Mathematics • Year 8 • Unit 3 • Lesson 1

Pythagoras' Theorem

Build fluency with c² = a² + b² for finding the hypotenuse. One fully worked example, one guided example with blanks, then eight independent problems ramping from clean triples to decimal answers.

Build · I Do / We Do / You Do

1. I do — fully worked example

Read every line. Each step has a short reason so you can see why, not just what.

Problem. A right-angled triangle has legs a = 3 cm and b = 4 cm. Find the hypotenuse c.

Step 1 — Identify the hypotenuse.

c is opposite the right angle (the longest side).

Reason: the theorem only works when c is the hypotenuse, never a leg.

Step 2 — Write the formula.

c² = a² + b²

Reason: this is the theorem. Always write it first so you don't confuse the operations.

Step 3 — Substitute and calculate.

c² = 3² + 4² = 9 + 16 = 25

Reason: square each leg first, then add. Never add the legs before squaring.

Step 4 — Take the square root.

c = √25 = 5 cm

Reason: c² = 25 means c = √25. The square root is always the LAST step.

Answer: c = 5 cm (a 3-4-5 Pythagorean triple).

Stuck? Revisit lesson § Card 7 — the four steps: Identify → Square → Add → Square root.

2. We do — fill in the missing steps

Same shape as Section 1, with the working faded. Fill in each blank. 4 marks

Problem. A right-angled triangle has legs a = 5 cm and b = 12 cm. Find the hypotenuse c.

Step 1 — Hypotenuse: c is the side opposite the ______ angle.

Step 2 — Write the formula:

c² = ______ + ______

Step 3 — Substitute and calculate:

c² = 5² + 12² = ______ + ______ = ______

Step 4 — Take the square root:

c = √______ = ______ cm

Stuck? Revisit lesson § Card 8 — 5-12-13 is one of the three core Pythagorean triples.

3. You do — independent practice

Show all working. The first four are foundation (clean triples). The middle two are standard (decimal or scaled triples). The last two are extension (apply the theorem in context).

Foundation — clean Pythagorean triples

3.1 a = 6 cm, b = 8 cm. Find c. 1 mark

3.2 a = 8 cm, b = 15 cm. Find c. 1 mark

3.3 a = 7 cm, b = 24 cm. Find c. 1 mark

3.4 a = 9 cm, b = 12 cm. Find c. (Hint: 9-12-15 is a ×3 multiple of the 3-4-5 triple.) 1 mark

Standard — decimals and scaled triples

3.5 a = 10 cm, b = 10 cm. Find c to 1 decimal place. 2 marks

3.6 a = 1.2 m, b = 0.9 m. Find c. (Hint: this is a 3-4-5 triple scaled by 0.3.) 2 marks

Extension — apply to rectangles and ladders

3.7 A rectangle is 24 cm long and 7 cm wide. Find the length of its diagonal. (Hint: the diagonal cuts the rectangle into a right triangle with legs 24 and 7.) 2 marks

3.8 A right-angled triangle has legs 11 cm and 60 cm. Find the hypotenuse, then state whether 11-60-61 is a Pythagorean triple. 2 marks

Stuck on 3.7 / 3.8? The diagonal/hypotenuse is always the LONGEST side. Calculate c² = a² + b², then square root.

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Answers — Do not peek before attempting

Section 2 — We do (faded 5-12-?)

Step 1: opposite the right angle.
Step 2: c² = a² + b².
Step 3: c² = 5² + 12² = 25 + 144 = 169.
Step 4: c = √169 = 13 cm.

3.1 — a = 6, b = 8

c² = 36 + 64 = 100, so c = √100 = 10 cm. (6-8-10 is a ×2 multiple of 3-4-5.)

3.2 — a = 8, b = 15

c² = 64 + 225 = 289, so c = √289 = 17 cm. (8-15-17 is a core triple.)

3.3 — a = 7, b = 24

c² = 49 + 576 = 625, so c = √625 = 25 cm. (7-24-25 is a triple.)

3.4 — a = 9, b = 12

c² = 81 + 144 = 225, so c = √225 = 15 cm. Confirmed: 9-12-15 = 3-4-5 × 3.

3.5 — a = 10, b = 10

c² = 100 + 100 = 200, so c = √200 ≈ 14.1 cm (to 1 d.p.).

3.6 — a = 1.2, b = 0.9

c² = 1.44 + 0.81 = 2.25, so c = √2.25 = 1.5 m. (3-4-5 × 0.3 = 0.9-1.2-1.5.)

3.7 — Rectangle 24 × 7

Diagonal: c² = 24² + 7² = 576 + 49 = 625, so diagonal = 25 cm. (7-24-25 triple again.)

3.8 — Legs 11 and 60

c² = 121 + 3600 = 3721, so c = √3721 = 61 cm. Yes — 11-60-61 IS a Pythagorean triple because 11² + 60² = 61² exactly.