Pythagorean Theorem Lesson Plan (Two Days)

Two ready-to-teach 50-minute lessons. On day 1 students discover a2 + b2 = c2 by counting squares and see why it is true; on day 2 they find missing legs, test the converse and measure distances on a grid.

By the PythagoreanTheorem.net teamUpdated 11 min read

This Pythagorean theorem lesson plan covers two 50-minute periods for grade 8. On day 1, students count squares on grid paper, discover that a2 + b2 = c2 in right triangles and follow a rearrangement proof. On day 2, they find missing legs, use the converse and calculate distances on the coordinate plane. Every task below includes its answers.

The plan follows the progression in the teaching guide: discover, then prove, then practice. The day 1 warm-up checks the prerequisites.

Lesson overview

Grade and length
Grade 8 (also works as a review in Algebra 1 or Geometry). Two 50-minute lessons.

Standards (Common Core, grade 8 geometry)

  • 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.
  • Practices: MP3 (construct viable arguments) and MP7 (look for and make use of structure).

Objectives. Students will be able to:

  1. Identify the legs and the hypotenuse of a right triangle in any position.
  2. Explain with areas why a2 + b2 = c2 in every right triangle.
  3. Find a missing side, giving the exact answer and a rounded decimal.
  4. Use the converse to decide whether a triangle is right.
  5. Find the distance between two points on the coordinate plane.

Vocabulary: right angle, leg, hypotenuse, square of a number, square root, converse.

Materials: 1 cm grid paper or the handout below, rulers, blue, orange and green pencils, card and scissors (four congruent right triangles per pair, or one teacher set), calculators, mini whiteboards. For the proof you can also project the animation on the rearrangement proof page.

Timing at a glance

Day 1: discover and proveMinDay 2: applyMin
Warm-up: squares and roots5Warm-up: fix the exit-ticket errors5
Launch: legs and hypotenuse5Finding a missing leg7
Discovery with grid squares15Converse investigation8
Share data, write the conjecture5Distance on the coordinate plane8
Rearrangement proof8Guided practice15
Guided practice8Exit ticket5
Exit ticket4Homework launch2
Total50Total50

Day 1: discover the pattern and prove it

Warm-up (5 minutes)

Students answer on mini whiteboards: 62, 122, √81, "between which two whole numbers is √50?" and "what is the area of a right triangle with legs 3 and 4?"

Answers

36; 144; 9; between 7 and 8 (it is about 7.07); 6 square units. If several students write 62 = 12, spend two extra minutes on what squaring means before moving on. Students who struggle can keep the square roots page open as a reference.

Launch (5 minutes)

Draw a right triangle and name its parts: the two legs form the right angle and the hypotenuse is opposite it. Then project three triangles turned in different directions and ask students to trace each hypotenuse on their whiteboards. The rule to repeat: find the right angle first; the hypotenuse is the side that doesn't touch it. More on this on the parts of a right triangle page.

Pose the question for the day: "If you know two sides of a right triangle, is there a rule that gives you the third?"

Discovery with grid squares (15 minutes)

916?abc2566667 × 7 − 4 × 6 = 25surround the tilted square, subtract the corners9 + 16 = 25
The handout. The squares on the legs can be counted directly. The tilted square on the hypotenuse is found by surrounding it with a 7 × 7 square and subtracting the four corner triangles.

In pairs, students draw right triangles on grid paper with the legs listed in the table, then draw a square outward on each side. They shade the leg squares blue and orange and the hypotenuse square green.

  1. Count the leg squares. These sit on the grid lines, so counting is easy.
  2. Find the tilted square. Students will try to count part-squares and get stuck. Let them try for a minute, then show the trick: draw the smallest grid-aligned square around it and subtract the four corner triangles.
  3. Record the results and look for a pattern before anyone says it out loud.
LegsSquare on aSquare on bSquare on cHow the c square was found
1 and 11122 × 2 − 4 × ½ = 2
1 and 21453 × 3 − 4 × 1 = 5
2 and 24484 × 4 − 4 × 2 = 8
1 and 319104 × 4 − 4 × 1.5 = 10
2 and 349135 × 5 − 4 × 3 = 13
3 and 4916257 × 7 − 4 × 6 = 25

Early finishers: draw the triangle with corners at (0, 0), (4, 0) and (−2, 3), which has an obtuse angle at (0, 0). Its squares have areas 16, 13 and 45, and 16 + 13 = 29, not 45. That shows the pattern needs the right angle, and sets up day 2's converse.

Share the data and write the conjecture (5 minutes)

Collect a few rows on the board and ask what students notice. Push for a sentence before any symbols: "In a right triangle, the square on the hypotenuse has the same area as the two squares on the legs put together." Then write it as a2 + b2 = c2 and give it its name. Ask the class to predict, without drawing, the hypotenuse square for legs 5 and 12 (169, so the hypotenuse is 13).

Why it is always true: the rearrangement proof (8 minutes)

c²a²b²=uncovered: c²uncovered: a² + b²
The same four triangles inside the same square. Whatever is left uncovered must have the same area in both pictures, so c2 = a2 + b2.

Six examples are evidence, not proof. Show two equal squares of side a + b. In the first, place the four triangles so they leave a tilted square of side c uncovered. In the second, slide them into two rectangles, leaving a square of side a and a square of side b. Ask: "What stayed the same? What does that tell us?" Students should say that the frame and the triangles are identical, so the uncovered areas must be equal. The hands-on version is activity 1 on the activities page.

Guided practice (8 minutes)

Find the hypotenuse. Give exact answers and round to the nearest hundredth where needed.

  1. Legs 6 and 8
  2. Legs 5 and 12
  3. Legs 4 and 7
  4. Legs 5 and 5
  5. A TV screen is 32 inches wide and 18 inches tall. How long is its diagonal?
Answers
  1. c = √36 + 64 = √100 = 10
  2. c = √25 + 144 = √169 = 13
  3. c = √16 + 49 = √65 ≈ 8.06
  4. c = √25 + 25 = √50 = 5√2 ≈ 7.07
  5. √1024 + 324 = √1348 ≈ 36.72 inches

Exit ticket (4 minutes)

  1. In triangle PQR the right angle is at Q. Which side is the hypotenuse?
  2. The legs of a right triangle are 9 and 12. Find the hypotenuse.
  3. The legs are 2 and 5. Find the hypotenuse, exact and to the nearest hundredth.
  4. Sam says the hypotenuse of a triangle with legs 3 and 4 is 7. Explain his mistake.
Answers
  1. PR, the side that does not touch Q.
  2. √81 + 144 = √225 = 15
  3. √4 + 25 = √29 ≈ 5.39
  4. He added the lengths instead of their squares. 32 + 42 = 25, so the hypotenuse is √25 = 5.

Sort the tickets into three piles: all correct, setup errors (items 1 and 4) and arithmetic errors. The setup pile is your day 2 warm-up.

Day 2: missing legs, the converse and distance

Warm-up: fix the errors (5 minutes)

Project two anonymous wrong answers from yesterday's tickets, such as "c = 3 + 4 = 7" and "c = 29" for legs 2 and 5. Students identify each mistake and correct it with a partner.

Finding a missing leg (7 minutes)

Model one example and insist on the same setup every time: write the theorem, put the hypotenuse alone, substitute, then solve. A ladder 13 ft long leans against a wall with its foot 5 ft from the wall. How high does it reach?

c2= a2 + b2132= 52 + b2169= 25 + b2b2= 144b= 12

The ladder reaches 12 ft up the wall. Point out the check: a leg must be shorter than the hypotenuse, and 12 is shorter than 13. If a student gets √194, they added. The missing side lesson has more examples.

Converse investigation (8 minutes)

Give each group three sets of lengths: 6, 8, 10; 5, 6, 8; and 4, 5, 6. For each, they compare the sum of the squares of the two shorter sides with the square of the longest. If you have straws or strips of card, let them build the triangles and look at the largest angle.

Results
  • 6, 8, 10: 36 + 64 = 100 = 102, a right triangle.
  • 5, 6, 8: 25 + 36 = 61, less than 82 = 64, an obtuse triangle.
  • 4, 5, 6: 16 + 25 = 41, more than 62 = 36, an acute triangle.

State the converse: if a2 + b2 = c2, the triangle has a right angle opposite c. Mention that builders use exactly this to check that a corner is square. The comparison test for acute and obtuse triangles is a natural extension for fast groups.

Distance on the coordinate plane (8 minutes)

246810246810(1, 2)(7, 10)7 − 1 = 610 − 2 = 8dd² = 6² + 8²d² = 36 + 64 = 100d = √100 = 10
Any slanted segment on a grid is the hypotenuse of a right triangle whose legs are the differences of the coordinates.

Plot (1, 2) and (7, 10) and ask how far apart they are. Students can't count along a slanted line, so draw the right triangle underneath it. The horizontal leg is 7 − 1 = 6, the vertical leg is 10 − 2 = 8, and the distance is √62 + 82 = 10. Hold off on the symbolic distance formula until students have done a few by drawing the triangle; then show that the formula is the same three steps written with letters.

Guided practice (15 minutes)

  1. Hypotenuse 25, one leg 7. Find the other leg.
  2. Hypotenuse 10, one leg 6. Find the other leg.
  3. Is a triangle with sides 20, 21 and 29 a right triangle?
  4. Is a triangle with sides 6, 7 and 9 a right triangle? If not, is it acute or obtuse?
  5. Find the distance between (−3, 1) and (5, 7).
  6. Find the distance between (2, −1) and (5, 3).
  7. A rectangular park is 120 m by 50 m. How much shorter is the diagonal path than walking along two sides?
  8. Extension (3D): a box is 3 cm by 4 cm by 12 cm. Find the length of the longest straight rod that fits inside.
Answers
  1. √625 − 49 = √576 = 24
  2. √100 − 36 = √64 = 8
  3. Yes: 400 + 441 = 841 = 292.
  4. No: 36 + 49 = 85 but 92 = 81. Since 85 > 81, it is acute.
  5. Legs 8 and 6, distance √64 + 36 = 10.
  6. Legs 3 and 4, distance 5.
  7. The diagonal is √14,400 + 2,500 = √16,900 = 130 m; walking two sides is 170 m, so the path is 40 m shorter.
  8. The base diagonal is √9 + 16 = 5, then the rod is √25 + 144 = 13 cm. The 3D worksheet has twelve more problems like this.

Exit ticket (5 minutes)

  1. A right triangle has hypotenuse 17 and a leg of 8. Find the other leg.
  2. Do the lengths 9, 12 and 16 form a right triangle? Show how you know.
  3. Find the distance between (−2, 1) and (4, 9).
Answers
  1. √289 − 64 = √225 = 15
  2. No: 81 + 144 = 225, but 162 = 256. (Since 225 < 256, the triangle is obtuse.)
  3. Legs 6 and 8, so the distance is √36 + 64 = 10.

Homework

Eight problems that mix both days. Round to the nearest hundredth where needed.

  1. Legs 7 and 24. Find the hypotenuse.
  2. Legs 3 and 6. Find the hypotenuse in simplest radical form.
  3. Hypotenuse 15, leg 9. Find the other leg.
  4. Hypotenuse 8, leg 5. Find the other leg.
  5. Is 11, 60, 61 a right triangle?
  6. Find the distance between (−4, −2) and (1, 10).
  7. A 10 ft ladder reaches 8 ft up a wall. How far is its foot from the wall?
  8. Find the longest diagonal of a 2 × 3 × 6 box.
Answers
  1. √49 + 576 = √625 = 25
  2. √9 + 36 = √45 = 3√5 ≈ 6.71
  3. √225 − 81 = √144 = 12
  4. √64 − 25 = √39 ≈ 6.24
  5. Yes: 121 + 3,600 = 3,721 = 612.
  6. Legs 5 and 12, distance 13.
  7. √100 − 64 = √36 = 6 ft
  8. √4 + 9 + 36 = √49 = 7

For extra practice, assign one of the printable sets: the hypotenuse worksheet after day 1, and the missing side, converse or distance formula worksheets after day 2. Students can check their own answers with the step-by-step calculator.

Differentiation

Support

  • Hand out the discovery triangles already drawn, so the time goes into finding areas, not drawing.
  • Give a table of squares from 12 to 252.
  • Use only whole-number answers in the first practice set, then add radicals.
  • Pair roles: one student counts, the other records and checks.
  • Allow calculators for arithmetic; grade the setup.

Extension

  • Prove the converse: build a right triangle with legs a and b and compare it with the original (SSS).
  • Find a tilted square of area 10 on the grid. Is there one with area 3?
  • Find every right triangle with whole-number sides and a hypotenuse up to 30.
  • Replace the rearrangement proof with the similar triangles proof for geometry students.

Frequently asked questions

Can I teach this Pythagorean theorem lesson in a single day?

Day 1 works on its own as an introduction to the theorem. Day 2 then becomes the next lesson, or you can split it: missing legs and distance in one lesson, the converse in another. Trying to fit both days into one period usually means cutting the discovery, which is the part students remember.

How do I adapt the plan for a 90-minute block?

Teach day 1 in full, use its exit ticket as a mid-block check, then run day 2 from the missing-leg example onward. Keep the converse investigation and drop two or three guided practice items. Leave the day 2 exit ticket for the start of the next class.

Does this lesson plan meet the grade 8 standards?

Yes. Day 1 covers explaining a proof (8.G.B.6), day 2 covers the converse (8.G.B.6), missing sides in 2D and 3D (8.G.B.7) and distance between points (8.G.B.8). The 8th grade page explains each standard in detail.

How can I check what students learned after the two days?

Give the online quiz as a self-marking check, or collect the homework and look for the same three error types as on the exit tickets: wrong hypotenuse, adding instead of subtracting, and a missing square root.

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