LESSON PLAN

The Pythagorean Theorem

Apothem Team

Grade 8 · Geometry

LESSON AT A GLANCE

Warm-up

5 min

Explore

15 min

Formalize

10 min

Practice

12 min

Exit ticket

3 min

Warm-up

Hand every pair a loop of string knotted into 12 equal segments — the ancient Egyptian rope. Challenge: "Pull it into a triangle with sides 3, 4, and 5 segments. What do you notice about the biggest angle?" A perfect right angle appears, no protractor involved.

Rope-stretchers squared the pyramids' corners this way four millennia ago. The question of the unit: WHY does 3-4-5 force a right angle — and what's the general law hiding behind it?

Explore

Square-counting discovery: on grid paper, draw a right triangle with legs 3 and 4, then build a literal square on each side (the 5-side square drawn tilted, its area found by surrounding-square-minus-triangles: $49 - 4 \times 6 = 25$ ). Tally: $9 + 16 = 25$ — the leg-squares EXACTLY fill the hypotenuse-square.

Test a second triangle (legs 6, 8: $36 + 64 = 100 = 10^2$ ✓) and a NON-right triangle (the areas refuse to balance). Then one visual proof for conviction: the two 7×7 rearrangement squares — four copies of the triangle arranged two ways leave $a^2 + b^2$ in one square and $c^2$ in the other. Same four triangles, same big square: the leftovers must match.

Formalize

Formalize the theorem and its converse — both directions earn their keep:

$a^2 + b^2 = c^2 \quad \text{(right triangle, } c \text{ the hypotenuse)} \qquad \text{converse: if } a^2+b^2=c^2 \text{, the angle is right}$

The hypotenuse is the side OPPOSITE the right angle — always the longest, never a leg. Finding a leg rearranges: $b^2 = c^2 - a^2$ (subtract, students who always add are answering a different triangle). Non-perfect-square answers stay exact as roots ( $\sqrt{41}$ ) or round with the ≈ flag.

Practice

Practice: two find-the-hypotenuse, two find-a-leg, one converse test (is 5-12-13 right? is 6-7-9?), one applied diagonal (does a 2.4 m board fit through a 2.1 × 0.9 m door?).

Exit ticket: a ladder leans 2 m from a wall, reaching 6 m up. Ladder length, exact and rounded. ( $\sqrt{40} = 2\sqrt{10} \approx 6.3$ m.)

Exit ticket

Practice: two find-the-hypotenuse, two find-a-leg, one converse test (is 5-12-13 right? is 6-7-9?), one applied diagonal (does a 2.4 m board fit through a 2.1 × 0.9 m door?).

Exit ticket: a ladder leans 2 m from a wall, reaching 6 m up. Ladder length, exact and rounded. ( $\sqrt{40} = 2\sqrt{10} \approx 6.3$ m.)

TIP The theorem is about AREAS of squares, not a letter-chant. Students who've counted the 25 squares themselves never write

a + b = c

; students who met it as "a-squared plus b-squared" sometimes do.

WORKED EXAMPLES

Example 1 — The classic ladder: find the hypotenuse

The scene: a ladder's foot sits 1.5 m from the wall; its top touches 3.6 m up. How long is the ladder?

Step 1: Certify the right angle: wall meets ground squarely — licence granted.

Step 2: Identify parts: legs 1.5 and 3.6; the ladder is the hypotenuse.

Step 3: Apply: $c^2 = 1.5^2 + 3.6^2 = 2.25 + 12.96 = 15.21$ .

Step 4: Root: $c = \sqrt{15.21} = 3.9$ m (perfect: $3.9^2 = 15.21$ ).

Step 5: Sanity: the hypotenuse (3.9) exceeds both legs but is less than their sum (5.1) ✓ — both bounds every right triangle must respect.

Example 2 — Find the leg, resist the add: the zipline anchor

The setup: a 25 m zipline runs from a platform to an anchor 20 m away horizontally. How high is the platform?

Step 1: Cast the parts: the zipline is the HYPOTENUSE (25); the ground run is a leg (20); the height is the missing LEG.

Step 2: Rearrange before computing: $h^2 = 25^2 - 20^2 = 625 - 400 = 225$ .

Step 3: Root: $h = 15$ m.

Step 4: The error to inoculate: adding gives $\sqrt{1025} \approx 32$ — a "height" LONGER than the zipline itself, geometrically absurd. The pre-computation size-check (legs < hypotenuse) catches the wrong operation before the calculator even warms up.

Step 5: Pattern note: 15-20-25 is 3-4-5 scaled by 5 — Pythagorean triples come in families, and spotting the scale saves the whole computation.

Example 3 — The converse at work: is the deck square?

The job: a builder's rectangular deck frame measures 2.4 m by 3.2 m, and the diagonal tapes at 4.1 m. Is the corner square?

Step 1: What SHOULD the diagonal be if the corner is right? $d^2 = 2.4^2 + 3.2^2 = 5.76 + 10.24 = 16$ → $d = 4.0$ m exactly.

Step 2: Compare with reality: taped 4.1 m — one centimetre-decimetre… 10 cm too long. The corner is NOT square: the frame has splayed wider than 90°.

Step 3: The fix direction: pull the corners to shorten the diagonal toward 4.0 (a diagonal too long means the angle opened past 90°; too short means pinched).

Step 4: Name what was used: the CONVERSE — measuring all three sides to test the angle. Builders call it the 3-4-5 check and use it weekly; mathematics calls it the converse of Pythagoras and proved it once, for all decks, forever.

MATERIALS

•Knotted rope loops (12 segments)

•Grid paper

•Rearrangement proof puzzle pieces

•Rulers

•Practice set (PDF)

WATCH FOR

!The formula applied to non-right triangles. The theorem carries a licence condition: right angle required (the converse is the licence-checker).

!Hypotenuse misidentified as "the slanted-looking side." It's opposite the right angle — orientation is costume.

!Adding leg² and hypotenuse² when hunting a leg. Estimate first: a leg must be SHORTER than the hypotenuse; a too-long answer self-reports.