8.G.B.6: Explaining a Proof of the Pythagorean Theorem and Its Converse
In plain English: 8.G.B.6 is the Common Core grade 8 math standard that asks students to explain a proof of the Pythagorean Theorem, a² + b² = c² for the legs a and b and the hypotenuse c of a right triangle, and a proof of its converse: a triangle whose sides fit a² + b² = c² has a right angle. It comes just before students apply the theorem in 8.G.B.7.
Explain a proof of the Pythagorean Theorem and its converse.
Common Core State Standards for Mathematics · Domain: Geometry (G) · Cluster: Understand and apply the Pythagorean Theorem. Also written as 8.G.6 · Official standard
In a right triangle, the two sides that form the right angle are the legs, called a and b, and the longest side, across from the right angle, is the hypotenuse, called c. The Pythagorean Theorem says that a² + b² = c². A theorem is a statement that has been proved, and a proof is an argument that shows a statement is true in every case, not only in the examples someone checked. This standard asks students to explain a proof, step by step, and to say why each step is true.
The lesson uses a rearrangement proof: the same big square holds four copies of a right triangle in two different ways, and the space left over is c² in one and a² + b² in the other. Then it turns to the converse, the statement with the "if" and "then" parts swapped: if the sides of a triangle fit a² + b² = c², the triangle has a right angle. Students explain its proof by building a right triangle with the same legs and showing that the two triangles are copies of each other.
Learning Objectives
By the end of this lesson, students will be able to:
Explain each step of a rearrangement proof of the Pythagorean Theorem, including where the right angle is used
Explain why checking numerical examples is not a proof, and why a proof with the letters a, b and c covers every right triangle
State the converse of the Pythagorean Theorem and explain its proof by building a right triangle with the same legs
Use the converse to decide whether a triangle with given side lengths is a right triangle, and explain the decision
Prior Knowledge Required
Students should already be comfortable with:
The area of a triangle is half the area of the rectangle around it: ½ × base × height 6.G.A.1
Square roots: the square root of 49 is 7, because 7² = 49 8.EE.A.2
Three side lengths make only one triangle shape, so two triangles with the same three sides are copies of each other 7.G.A.2
Congruent figures (exact copies, the same shape and size) have equal angles, and the angles of a triangle add up to 180° (8.G.A.2, 8.G.A.5)
Students draw a right triangle on grid paper with legs of 3 and 4 units along the grid lines, and a second one with legs of 6 and 8 units. They measure each hypotenuse with a ruler, in grid units.
Warm-Up Prompt
"Square each leg and each hypotenuse you measured. What do you notice? Do you think it will happen for every right triangle? How many triangles would you need to check to be sure?"
Students find hypotenuses of 5 and 10 units and notice that 9 + 16 = 25 and 36 + 64 = 100. Push on the last question: there are infinitely many right triangles, so no number of checks is enough. That is why mathematicians need a proof, an argument that works with the letters a, b and c for every right triangle at once.
Direct Instruction20 minutes
Walk through the proof with Diagram 1 and paper cutouts of four copies of one right triangle. Students copy each step and the reason for it:
Start with any right triangle, with legs a and b and hypotenuse c. Make four copies of it (congruent triangles, exact copies of each other).
Arrangement 1: place the four triangles inside a square with side a + b, one in each corner, as in Diagram 1. Each side of the big square is a leg a plus a leg b.
Why the middle is a square: its four sides are hypotenuses, so each has length c. At each point where two triangles meet, the two acute angles of the triangle (the angles smaller than 90°) and the corner of the middle figure fill a straight line. The acute angles add up to 180° - 90° = 90°, so the corner is 180° - 90° = 90°. So the middle figure is a square with area c².
Arrangement 2: slide the same four triangles into two rectangles, each a by b, inside a second square with side a + b. The space left over is a square with side a and a square with side b, with areas a² and b².
Compare: both big squares have the same area, (a + b)², and both hold the same four triangles. So the uncovered areas are equal: c² = a² + b².
The converse: suppose a triangle has sides a, b and c with a² + b² = c². Build a new right triangle with legs a and b. By the theorem, its hypotenuse h has h² = a² + b² = c², so h = c. The two triangles have the same three side lengths, so they are copies of each other, and the angle across from c in the first triangle equals the right angle of the new one. So the first triangle is a right triangle.
When the equation fails: if a² + b² does not equal c², the triangle cannot be a right triangle with c as its longest side, because the theorem would force a² + b² = c².
The rearrangement proof (Diagram 1)
Explain why the uncovered area in Arrangement 1 equals the uncovered area in Arrangement 2, and what that proves.
Equation: Both big squares have side a + b, so they have the same area. Both hold the same four copies of one triangle, so the triangles cover the same area in each. What is left must be equal: c² in Arrangement 1 and a² + b² in Arrangement 2. So a² + b² = c² for every right triangle, since a, b and c can be any legs and hypotenuse.
The proof with numbers
Follow the proof for a right triangle with legs 8 and 15.
Equation: The big square has side 8 + 15 = 23 and area 23² = 529. The four triangles have area 4 × ½ × 8 × 15 = 240. Arrangement 1 leaves 529 - 240 = 289 = c², so c = 17. Arrangement 2 leaves 8² + 15² = 64 + 225 = 289. The areas agree, as the proof says.
The converse (Diagram 2)
Triangle 1 has sides a, b and c with a² + b² = c². Explain why the angle at C, across from side c, is a right angle.
Equation: Build Triangle 2 with legs a and b and a right angle at F. By the Pythagorean Theorem, its hypotenuse squared is a² + b², which equals c², so its hypotenuse is c. Both triangles have sides a, b and c, so they are copies of each other, and the angle at C equals the right angle at F.
Using the converse
A triangle has sides 28 cm, 45 cm and 53 cm. Is it a right triangle? If so, where is the right angle?
Equation: Square the two shorter sides and the longest: 28² + 45² = 784 + 2,025 = 2,809, and 53² = 2,809. They are equal, so by the converse the triangle is a right triangle. The right angle is across from the 53 cm side, between the 28 cm and 45 cm sides.
When the converse says no
A triangle has sides 6, 9 and 11. Is it a right triangle?
Equation: 6² + 9² = 36 + 81 = 117, but 11² = 121. If the triangle had a right angle, the theorem would make the two numbers equal, so it has no right angle. (Its largest angle is a little more than 90°, since 121 is larger than 117.)
Point out that the proof never uses particular numbers. The numbers in the second example only illustrate it. For the converse, stress which side is c: it must be the longest side, because the hypotenuse is always the longest side of a right triangle.
Guided Practice15 minutes
Pairs work through each problem. One partner computes, and the other says which step of the proof or of the converse the computation matches; they switch for the next problem.
Guided practice problems with answers
Problem
Answer
Follow the rearrangement proof for legs 5 and 12: find the area of the big square, the four triangles and the uncovered square.
Big square 17² = 289. Triangles 4 × 30 = 120. Uncovered 289 - 120 = 169 = 25 + 144, so c = 13
A triangle has sides 18, 24 and 30. Is it a right triangle?
18² + 24² = 324 + 576 = 900 = 30², so yes, with the right angle across from the 30 side
A triangle has sides 5, 6 and 8. Is it a right triangle?
25 + 36 = 61, but 8² = 64, so no
On grid paper, draw a right triangle with legs 2 and 3 units along the grid lines, and draw the square on its hypotenuse. Find the area of that square by boxing it in a 5 by 5 square and subtracting the four corner triangles.
25 - 4 × 3 = 13, which equals 2² + 3² = 4 + 9
Watch for students who square the longest side and add it to a shorter one, and for students who add the side lengths without squaring. Ask: "Which side is across from the right angle?"
Independent Practice15 minutes
Students work alone, then compare with a partner, explaining one problem each in full sentences.
Independent practice problems with answers
Problem
Answer
Follow the rearrangement proof for legs 12 and 16.
Big square 28² = 784. Triangles 4 × 96 = 384. Uncovered 400 = 144 + 256, so c = 20
Is a triangle with sides 11, 60 and 61 a right triangle?
121 + 3,600 = 3,721 = 61², so yes
Is a triangle with sides 8, 11 and 14 a right triangle?
64 + 121 = 185, but 14² = 196, so no
Use the grid method to find the area of the square on the hypotenuse of a right triangle with legs 1 and 2 units.
3 by 3 box: 9 - 4 × 1 = 5 = 1 + 4
A triangle has sides 13, 84 and 85. Which side has to play the role of c when you test it, and why? Is it a right triangle?
85, the longest side, because a hypotenuse is always the longest side. 169 + 7,056 = 7,225 = 85², so yes
Closure5-10 minutes
Exit ticket: (1) A triangle has sides 14, 48 and 50. Use the converse to decide whether it is a right triangle. (Yes: 196 + 2,304 = 2,500 = 50².) (2) In one sentence, explain why Arrangement 1 and Arrangement 2 leave the same uncovered area. (3) In one sentence, explain the difference between the Pythagorean Theorem and its converse.
Differentiation Strategies
For Struggling Students
Give students the four triangles and the two square frames already cut out, so they can focus on the argument instead of the cutting
Provide a proof frame: "Both big squares have area ____. Both hold four triangles with total area ____. So the leftover areas ____ and ____ are equal."
For the converse, have students list the three sides in order from shortest to longest before squaring
For Advanced Students
Ask students to follow the proof with a = b, when the right triangle has two equal legs, and to say what the two arrangements look like
Have students write the area of each big square as an equation, once from the pieces of Arrangement 1 and once from the pieces of Arrangement 2, and explain how the two equations prove the theorem
Extension (beyond this standard): test triangles where a² + b² is more or less than c², measure the largest angle, and describe the pattern (less than 90° when a² + b² is larger, more than 90° when it is smaller)
Assessment Guidance
What to Look For
A complete explanation of the proof says that the two big squares are equal, that they hold the same four triangles, and that the leftover areas are c² and a² + b². Strong answers also say why the middle figure in Arrangement 1 is a square. For the converse, look for the three steps: build a right triangle with legs a and b, use the theorem to show its hypotenuse is c, and conclude that the triangles are copies. Watch for students who think checking a few triangles proves the theorem, who use the theorem when they mean the converse, and who test the wrong side as c.
02
Classroom Activities
3 Activities
1
Cut and Rearrange
20 minPairs
Pairs cut out eight copies of a right triangle with legs 4 cm and 7 cm and two card-stock squares with sides of 11 cm, then build both arrangements of Diagram 1 and explain the proof in their own words.
Procedure
Draw one right triangle with legs 4 cm and 7 cm on card stock, cut it out, and trace it to make eight copies in all
Lay four triangles on the first 11 cm square as in Arrangement 1, and the other four on the second square as in Arrangement 2
Measure the side of the uncovered middle square in Arrangement 1, and find the areas of all the uncovered pieces
Write the proof in four sentences, one for each of these words: square, triangles, leftover, so
Answer Key
Each big square has area 11² = 121 cm², and each set of four triangles covers 4 × 14 = 56 cm². Arrangement 1 leaves 121 - 56 = 65 cm², and Arrangement 2 leaves 4² + 7² = 16 + 49 = 65 cm². The middle square measures about 8.1 cm on a side, since the square root of 65 is about 8.06.
Discussion Questions
The hypotenuse here is not a whole number. Does the proof still work? Why?
Where did you use the fact that each triangle has a right angle?
Modification for Distance Learning
Students use free geometry software to draw the two arrangements, drag the vertices to change a and b, and watch the measured areas stay equal.
2
Squares on the Sides
15 minPairs
Pairs draw four right triangles on grid paper with legs along the grid lines, draw the square on each side, and find the area of each tilted square on the hypotenuse by boxing it in a larger square and subtracting four corner triangles.
The Four Triangles
Triangle 1: legs 1 and 1
Triangle 2: legs 2 and 2
Triangle 3: legs 1 and 4
Triangle 4: legs 3 and 4
Answer Key
Triangle 1: 2 by 2 box, 4 - 4 × 0.5 = 2, and 1 + 1 = 2
Triangle 2: 4 by 4 box, 16 - 4 × 2 = 8, and 4 + 4 = 8
Triangle 3: 5 by 5 box, 25 - 4 × 2 = 17, and 1 + 16 = 17
Triangle 4: 7 by 7 box, 49 - 4 × 6 = 25, and 9 + 16 = 25
Discussion Questions
The box around each tilted square has side a + b. How is this box the same as Arrangement 1 of Diagram 1?
Only Triangle 4 has a whole-number hypotenuse. Why does the area method work even when the hypotenuse is not a whole number?
Challenge Variation
A proof published in 1876 by James Garfield, later a U.S. president, uses a trapezoid made of two copies of the triangle and a half square. Students draw it for legs a and b, write its area in two ways, and explain why the two areas give a² + b² = c².
3
Is It a Right Triangle?
20 minGroups of 3
Each group gets four sets of three paper strips, 12 strips in all, and builds a triangle from each set. They test each corner with the square corner of an index card and with the converse, and compare the two tests.
The Four Sets (in cm)
Set A: 6, 8 and 10
Set B: 7, 7 and 10
Set C: 5, 7 and 9
Set D: 4, 7.5 and 8.5
Procedure
Tape each set into a triangle, with the ends of the strips meeting exactly
Fit the corner of an index card into the largest angle and record whether it looks square
Compute a² + b² for the two shorter sides and c² for the longest side
Decide with the converse, and compare the result with the index-card test
Answer Key
Set A: 36 + 64 = 100 = 10², a right triangle
Set B: 49 + 49 = 98, less than 10² = 100, not a right triangle (its largest angle is about 91°)
Set C: 25 + 49 = 74, less than 9² = 81, not a right triangle (its largest angle is about 96°)
Set D: 16 + 56.25 = 72.25 = 8.5², a right triangle
Discussion Questions
Set B looks almost square at its largest corner. Which test is more reliable, the index card or the converse? Why?
Set D has a decimal side. Does the converse still apply?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: The Rearrangement Proof
Two squares with side a + b, each holding four copies of the same right triangle with legs a and b and hypotenuse c (drawn to scale for a = 3 and b = 5 units, but the argument uses only the letters). Arrangement 1 leaves a tilted square with area c². Arrangement 2 leaves two squares with areas a² and b². The big squares and the triangles are the same, so c² = a² + b².
Diagram 2: Proving the Converse
Triangle 1 has sides a, b and c with a² + b² = c², and the angle at C is unknown. Triangle 2 is built with legs a and b and a right angle at F. By the Pythagorean Theorem its hypotenuse is c, so the two triangles have the same three sides, and the angle at C is a right angle.
04
Homework Assignment
~30 min
8.G.B.6 Homework: Proving the Pythagorean Theorem and Its Converse
Directions: Show every computation, draw a sketch for each problem, and explain your reasoning in full sentences. For the converse problems, always test the longest side as c.
Part 1: The Pythagorean Theorem (Problems 1-3)
A right triangle has legs 9 and 12. (a) Follow Arrangement 1 of the rearrangement proof: find the area of the big square, the total area of the four triangles and the area of the uncovered square. (b) Follow Arrangement 2: find the areas of the two uncovered squares. (c) Explain why the two answers had to be equal.
On grid paper, draw a right triangle with legs 2 and 5 units along the grid lines. (a) Draw the square on its hypotenuse and find its area by boxing it in and subtracting the four corner triangles. (b) Compare your answer with 2² + 5². (c) Between which two whole numbers is the length of the hypotenuse?
Priya says: "I checked a² + b² = c² for three right triangles, so I proved the Pythagorean Theorem." (a) Explain why her checks are not a proof. (b) Explain where the rearrangement proof uses the fact that the triangle has a right angle.
Part 2: The Converse (Problems 4-6)
A triangle has sides 16, 30 and 34. (a) Use the converse to decide whether it is a right triangle. (b) If it is, name the two sides that form the right angle.
A triangle has sides 10, 11 and 15. (a) Is it a right triangle? (b) Explain your answer using the Pythagorean Theorem.
A builder frames a rectangular garden bed. Along one corner, she marks 80 cm on one board and 150 cm on the other, and the distance between the two marks is 170 cm. At a second corner, the same marks are 168 cm apart. (a) Which corners are square (90°)? (b) Explain, step by step as in the proof of the converse, why a 170 cm distance guarantees a square corner.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Computations
All areas and squares correct
One or two computing errors
Many errors
Proof of the Theorem
Explains the equal squares, the equal triangles and the equal leftovers
Explanation missing one of the three parts
No explanation
Converse
Tests the longest side as c and explains the conclusion with the converse
Correct decision without a reason, or the wrong side tested
Missing or incorrect
Communication
Sketches labeled and sentences clear
Some sketches or labels missing
No sketches or sentences
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Question 6 uses Diagram 1. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
In the rearrangement proof, two squares with side a + b each hold four copies of the same right triangle. Why must the uncovered areas in the two squares be equal?
Answer: C
The big squares have the same area, (a + b)², and the same four triangles are taken away from each, so what is left is equal. Choice A is false: the second square leaves two squares with sides a and b. Choice B is false, since the triangles are the same four in both squares. Choice D is false: the big squares have side a + b.
Question 2 of 20 · Multiple Choice
Follow the rearrangement proof for a right triangle with legs 7 and 24. What is the area of the uncovered square in Arrangement 1?
Answer: B
The big square has area (7 + 24)² = 31² = 961. The four triangles have area 4 × ½ × 7 × 24 = 336. So the uncovered square is 961 - 336 = 625, and c = 25. Choice A forgets to subtract the triangles. Choice C uses 7 × 24 = 168 for each triangle, forgetting the ½. Choice D subtracts only one triangle.
Question 3 of 20 · Multiple Choice
Where does the rearrangement proof use the fact that each triangle has a right angle?
Answer: B
At each side of the big square, the two acute angles of the triangle add up to 90°, so the corner of the middle figure is 180° - 90° = 90°, and the middle figure is a square with area c². The right angles also make the big figure's corners square. Choice A is true for any four copies of a triangle. Choice C is false: the big square has side a + b. Choice D is false: the legs can be any lengths.
Question 4 of 20 · Multiple Choice
On grid paper, a right triangle has legs 1 and 3 units. The square on its hypotenuse fits in a 4 by 4 box with four corner triangles cut off. What is the area of the square on the hypotenuse?
Answer: D
The box has area 16, and each corner triangle has area ½ × 1 × 3 = 1.5, so the square has area 16 - 4 × 1.5 = 10, which equals 1² + 3². Choice A forgets to subtract the corner triangles. Choice B adds the legs, 1 + 3. Choice C subtracts only two of the corner triangles.
Question 5 of 20 · Multiple Choice
In Arrangement 1 of Diagram 1, each of the four triangles has legs a and b. What is the total area of the four triangles?
Answer: D
Each right triangle is half of an a by b rectangle, so its area is ½ab, and four of them cover 4 × ½ab = 2ab. Choice A forgets the ½ and uses ab for each triangle. Choice B is the area of only two triangles. Choice C uses the hypotenuse c as if it were both the base and the height of each triangle.
Question 6 of 20 · Multiple Choice
In Arrangement 2 of Diagram 1, the big square is split into two squares and two rectangles. Which equation describes its area?
Answer: D
The pieces are a square with area a², a square with area b², and two rectangles with area ab each, so (a + b)² = a² + b² + 2ab. Choice A leaves out the two rectangles. Choice B counts only one rectangle. Choice C is the perimeter of the big square, not its area.
Question 7 of 20 · Multiple Choice
Which statement is the converse of the Pythagorean Theorem?
Answer: B
The theorem says: right angle, then a² + b² = c². The converse swaps the two parts: a² + b² = c², then right angle. Choice A is the theorem itself. Choice C is a different fact about right triangles. Choice D is false: when the equation fails, the triangle has no right angle across from c.
Question 8 of 20 · Multiple Choice
A triangle has sides 12, 35 and 37. What does the converse of the Pythagorean Theorem tell you?
Answer: C
12² + 35² = 144 + 1,225 = 1,369 = 37², so by the converse it is a right triangle. Choice A only shows that the sides can make a triangle. Choice B adds the sides without squaring them. Choice D tests 35 as c, but c must be the longest side.
Question 9 of 20 · Multiple Choice
A triangle has sides 10, 13 and 16. Which answer uses the converse correctly?
Answer: A
100 + 169 = 269, which is not 256, so the triangle has no right angle: if it did, the theorem would make the two numbers equal. Choice B only shows the sides make a triangle. Choice C is false: 256 - 169 = 87, not 100. Choice D reaches the right conclusion with the wrong test, using 13 as c instead of the longest side, 16.
Question 10 of 20 · Multiple Choice
In the proof of the converse, a triangle has sides a, b and c with a² + b² = c². What do you build first?
Answer: C
The proof builds a right triangle with legs a and b, then shows that its hypotenuse is c. Choice A is not a triangle at all, since a + b is exactly the sum of the other two sides. Choice B belongs to the proof of the theorem, not the converse. Choice D uses the wrong legs, so its hypotenuse would not be c.
Question 11 of 20 · Multiple Choice
In the proof of the converse, why is the hypotenuse of the new right triangle equal to c?
Answer: A
The new triangle is a right triangle, so the Pythagorean Theorem gives hypotenuse² = a² + b². We know a² + b² = c², so the hypotenuse is c. Choice B is false: the hypotenuse is shorter than the two legs together. Choice C is a measurement, not a proof. Choice D is false: the third side depends on the angle between a and b.
Question 12 of 20 · Multiple Choice
In the proof of the converse, why can you conclude that the original triangle has a right angle?
Answer: D
Three side lengths make only one triangle shape, so the original triangle is a copy of the new right triangle, and the angle across from c is 90°. Choice A is false: most triangles have no right angle. Choice B is false: a 180° sum does not force a 90° angle, as a triangle with angles 60°, 60° and 60° shows. Choice C is true for many triangles that are not right triangles.
Question 13 of 20 · Multiple Choice
Which set of side lengths makes a right triangle?
Answer: B
10² + 24² = 100 + 576 = 676 = 26². Choice A adds the lengths, 5 + 12 = 17, instead of the squares, and those lengths cannot even make a triangle. Choice C gives 36 + 81 = 117, not 144. Choice D gives 49 + 64 = 113, not 100.
Question 14 of 20 · Multiple Choice
A window frame is 36 in wide and 48 in tall. A carpenter measures the diagonal and gets 61 in. Is the corner square (90°)?
Answer: A
For a square corner, the diagonal d would have d² = 36² + 48² = 1,296 + 2,304 = 3,600, so d = 60 in. Since 61² = 3,721 is not 3,600, the converse test fails and the corner is not square. Choice B only shows the lengths make a triangle. Choice C ignores the fact that a 1 in difference means a corner that is not square. Choice D adds 36 and 48.
Question 15 of 20 · Short Answer
Follow the rearrangement proof for a right triangle with legs 20 and 21. Find the area of the big square, the four triangles and the uncovered square in Arrangement 1, and the length of the hypotenuse.
Big square: (20 + 21)² = 41² = 1,681. Four triangles: 4 × ½ × 20 × 21 = 840. Uncovered square: 1,681 - 840 = 841, which equals 20² + 21² = 400 + 441. The hypotenuse is 29, since 29² = 841.
Question 16 of 20 · Short Answer
On grid paper, a right triangle has legs 1 and 5 units along the grid lines. Use the grid method to find the area of the square on its hypotenuse, and explain how your answer fits the Pythagorean Theorem.
Box the tilted square in a 6 by 6 square with area 36. Each corner triangle has area ½ × 1 × 5 = 2.5, so the tilted square has area 36 - 4 × 2.5 = 26. That equals 1² + 5² = 1 + 25, as the theorem says.
Question 17 of 20 · Short Answer
A triangle has sides 15, 36 and 39. Use the converse to decide whether it is a right triangle, and say where the right angle is.
15² + 36² = 225 + 1,296 = 1,521, and 39² = 1,521. They are equal, so it is a right triangle. The right angle is across from the longest side, 39, between the sides of length 15 and 36.
Question 18 of 20 · Short Answer
A triangle has sides 7, 9 and 12. Is it a right triangle? Explain using the Pythagorean Theorem.
No. 7² + 9² = 49 + 81 = 130, but 12² = 144. If the triangle had a right angle, its longest side would be the hypotenuse, and the theorem would force 7² + 9² = 12². The numbers are not equal, so it has no right angle.
Question 19 of 20 · Short Answer
Explain the rearrangement proof of the Pythagorean Theorem in your own words, using the letters a, b and c.
Take four copies of a right triangle with legs a and b and hypotenuse c. In a square with side a + b, place them in the corners: the middle is a square with side c, so the leftover area is c². In a second square with side a + b, place them as two rectangles: the leftover is a square with side a and a square with side b, with area a² + b². Both big squares are equal and hold the same four triangles, so the leftovers are equal: a² + b² = c². Because the letters stand for any right triangle, this proves it for all of them.
Question 20 of 20 · Short Answer
A triangle has sides 2 m, 4.8 m and 5.2 m. Explain, step by step as in the proof of the converse, why it must have a right angle.
First, 2² + 4.8² = 4 + 23.04 = 27.04 = 5.2². Build a right triangle with legs 2 m and 4.8 m. By the theorem, its hypotenuse squared is 27.04, so its hypotenuse is 5.2 m. The two triangles have the same three sides, 2 m, 4.8 m and 5.2 m, so they are copies, and the angle across from the 5.2 m side is 90°.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does 8.G.B.6 mean?
8.G.B.6 means students can explain why the Pythagorean Theorem is true and why its converse is true. They follow a proof step by step and give the reason for each step, instead of only using the formula a² + b² = c². Applying the theorem to find lengths comes next, in 8.G.B.7 and 8.G.B.8.
What is the converse of the Pythagorean Theorem?
It says that if the sides of a triangle fit a² + b² = c², with c the longest side, then the triangle has a right angle. The theorem goes from a right angle to the equation; the converse goes from the equation to a right angle. Builders use it to check that corners are square.
Which proof of the Pythagorean Theorem do grade 8 students learn?
The standard does not name one, so teachers choose. Many classes use a rearrangement proof like the one on this page, because students can cut out the pieces and see it. Other common choices are the grid method with squares on the sides, a proof with similar triangles (triangles with the same shape, possibly different sizes), and a trapezoid proof published in 1876 by James Garfield.
Why isn't checking examples a proof?
Because examples only show the fact for the triangles someone checked, and there are infinitely many right triangles. A proof uses letters such as a, b and c that stand for any right triangle, so its argument covers every case at once. Examples are still useful to test a proof and to see why it works.
Why does the rearrangement proof need a right angle?
The right angle makes the middle figure a square. The two acute angles of a right triangle add up to 90°, so the corner between them in Arrangement 1 is 90°. For a triangle with no right angle, the middle figure would not be a square, and its area would not be c².
How do you prove the converse of the Pythagorean Theorem?
Start with a triangle whose sides fit a² + b² = c². Build a right triangle with legs a and b; by the theorem its hypotenuse is c. The two triangles have the same three sides, so they are copies of each other, and the angle across from c in the first triangle is also a right angle.
Do students have to invent a proof for 8.G.B.6?
No. The standard says "explain a proof", so students must understand one proof well enough to explain each step and why it is true. Many teachers still ask students to rebuild the proof with cutouts, which helps them remember the steps. Writing an original proof is not required.
What mistakes do students make with 8.G.B.6?
A common mistake is testing the converse with the wrong side as c; c must be the longest side. Others add the side lengths instead of their squares, confuse the theorem with its converse, or think that checking a few triangles proves the theorem. Some forget the ½ when finding the area of the triangles in the proof.
How does 8.G.B.6 connect to 8.G.B.7 and 8.G.B.8?
8.G.B.6 explains why the theorem is true. 8.G.B.7 uses it to find unknown side lengths in two and three dimensions, and 8.G.B.8 uses it to find the distance between two points on a coordinate grid. In high school, HSG.SRT.B.4 proves the theorem again with similar triangles.
How can parents help with 8.G.B.6 at home?
Cut four copies of a right triangle from cereal-box cardboard, and ask your child to arrange them in a square frame in two ways and explain why the leftover areas match. You can also measure a corner of a room with a tape measure: marks at 30 cm and 40 cm should be 50 cm apart if the corner is square.
07
Related Standards
6 standards
These standards connect to 8.G.B.6: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
6.G.A.1Prerequisite
Find the area of triangles and other polygons by composing and decomposing shapes
