Mathematics Standard • Year 11 • Module 2 • Lesson 3

Pythagoras' Theorem

Build fluency in finding any side of a right-angled triangle — hypotenuse first, then shorter side — keeping working clean and exact.

Build · Skill Drill

1. Quick recall

Answer each question in the space provided. 1 mark each

Q1.1 Write Pythagoras' theorem in symbol form:

______ + ______ = ______ where c is the ____________ side.

Q1.2 Complete the rearrangements:

To find the HYPOTENUSE c: c = √( ______ + ______ )

To find a SHORTER side a: a = √( ______ − ______ )

Q1.3 Circle TWO Pythagorean triples (sets of three sides that form a right-angled triangle):

2 – 3 – 4 3 – 4 – 5 5 – 12 – 13 6 – 7 – 8 7 – 8 – 9

Stuck? Revisit lesson § The Theorem and § Common Pythagorean Triples diagram.

2. Worked example — finding the hypotenuse

Follow each line of working — every step earns a method mark.

Problem. A right-angled triangle has shorter sides 6 cm and 8 cm. Find the hypotenuse.

Step 1 — Decide: hypotenuse or shorter side?

Finding c (hypotenuse) → ADD the squares.

Reason: hypotenuse → add; shorter side → subtract. Commit before substituting.

Step 2 — Write the formula and substitute.

c² = a² + b² = 6² + 8² = 36 + 64 = 100

Reason: square each side first, then add.

Step 3 — Square root and state the answer.

c = √100 = 10 cm

Reason: 6-8-10 is a Pythagorean triple (a scaled 3-4-5). Always write the unit.

3. Faded example — finding a shorter side

A right-angled triangle has hypotenuse 15 m and one shorter side 9 m. Find the other shorter side. Fill in every blank. 3 marks

Step 1 — Decide: Finding a __________ side, so SUBTRACT the squares.

Step 2 — Rearranged form: a² = c² − b² = ______² − ______²

Step 3 — Square and subtract: a² = ______ − ______ = ______

Step 4 — Square root: a = √______ = ______ m

Sense check: the answer must be smaller than the hypotenuse 15 m ✓

Stuck? Revisit lesson § Worked Example 2 — Finding a Shorter Side. 9-12-15 is a scaled 3-4-5.

4. Graduated practice — find the missing side

Write the formula, substitute, then state the answer with the unit. Round to 2 d.p. unless answer is exact.

Foundation — clean triples (4 questions)

Q	Problem	Answer (with unit)
4.1 1	Find c: shorter sides 3 cm and 4 cm.
4.2 1	Find c: shorter sides 5 m and 12 m.
4.3 1	Find shorter side: hypotenuse 13 cm, other side 5 cm.
4.4 1	Find shorter side: hypotenuse 17 m, other side 8 m.

Standard — typical HSC difficulty (6 questions)

4.5 Find c to 2 d.p.: shorter sides 7 cm and 9 cm. 2 marks

4.6 Find c to 2 d.p.: shorter sides 1.5 m and 2 m. 2 marks

4.7 Find the shorter side to 2 d.p.: hypotenuse 10 cm, other side 4 cm. 2 marks

4.8 Find the shorter side: hypotenuse 7.5 m, other side 4.5 m. 2 marks

4.9 A 5 m ladder leans against a wall. The base is 2 m from the wall. How high up the wall does the ladder reach (to 2 d.p.)? 3 marks

4.10 A rectangular room is 9 m long and 5 m wide. Find the length of the diagonal of the floor to 2 d.p. 2 marks

Extension — slant height + multi-step (2 questions)

4.11 A cone has vertical height 12 cm and base radius 5 cm. Find the slant height to 2 d.p. 3 marks

4.12 A 4 m rectangular gate is 1.8 m tall. Find the length of a diagonal brace from corner to corner, then determine the cost of timber for the brace at $12.50 per metre, rounded up to the nearest 10 cm. 3 marks

Stuck on 4.11? Inside any cone there is a right triangle: r and h are the legs; the slant height is the hypotenuse.

5. Self-check the easy 3

Tick the first three once you've checked your method.

For 4.1 (3, 4) I ADDED the squares because I was finding the hypotenuse.

For 4.3 (hyp 13, side 5) I SUBTRACTED the squares because I was finding a shorter side, and my answer (12) is less than 13 ✓.

For 4.7 (hyp 10, side 4) my answer is less than 10 — if it were larger, I'd have added when I should have subtracted.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — Do not peek before attempting

Q1.1 — The theorem

a² + b² = c², where c is the hypotenuse (longest) side.

Q1.2 — Rearrangements

c = √(a² + b²); a = √(c² − b²).

Q1.3 — Pythagorean triples

3 – 4 – 5 and 5 – 12 – 13. Quick check: 3² + 4² = 25 = 5²; 5² + 12² = 169 = 13².

Q3 — Faded shorter-side example

Step 1: shorter side, so subtract. Step 2: a² = 15² − 9². Step 3: a² = 225 − 81 = 144. Step 4: a = √144 = 12 m.

Q4.1 — Hyp from 3 and 4

c² = 9 + 16 = 25, c = 5 cm.

Q4.2 — Hyp from 5 and 12

c² = 25 + 144 = 169, c = 13 m.

Q4.3 — Shorter, hyp 13, side 5

a² = 169 − 25 = 144, a = 12 cm.

Q4.4 — Shorter, hyp 17, side 8

a² = 289 − 64 = 225, a = 15 m.

Q4.5 — Hyp from 7 and 9

c² = 49 + 81 = 130, c = √130 = 11.40 cm (to 2 d.p.).

Q4.6 — Hyp from 1.5 and 2

c² = 2.25 + 4 = 6.25, c = √6.25 = 2.50 m.

Q4.7 — Shorter, hyp 10, side 4

a² = 100 − 16 = 84, a = √84 = 9.17 cm (to 2 d.p.).

Q4.8 — Shorter, hyp 7.5, side 4.5

a² = 56.25 − 20.25 = 36, a = 6.00 m.

Q4.9 — Ladder against a wall

Ladder = hypotenuse 5, ground side = 2. Wall height² = 25 − 4 = 21. Wall = √21 = 4.58 m (to 2 d.p.).

Q4.10 — Room diagonal

d² = 81 + 25 = 106, d = √106 = 10.30 m (to 2 d.p.).

Q4.11 — Cone slant height

ℓ² = r² + h² = 25 + 144 = 169. ℓ = 13.00 cm. (5-12-13 triple.)

Q4.12 — Gate diagonal and cost

d² = 4² + 1.8² = 16 + 3.24 = 19.24, d = √19.24 = 4.386... m. Round UP to nearest 10 cm → 4.4 m. Cost = 4.4 × $12.50 = $55.00. Marking note: rounding up (not down) is required because timber must reach corner-to-corner; rounding to 4.3 m would leave it short.