππππ = ππππ ππππππ + ππππ = ππππ
ππππππ = ππππ ππππ = ππ
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8β’2 Lesson 15
9.
10.
11.
12.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ ππππ+ ππ = ππππ
ππππ = ππππ
ππππ + ππππ = ππππ ππππππ + ππππ = ππππ ππππππ+ ππππ = ππππ
ππππππ = ππππ
ππππ + ππππ = ππππ ππππππ + ππππ = ππππ ππππππ+ ππππ = ππππ
ππππππ = ππππ
ππππ + ππππ = ππππ ππππππ + ππππ = ππππ ππππππ+ ππππ = ππππ
ππππππ = ππππ ππππ = ππ
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8β’2 Lesson 16
Lesson 16: Applications of the Pythagorean Theorem
Student Outcomes
Students use the Pythagorean theorem to determine missing side lengths of right triangles.
Lesson Notes Since 8.G.B.6 and 8.G.B.7 are post-test standards, this lesson is designated as an extension lesson for this module. However, the content within this lesson is prerequisite knowledge for Module 7. If this lesson is not used with students as part of the work within Module 2, it must be used with students prior to beginning work on Module 7. Please realize that many mathematicians agree that the Pythagorean theorem is the most important theorem in geometry and has immense implications in much of high school mathematics in general (e.g., especially when studying quadratics and trigonometry). It is crucial that students see the teacher explain several proofs of the Pythagorean Theorem and practice using it before being expected to produce a proof on their own.
Classwork
Example 1 (4 minutes)
Pythagorean theorem as it applies to missing side lengths of triangles:
Example 1
Given a right triangle with a hypotenuse with length ππππ units and a leg with length ππ units, as shown, determine the length of the other leg.
The length of the leg is ππππ units.
Let ππ represent the missing leg of the right triangle; then, by the Pythagorean theorem: 52 + ππ2 = 132.
If we let ππ represent the missing leg of the right triangle, then by the Pythagorean theorem: ππ2 + 52 = 132.
ππππ + ππππ = ππππππ ππππ β ππππ + ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππ ππππ = ππππππ β ππππ ππππ = ππππππ ππ = ππππ
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8β’2 Lesson 16
Which of these two equations is correct: 52 + ππ2 = 132 or ππ2 + 52 = 132? It does not matter which equation we use as long as we are showing the sum of
the squares of the legs as equal to the square of the hypotenuse.
Using the first of our two equations, 52 + ππ2 = 132, what can we do to solve for ππ in the equation?
We need to subtract 52 from both sides of the equation. 52 + ππ2 = 132
52 β 52 + ππ2 = 132 β 52ππ2 = 132 β 52
Point out to students that we are looking at the Pythagorean theorem in a form that allows us to find the length of one of the legs of the right triangle. That is, ππ2 = ππ2 β ππ2.
ππ2 = 132 β 52ππ2 = 169 β 25ππ2 = 144ππ = 12
The length of the leg of the right triangle is 12 units.
Example 2 (4 minutes)
Pythagorean theorem as it applies to missing side lengths of triangles in a real-world problem:
Suppose you have a ladder of length 13 feet. Suppose that to make it sturdy enough to climb, you must place the ladder exactly 5 feet from the wall of a building. You need to post a banner on the building 10 feet above the ground. Is the ladder long enough for you to reach the location you need to post the banner?
The ladder against the wall forms a right angle. For that reason, we can use the Pythagorean theorem to find out how far up the wall the ladder will reach. If we let β represent the height the ladder can reach, what equation will represent this problem?
52 + β2 = 132 or β2 = 132 β 52
Scaffolding: If students do not believe that we could use either equation, solve each of them and show that the answer is the same.
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8β’2 Lesson 16
Using either equation, we see that this is just like Example 1. We know that the missing side of the triangle is 12 feet. Is the ladder long enough for you to reach the 10-foot banner location?
Yes, the ladder allows us to reach 12 feet up the wall.
Example 3 (3 minutes)
Pythagorean theorem as it applies to missing side lengths of a right triangle:
Given a right triangle with a hypotenuse of length 15 units and a leg of length 9, what is the length of the other leg?
If we let the length of the missing leg be represented by ππ, what equation will allow us to determine its value?
ππ2 + 92 = 152 or ππ2 = 152 β 92. Finish the computation:
ππ2 = 225 β 81ππ2 = 144ππ = 12
The length of the missing leg of this triangle is 12 units.
Exercises 1β2 (5 minutes)
Students work on Exercises 1 and 2 independently.
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8β’2 Lesson 16
Exercises 1β2
1. Use the Pythagorean theorem to find the missing length of the leg in the right triangle.
Let ππ represent the missing leg length; then,
ππππππ + ππππ = ππππππππππππ β ππππππ = ππππππ β ππππππ
ππππ = ππππππ β ππππππ= ππππππ= ππππ.
The length of the leg is ππππ units.
2. You have a ππππ-foot ladder and need to reach exactly ππ feet up the wall. How far away from the wall should you place the ladder so that you can reach your desired location?
Let ππ represent the distance the ladder must be placed from the wall; then,
ππππ + ππππ = ππππππππππ + ππππ β ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππππππ = ππππππππ = ππππ.
The ladder must be placed exactly ππππ feet from the wall.
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8β’2 Lesson 16
Example 4 (5 minutes)
Pythagorean theorem as it applies to distances on a coordinate plane:
We want to find the length of the segment π΄π΄π΄π΄ on the coordinate plane, as shown.
If we had a right triangle, then we could use the Pythagorean theorem to determine the length of the segment. Letβs draw a line parallel to the π¦π¦-axis through point π΄π΄. We will also draw a line parallel to the π₯π₯-axis through point π΄π΄.
How can we be sure we have a right triangle?
The coordinate plane is set up so that the intersection of the π₯π₯-axis and π¦π¦-axis are perpendicular. The line parallel to the π¦π¦-axis through π΄π΄ is just a translation of the π¦π¦-axis. Similarly, the line parallel to the π₯π₯-axis through π΄π΄ is a translation of the π₯π₯-axis. Since translations preserve angle measure, the intersection of the two red lines are also perpendicular, meaning we have a 90Β° angle and a right triangle.
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8β’2 Lesson 16
Now that we are sure we can use the Pythagorean theorem, we need to know the lengths of the legs. Count the units from point π΄π΄ to the right angle and point π΄π΄ to the right angle. What are those lengths?
The base of the triangle is 6 units, and the height of the triangle is 3 units.
What equation can we use to find the length of the segment π΄π΄π΄π΄? Letβs represent that length by ππ.
32 + 62 = ππ2 The length of ππ is
32 + 62 = ππ29 + 36 = ππ2
45 = ππ2.
We cannot get a precise answer, so we will leave the length of ππ as ππ2 = 45.
Example 5 (3 minutes)
Pythagorean Theorem as it applies to the length of a diagonal in a rectangle:
Given a rectangle with side lengths of 8 cm and 2 cm, as shown, what is the length of the diagonal?
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8β’2 Lesson 16
If we let the length of the diagonal be represented by ππ, what equation can we use to find its length?
22 + 82 = ππ2 The length of ππ is
22 + 82 = ππ24 + 64 = ππ2
68 = ππ2.
We cannot get a precise answer, so we will leave the length of ππ as ππ2 = 68.
Exercises 3β6 (11 minutes)
Students work independently on Exercises 3β6.
Exercises 3β6
3. Find the length of the segment π¨π¨π¨π¨, if possible.
If we let the length of π¨π¨π¨π¨ be represented by ππ, then
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ ππ = ππ
The length of segment π¨π¨π¨π¨ is ππ units.
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8β’2 Lesson 16
4. Given a rectangle with dimensions ππ cm and ππππ cm, as shown, find the length of the diagonal, if possible.
Let ππ represent the length of the diagonal; then,
ππππ = ππππ + ππππππ= ππππ+ ππππππ= ππππππ.
We cannot find a precise answer for ππ, so the length is (ππππ = ππππππ) cm.
5. A right triangle has a hypotenuse of length ππππ in. and a leg with length ππ in. What is the length of the other leg?
If we let ππ represent the length of the other leg, then
ππππ + ππππ = ππππππππππ + ππππ β ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππ= ππππππ β ππππ= ππππππ.
We cannot find a precise length for ππ, so the leg is (ππππ = ππππππ) in.
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8β’2 Lesson 16
6. Find the length of ππ in the right triangle below, if possible.
By the Pythagorean theorem,
ππππ + ππππ = ππππππππππ β ππππ + ππππ = ππππππ β ππππ
= ππππππ β ππππ= ππππππ.
A precise length for side ππ cannot be found, so ππππ = ππππππ.
Closing (5 minutes)
Summarize, or have students summarize, the lesson.
We know how to use the Pythagorean theorem to find the length of a missing side of a right triangle whether it be one of the legs or the hypotenuse.
We know how to apply the Pythagorean theorem to a real life problem like how high a ladder will reach along a wall.
We know how to find the length of a diagonal of a rectangle.
We know how to determine the length of a segment that is on the coordinate plane.
Exit Ticket (5 minutes)
Lesson Summary
The Pythagorean theorem can be used to find the unknown length of a leg of a right triangle.
An application of the Pythagorean theorem allows you to calculate the length of a diagonal of a rectangle, the distance between two points on the coordinate plane, and the height that a ladder can reach as it leans against a wall.
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8β’2 Lesson 16
Name Date
Lesson 16: Applications of the Pythagorean Theorem
Exit Ticket 1. Find the length of the missing side of the rectangle shown below, if possible.
2. Find the length of all three sides of the right triangle shown below, if possible.
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8β’2 Lesson 16
Exit Ticket Sample Solutions
1. Find the length of the missing side of the rectangle shown below, if possible.
Let ππ represent the length of the unknown leg. Then,
ππππ + ππππ = ππππππππππ + ππππ β ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππππππ = ππππππ β ππππππππ = ππππ.
The precise length of the side cannot be found, but ππππ = ππππ units.
2. Find the length of all three sides of the right triangle shown below, if possible.
The two legs are each ππ units in length. The hypotenuse is
ππππ + ππππ = ππππππππ + ππππ = ππππ
ππππ = ππππ.
The precise length of the hypotenuse cannot be found, but ππππ = ππππ units.
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8β’2 Lesson 16
Problem Set Sample Solutions Students practice using the Pythagorean theorem to find missing lengths in right triangles.
1. Find the length of the segment π¨π¨π¨π¨ shown below, if possible.
If we let the length of π¨π¨π¨π¨ be represented by ππ, then by the Pythagorean theorem
ππππ + ππππ = ππππππππ + ππππ = ππππππ
ππππππ = ππππππππ = ππ.
The length of the segment π¨π¨π¨π¨ is ππππ units.
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8β’2 Lesson 16
2. A ππππ-foot ladder is placed ππππ feet from the wall, as shown. How high up the wall will the ladder reach?
Let ππ represent the height up the wall that the ladder will reach. Then,
ππππ + ππππππ = ππππππππππ + ππππππ β ππππππ = ππππππ β ππππππ
ππππ = ππππππ β ππππππππππ = ππππππ β ππππππππππ = ππππππππ = ππππ.
The ladder will reach ππππ feet up the wall.
3. A rectangle has dimensions ππ in. by ππππ in. What is the length of the diagonal of the rectangle?
If we let ππ represent the length of the diagonal, then
ππππ + ππππππ = ππππππππ + ππππππ = ππππ
ππππππ = ππππ.
A precise answer cannot be determined for the length of the diagonal, so we say that (ππππ = ππππππ) in.
Use the Pythagorean theorem to find the missing side lengths for the triangles shown in Problems 4β8.
4. Determine the length of the missing side, if possible.
ππππππ + ππππ = ππππππππππππ β ππππππ + ππππ = ππππππ β ππππππ
ππππ = ππππππ β ππππππππππ = ππππππ β ππππππππππ = ππππππ = ππ
The length of the missing side is ππ units.
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8β’2 Lesson 16
5. Determine the length of the missing side, if possible.
6. Determine the length of the missing side, if possible.
7. Determine the length of the missing side, if possible.
8. Determine the length of the missing side, if possible.
ππππ + ππππ = ππππππππ + ππππ β ππππ = ππππ β ππππ
ππππ = ππππ β ππππππππ = ππππ β ππππππ = ππππ
We cannot get a precise answer, but ππππ = ππππ.
ππππ + ππππ = ππππππππππ + ππππ β ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππππππ = ππππππ β ππππππππ = ππππππ
We cannot get a precise answer, but ππππ = ππππππ.
ππππ + ππππ = ππππππππ + ππππ β ππππ = ππππ β ππππ
ππππ = ππππ β ππππππππ = ππππ β ππππππ = ππππ
We cannot get a precise answer, but ππππ = ππππ.
ππππ + ππππ = ππππππππππ β ππππ + ππππ = ππππππ β ππππ
ππππ = ππππππ β ππππππππ = ππππππ β ππππππππ = ππππ
We cannot get a precise answer, but ππππ = ππππ.
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GRADE 8 β’ MODULE 3
Topic C:
The Pythagorean Theorem
8.G.B.6, 8.G.B.7
Focus Standard: 8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse.
8.G.B.7 Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
Instructional Days: 2
Lesson 13: Proof of the Pythagorean Theorem (S)1
Lesson 14: Converse of the Pythagorean Theorem (P)
It is recommended that students have some experience with the lessons in Topic D from Module 2 before beginning these lessons. Lesson 13 of Topic C, presents students with a general proof that uses the angle-angle criterion. In Lesson 14, students are presented with a proof of the converse of the Pythagorean Theorem. Also in Lesson 14, students apply their knowledge of the Pythagorean Theorem (i.e., given a right triangle with sides ππ, ππ, and ππ, where ππ is the hypotenuse, then ππ2 + ππ2 = ππ2) to determine unknown side lengths in right triangles. Students also use the converse of the theorem (i.e., given a triangle with lengths ππ, ππ and ππ, so that ππ2 + ππ2 = ππ2, then the triangle is a right triangle with hypotenuse ππ) to determine if a given triangle is in fact a right triangle.
1 Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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8β’3 Lesson 13
Lesson 13: Proof of the Pythagorean Theorem
Student Outcomes
Students practice applying the Pythagorean Theorem to find lengths of sides of right triangles in two dimensions.
Lesson Notes Since 8.G.B.6 and 8.G.B.7 are post-test standards, this lesson is designated as an extension lesson for this module. However, the content within this lesson is prerequisite knowledge for Module 7. If this lesson is not used with students as part of the work within Module 3, it must be used with students prior to beginning work on Module 7. Please realize that many mathematicians agree that the Pythagorean Theorem is the most important theorem in geometry and has immense implications in much of high school mathematics in general (e.g., learning of quadratics, trigonometry, etc.). It is crucial that students see the teacher explain several proofs of the Pythagorean Theorem and practice using it before being expected to produce a proof on their own.
Classwork
Discussion (20 minutes)
The following proof of the Pythagorean Theorem is based on the fact that similarity is transitive. It begins with the right triangle, shown on the next page, split into two other right triangles. The three triangles are placed in the same orientation, and students verify that a pair of triangles is similar using the AA criterion, then a second pair of triangles is shown to be similar using the AA criterion, and then finally all three triangle pairs are shown to be similar by the fact that similarity is transitive. Once it is shown that all three triangles are in fact similar, the theorem is proved by comparing the ratios of corresponding side lengths. Because some of the triangles share side lengths that are the same (or sums of lengths), then the formula ππ2 + ππ2 = ππ2 is derived. Symbolic notation is used explicitly for the lengths of sides. Therefore, it may be beneficial to do this proof simultaneously with triangles that have concrete numbers for side lengths. Another option to prepare students for the proof is showing the video presentation first, and then working through this Discussion.
The concept of similarity can be used to prove one of the great theorems in mathematics, the Pythagorean Theorem. What do you recall about the Pythagorean Theorem from our previous work? The Pythagorean Theorem is a theorem about the lengths of the legs and the hypotenuse of right
triangles. Specifically, if ππ and ππ are the lengths of legs of a right triangle and ππ is the length of the hypotenuse, then ππ2 + ππ2 = ππ2. The hypotenuse is the longest side of the triangle, and it is opposite the right angle.
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8β’3 Lesson 13
In this lesson we are going to take a right triangle, β³ π΄π΄π΄π΄π΄π΄, and use what we know about similarity of triangles to prove ππ2 + ππ2 = ππ2.
For the proof, we will draw a line from vertex π΄π΄ to a point π·π· so that the line is perpendicular to side π΄π΄π΄π΄.
We draw this particular line, line π΄π΄π·π·, because it divides the original triangle into three similar triangles. Before
we move on, can you name the three triangles?
The three triangles are β³ π΄π΄π΄π΄π΄π΄, β³ π΄π΄π΄π΄π·π·, and β³ π΄π΄π΄π΄π·π·. Letβs look at the triangles in a different orientation in order to see why they are similar. We can use our basic
rigid motions to separate the three triangles. Doing so ensures that the lengths of segments and measures of angles are preserved.
To have similar triangles by the AA criterion, the triangles must have two common angles. Which angles prove
that β³ π΄π΄π·π·π΄π΄ and β³ π΄π΄π΄π΄π΄π΄ similar?
It is true that β³ π΄π΄π·π·π΄π΄ β½ β³ π΄π΄π΄π΄π΄π΄ because they each have a right angle and β π΄π΄, which is not the right angle, is common to both triangles.
What does that tell us about β π΄π΄ from β³ π΄π΄π·π·π΄π΄ and β π΄π΄ from β³ π΄π΄π΄π΄π΄π΄?
It means that the angles correspond and must be equal in measure because of the triangle sum theorem.
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8β’3 Lesson 13
Which angles prove that β³ π΄π΄π΄π΄π΄π΄ and β³ π΄π΄π·π·π΄π΄ are similar?
It is true that β³ π΄π΄π΄π΄π΄π΄ β½ β³ π΄π΄π·π·π΄π΄ because they each have a right angle and β π΄π΄, which is not the right angle, is common to both triangles.
What does that tell us about β π΄π΄ from β³ π΄π΄π΄π΄π΄π΄ and β π΄π΄ from β³ π΄π΄π·π·π΄π΄?
The angles correspond and must be equal in measure because of the triangle sum theorem. If β³ π΄π΄π·π·π΄π΄ β½ β³ π΄π΄π΄π΄π΄π΄ and β³ π΄π΄π΄π΄π΄π΄ β½ β³ π΄π΄π·π·π΄π΄, is it true that β³ π΄π΄π·π·π΄π΄ β½ β³ π΄π΄π·π·π΄π΄? How do you know?
Yes, because similarity is a transitive relation.
When we have similar triangles, we know that their side lengths are proportional. Therefore, if we consider β³ π΄π΄π·π·π΄π΄ and β³ π΄π΄π΄π΄π΄π΄, we can write
|π΄π΄π΄π΄||π΄π΄π΄π΄| =
|π΄π΄π·π·||π΄π΄π΄π΄|.
By the cross-multiplication algorithm, |π΄π΄π΄π΄|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|.
By considering β³ π΄π΄π΄π΄π΄π΄ and β³ π΄π΄π·π·π΄π΄, we can write |π΄π΄π΄π΄||π΄π΄π΄π΄| =
|π΄π΄π΄π΄||π΄π΄π·π·|.
Which again by the cross-multiplication algorithm, |π΄π΄π΄π΄|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|.
If we add the two equations together, we get |π΄π΄π΄π΄|2 + |π΄π΄π΄π΄|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| + |π΄π΄π΄π΄| β |π΄π΄π·π·|.
By the distributive property, we can rewrite the right side of the equation because there is a common factor of |π΄π΄π΄π΄|. Now we have
|π΄π΄π΄π΄|2 + |π΄π΄π΄π΄|2 = |π΄π΄π΄π΄|(|π΄π΄π·π·| + |π΄π΄π·π·|). Keeping our goal in mind, we want to prove that ππ2 + ππ2 = ππ2; letβs see how close we are.
Using our diagram where three triangles are within one, (shown below), what side lengths are represented by |π΄π΄π΄π΄|2 + |π΄π΄π΄π΄|2?
π΄π΄π΄π΄ is side length ππ, and π΄π΄π΄π΄ is side length ππ, so the left side of our equation represents ππ2 + ππ2.
Now letβs examine the right side of our equation: |π΄π΄π΄π΄|(|π΄π΄π·π·| + |π΄π΄π·π·|). We want this to be equal to ππ2; is it? If we add the side length π΄π΄π·π· and side length π΄π΄π·π·, we get the entire side length of π΄π΄π΄π΄; therefore, we
have |π΄π΄π΄π΄|(|π΄π΄π·π·| + |π΄π΄π·π·|) = |π΄π΄π΄π΄| β |π΄π΄π΄π΄| = |π΄π΄π΄π΄|2 = ππ2. We have just proven the Pythagorean Theorem using what we learned about similarity. At this point we have
seen the proof of the theorem in terms of congruence and now similarity.
Scaffolding: Use concrete numbers to quickly convince students that adding two equations together leads to another true equation. For example: 5 = 3 + 2 and 8 = 4 + 4; therefore, 5 + 8 =3 + 2 + 4 + 4.
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8β’3 Lesson 13
Video Presentation (7 minutes)
The video located at the following link is an animation1 of the preceding proof using similar triangles: http://www.youtube.com/watch?v=QCyvxYLFSfU.
Exercises 1β3 (8 minutes)
Students work independently to complete Exercises 1β3.
Exercises 1β3
Use the Pythagorean Theorem to determine the unknown length of the right triangle.
1. Determine the length of side ππ in each of the triangles below.
a.
ππππ + ππππππ = ππππ
ππππ+ ππππππ = ππππ
ππππππ = ππππ
ππππ = ππ
b.
ππ.ππππ + ππ.ππππ = ππππ
ππ.ππππ+ ππ.ππππ = ππππ
ππ.ππππ = ππππ ππ.ππ = ππ
2. Determine the length of side ππ in each of the triangles below.
a.
ππππ + ππππ = ππππ
ππππ+ ππππ = ππππ
ππππβ ππππ+ ππππ = ππππ β ππππ
ππππ = ππ ππ = ππ
b.
ππ.ππππ + ππππ = ππ.ππππ
ππ.ππ + ππππ = ππ.ππ
ππ.ππ β ππ.ππ + ππππ = ππ.ππ β ππ.ππ
ππππ = ππ.ππ ππ = ππ.ππ
1 Animation developed by Larry Francis.
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8β’3 Lesson 13
3. Determine the length of πΈπΈπΈπΈ. (Hint: Use the Pythagorean Theorem twice.)
ππππππ + |πΈπΈπΈπΈ|ππ = ππππππ
ππππππ + |πΈπΈπΈπΈ|ππ = ππππππ
ππππππ β ππππππ + |πΈπΈπΈπΈ|ππ = ππππππ β ππππππ
|πΈπΈπΈπΈ|ππ = ππππ
|πΈπΈπΈπΈ| = ππ
ππππππ + |πΈπΈπΈπΈ|ππ = ππππππ
ππππππ+ |πΈπΈπΈπΈ|ππ = ππππππ
ππππππ β ππππππ+ |πΈπΈπΈπΈ|ππ = ππππππ β ππππππ
|πΈπΈπΈπΈ|ππ = ππππππ
|πΈπΈπΈπΈ| = ππππ
Since |πΈπΈπΈπΈ| + |πΈπΈπΈπΈ| = |πΈπΈπΈπΈ|, then the length of side πΈπΈπΈπΈ is ππ + ππππ, which is ππππ.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson.
We have now seen another proof of the Pythagorean Theorem, but this time we used what we knew about similarity, specifically similar triangles.
We practiced using the Pythagorean Theorem to find unknown lengths of right triangles.
Exit Ticket (5 minutes)
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8β’3 Lesson 13
Name ___________________________________________________ Date____________________
Lesson 13: Proof of the Pythagorean Theorem
Exit Ticket Determine the length of side π΄π΄π·π· in the triangle below.
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8β’3 Lesson 13
Exit Ticket Sample Solutions
Determine the length of side π©π©π©π© in the triangle below.
First determine the length of side π©π©π©π©. ππππππ + π©π©π©π©ππ = ππππππ
ππππππ+ π©π©π©π©ππ = ππππππ π©π©π©π©ππ = ππππππ β ππππππ π©π©π©π©ππ = ππππ π©π©π©π© = ππ
Then determine the length of side π©π©π©π©. ππππππ + π©π©π©π©ππ = ππππππ
ππππππ+ π©π©π©π©ππ = ππππππ π©π©π©π©ππ = ππππππ β ππππππ π©π©π©π©ππ = ππππ π©π©π©π© = ππ
Adding the lengths of sides π©π©π©π© and π©π©π©π© will determine the length of side π©π©π©π©; therefore, ππ + ππ = ππππ. π©π©π©π© has a length of ππππ.
Problem Set Sample Solutions Students practice using the Pythagorean Theorem to find unknown lengths of right triangles.
Use the Pythagorean Theorem to determine the unknown length of the right triangle.
1. Determine the length of side ππ in each of the triangles below.
a.
ππππ + ππππ = ππππ
ππππ + ππππ = ππππ
ππππππ = ππππ
ππππ = ππ
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8β’3 Lesson 13
b.
ππ.ππππ + ππ.ππππ = ππππ
ππ.ππππ + ππ.ππππ = ππππ
ππ = ππππ ππ = ππ
2. Determine the length of side ππ in each of the triangles below.
a.
ππππ + ππππ = ππππππ
ππππ + ππππ = ππππππ
ππππ + ππππ β ππππ = ππππππ β ππππ
ππππ = ππππππ ππ = ππππ
b.
ππππ + ππ.ππππ = ππ.ππππ
ππππ + ππ.ππππ = ππ.ππππ
ππππ + ππ.ππππ β ππ.ππππ = ππ.ππππ β ππ.ππππ
ππππ = ππ.ππππ ππ = ππ.ππ
3. Determine the length of side ππ in each of the triangles below.
a.
ππππππ + ππππ = ππππππ
ππππππ + ππππ = ππππππ
ππππππ β ππππππ + ππππ = ππππππ β ππππππ
ππππ = ππππππ ππ = ππππ
b.
ππππ + ππππ = ππ.ππππ
ππ+ ππππ = ππ.ππππ
ππ β ππ+ ππππ = ππ.ππππ β ππ
ππππ = ππ.ππππ ππ = ππ.ππ
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8β’3 Lesson 13
4. Determine the length of side ππ in each of the triangles below.
a.
ππππ + ππππππ = ππππππ
ππππ + ππππππ = ππππππ
ππππ + ππππππ β ππππππ = ππππππ β ππππππ
ππππ = ππππππ ππ = ππππ
b.
ππππ + ππ.ππππ = ππππ
ππππ + ππ.ππππ = ππ
ππππ + ππ.ππππ β ππ.ππππ = ππ β ππ.ππππ
ππππ = ππ.ππππ ππ = ππ.ππ
5. What did you notice in each of the pairs of Problems 1β4? How might what you noticed be helpful in solving problems like these?
In each pair of problems, the problems and solutions were similar. For example, in Problem 1, part (a) showed the sides of the triangle were ππ, ππ, and ππππ, and in part (b), they were ππ.ππ, ππ.ππ, and ππ. The side lengths in part (b) were a tenth of the value of the lengths in part (a). The same could be said about parts (a) and (b) of Problems 2β4. This might be helpful for solving problems in the future. If I am given sides lengths that are decimals, then I could multiply them by a factor of ππππ to make whole numbers, which are easier to work with. Also, if I know common numbers that satisfy the Pythagorean Theorem, like side lengths of ππ, ππ, and ππ, then I will recognize them more easily in their decimal forms, i.e., ππ.ππ, ππ.ππ, and ππ.ππ.
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8β’3 Lesson 14
Lesson 14: The Converse of the Pythagorean Theorem
Student Outcomes
Students illuminate the converse of the Pythagorean Theorem through computation of examples and counterexamples.
Students apply the theorem and its converse to solve problems.
Lesson Notes Since 8.G.B.6 and 8.G.B.7 are post-test standards, this lesson is designated as an extension lesson for this module. However, the content within this lesson is prerequisite knowledge for Module 7. If this lesson is not used with students as part of the work within Module 3, it must be used with students prior to beginning work on Module 7. Please realize that many mathematicians agree that the Pythagorean Theorem is the most important theorem in geometry and has immense implications in much of high school mathematics in general (e.g., learning of quadratics and trigonometry). It is crucial that students see the teacher explain several proofs of the Pythagorean Theorem and practice using it before being expected to produce a proof on their own.
Classwork
Concept Development (8 minutes)
So far, you have seen two different proofs of the Pythagorean Theorem:
If the lengths of the legs of a right triangle are ππ and ππ, and the length of the hypotenuse is ππ, then ππ2 + ππ2 = ππ2.
This theorem has a converse:
If the lengths of three sides of a triangle, ππ, ππ, and ππ, satisfy ππ2 = ππ2 + ππ2, then the triangle is a right triangle, and furthermore, the side of length ππ is opposite the right angle.
Consider an exercise in which students attempt to draw a triangle on graph paper that satisfies ππ2 = ππ2 + ππ2 but is not a right triangle. Students should have access to rulers for this. Activities of this type may be sufficient to develop conceptual understanding of the converse; a formal proof by contradiction follows that may also be used.
The following is a proof of the converse. Assume we are given a triangle π΄π΄π΄π΄π΄π΄ with sides ππ, ππ, and ππ. We want to show that β π΄π΄π΄π΄π΄π΄ is a right angle. To do so, we will assume that β π΄π΄π΄π΄π΄π΄ is not a right angle. Then |β π΄π΄π΄π΄π΄π΄| > 90Β° or |β π΄π΄π΄π΄π΄π΄| < 90Β°. For brevity, we will only show the case for when |β π΄π΄π΄π΄π΄π΄| > 90Β° (the proof of the other case is similar). In the diagram below, we extend π΄π΄π΄π΄ to a ray π΄π΄π΄π΄ and let the perpendicular from π΄π΄ meet the ray at point π·π·.
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A STORY OF RATIOS
8β’3 Lesson 14
Let ππ = |π΄π΄π·π·| and ππ = |π΄π΄π·π·|.
Then applying the Pythagorean Theorem to β³ π΄π΄π΄π΄π·π· and β³ π΄π΄π΄π΄π·π· results in
ππ2 = ππ2 + ππ2 and ππ2 = (ππ + ππ)2 + ππ2.
Since we know what ππ2 and ππ2 are from the above equations, we can substitute those values into ππ2 = ππ2 + ππ2 to get
(ππ + ππ)2 + ππ2 = ππ2 + ππ2 + ππ2.
Since (ππ + ππ)2 = (ππ + ππ)(ππ + ππ) = ππ2 + ππππ + ππππ + ππ2 = ππ2 + 2ππππ + ππ2, then we have
ππ2 + 2ππππ + ππ2 + ππ2 = ππ2 + ππ2 + ππ2.
We can subtract the terms ππ2, ππ2, and ππ2 from both sides of the equal sign. Then we have 2ππππ = 0.
But this cannot be true because 2ππππ is a length; therefore, it cannot be equal to zero. Which means our assumption that |β π΄π΄π΄π΄π΄π΄| > 90Β° cannot be true. We can write a similar proof to show that |β π΄π΄π΄π΄π΄π΄| < 90Β° cannot be true either. Therefore, |β π΄π΄π΄π΄π΄π΄| = 90Β°.
Example 1 (7 minutes)
To maintain the focus of the lesson, allow the use of calculators in order to check the validity of the right angle using the Pythagorean Theorem.
The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle?
To find out, we need to put these numbers into the Pythagorean Theorem. Recall that side ππ is always the
longest side. Since 610 is the largest number, it is representing the ππ in the Pythagorean Theorem. To determine if this triangle is a right triangle, then we need to verify the computation.
?2722 + 5462 = 6102
Scaffolding: You may need to demonstrate how to use the squared button on a calculator.
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8β’3 Lesson 14
Find the value of the left side of the equation: 2722 + 5462 = 372,100. Then, find the value of the right side of the equation: 6102 = 372,100. Since the left side of the equation is equal to the right side of the equation, then we have a true statement, i.e., 2722 + 5462 = 6102. What does that mean about the triangle?
It means that the triangle with side lengths of 272, 546, and 610 is a right triangle.
Example 2 (5 minutes)
The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle?
What do we need to do to find out if this is a right triangle?
We need to see if it makes a true statement when we replace ππ, ππ, and ππ with the numbers using the Pythagorean Theorem.
Which number is ππ? How do you know?
The longest side is 12; therefore, ππ = 12. Use your calculator to see if it makes a true statement. (Give students a minute to calculate.) Is it a right
triangle? Explain.
No, it is not a right triangle. If it were a right triangle the equation 72 + 92 = 122 would be true. But the left side of the equation is equal to 130, and the right side of the equation is equal to 144. Since 130 β 144, then these lengths do not form a right triangle.
Exercises 1β7 (15 minutes)
Students complete Exercises 1β4 independently. Use of calculators is recommended.
Exercises 1β7
1. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππππππ. The right side of the equation is equal to ππππππ. That means ππππ + ππππππ = ππππππ is true, and the triangle shown is a right triangle, by the converse of the Pythagorean Theorem.
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8β’3 Lesson 14
2. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππ.ππππ + ππ.ππππ = ππ.ππππ is a true statement. The left side of the equation is equal to ππππ.ππππ. The right side of the equation is equal to ππππ.ππππ. That means ππ.ππππ + ππ.ππππ = ππ.ππππ is not true, and the triangle shown is not a right triangle.
3. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππππππ = ππππππππ is a true statement. The left side of the equation is equal to ππππ,ππππππ. The right side of the equation is equal to ππππ,ππππππ. That means ππππππ + ππππππππ = ππππππππ is true, and the triangle shown is a right triangle by the converse of the Pythagorean Theorem.
4. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππ,ππππππ. The right side of the equation is equal to ππ,ππππππ. That means ππππ + ππππππ = ππππππ is true, and the triangle shown is a right triangle by the converse of the Pythagorean Theorem.
5. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππ,ππππππ. The right side of the equation is equal to ππ,ππππππ. That means ππππππ + ππππππ = ππππππ is not true, and the triangle shown is not a right triangle.
6. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππ + ππππ = ππππ is a true statement. The left side of the equation is equal to ππππ. The right side of the equation is equal to ππππ. That means ππππ + ππππ = ππππ is not true, and the triangle shown is not a right triangle.
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8β’3 Lesson 14
7. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππ.ππππ + ππππ = ππ.ππππ is a true statement. The left side of the equation is equal to ππππ.ππππ. The right side of the equation is equal to ππππ.ππππ. That means ππ.ππππ + ππππ = ππ.ππππ is true, and the triangle shown is a right triangle by the converse of the Pythagorean Theorem.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson.
We know the converse of the Pythagorean Theorem states that if side lengths of a triangle ππ, ππ, ππ, satisfy ππ2 + ππ2 = ππ2, then the triangle is a right triangle.
We know that if the side lengths of a triangle ππ, ππ, ππ, do not satisfy ππ2 + ππ2 = ππ2, then the triangle is not a right triangle.
Exit Ticket (5 minutes)
Lesson Summary
The converse of the Pythagorean Theorem states that if side lengths of a triangle ππ, ππ, ππ, satisfy ππππ + ππππ = ππππ, then the triangle is a right triangle.
If the side lengths of a triangle ππ, ππ, ππ, do not satisfy ππππ + ππππ = ππππ, then the triangle is not a right triangle.
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8β’3 Lesson 14
Name ___________________________________________________ Date____________________
Lesson 14: The Converse of the Pythagorean Theorem
Exit Ticket 1. The numbers in the diagram below indicate the lengths of the sides of the triangle. Bernadette drew the following
triangle and claims it a right triangle. How can she be sure?
2. Will the lengths 5, 9, and 14 form a right triangle? Explain.
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8β’3 Lesson 14
Exit Ticket Sample Solutions
1. The numbers in the diagram below indicate the units of length of each side of the triangle. Bernadette drew the following triangle and claims it a right triangle. How can she be sure?
Since ππππ is the longest side, if this triangle was a right triangle, ππππ would have to be the hypotenuse (or ππ). Now she needs to check to see if ππππππ + ππππππ = ππππππ is a true statement. The left side is ππ,ππππππ, and the right side is ππ,ππππππ. That means ππππππ + ππππππ = ππππππ is true, and this is a right triangle.
2. Will the lengths ππ, ππ, and ππππ form a right triangle? Explain.
No, lengths of ππ, ππ, and ππππ will not form a right triangle. If they did, then the equation ππππ + ππππ = ππππππ would be a true statement. However, the left side equals ππππππ, and the right side equals ππππππ. Therefore, these lengths do not form a right triangle.
Problem Set Sample Solutions Students practice using the converse of the Pythagorean Theorem and identifying common errors in computations.
1. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππππππ. The right side of the equation is equal to ππππππ. That means ππππππ + ππππππ = ππππππ is true, and the triangle shown is a right triangle.
2. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππ,ππππππ. The right side of the equation is equal to ππ,ππππππ. That means ππππππ + ππππππ = ππππππ is not true, and the triangle shown is not a right triangle.
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8β’3 Lesson 14
3. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππ,ππππππ. The right side of the equation is equal to ππ,ππππππ. That means ππππππ + ππππππ = ππππππ is true, and the triangle shown is a right triangle.
4. The numbers in the diagram below indicate the units of length of each side of the triangle. Sam said that the following triangle is a right triangle. Explain to Sam what he did wrong to reach this conclusion and what the correct solution is.
Sam forgot to square each of the side lengths. In other words, he said ππ+ ππππ =ππππ, which is a true statement. However, to show that a triangle is a right triangle, you have to use the Pythagorean Theorem, which is ππππ + ππππ = ππππ. Using the Pythagorean Theorem, the left side of the equation is equal to ππ,ππππππ, and the right side is equal to ππ,ππππππ. Since ππ,ππππππ β ππ,ππππππ, then the triangle is not a right triangle.
5. The numbers in the diagram below indicate the units of length of each side of the triangle. Is the triangle shown below a right triangle? Show your work, and answer in a complete sentence.
We need to check if ππππππ + ππππ = ππππππ is a true statement. The left side of the equation is equal to ππππππ. The right side of the equation is equal to ππππππ. That means ππππππ + ππππ = ππππππ is true, and the triangle shown is a right triangle.
6. Jocelyn said that the triangle below is not a right triangle. Her work is shown below. Explain what she did wrong, and show Jocelyn the correct solution.
We need to check if ππππππ + ππππππ = ππππππ is a true statement. The left side of the equation is equal to ππ,ππππππ. The right side of the equation is equal to ππ,ππππππ. That means ππππππ + ππππππ = ππππππ is not true, and the triangle shown is not a right triangle.
Jocelyn made the mistake of not putting the longest side of the triangle in place of ππ in the Pythagorean Theorem, ππππ + ππππ = ππππ. Specifically, she should have used ππππ for ππ, but instead she used ππππ for ππ. If she had done that part correctly, she would have seen that, in fact, ππππππ + ππππππ = ππππππ is a true statement because both sides of the equation equal ππ,ππππππ. That means that the triangle is a right triangle.
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8β’7 Lesson 1
Lesson 1: The Pythagorean Theorem
Student Outcomes
Students know that they can estimate the length of a side of a right triangle as a number between two integers and identify the integer to which the length is closest.
Lesson Notes Before beginning this lesson, it is imperative that students are familiar with the lessons in Modules 2 and 3 that relate to the Pythagorean theorem. This lesson assumes knowledge of the theorem and its basic applications. Students should not use calculators during this lesson.
In this lesson, students are exposed to expressions that involve irrational numbers, but they will not learn the definition of an irrational number until Topic B. It is important for students to understand that these irrational numbers can be approximated, but it is not yet necessary that they know the definition.
Classwork
Opening (5 minutes)
Show students the three triangles below. Give students the direction to determine as much as they can about the triangles. If necessary, give the direction to apply the Pythagorean theorem, in particular. Then, have a discussion with students about their recollection of the theorem. Basic points should include the theorem, the converse of the theorem, and the fact that when the theorem leads them to an answer of the form ππ2 = π₯π₯2, then ππ = π₯π₯ (perfect squares).
In the first triangle, students are required to use the Pythagorean theorem to determine the unknown side length. Let us call the unknown side length π₯π₯ cm. Then, π₯π₯ is 8 cm because, by the Pythagorean theorem, 172 β 152 = π₯π₯2, and 64 =π₯π₯2. Since 64 is a perfect square, then students should identify the length of π₯π₯ as 8 cm. In the second triangle, students are required to use the converse of the Pythagorean theorem to determine that it is a right triangle. In the third triangle, students are required to use the converse of the Pythagorean theorem to determine that it is not a right triangle.
Scaffolding: In teaching about right triangles and guiding students in learning to identify the hypotenuse, it may be necessary to provide additional support in addressing the differences among the terms long, longer, and longest, as comparative words like these (with the same root) may not yet be familiar to English language learners.
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8β’7 Lesson 1
Example 1 (3 minutes)
Recall the Pythagorean theorem and its converse for right triangles.
The Pythagorean theorem states that a right triangle with leg lengths ππ and ππ and hypotenuse ππ will satisfy ππ2 + ππ2 = ππ2. The converse of the theorem states that if a triangle with side lengths ππ, ππ, and ππ satisfies the equation ππ2 + ππ2 = ππ2, then the triangle is a right triangle.
Example 1
Write an equation that will allow you to determine the length of the unknown side of the right triangle.
Write an equation that will allow you to determine the length of the unknown side of the right triangle.
Note: Students may use a different symbol to represent the unknown side length.
Let ππ represent the unknown side length. Then, 52 + ππ2 = 132.
Verify that students wrote the correct equation; then, allow them to solve it. Ask them how they knew the correct answer was 12. They should respond that 132 β 52 = 144, and since 144 is a perfect square, they knew that the unknown side length must be 12 cm.
Example 2 (5 minutes)
Example 2
Write an equation that will allow you to determine the length of the unknown side of the right triangle.
Write an equation that will allow you to determine the length of the unknown side of the right triangle.
Let ππ represent the length of the hypotenuse. Then, 42 + 92 = ππ2.
Scaffolding: Consider using cutouts of the triangles in this lesson to further illustrate the difference between triangles with whole number hypotenuses and those without. Then, call on students to measure the lengths directly for Examples 1β3. Cutouts drawn to scale are provided at the end of the lesson.
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8β’7 Lesson 1
There is something different about this triangle. What is the length of the missing side? If you cannot find the length of the missing side exactly, then find a good approximation.
Provide students time to find an approximation for the length of the unknown side. Select students to share their answers and explain their reasoning. Use the points below to guide their thinking as needed.
How is this problem different from the last one?
The answer is ππ2 = 97. Since 97 is not a perfect square, the exact length cannot be represented as an integer.
Since 97 is not a perfect square, we cannot determine the exact length of the hypotenuse as an integer; however, we can make an estimate. Think about all of the perfect squares we have seen and calculated in past discussions. The number 97 is between which two perfect squares?
The number 97 is between 81 and 100.
If the length of the hypotenuse were ππ2 = 81, what would be the length of the hypotenuse? The length would be 9 cm.
If the length of the hypotenuse were ππ2 = 100, what would be the length of the hypotenuse? The length would be 10 cm.
At this point, we know that the length of the hypotenuse is somewhere between 9 cm and 10 cm. Think about the length to which it is closest. The actual length of the hypotenuse is ππ2 = 97. To which perfect square number, 100 or 81, is 97 closer?
The number 97 is closer to the perfect square 100 than to the perfect square 81.
Now that we know that the length of the hypotenuse of this right triangle is between 9 cm and 10 cm, but closer to 10 cm, letβs try to get an even better estimate of the length. Choose a number between 9 and 10 but closer to 10. Square that number. Do this a few times to see how close you can get to the number 97.
Provide students time to check a few numbers between 9 and 10. Students should see that the length is somewhere between 9.8 and 9.9 because 9.82 = 96.04 and 9.92 = 98.01. Some students may even check 9.85; 9.852 = 97.0225. This activity will show students that an estimation of the length being between 9 cm and 10 cm is indeed accurate, and it will help students develop an intuitive sense of how to estimate square roots.
Example 3 (4 minutes)
Example 3
Write an equation to determine the length of the unknown side of the right triangle.
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8β’7 Lesson 1
Write an equation to determine the length of the unknown side of the right triangle.
Let ππ represent the length of the hypotenuse. Then, 32 + 82 = ππ2.
Verify that students wrote the correct equation, and then allow them to solve it. Instruct them to estimate the length, if necessary. Then, let them continue to work. When most students have finished, ask the questions below.
Could you determine an answer for the length of the hypotenuse as an integer?
No. The square of the length of the hypotenuse, ππ2 = 73, is not a perfect square.
Optionally, you can ask, βCan anyone find the exact length of side ππ, as a rational number?β It is important that students recognize that no one can determine the exact length of the hypotenuse as a rational number at this point.
Since 73 is not a perfect square, we cannot determine the exact length of the hypotenuse as a whole number.Letβs estimate the length. Between which two whole numbers is the length of the hypotenuse? Explain.
Since 73 is between the two perfect squares 64 and 81, we know the length of the hypotenuse must be between 8 cm and 9 cm.
Is the length closer to 8 cm or 9 cm? Explain. The length is closer to 9 cm because 73 is closer to 81 than it is to 64.
The length of the hypotenuse is between 8 cm and 9 cm but closer to 9 cm.
Example 4 (8 minutes)
Example 4
In the figure below, we have an equilateral triangle with a height of ππππ inches. What do we know about an equilateral triangle?
In the figure below, we have an equilateral triangle with a height of 10 inches. What do we know about anequilateral triangle?
Equilateral triangles have sides that are all of the same length and angles that are all of the same degree, namely 60Β°.
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8β’7 Lesson 1
Letβs say the length of the sides is π₯π₯ inches. Determine the approximate length of the sides of the triangle.
What we actually have here are two congruent right triangles.
Trace one of the right triangles on a transparency, and reflect across the line representing the height of the triangle to convince students that an equilateral triangle is composed of two congruent right triangles.
With this knowledge, we need to determine the length of the base of one of the right triangles. If we know that the length of the base of the equilateral triangle is π₯π₯, then what is the length of the base of one of the right triangles? Explain.
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8β’7 Lesson 1
The length of the base of one of the right triangles must be 12π₯π₯ because the equilateral triangle has a
base of length π₯π₯. Since the equilateral triangle is composed of two congruent right triangles, we know that the base of each of the right triangles is of the same length (reflections preserve lengths of segments). Therefore, each right triangle has a base length of 1
2π₯π₯.
Now that we know the length of the base of the right triangle, write an equation for this triangle using thePythagorean theorem.
οΏ½12π₯π₯οΏ½
2+ 102 = π₯π₯2
Verify that students wrote the correct equation, and then ask students to explain the meaning of each term of the equation. Allow students time to solve the equation in pairs or small groups. Instruct them to make an estimate of the length, if necessary. Then, let them continue to work. When most students have finished, continue with the discussion below.
Explain your thinking about this problem. What did you do with the equation οΏ½12π₯π₯οΏ½
2+ 102 = π₯π₯2?
If students are stuck, ask them questions that help them work through the computations below. For example, you can
ask them what they recall about the laws of exponents to simplify the term οΏ½12π₯π₯οΏ½
2 or how to use the properties of
equality to get the answer in the form of π₯π₯2 equal to a constant.
We had to solve for π₯π₯:
οΏ½12π₯π₯οΏ½
2
+ 102 = π₯π₯2
14π₯π₯2 + 100 = π₯π₯2
14π₯π₯2 β
14 π₯π₯
2 + 100 = π₯π₯2 β14 π₯π₯
2
100 =34π₯π₯2
4003
= π₯π₯2
133.3 β π₯π₯2
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8β’7 Lesson 1
Now that we know that π₯π₯2 β 133.3, find a whole number estimate for the length of π₯π₯. Explain your thinking. The length of π₯π₯ is approximately 12 in. The number 133.3 is between the perfect squares 121 and 144.
Since 133.3 is closer to 144 than 121, we know that the value of π₯π₯ is between 11 and 12 but closer to 12.
Exercises 1β3 (7 minutes)
Students complete Exercises 1β3 independently.
Exercises
1. Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the righttriangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ = ππππ
The length of the unknown side of the triangle is approximately ππ ππππ. The number ππππ is between the perfect squares ππππ and ππππππ. Since ππππ is closer to ππππ than ππππππ, then the length of the unknown side of the triangle is closer to ππ than it is to ππππ.
2. Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the righttriangle. Explain why your estimate makes sense.
Let ππ be the length of the hypotenuse.
ππππ + ππππππ = ππππ ππππ + ππππππ = ππππ
ππππππ = ππππ
The length of the hypotenuse is approximately ππππ in. The number ππππππ is between the perfect squares ππππππ and ππππππ. Since ππππππ is closer to ππππππ than ππππππ, then the length of the unknown side of the triangle is closer to ππππ than it is to ππππ.
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8β’7 Lesson 1
3. Use the Pythagorean theorem to find a whole number estimate of the length of the unknown side of the righttriangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ = ππππ
The length of the hypotenuse is approximately ππ ππππ. The number ππππ is between the perfect squares ππππ and ππππ. Since ππππ is closer to ππππ than ππππ, then the length of the unknown side of the triangle is closer to ππ than it is to ππ.
Discussion (3 minutes)
Our estimates for the lengths in the problems in this lesson are acceptable, but we can do better. Instead ofsaying that a length is between two particular whole numbers and closer to one compared to the other, we willsoon learn how to make more precise estimates.
Obviously, since the lengths have been between two integers (e.g., between 8 and 9), we will need to look atthe numbers between the integers: the rational numbers (fractions). That means we will need to learn moreabout rational numbers and all numbers between the integers on the number line, in general.
The examination of those numbers will be the focus of the next several lessons.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
We know what a perfect square is.
We know that when the square of the length of an unknown side of a triangle is not equal to a perfect square,we can estimate the side length by determining which two perfect squares the square of the length is between.
We know that we will need to look more closely at the rational numbers in order to make better estimates ofthe lengths of unknown sides of a right triangle.
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8β’7 Lesson 1
Exit Ticket (5 minutes)
Lesson Summary
Perfect square numbers are those that are a product of an integer factor multiplied by itself. For example, the number ππππ is a perfect square number because it is the product of ππ multiplied by ππ.
When the square of the length of an unknown side of a right triangle is not equal to a perfect square, you can estimate the length as a whole number by determining which two perfect squares the square of the length is between.
Example:
Let ππ represent the length of the hypotenuse. Then,
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ.
The number ππππ is not a perfect square, but it is between the perfect squares ππππ and ππππ. Therefore, the length of the hypotenuse is between ππ and ππ but closer to ππ because ππππ is closer to the perfect square ππππ than it is to the perfect square ππππ.
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8β’7 Lesson 1
Name Date
Lesson 1: The Pythagorean Theorem
Exit Ticket
1. Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, thendetermine which two integers the length is between and the integer to which it is closest.
2. Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, thendetermine which two integers the length is between and the integer to which it is closest.
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8β’7 Lesson 1
Exit Ticket Sample Solutions
1. Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, thendetermine which two integers the length is between and the integer to which it is closest.
Let ππ be the length of the unknown side.
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ = ππππππ ππ = ππππ
The length of the unknown side is ππππ π’π’π’π’. The Pythagorean theorem led me to the fact that the square of the unknown side is ππππππ. We know ππππππ is a perfect square, and ππππππ is equal to ππππππ; therefore, ππ = ππππ, and the unknown length of the triangle is ππππ π’π’π’π’.
2. Determine the length of the unknown side of the right triangle. If you cannot determine the length exactly, thendetermine which two integers the length is between and the integer to which it is closest.
Let ππ be the length of the unknown side.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
The number ππππ is between the perfect squares ππππ and ππππ. Since ππππ is closer to ππππ than ππππ, then the length of the unknown side of the triangle is closer to ππ than ππ.
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8β’7 Lesson 1
Problem Set Sample Solutions
1. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππππ + ππππ = ππππππ ππππππ + ππππ = ππππππ
ππππ = ππππ
The number ππππ is between the perfect squares ππππ and ππππ. Since ππππ is closer to ππππ than it is to ππππ, then the length of the unknown side of the triangle is closer to ππ than it is to ππ.
2. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππ + ππππππ = ππππππ ππππ + ππππππ = ππππππ
ππππ = ππππ ππ = ππ
The length of the unknown side is ππ ππππ. The Pythagorean theorem led me to the fact that the square of the unknown side is ππππ. Since ππππ is a perfect square, ππππ is equal to ππππ; therefore, ππ = ππ, and the unknown length of the triangle is ππ ππππ.
3. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the hypotenuse.
ππππ + ππππππ = ππππ ππππ + ππππππ = ππππ
ππππππ = ππππ
The number ππππππ is between the perfect squares ππππππ and ππππππ. Since ππππππ is closer to ππππππ than it is to ππππππ, then the length of the hypotenuse of the triangle is closer to ππππ than it is to ππππ.
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8β’7 Lesson 1
4. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππ + ππππππ = ππππππ ππππ + ππππππ = ππππππ
ππππ = ππππ
The number ππππ is between the perfect squares ππππ and ππππ. Since ππππ is closer to ππππ than it is to ππππ, then the length of the unknown side of the triangle is closer to ππ than it is to ππ.
5. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the hypotenuse.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππππ = ππππ ππππ = ππ
The length of the hypotenuse is ππππ π’π’π’π’. The Pythagorean theorem led me to the fact that the square of the unknown side is ππππππ. We know ππππππ is a perfect square, and ππππππ is equal to ππππππ; therefore, ππ = ππππ, and the length of the hypotenuse of the triangle is ππππ π’π’π’π’.
6. Determine the length of the unknown side of the right triangle. Explain how you know your answer is correct.
Let ππ be the length of the hypotenuse.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
The number ππππ is between the perfect squares ππππ and ππππ. Since ππππ is closer to ππππ than it is to ππππ, then the length of the hypotenuse of the triangle is closer to ππ than it is to ππ.
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8β’7 Lesson 1
7. Use the Pythagorean theorem to estimate the length of the unknown side of the right triangle. Explain why your estimate makes sense.
Let ππ be the length of the unknown side.
ππππ + ππππ = ππππππ ππ + ππππ = ππππππ
ππππ = ππππππ
The number ππππππ is between the perfect squares ππππππ and ππππππ. Since ππππππ is closer to ππππππ than it is to ππππππ, then the length of the unknown side of the triangle is closer to ππππ than it is to ππππ.
8. The triangle below is an isosceles triangle. Use what you know about the Pythagorean theorem to determine the approximate length of the base of the isosceles triangle.
Let ππ represent the length of the base of one of the right triangles of the isosceles triangle.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
Since ππππ is between the perfect squares ππππ and ππππ but closer to ππππ, the approximate length of the base of the right triangle is ππ ππππ. Since there are two right triangles, then the length of the base of the isosceles triangle is approximately ππππ ππππ.
9. Give an estimate for the area of the triangle shown below. Explain why it is a good estimate.
Let ππ represent the length of the base of the right triangle.
ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
Since ππππ is between the perfect squares ππππ and ππππ but closer to ππππ, the approximate length of the base is ππ ππππ. So, the approximate area of the triangle is
ππ(ππ)ππ
= ππ.
ππ ππππππ is a good estimate because of the approximation of the length of the base. Further, since the hypotenuse is the longest side of the right triangle, approximating the length of the base as ππ ππππ makes mathematical sense because it has to be shorter than the hypotenuse.
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8β’7 Lesson 1
Example 1
Example 2
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8β’7 Lesson 1
Example 3
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8
G R A D E
Mathematics Curriculum GRADE 8 β’ MODULE 7
Topic C: The Pythagorean Theorem
207
Topic C:
The Pythagorean Theorem
8.G.B.6, 8.G.B.7, 8.G.B.8
Focus Standards: 8.G.B.6 Explain a proof of the Pythagorean Theorem and its converse.
8.G.B.7
Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
8.G.B.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
Instructional Days: 4
Lesson 15: Pythagorean Theorem, Revisited (S)1
Lesson 16: Converse of the Pythagorean Theorem (S)
Lesson 17: Distance on the Coordinate Plane (P)
Lesson 18: Applications of the Pythagorean Theorem (E)
In Lesson 15, students engage with another proof of the Pythagorean theorem. This time, students compare the areas of squares that are constructed along each side of a right triangle in conjunction with what they know about similar triangles. Now that students know about square roots, students can determine the approximate length of an unknown side of a right triangle even when the length is not a whole number. Lesson 16 shows students another proof of the converse of the Pythagorean theorem based on the notion of congruence. Students practice explaining proofs in their own words in Lessons 15 and 16 and apply the converse of the theorem to make informal arguments about triangles as right triangles. Lesson 17 focuses on the application of the Pythagorean theorem to calculate the distance between two points on the coordinate plane. Lesson 18 gives students practice applying the Pythagorean theorem in a variety of mathematical and real-world scenarios. Students determine the height of isosceles triangles, determine the length of the diagonal of a rectangle, and compare lengths of paths around a right triangle.
1 Lesson Structure Key: P-Problem Set Lesson, M-Modeling Cycle Lesson, E-Exploration Lesson, S-Socratic Lesson
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8β’7 Lesson 15
Lesson 15: Pythagorean Theorem, Revisited
Student Outcomes
Students know that the Pythagorean theorem can be interpreted as a statement about the areas of similar geometric figures constructed on the sides of a right triangle.
Students explain a proof of the Pythagorean theorem.
Lesson Notes The purpose of this lesson is for students to review and practice presenting the proof of the Pythagorean theorem using similar triangles. Then, students will apply this knowledge to another proof that uses areas of similar figures such as squares.
Classwork
Discussion (20 minutes)
This discussion is an opportunity for students to practice explaining a proof of the Pythagorean theorem using similar triangles. Instead of leading the discussion, consider posing the questions, one at a time, to small groups of students and allow time for discussions. Then, have select students share their reasoning while others critique.
To prove the Pythagorean theorem, ππ2 + ππ2 = ππ2, use a right triangle, shown below. Begin by drawing a segment from the right angle, perpendicular to side π΄π΄π΄π΄ through point πΆπΆ. Label the intersection of the segments point π·π·.
Proof of the Pythagorean Theorem
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8β’7 Lesson 15
Using one right triangle, we created 3 right triangles. Name those triangles. The three triangles are β³ π΄π΄π΄π΄πΆπΆ, β³ π΄π΄πΆπΆπ·π·, and β³π΄π΄πΆπΆπ·π·.
We can use our basic rigid motions to reorient the triangles so they are easier to compare, as shown below.
The next step is to show that these triangles are similar. Begin by showing that β³ π΄π΄π·π·πΆπΆ β½ β³ π΄π΄πΆπΆπ΄π΄. Discuss in your group.
β³ π΄π΄π·π·πΆπΆ and β³ π΄π΄πΆπΆπ΄π΄ are similar because they each have a right angle, and they each share β π΄π΄. Then, by the AA criterion for similarity, β³ π΄π΄π·π·πΆπΆ β½ β³ π΄π΄πΆπΆπ΄π΄.
Now, show that β³ π΄π΄πΆπΆπ΄π΄ β½ β³ πΆπΆπ·π·π΄π΄. Discuss in your group.
β³ π΄π΄πΆπΆπ΄π΄ β½ β³ πΆπΆπ·π·π΄π΄ because they each have a right angle, and they each share β π΄π΄. Then, by the AA criterion for similarity, β³ π΄π΄πΆπΆπ΄π΄ β½ β³ πΆπΆπ·π·π΄π΄.
Are β³ π΄π΄π·π·πΆπΆ and β³ πΆπΆπ·π·π΄π΄ similar? Discuss in your group.
We know that similarity has the property of transitivity; therefore, since β³ π΄π΄π·π·πΆπΆ β½ β³ π΄π΄πΆπΆπ΄π΄, and β³ π΄π΄πΆπΆπ΄π΄ β½ β³ πΆπΆπ·π·π΄π΄, then β³ π΄π΄π·π·πΆπΆ β½ β³ πΆπΆπ·π·π΄π΄.
Scaffolding: You may also consider showing a concrete example, such as a 6-8-10 triangle, along with the general proof.
You can have students verify similarity using a protractor to compare corresponding angle measures. There is a reproducible available at the end of the lesson.
Scaffolding: A good hands-on visual
that can be used here requires a 3 Γ 5 notecard. Have students draw the diagonal and then draw the perpendicular line from πΆπΆ to side π΄π΄π΄π΄.
Make sure students label
all of the parts to match the triangle to the left. Next, have students cut out the three triangles. Students will then have a notecard version of the three triangles shown below and can use them to demonstrate the similarity among them.
The next scaffolding box shows a similar diagram for the concrete case of a 6-8-10 triangle.
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8β’7 Lesson 15
If we consider β³ π΄π΄π·π·πΆπΆ and β³ π΄π΄πΆπΆπ΄π΄, we can write a statement about corresponding sides being equal in ratio that will help us reach our goal of showing ππ2 + ππ2 = ππ2. Discuss in your group. Using β³ π΄π΄π·π·πΆπΆ and β³ π΄π΄πΆπΆπ΄π΄, we can write
|π΄π΄πΆπΆ||π΄π΄π΄π΄| =
|π΄π΄π·π·||π΄π΄πΆπΆ|
which is equal to |π΄π΄πΆπΆ|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|.
Since |π΄π΄πΆπΆ| = ππ, we have
ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|. Consider that β³ π΄π΄πΆπΆπ΄π΄ and β³ πΆπΆπ·π·π΄π΄ will give us another piece that we need. Discuss in your group.
Using β³ π΄π΄πΆπΆπ΄π΄ and β³ πΆπΆπ·π·π΄π΄, we can write |π΄π΄π΄π΄||π΄π΄πΆπΆ| =
|π΄π΄πΆπΆ||π΄π΄π·π·|
which is equal to |π΄π΄πΆπΆ|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|.
Since |π΄π΄πΆπΆ| = ππ, we have
ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|. The two equations ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| and ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| are all that we need to finish the proof. Discuss in
your group. By adding the equations together, we have
ππ2 + ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| + |π΄π΄π΄π΄| β |π΄π΄π·π·|. The length |π΄π΄π΄π΄| = |π΄π΄π΄π΄| = ππ, so by substitution we have
ππ2 + ππ2 = ππ β |π΄π΄π·π·| + ππ β |π΄π΄π·π·|. Using the distributive property
ππ2 + ππ2 = ππ β (|π΄π΄π·π·| + |π΄π΄π·π·|). The length |π΄π΄π·π·| + |π΄π΄π·π·| = ππ, so by substitution
ππ2 + ππ2 = ππ β ππ ππ2 + ππ2 = ππ2.
Discussion (15 minutes)
Now, letβs apply this knowledge to another proof of the Pythagorean theorem. Compare the area of similar figures drawn from each side of a right triangle. We begin with a right triangle:
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8β’7 Lesson 15
Next, we will construct squares off of each side of the right triangle in order to compare the areas of similar figures. However, are all squares similar? Discuss in your group.
Yes, all squares are similar. Assume you have a square with side length equal to 1 unit. You can dilate from a center by any scale factor to make a square of any size similar to the original one.
What would it mean, geometrically, for ππ2 + ππ2 to equal ππ2? It means that the sum of the areas of ππ2 and ππ2 is equal to the area ππ2.
There are two possible ways to continue; one way is by examining special cases on grid paper, as mentioned in the scaffolding box above, and showing the relationship between the squares physically. The other way is by using the algebraic proof of the general case that continues below.
This is where the proof using similar triangles will be helpful.
Scaffolding: Consider using concrete values for the sides of the right triangle (e.g., 3, 4, 5, then 5, 12, 13, then 6, 8, 10) and then moving to the general triangle with sides ππ, ππ, ππ. Given a triangle with sides 3, 4, 5 drawn on grid paper, students will be able to count squares or cut them out and physically place them on top of the larger square to compare the areas. An example of this is shown at the end of the lesson.
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8β’7 Lesson 15
When we compared β³ π΄π΄πΆπΆπ΄π΄ and β³ πΆπΆπ·π·π΄π΄, we wrote a statement about their corresponding side lengths, |π΅π΅π΅π΅||π΅π΅π΅π΅|
=|π΅π΅π΅π΅||π΅π΅π΅π΅|
, leading us to state that |π΄π΄πΆπΆ|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| and ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|. How might this information
be helpful in leading us to show that the areas of ππ2 + ππ2 are equal to the area of ππ2? Discuss in your group.
Since |π΄π΄π΄π΄| = ππ, then we have ππ2 = ππ β |π΄π΄π·π·|, which is part of the area of ππ2 that we need. Explain the statement ππ2 = ππ β |π΄π΄π·π·| in terms of the diagram below.
The square built from the leg of length ππ is equal in area to the rectangle built from segment π΄π΄π·π·, with length ππ. This is part of the area of the square with side ππ.
MP.2
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8β’7 Lesson 15
Now, we must do something similar with the area of ππ2. Discuss in your group. Using β³ π΄π΄π·π·πΆπΆ and β³ π΄π΄πΆπΆπ΄π΄, we wrote
|π΄π΄πΆπΆ||π΄π΄π΄π΄| =
|π΄π΄π·π·||π΄π΄πΆπΆ| ,
which is equal to |π΄π΄πΆπΆ|2 = |π΄π΄π΄π΄| β |π΄π΄π·π·|.
By substitution
ππ2 = |π΄π΄π΄π΄| β |π΄π΄π·π·| ππ2 = ππ β |π΄π΄π·π·|.
Explain the statement ππ2 = ππ β |π΄π΄π·π·| in terms of the diagram below.
The square built from the leg of length ππ is equal, in area, to the rectangle built from segment π΄π΄π·π·, with
length ππ. This is the other part of the area of the square with side ππ. Our knowledge of similar figures, as well as our understanding of the proof of the Pythagorean theorem using
similar triangles, led us to another proof where we compared the areas of similar figures constructed off the sides of a right triangle. In doing so, we have shown that ππ2 + ππ2 = ππ2, in terms of areas.
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8β’7 Lesson 15
Explain how the diagram shows that the Pythagorean theorem is true.
The Pythagorean theorem states that given a right triangle with lengths ππ,ππ, ππ that ππ2 + ππ2 = ππ2. The diagram shows that the area of the squares off of the legs ππ and ππ are equal to the area off of the hypotenuse ππ. Since the area of a square is found by multiplying a side by itself, then the area of a square with length ππ is ππ2, ππ is ππ2, and ππ is ππ2. The diagram shows that the areas ππ2 + ππ2 are equal to the area of ππ2, which is exactly what the theorem states.
To solidify student understanding of the proof, consider showing the six-minute video to students located at http://www.youtube.com/watch?v=QCyvxYLFSfU. If you have access to multiple computers or tablets, have small groups of students watch the video together, so they can pause and replay parts of the proof as necessary.
Scaffolding: The geometric illustration of the proof, shown to the left, can be used as further support or as an extension to the claim that the sum of the areas of the smaller squares is equal to the area of the larger square.
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8β’7 Lesson 15
Another short video that demonstrates ππ2 + ππ2 = ππ2 using area is at the following link: http://9gag.com/gag/aOqPoMD/cool-demonstration-of-the-pythagorean-theorem.
It does not explain the proof but merely shows that it is true.
Closing (5 minutes)
Consider having students explain how to show the Pythagorean theorem, using area, for a triangle with legs of length 40 units and 9 units and a hypotenuse of 41 units. Have students draw a diagram to accompany their explanation.
Summarize, or ask students to summarize, the main points from the lesson:
We know the proof of the Pythagorean theorem using similarity better than before.
We can prove the Pythagorean theorem using what we know about similar figures, generally, and what we know about similar triangles, specifically.
We know a proof for the Pythagorean theorem that uses area.
Exit Ticket (5 minutes)
Lesson Summary
The Pythagorean theorem can be proven by showing that the sum of the areas of the squares constructed off of the legs of a right triangle is equal to the area of the square constructed off of the hypotenuse of the right triangle.
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8β’7 Lesson 15
Name Date
Lesson 15: Pythagorean Theorem, Revisited
Exit Ticket Explain a proof of the Pythagorean theorem in your own words. Use diagrams and concrete examples, as necessary, to support your explanation.
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8β’7 Lesson 15
Exit Ticket Sample Solutions
Explain a proof of the Pythagorean theorem in your own words. Use diagrams and concrete examples, as necessary, to support your explanation.
Proofs will vary. The critical parts of the proof that demonstrate proficiency include an explanation of the similar triangles β³ π¨π¨π¨π¨π¨π¨, β³ π¨π¨π¨π¨π¨π¨, and β³ π¨π¨π¨π¨π¨π¨, including a statement about the ratio of their corresponding sides being equal, leading to the conclusion of the proof.
Problem Set Sample Solutions Students apply the concept of similar figures to show the Pythagorean theorem is true.
1. For the right triangle shown below, identify and use similar triangles to illustrate the Pythagorean theorem.
First, I must draw a segment that is perpendicular to side π¨π¨π¨π¨ that goes through point π¨π¨. The point of intersection of that segment and side π¨π¨π¨π¨ will be marked as point π¨π¨.
Then, I have three similar triangles, β³ π¨π¨π¨π¨π¨π¨, β³ π¨π¨π¨π¨π¨π¨, and β³ π¨π¨π¨π¨π¨π¨, as shown below.
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8β’7 Lesson 15
β³ π¨π¨π¨π¨π¨π¨ and β³ π¨π¨π¨π¨π¨π¨ are similar because each one has a right angle, and they both share β π¨π¨. By AA criterion, β³ π¨π¨π¨π¨π¨π¨ ~ β³ π¨π¨π¨π¨π¨π¨. β³ π¨π¨π¨π¨π¨π¨ and β³ π¨π¨π¨π¨π¨π¨ are similar because each one has a right angle, and they both share β π¨π¨. By AA criterion, β³ π¨π¨π¨π¨π¨π¨ ~ β³ π¨π¨π¨π¨π¨π¨. By the transitive property, we also know that β³ π¨π¨π¨π¨π¨π¨ ~ β³ π¨π¨π¨π¨π¨π¨.
Since the triangles are similar, they have corresponding sides that are equal in ratio. β³ π¨π¨π¨π¨π¨π¨ and β³ π¨π¨π¨π¨π¨π¨,
ππππππ
=|π¨π¨π¨π¨|ππ
,
which is the same as ππππ = ππππ(|π¨π¨π¨π¨|).
For β³ π¨π¨π¨π¨π¨π¨ and β³ π¨π¨π¨π¨π¨π¨,
ππππππππ
=|π¨π¨π¨π¨|ππππ
,
which is the same as ππππππ = ππππ(|π¨π¨π¨π¨|).
Adding these two equations together I get
ππππ + ππππππ = ππππ(|π¨π¨π¨π¨|) + ππππ(|π¨π¨π¨π¨|).
By the distributive property,
ππππ + ππππππ = ππππ(|π¨π¨π¨π¨| + |π¨π¨π¨π¨|);
however, |π¨π¨π¨π¨| + |π¨π¨π¨π¨| = |π¨π¨π¨π¨| = ππππ. Therefore,
ππππ + ππππππ = ππππ(ππππ) ππππ + ππππππ = ππππππ.
2. For the right triangle shown below, identify and use squares formed by the sides of the triangle to illustrate the Pythagorean theorem.
The sum of the areas of the smallest squares is ππππππ + ππππππ = ππππππ πππ¦π¦ππ. The area of the largest square is ππππππ = ππππππ πππ¦π¦ππ. The sum of the areas of the squares off of the legs is equal to the area of the square off of the hypotenuse; therefore, ππππ + ππππ = ππππ.
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8β’7 Lesson 15
3. Reese claimed that any figure can be drawn off the sides of a right triangle and that as long as they are similar figures, then the sum of the areas off of the legs will equal the area off of the hypotenuse. She drew the diagram by constructing rectangles off of each side of a known right triangle. Is Reeseβs claim correct for this example? In order to prove or disprove Reeseβs claim, you must first show that the rectangles are similar. If they are, then you can use computations to show that the sum of the areas of the figures off of the sides ππ and ππ equals the area of the figure off of side ππ.
The rectangles are similar because their corresponding side lengths are equal in scale factor. That is, if we compare the longest side of the rectangle to the side with the same length as the right triangle sides, we get the ratios
ππ. ππππ
=ππ.ππππ
=ππππ
= ππ.ππ
Since the corresponding sides were all equal to the same constant, then we know we have similar rectangles. The areas of the smaller rectangles are ππππ.ππ πππ¦π¦ππ and ππππ. ππ πππ¦π¦ππ, and the area of the largest rectangle is ππππ πππ¦π¦ππ. The sum of the smaller areas is equal to the larger area:
ππππ.ππ + ππππ.ππ = ππππ ππππ = ππππ
Therefore, we have shown that the sum of the areas of the two smaller rectangles is equal to the area of the larger rectangle, and Reeseβs claim is correct.
4. After learning the proof of the Pythagorean theorem using areas of squares, Joseph got really excited and tried explaining it to his younger brother. He realized during his explanation that he had done something wrong. Help Joseph find his error. Explain what he did wrong.
Based on the proof shown in class, we would expect the sum of the two smaller areas to be equal to the sum of the larger area, i.e., ππππ + ππππ should equal ππππ. However, ππππ + ππππ = ππππ. Joseph correctly calculated the areas of each square, so that was not his mistake. His mistake was claiming that a triangle with sides lengths of ππ, ππ, and ππ was a right triangle. We know that the Pythagorean theorem only works for right triangles. Considering the converse of the Pythagorean theorem, when we use the given side lengths, we do not get a true statement.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ β ππππ
Therefore, the triangle Joseph began with is not a right triangle, so it makes sense that the areas of the squares were not adding up like we expected.
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8β’7 Lesson 15
5. Draw a right triangle with squares constructed off of each side that Joseph can use the next time he wants to show his younger brother the proof of the Pythagorean theorem.
Answers will vary. Verify that students begin, in fact, with a right triangle and do not make the same mistake that Joseph did. Consider having students share their drawings and explanations of the proof in future class meetings.
6. Explain the meaning of the Pythagorean theorem in your own words.
If a triangle is a right triangle, then the sum of the squares of the legs will be equal to the square of the hypotenuse. Specifically, if the leg lengths are ππ and ππ, and the hypotenuse is length ππ, then for right triangles ππππ + ππππ = ππππ.
7. Draw a diagram that shows an example illustrating the Pythagorean theorem.
Diagrams will vary. Verify that students draw a right triangle with side lengths that satisfy the Pythagorean theorem.
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8β’7 Lesson 15
Diagrams referenced in scaffolding boxes can be reproduced for student use.
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8β’7 Lesson 16
Lesson 16: Converse of the Pythagorean Theorem
Student Outcomes
Students explain a proof of the converse of the Pythagorean theorem. Students apply the theorem and its converse to solve problems.
Lesson Notes Students had their first experience with the converse of the Pythagorean theorem in Module 3, Lesson 14. In that lesson, students learned the proof of the converse by contradiction. That is, students were asked to draw a triangle with sides ππ, ππ, ππ, where the angle between side ππ and ππ is greater than 90Β°. The proof using the Pythagorean theorem led students to an expression that was not possible; that is, two times a length was equal to zero. This contradiction meant that the angle between sides ππ and ππ was in fact 90Β°. In this lesson, students are given two triangles with base and height dimensions of ππ and ππ. They are told that one of the triangles is a right triangle and has lengths that satisfy the Pythagorean theorem. Students must use computation and their understanding of the basic rigid motions to show that the triangle with an unmarked angle is also a right triangle. The proof is subtle, so it is important from the beginning that students understand the differences between the triangles used in the discussion of the proof of the converse.
Classwork
Discussion (20 minutes)
So far you have seen three different proofs of the Pythagorean theorem:
THEOREM: If the lengths of the legs of a right triangle are ππ and ππ, and the length of the hypotenuse is ππ, then ππ2 + ππ2 = ππ2.
Provide students time to explain to a partner a proof of the Pythagorean theorem. Allow them to choose any one of the three proofs they have seen. Remind them of the proof from Module 2 that was based on congruent triangles, knowledge about angle sum of a triangle, and angles on a line. Also remind them of the proof from Module 3 that was based on their knowledge of similar triangles and corresponding sides being equal in ratio. Select students to share their proofs with the class. Encourage other students to critique the reasoning of the student providing the proof.
What do you recall about the meaning of the word converse?
Consider pointing out the hypothesis and conclusion of the Pythagorean theorem and then asking students to describe the converse in those terms.
The converse is when the hypothesis and conclusion of a theorem are reversed.
Scaffolding: Provide students samples of converses (and note that converses are not always true):
If it is a right angle, then the angle measure is 90Β°. Converse: If the angle measure is 90Β°, then it is a right angle.
If it is raining, I will study inside the house. Converse: If I study inside the house, it is raining.
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You have also seen one proof of the converse:
If the lengths of three sides of a triangle ππ, ππ, and ππ satisfy ππ2 = ππ2 + ππ2, then the triangle is a right triangle, and furthermore, the side of length ππ is opposite the right angle.
The following is another proof of the converse. Assume we are given a triangle π΄π΄π΄π΄π΄π΄ so that the sides ππ, ππ, and ππ satisfy ππ2 = ππ2 + ππ2. We want to show that β π΄π΄π΄π΄π΄π΄ is a right angle. To do so, we construct a right triangle π΄π΄β²π΄π΄β²π΄π΄β² with leg lengths of ππ and ππ and right angle β π΄π΄β²π΄π΄β²π΄π΄β².
Proof of the Converse of the Pythagorean Theorem
What do we know or not know about each of these triangles?
In the first triangle, π΄π΄π΄π΄π΄π΄, we know that ππ2 + ππ2 = ππ2. We do not know if angle π΄π΄ is a right angle. In the second triangle, π΄π΄β²π΄π΄β²π΄π΄β², we know that it is a right triangle.
What conclusions can we draw from this? By applying the Pythagorean theorem to β³ π΄π΄β²π΄π΄β²π΄π΄β², we get |π΄π΄β²π΄π΄β²|2 = ππ2 + ππ2. Since we are given
ππ2 = ππ2 + ππ2, then by substitution, |π΄π΄β²π΄π΄β²|2 = ππ2, and then |π΄π΄β²π΄π΄β²| = ππ. Since ππ is also |π΄π΄π΄π΄|, then |π΄π΄β²π΄π΄β²| = |π΄π΄π΄π΄|. That means that both triangles have sides ππ, ππ, and ππ that are the exact same lengths. Therefore, if we translated one triangle along a vector (or applied any required rigid motion(s)), it would map onto the other triangle showing a congruence. Any congruence preserves the degree of angles. Which means that β π΄π΄π΄π΄π΄π΄ is a right angle, or 90Β° = β π΄π΄β²π΄π΄β²π΄π΄β² = β π΄π΄π΄π΄π΄π΄.
Provide students time to explain to a partner a proof of the converse of the Pythagorean theorem. Allow them to choose either proof that they have seen. Remind them of the proof from Module 3 that was a proof by contradiction, where we assumed that the triangle was not a right triangle and then showed that the assumption was wrong. Select students to share their proofs with the class. Encourage other students to critique the reasoning of the student providing the proof.
Exercises 1β7 (15 minutes)
Students complete Exercises 1β7 independently. Remind students that since each of the exercises references the side length of a triangle, we need only consider the positive square root of each number, because we cannot have a negative length.
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8β’7 Lesson 16
Exercises 1β7
1. Is the triangle with leg lengths of ππ π¦π¦π¦π¦., ππ π¦π¦π¦π¦., and hypotenuse of length βππππ π¦π¦π¦π¦. a right triangle? Show your work, and answer in a complete sentence.
ππππ + ππππ = οΏ½βπππποΏ½ππ
ππ + ππππ = ππππ ππππ = ππππ
Yes, the triangle with leg lengths of ππ π¦π¦π¦π¦., ππ π¦π¦π¦π¦., and hypotenuse of length βππππ π¦π¦π¦π¦. is a right triangle because it satisfies the Pythagorean theorem.
2. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the hypotenuse of the triangle.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ ππ.ππ β ππ
The length of the hypotenuse of the right triangle is exactly βππππ inches and approximately ππ.ππ inches.
3. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the hypotenuse of the triangle.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ βππππ = ππ
οΏ½ππππ Γ βππ = ππ
οΏ½ππππ Γ βππ Γ βππ = ππ
ππβππππ = ππ
The length of the hypotenuse of the right triangle is exactly ππβππππ π¦π¦π¦π¦ and approximately ππ.ππ π¦π¦π¦π¦.
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8β’7 Lesson 16
4. Is the triangle with leg lengths of ππ π¦π¦π’π’., ππ π¦π¦π’π’., and hypotenuse of length βππππππ π¦π¦π’π’. a right triangle? Show your work, and answer in a complete sentence.
ππππ + ππππ = οΏ½βπππππποΏ½ππ
ππππ + ππππ = ππππππ
ππππππ β ππππππ
No, the triangle with leg lengths of ππ π¦π¦π’π’., ππ π¦π¦π’π’., and hypotenuse of length βππππππ π¦π¦π’π’. is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
5. Is the triangle with leg lengths of βππππ πππ¦π¦, ππ πππ¦π¦, and hypotenuse of length ππ πππ¦π¦ a right triangle? Show your work, and answer in a complete sentence.
οΏ½βπππποΏ½ππ
+ ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
Yes, the triangle with leg lengths of βππππ πππ¦π¦, ππ πππ¦π¦, and hypotenuse of length ππ πππ¦π¦ is a right triangle because the lengths satisfy the Pythagorean theorem.
6. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence.
Let ππ represent the hypotenuse of the triangle.
ππππ + οΏ½βπππποΏ½ππ
= ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = οΏ½ππππ ππ = ππ
The length of the hypotenuse of the right triangle is ππ ππππ.
7. The triangle shown below is an isosceles right triangle. Determine the length of the legs of the triangle. Show your work, and answer in a complete sentence.
Let ππ represent the length of the side of the isosceles triangle.
ππππ + ππππ = οΏ½βπππποΏ½ππ
ππππππ = ππππ ππππππ
ππ=ππππππ
ππππ = ππ
οΏ½ππππ = βππ ππ = ππ
The leg lengths of the isosceles triangle are ππ πππ¦π¦.
Lesson 16: Converse of the Pythagorean Theorem
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Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
The converse of the Pythagorean theorem states that if side lengths of a triangle ππ, ππ, ππ satisfy ππ2 + ππ2 = ππ2, then the triangle is a right triangle.
If the side lengths of a triangle ππ, ππ, ππ do not satisfy ππ2 + ππ2 = ππ2, then the triangle is not a right triangle. We know how to explain a proof of the Pythagorean theorem and its converse.
Exit Ticket (5 minutes)
Lesson Summary
The converse of the Pythagorean theorem states that if a triangle with side lengths ππ, ππ, and ππ satisfies ππππ + ππππ = ππππ, then the triangle is a right triangle.
The converse can be proven using concepts related to congruence.
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8β’7 Lesson 16
Name ___________________________________________________ Date____________________
Lesson 16: The Converse of the Pythagorean Theorem
Exit Ticket 1. Is the triangle with leg lengths of 7 mm, 7 mm, and a hypotenuse of length 10 mm a right triangle? Show your
work, and answer in a complete sentence.
2. What would the length of hypotenuse need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
3. If one of the leg lengths is 7 mm, what would the other leg length need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
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8β’7 Lesson 16
Exit Ticket Sample Solutions
1. Is the triangle with leg lengths of ππ π¦π¦π¦π¦, ππ π¦π¦π¦π¦, and a hypotenuse of length ππππ π¦π¦π¦π¦ a right triangle? Show your work, and answer in a complete sentence.
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ β ππππππ
No, the triangle with leg lengths of ππ π¦π¦π¦π¦, ππ π¦π¦π¦π¦, and hypotenuse of length ππππ π¦π¦π¦π¦ is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
2. What would the length of the hypotenuse need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
Let ππ represent the length of the hypotenuse in millimeters.
Then,
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ
The hypotenuse would need to be βππππ π¦π¦π¦π¦ for the triangle with sides of ππ π¦π¦π¦π¦ and ππ π¦π¦π¦π¦ to be a right triangle.
3. If one of the leg lengths is ππ π¦π¦π¦π¦, what would the other leg length need to be so that the triangle in Problem 1 would be a right triangle? Show work that leads to your answer.
Let ππ represent the length of one leg in millimeters.
Then,
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ + ππππ β ππππ = ππππππ β ππππ ππππ = ππππ
ππ = βππππ
The leg length would need to be βππππ π¦π¦π¦π¦ so that the triangle with one leg length of ππ π¦π¦π¦π¦ and the hypotenuse of ππππ π¦π¦π¦π¦ is a right triangle.
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8β’7 Lesson 16
Problem Set Sample Solutions
1. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the length of the hypotenuse of the triangle in centimeters.
ππππ + ππππ = ππππ ππ + ππ = ππππ
ππ = ππππ
βππ = οΏ½ππππ ππ. ππ β ππ
The length of the hypotenuse is exactly βππ πππ¦π¦ and approximately ππ.ππ πππ¦π¦.
2. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the unknown length of the triangle in feet.
ππππ + ππππ = ππππππ ππππ + ππππ = ππππππ
ππππ β ππππ + ππππ = ππππππ β ππππ ππππ = ππππ
οΏ½ππππ = βππππ
ππ = οΏ½ππππ Γ βππ Γ οΏ½ππππ
ππ = ππβππ ππ β ππ.ππ
The length of the unknown side of the triangle is exactly ππβππ ππππ. and approximately ππ.ππ ππππ.
3. Is the triangle with leg lengths of βππ πππ¦π¦, ππ πππ¦π¦, and hypotenuse of length βππππ πππ¦π¦ a right triangle? Show your work, and answer in a complete sentence.
οΏ½βπποΏ½ππ
+ ππππ = οΏ½βπππποΏ½ππ
ππ + ππππ = ππππ
ππππ = ππππ
Yes, the triangle with leg lengths of βππ πππ¦π¦, ππ πππ¦π¦, and hypotenuse of length βππππ πππ¦π¦ is a right triangle because the lengths satisfy the Pythagorean theorem.
4. Is the triangle with leg lengths of βππ π€π€π¦π¦, ππ π€π€π¦π¦, and hypotenuse of length βππππ π€π€π¦π¦ a right triangle? Show your work, and answer in a complete sentence.
οΏ½βπποΏ½ππ
+ ππππ = οΏ½βπππποΏ½ππ
ππ + ππππ = ππππ
ππππ β ππππ
No, the triangle with leg lengths of βππ π€π€π¦π¦, ππ π€π€π¦π¦, and hypotenuse of length βππππ π€π€π¦π¦ is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
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8β’7 Lesson 16
5. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the length of the hypotenuse of the triangle in millimeters.
ππππ + ππππππ = ππππ ππππ + ππππππ = ππππ
ππππππ = ππππ
βππππππ = οΏ½ππππ
οΏ½ππππ = ππ
οΏ½ππππ Γ βππ = ππ
ππβππ = ππ ππππ.ππ β ππ
The length of the hypotenuse is exactly ππβππ π¦π¦π¦π¦ and approximately ππππ.ππ π¦π¦π¦π¦.
6. Is the triangle with leg lengths of ππ, ππ, and hypotenuse of length βππππ a right triangle? Show your work, and answer in a complete sentence.
ππππ + ππππ = οΏ½βπππποΏ½ππ
ππ + ππππ = ππππ ππππ = ππππ
Yes, the triangle with leg lengths of ππ and ππ and hypotenuse of length βππππ is a right triangle because the lengths satisfy the Pythagorean theorem.
7. What is the length of the unknown side of the right triangle shown below? Show your work, and answer in a complete sentence. Provide an exact answer and an approximate answer rounded to the tenths place.
Let ππ represent the unknown side length of the triangle in inches.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππ β ππ + ππππ = ππππ β ππ ππππ = ππππ
οΏ½ππππ = βππππ
ππ = οΏ½ππππ Γ βππ Γ βππ
ππ = ππβππππ ππ β ππ.ππ
The length of the unknown side of the triangle is exactly ππβππππ inches and approximately ππ.ππ inches.
8. Is the triangle with leg lengths of ππ and βππ and hypotenuse of length ππ a right triangle? Show your work, and answer in a complete sentence.
ππππ + οΏ½βππ οΏ½ππ
= ππππ ππ + ππ = ππ
ππ = ππ
Yes, the triangle with leg lengths of ππ and βππ and hypotenuse of length ππ is a right triangle because the lengths satisfy the Pythagorean theorem.
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9. Corey found the hypotenuse of a right triangle with leg lengths of ππ and ππ to be βππππ. Corey claims that since βππππ = ππ. ππππ when estimating to two decimal digits, that a triangle with leg lengths of ππ and ππ and a hypotenuse of ππ.ππππ is a right triangle. Is he correct? Explain.
No, Corey is not correct.
ππππ + ππππ = (ππ.ππππ)ππ ππ + ππ = ππππ.ππππππππ ππππ β ππππ.ππππππππ
No, the triangle with leg lengths of ππ and ππ and hypotenuse of length ππ.ππππ is not a right triangle because the lengths do not satisfy the Pythagorean theorem.
10. Explain a proof of the Pythagorean theorem.
Consider having students share their proof with a partner while their partner critiques their reasoning. Accept any of the three proofs that the students have seen.
11. Explain a proof of the converse of the Pythagorean theorem.
Consider having students share their proof with a partner while their partner critiques their reasoning. Accept either of the proofs that the students have seen.
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Lesson 17: Distance on the Coordinate Plane
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8β’7 Lesson 17
Lesson 17: Distance on the Coordinate Plane
Student Outcomes
Students determine the distance between two points on a coordinate plane using the Pythagorean theorem.
Lesson Notes Calculators will be helpful in this lesson for determining values of radical expressions.
Classwork
Example 1 (6 minutes)
Example 1
What is the distance between the two points π¨π¨ and π©π© on the coordinate plane?
What is the distance between the two points π΄π΄ and π΅π΅ on the coordinate plane?
The distance between points π΄π΄ and π΅π΅ is 6 units.
Scaffolding: Students may benefit from physically measuring lengths to understand finding distance. A reproducible of cut-outs for this example has been included at the end of the lesson.
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What is the distance between the two points π¨π¨ and π©π© on the coordinate plane?
What is the distance between the two points π΄π΄ and π΅π΅ on the coordinate plane?
The distance between points π΄π΄ and π΅π΅ is 2 units.
What is the distance between the two points π¨π¨ and π©π© on the coordinate plane? Round your answer to the tenths place.
What is the distance between the two points π΄π΄ and π΅π΅ on the coordinate plane? Round your answer to the tenths place.
Provide students time to solve the problem. Have students share their work and estimations of the distance between the points. The questions below can be used to guide studentsβ thinking.
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We cannot simply count units between the points because the line that connects π΄π΄ to π΅π΅ is not horizontal or vertical. What have we done recently that allowed us to find the length of an unknown segment?
The Pythagorean theorem allows us to determine the length of an unknown side of a right triangle.
Use what you know about the Pythagorean theorem to determine the distance between points π΄π΄ and π΅π΅.
Provide students time to solve the problem now that they know that the Pythagorean theorem can help them. If necessary, the questions below can guide studentsβ thinking.
We must draw a right triangle so that |π΄π΄π΅π΅| is the hypotenuse. How can we construct the right triangle that we need?
Draw a vertical line through π΅π΅ and a horizontal line through π΄π΄. Or, draw a vertical line through π΄π΄ and a horizontal line through π΅π΅.
MP.1 &
MP.7
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Letβs mark the point of intersection of the horizontal and vertical lines we drew as point πΆπΆ. What is the length of |π΄π΄πΆπΆ|? |π΅π΅πΆπΆ|?
The length of |π΄π΄πΆπΆ| = 6 units, and the length of |π΅π΅πΆπΆ| = 2 units.
Now that we know the lengths of the legs of the right triangle, we can determine the length of |π΄π΄π΅π΅|.
Remind students that because we are finding a length, we need only consider the positive value of the square root because a negative length does not make sense. If necessary, remind students of this fact throughout their work in this lesson.
Let ππ be the length of π΄π΄π΅π΅.
22 + 62 = ππ2 4 + 36 = ππ2
40 = ππ2
β40 = ππ 6.3 β ππ
The distance between points π΄π΄ and π΅π΅ is approximately 6.3 units.
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Example 2 (6 minutes)
Example 2
Given two points π¨π¨ and π©π© on the coordinate plane, determine the distance between them. First, make an estimate; then, try to find a more precise answer. Round your answer to the tenths place.
Provide students time to solve the problem. Have students share their work and estimations of the distance between the points. The questions below can be used to guide studentsβ thinking.
We know that we need a right triangle. How can we draw one?
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Draw a vertical line through π΅π΅ and a horizontal line through π΄π΄. Or draw a vertical line through π΄π΄ and a horizontal line through π΅π΅.
Mark the point πΆπΆ at the intersection of the horizontal and vertical lines. What do we do next?
Count units to determine the lengths of the legs of the right triangle, and then use the Pythagorean theorem to find |π΄π΄π΅π΅|.
Show the last diagram, and ask a student to explain the answer.
The length |π΄π΄πΆπΆ| = 3 units, and the length |π΅π΅πΆπΆ| = 3 units. Let ππ be |π΄π΄π΅π΅|. 32 + 32 = ππ2
9 + 9 = ππ2 18 = ππ2
β18 = ππ 4.2 β ππ
The distance between points π΄π΄ and π΅π΅ is approximately 4.2 units.
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Exercises 1β4 (12 minutes)
Students complete Exercises 1β4 independently.
Exercises 1β4
For each of the Exercises 1β4, determine the distance between points π¨π¨ and π©π© on the coordinate plane. Round your answer to the tenths place.
1.
Let ππ represent |π¨π¨π©π©|.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
2.
Let ππ represent |π¨π¨π©π©|. ππππππ + ππππ = ππππ ππππππ + ππππ = ππππ
ππππππ = ππππ
βππππππ = ππ ππππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππππ.ππ units.
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3.
Let ππ represent |π¨π¨π©π©|. ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
4.
Let ππ represent |π¨π¨π©π©|. ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
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Example 3 (14 minutes)
Is the triangle formed by the points π΄π΄, π΅π΅, πΆπΆ a right triangle?
Provide time for small groups of students to discuss and determine if the triangle formed is a right triangle. Have students share their reasoning with the class. If necessary, use the questions below to guide their thinking.
Example 3
Is the triangle formed by the points π¨π¨, π©π©, πͺπͺ a right triangle?
How can we verify if a triangle is a right triangle?
Use the converse of the Pythagorean theorem.
What information do we need about the triangle in order to use the converse of the Pythagorean theorem, and how would we use it?
We need to know the lengths of all three sides; then, we can check to see if the side lengths satisfy the Pythagorean theorem.
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Clearly, the length of |π΄π΄π΅π΅| = 10 units. How can we determine |π΄π΄πΆπΆ|?
To find |π΄π΄πΆπΆ|, follow the same steps used in the previous problem. Draw horizontal and vertical lines to form a right triangle, and use the Pythagorean theorem to determine the length.
Determine |π΄π΄πΆπΆ|. Leave your answer in square root form unless it is a perfect square.
Let ππ represent |π΄π΄πΆπΆ|.
12 + 32 = ππ2 1 + 9 = ππ2
10 = ππ2
β10 = ππ
Now, determine |π΅π΅πΆπΆ|. Again, leave your answer in square root form unless it is a perfect square.
Let ππ represent |π΅π΅πΆπΆ|.
92 + 32 = ππ2 81 + 9 = ππ2
90 = ππ2
β90 = ππ
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The lengths of the three sides of the triangle are 10 units, β10 units, and β90 units. Which number represents the hypotenuse of the triangle? Explain.
The side π΄π΄π΅π΅ must be the hypotenuse because it is the longest side. When estimating the lengths of the other two sides, I know that β10 is between 3 and 4, and β90 is between 9 and 10. Therefore, the side that is 10 units in length is the hypotenuse.
Use the lengths 10, β10, and β90 to determine if the triangle is a right triangle. Sample Response
οΏ½β10οΏ½2
+ οΏ½β90οΏ½2
= 102
10 + 90 = 100 100 = 100
Therefore, the points π΄π΄, π΅π΅, πΆπΆ form a right triangle.
Closing (3 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
To find the distance between two points on the coordinate plane, draw a right triangle and use the Pythagorean theorem.
To verify if a triangle in the plane is a right triangle, use both the Pythagorean theorem and its converse.
Exit Ticket (4 minutes)
Lesson Summary
To determine the distance between two points on the coordinate plane, begin by connecting the two points. Then, draw a vertical line through one of the points and a horizontal line through the other point. The intersection of the vertical and horizontal lines forms a right triangle to which the Pythagorean theorem can be applied.
To verify if a triangle is a right triangle, use the converse of the Pythagorean theorem.
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Name Date
Lesson 17: Distance on the Coordinate Plane
Exit Ticket Use the following diagram to answer the questions below.
1. Determine |π΄π΄πΆπΆ|. Leave your answer in square root form unless it is a perfect square.
2. Determine |πΆπΆπ΅π΅|. Leave your answer in square root form unless it is a perfect square.
3. Is the triangle formed by the points π΄π΄, π΅π΅, πΆπΆ a right triangle? Explain why or why not.
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Exit Ticket Sample Solutions Use the following diagram to answer the questions below.
1. Determine |π¨π¨πͺπͺ|. Leave your answer in square root form unless it is a perfect square.
Let ππ represent |π¨π¨πͺπͺ|.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ
2. Determine |πͺπͺπ©π©|. Leave your answer in square root form unless it is a perfect square.
Let π π represent |πͺπͺπ©π©|.
ππππ + ππππ = π π ππ ππ+ ππππ = π π ππ
ππππ = π π ππ
βππππ = π π ππ = π π
3. Is the triangle formed by the points π¨π¨, π©π©, πͺπͺ a right triangle? Explain why or why not.
Using the lengths ππ,βππππ, and |π¨π¨π©π©| = ππ to determine if the triangle is a right triangle, I have to check to see if
ππππ + (βππππ)ππ = ππππ
ππππ + ππππ β ππππ
Therefore, the triangle formed by the points π¨π¨, π©π©, πͺπͺ is not a right triangle because the lengths of the triangle do not satisfy the Pythagorean theorem.
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Problem Set Sample Solutions For each of the Problems 1β4, determine the distance between points π¨π¨ and π©π© on the coordinate plane. Round your answer to the tenths place.
1.
Let ππ represent |π¨π¨π©π©|.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
2.
Let ππ represent |π¨π¨π©π©|.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
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3.
Let ππ represent |π¨π¨π©π©|.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ ππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππ.ππ units.
4.
Let ππ represent |π¨π¨π©π©|.
ππππππ + ππππ = ππππ ππππππ+ ππππ = ππππ
ππππππ = ππππ
βππππππ = ππ ππππ.ππ β ππ
The distance between points π¨π¨ and π©π© is about ππππ.ππ units.
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5. Is the triangle formed by points π¨π¨, π©π©, πͺπͺ a right triangle?
Let ππ represent |π¨π¨π©π©|.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ
Let ππ represent |π¨π¨πͺπͺ|.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ
Let ππ represent |π©π©πͺπͺ|.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππππ = ππππ
βππππ = ππ
οΏ½βπππποΏ½ππ
+ οΏ½βπππποΏ½ππ
= οΏ½βπππποΏ½ππ
ππππ + ππππ = ππππ ππππ β ππππ
No, the points do not form a right triangle.
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8β’7 Lesson 18
Lesson 18: Applications of the Pythagorean Theorem
Student Outcomes
Students apply the Pythagorean theorem to real-world and mathematical problems in two dimensions.
Lesson Notes It is recommended that students have access to a calculator as they work through the exercises. However, it is not recommended that students use calculators to answer the questions but only to check their work or estimate the value of an irrational number using rational approximation. Make clear to students that they can use calculators but that all mathematical work should be shown. This lesson includes a Fluency Exercise that will take approximately 10 minutes to complete. The Fluency Exercise is a white board exchange with problems on volume that can be found at the end of this lesson. It is recommended that the Fluency Exercise take place at the beginning of the lesson or after the discussion that concludes the lesson.
Classwork
Exploratory Challenge/Exercises 1β5 (20 minutes)
Students complete Exercises 1β5 in pairs or small groups. These problems are applications of the Pythagorean theorem and are an opportunity to remind students of Mathematical Practice 1: Make sense of problems and persevere in solving them. Students should compare their solutions and solution methods in their pairs, small groups, and as a class. If necessary, remind students that we are finding lengths, which means we need only consider the positive square root of a number.
Exercises
1. The area of the right triangle shown below is ππππ.ππππ π’π’π§π§ππ. What is the perimeter of the right triangle? Round your answer to the tenths place.
Let ππ represent the base of the triangle in inches where ππ = ππ.ππ.
π¨π¨ =ππππππ
ππππ.ππππ =ππ. ππππππ
ππππ.ππππ = ππ. ππππ ππππ.ππππππ.ππ
=ππ. ππππππ. ππ
ππ.ππ = ππ Let ππ represent the length of the hypotenuse in inches.
ππ.ππππ + ππ.ππππ = ππππ ππππ. ππππ + ππππ. ππππ = ππππ
ππππππ. ππππ = ππππ βππππππ. ππππ = οΏ½ππππ βππππππ. ππππ = ππ
The number βππππππ.ππππ is between ππππ and ππππ. When comparing with tenths, the number is actually equal to ππππ. ππ because ππππ.ππππ = ππππππ.ππππ. Therefore, the length of the hypotenuse is ππππ.ππ π’π’π§π§. The perimeter of the triangle is ππ.ππ π’π’π§π§. + ππ.ππ π’π’π§π§. + ππππ.ππ π’π’π§π§. = ππππ. ππ π’π’π§π§.
MP.1
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8β’7 Lesson 18
Note to Teacher: Check in with students to make sure they understand how TVs are measured in terms of their diagonal length, and what is meant by the ratio of 4: 3. To complete the problem, students must be clear that the size of a TV is not denoted by its length or width but by the length of the diagonal (hypotenuse). Also, students must have some sense that the ratio of length to width must be some multiple of the ratio 4: 3; otherwise, the TV would not give a good perspective. Consider showing students what a TV would look like with a ratio of 9: 12. They should notice that such dimensions yield a TV screen that is unfamiliar to them.
2. The diagram below is a representation of a soccer goal.
a. Determine the length of the bar, ππ, that would be needed to provide structure to the goal. Round your answer to the tenths place.
Let ππ represent the hypotenuse of the right triangle in feet.
ππππ + ππππ = ππππ ππππ + ππ = ππππ
ππππ = ππππ
βππππ = οΏ½ππππ
The number βππππ is between ππ and ππ. In the sequence of tenths, it is between ππ. ππ
and ππ. ππ because ππ.ππππ < οΏ½βπππποΏ½ππ
< ππ.ππππ. In the sequence of hundredths, the number
is between ππ.ππππ and ππ.ππππ because ππ.ππππππ < οΏ½βπππποΏ½ππ
< ππ.ππππππ. Since the number
βππππ is between ππ.ππππ and ππ.ππππ, it would round to ππ.ππ. The length of the bar that
provides structure for the goal is ππ.ππ ππππ.
b. How much netting (in square feet) is needed to cover the entire goal?
The area of the triangles are each ππππ ππππππ. The area of the rectangle in the back is ππππ ππππππ. The total area of netting required to cover the goal is ππππππ ππππππ.
3. The typical ratio of length to width that is used to produce televisions is ππ:ππ.
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8β’7 Lesson 18
a. A TV with those exact measurements would be quite small, so generally the size of the television is enlarged by multiplying each number in the ratio by some factor of ππ. For example, a reasonably sized television might have dimensions of ππ Γ ππ:ππ Γ ππ, where the original ratio ππ:ππ was enlarged by a scale factor of ππ. The size of a television is described in inches, such as a ππππ" TV, for example. That measurement actually refers to the diagonal length of the TV (distance from an upper corner to the opposite lower corner). What measurement would be applied to a television that was produced using the ratio of ππ Γ ππ: ππ Γ ππ?
Let ππ be the length of the diagonal in inches.
ππππππ + ππππππ = ππππ ππππππ + ππππππ = ππππ
ππππππ = ππππ
βππππππ = οΏ½ππππ ππππ = ππ
Since the TV has a diagonal length of ππππ inches, then it is a ππππ" TV.
b. A ππππ" TV was just given to your family. What are the length and width measurements of the TV?
Let ππ be the factor applied to the ratio ππ:ππ.
(ππππ)ππ + (ππππ)ππ = ππππππ ππππππ + ππππππππ = ππππππππ (ππ + ππππ)ππππ = ππππππππ
ππππππππ = ππππππππ ππππππππ
ππππ=ππππππππππππ
ππππ = ππππ. ππππ
οΏ½ππππ = βππππ. ππππ
ππ = βππππ. ππππ
The number βππππ.ππππ is between ππ and ππ. In working with the sequence of tenths, I realized the number βππππ.ππππ is actually equal to ππ.ππ because ππ. ππππ = ππππ.ππππ. Therefore, ππ = ππ.ππ, and the dimensions of the TV are ππ Γ ππ.ππ = ππππ. ππ inches and ππ Γ ππ.ππ = ππππ.ππ inches.
c. Check that the dimensions you got in part (b) are correct using the Pythagorean theorem.
ππππ.ππππ + ππππ.ππππ = ππππππ ππππππππ.ππππ + ππππππ.ππππ = ππππππππ
ππππππππ = ππππππππ
d. The table that your TV currently rests on is ππππ" in length. Will the new TV fit on the table? Explain.
The dimension for the length of the TV is ππππ.ππ inches. It will not fit on a table that is ππππ inches in length.
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4. Determine the distance between the following pairs of points. Round your answer to the tenths place. Use graph paper if necessary.
a. (ππ,ππ) and (βππ,βππ)
Let ππ represent the distance between the two points.
ππππππ + ππππ = ππππ ππππππ + ππππ = ππππ
ππππππ = ππππ
βππππππ = οΏ½ππππ
βππππππ = ππ
The number βππππππ is between ππππ and ππππ. In the sequence of tenths, it is between ππππ.ππ and ππππ. ππ because
ππππ.ππππ < οΏ½βπππππποΏ½ππ
< ππππ.ππππ. In the sequence of hundredths, it is between ππππ.ππππ and ππππ.ππππ, which means the number will round to ππππ. ππ. The distance between the two points is ππππ.ππ units.
b. (βππ,ππ) and (ππ,ππ)
Let ππ represent the distance between the two points.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = οΏ½ππππ
βππππ = ππ
The number βππππ is between ππ and ππ. In the sequence of tenths, it is between ππ. ππ and ππ because
ππ.ππππ < οΏ½βπππποΏ½ππ
< ππππ. In the sequence of hundredths, it is between ππ.ππππ and ππ.ππππ, which means it will round to ππ.ππ. The distance between the two points is ππ.ππ units.
c. Challenge: (ππππ,ππππ) and (ππππ ,ππππ). Explain your answer.
Note: Deriving the distance formula using the Pythagorean theorem is not part of the standard but does present an interesting challenge to students. Assign it only to students who need a challenge.
Let ππ represent the distance between the two points.
(ππππ β ππππ)ππ + (ππππ β ππππ)ππ = ππππ
οΏ½(ππππ β ππππ)ππ + (ππππ β ππππ)ππ = οΏ½ππππ
οΏ½(ππππ β ππππ)ππ + (ππππ β ππππ)ππ = ππ
I noticed that the dimensions of the right triangle were equal to the difference in ππ-values and difference in ππ-values. Using those expressions and what I knew about solving radical equations, I was able to determine the length of ππ.
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8β’7 Lesson 18
5. What length of ladder will be needed to reach a height of ππ feet along the wall when the base of the ladder is ππ feet from the wall? Round your answer to the tenths place.
Let ππ represent the length of the ladder in feet.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
βππππ = οΏ½ππππ βππππ = ππ
The number βππππ is between ππ and ππ. In the sequence of
tenths, it is between ππ and ππ.ππ because ππππ < οΏ½βπππποΏ½ππ
< ππ.ππππ. In the sequence of hundredths, it is between ππ. ππππ and ππ. ππππ, which means the number will round to ππ. ππ. The ladder must be ππ.ππ feet long to reach ππ feet up a wall when placed ππ feet from the wall.
Discussion (5 minutes)
This discussion provides a challenge question to students about how the Pythagorean theorem might be applied to a three-dimensional situation. The next lesson focuses on using the Pythagorean theorem to answer questions about cones and spheres.
The majority of our work with Pythagorean theorem has been in two dimensions. Can you think of any applications we have seen so far that are in three dimensions? The soccer goal is three-dimensional. A ladder propped up against a wall is three-dimensional.
What new applications of Pythagorean theorem in three dimensions do you think we will work on next?
Provide students time to think about this in pairs or small groups.
We have worked with solids this year, so there may be an application involving cones and spheres.
Fluency Exercise (10 minutes): Area and Volume II
RWBE: Refer to the Rapid White Board Exchanges section in the Module Overview for directions to administer a Rapid White Board Exchange.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:
We know some basic applications of the Pythagorean theorem in terms of measures of a television, length of a ladder, area and perimeter of right triangles, etc.
We know that there will be some three-dimensional applications of the theorem beyond what we have already seen.
Exit Ticket (5 minutes)
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
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Lesson 18: Applications of the Pythagorean Theorem
Exit Ticket Use the diagram of the equilateral triangle shown below to answer the following questions. Show work that leads to your answers.
a. What is the perimeter of the triangle?
b. What is the height, β, of the equilateral triangle? Write an exact answer using a square root and approximate answer rounded to the tenths place.
c. Using the approximate height found in part (b), estimate the area of the equilateral triangle.
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8β’7 Lesson 18
Exit Ticket Sample Solutions Use the diagram of the equilateral triangle shown below to answer the following questions. Show work that leads to your answers.
a. What is the perimeter of the triangle?
ππ + ππ + ππ = ππππ
The perimeter is ππππ π¦π¦π¦π¦.
b. What is the height, ππ, of the equilateral triangle? Write an exact answer using a square root and approximate answer rounded to the tenths place.
Using the fact that the height is one leg length of a right triangle, and I know the hypotenuse is ππ π¦π¦π¦π¦ and the other leg length is ππ π¦π¦π¦π¦, I can use the Pythagorean theorem to find ππ.
ππππ + ππππ = ππππ ππ + ππππ = ππππ
ππ β ππ + ππππ = ππππ β ππ ππππ = ππππ ππ = βππππ
ππ = βππ Γ ππ ππ = βππ Γ βππ
ππ = ππβππ
The number βππ is between ππ and ππ. In the sequence of tenths, it is between ππ. ππ and ππ.ππ because
ππ.ππππ < οΏ½βπποΏ½ππ
< ππ.ππππ. In the sequence of hundredths, it is between ππ. ππππ and ππ. ππππ, which means it would round to ππ. ππ. Then ππ Γ ππ.ππ = ππ. ππ π¦π¦π¦π¦ is the approximate length of the hypotenuse, and βππππ = ππβππ π¦π¦π¦π¦ is the exact length.
c. Using the approximate height found in part (b), estimate the area of the equilateral triangle.
π¨π¨ =ππππππ
π¨π¨ =ππ(ππ.ππ)ππ
π¨π¨ =ππππ.ππππ
π¨π¨ = ππ.ππ
The approximate area of the equilateral triangle is ππ.ππ π¦π¦π¦π¦ππ.
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
Problem Set Sample Solutions Students continue applying the Pythagorean theorem to solve real-world and mathematical problems.
1. A ππππ" TV is advertised on sale at a local store. What are the length and width of the television?
The TV is in the ratio of ππ:ππ and has measurements of ππππ:ππππ, where ππ is the scale factor of enlargement.
(ππππ)ππ + (ππππ)ππ = ππππππ ππππππ + ππππππππ = ππππππππ
ππππππππ = ππππππππ ππππππππ
ππππ=ππππππππππππ
ππππ = ππππππ
οΏ½ππππ = βππππππ ππ = ππππ
The length of the TV is ππ Γ ππππ = ππππ inches, and the width is ππ Γ ππππ = ππππ inches.
2. There are two paths that one can use to go from Sarahβs house to Jamesβs house. One way is to take C Street, and the other way requires you to use A Street and B Street. How much shorter is the direct path along C Street?
Let ππ represent the length of the hypotenuse of the right triangle in miles.
ππππ + ππ.ππππ = ππππ ππ + ππ. ππππ = ππππ
ππ. ππππ = ππππ
βππ. ππππ = οΏ½ππππ ππ. ππ = ππ
The path using A Street and B Street is ππ.ππ miles. The path along C Street is ππ.ππ miles. The path along C Street is exactly ππ mile shorter than the path along A Street and B Street.
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
3. An isosceles right triangle refers to a right triangle with equal leg lengths, ππ, as shown below.
What is the length of the hypotenuse of an isosceles right triangle with a leg length of ππ πππ¦π¦? Write an exact answer using a square root and an approximate answer rounded to the tenths place.
Let ππ be the length of the hypotenuse of the isosceles triangle in centimeters.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππππ = ππππ
βππππππ = οΏ½ππππ
βππππ Γ ππ = ππ βππππ Γ βππ = ππ
ππβππ = ππ
The number βππ is between ππ and ππ. In the sequence of tenths, it is between ππ. ππ and ππ. ππ because ππ.ππππ < οΏ½βπποΏ½ππ
<ππ.ππππ. Since the number ππ is closer to ππ.ππππ than ππ. ππππ, it would round to ππ.ππ. Then ππ Γ ππ. ππ = ππππ.ππ πππ¦π¦ is the approximate length of the hypotenuse, and ππβππ πππ¦π¦ is the exact length.
4. The area of the right triangle shown below is ππππ.ππ πππ¦π¦ππ.
a. What is the height of the triangle?
π¨π¨ =ππππππ
ππππ. ππ =ππ.ππππππ
ππππππ = ππ.ππππ ππππππππ.ππ
=ππ.ππππππ.ππ
ππππ = ππ
The height of the triangle is ππππ πππ¦π¦.
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
b. What is the perimeter of the right triangle? Round your answer to the tenths place.
Let ππ represent the length of the hypotenuse in centimeters.
ππ.ππππ + ππππππ = ππππ ππππ.ππππ + ππππππ = ππππ
ππππππ.ππππ = ππππ
βππππππ.ππππ = οΏ½ππππ
βππππππ.ππππ = ππ
The number βππππππ.ππππ is between ππππ and ππππ. In the sequence of tenths, the number is between ππππ. ππ and ππππ
because ππππ.ππππ < οΏ½βππππππ.πππποΏ½ππ
< ππππππ. Since ππππππ. ππππ is closer to ππππ.ππππ than ππππππ, then the approximate length of the hypotenuse is ππππ. ππ πππ¦π¦.
The perimeter of the triangle is ππ.ππ πππ¦π¦ + ππππ πππ¦π¦ + ππππ.ππ πππ¦π¦ = ππππ.ππ πππ¦π¦.
5. What is the distance between points (ππ,ππ) and (βππ,βππ)? Round your answer to the tenths place.
Let ππ represent the distance between the points.
ππππππ + ππππ = ππππ ππππππ + ππππ = ππππ
ππππππ = ππππ
βππππππ = οΏ½ππππ
βππππππ = ππ ππππ. ππ β ππ
The distance between the points is approximately ππππ.ππ units.
6. An equilateral triangle is shown below. Determine the area of the triangle. Round your answer to the tenths place.
Let ππ represent the height of the triangle in inches.
ππππ + ππππ = ππππ ππππ + ππππ = ππππ
ππππ = ππππ
οΏ½ππππ = βππππ
ππ = βππππ ππ β ππ.ππ
π¨π¨ =ππ(ππ. ππ)ππ
= ππ(ππ.ππ) = ππππ.ππ
The area of the triangle is ππππ.ππ π’π’π§π§ππ.
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
Area and Volume II 1. Find the area of the square shown below.
π¨π¨ = (ππ πππ¦π¦)ππ = ππππ πππ¦π¦ππ
2. Find the volume of the cube shown below.
π½π½ = (ππ πππ¦π¦)ππ = ππππππ πππ¦π¦ππ
3. Find the area of the rectangle shown below.
π¨π¨ = (ππ π¦π¦)(ππ π¦π¦) = ππππ π¦π¦ππ
4. Find the volume of the rectangular prism show below.
π½π½ = (ππππ π¦π¦ππ)(ππ π¦π¦) = ππππππ π¦π¦ππ
5. Find the area of the circle shown below.
π¨π¨ = (ππ π¦π¦)πππ π = πππππ π π¦π¦ππ
Lesson 18: Applications of the Pythagorean Theorem
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8β’7 Lesson 18
6. Find the volume of the cylinder show below.
π½π½ = (πππππ π π¦π¦ππ)(ππππ π¦π¦) = πππππππ π π¦π¦ππ
7. Find the area of the circle shown below.
π¨π¨ = (ππ π’π’π§π§.)πππ π = πππππ π π’π’π§π§ππ
8. Find the volume of the cone show below.
π½π½ = οΏ½πππποΏ½ (πππππ π π’π’π§π§ππ)(ππππ π’π’π§π§.)
= πππππππ π π’π’π§π§ππ
9. Find the area of the circle shown below.
π¨π¨ = (ππ π¦π¦π¦π¦)πππ π = πππππ π π¦π¦π¦π¦ππ
10. Find the volume of the sphere shown below.
π½π½ = οΏ½πππποΏ½π π (ππ π¦π¦π¦π¦)ππ
=ππππππ π¦π¦π¦π¦
πππ π
= πππππππ π π¦π¦π¦π¦ππ
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