Mathematics • Year 8 • Unit 3 • Lesson 3

Pythagoras Applications

Build fluency with drawing a right triangle from a word problem. One worked navigation example, one guided rectangle-diagonal example with blanks, then eight ramped problems on navigation, diagonals, grid distances and rafters.

Build · I Do / We Do / You Do

1. I do — fully worked example

The four-step routine for ANY application: Draw → Label → Identify the right angle → Apply Pythagoras.

Problem. A ship sails 12 km north, then turns 90° and sails 9 km east. How far is the ship from its starting point?

Step 1 — Draw a diagram.

Sketch: north leg = 12 km, east leg = 9 km, with a right angle at the turn. Hypotenuse = direct distance from start.

Reason: north and east are perpendicular compass directions — they automatically form a right angle.

Step 2 — Label the sides.

a = 12 km (north), b = 9 km (east), c = ? (direct distance, hypotenuse)

Reason: identify which side is the hypotenuse before writing any numbers.

Step 3 — Apply Pythagoras.

c² = 12² + 9² = 144 + 81 = 225

Reason: legs are 12 and 9; we want the hypotenuse, so ADD.

Step 4 — Square root, state units.

c = √225 = 15 km

Reason: don't forget units (km), and recognise 9-12-15 = 3-4-5 × 3.

Answer: The ship is 15 km from its starting point.

Stuck? Revisit lesson § Card 5 — clue words like "direct distance", "diagonal" and "straight-line" all signal the hypotenuse.

2. We do — fill in the missing steps

A rectangle diagonal problem with the working faded. Fill in each blank. 4 marks

Problem. A rectangular garden is 5 m wide and 12 m long. Find the diagonal length from one corner to the opposite corner.

Step 1 — Draw: sketch the rectangle and draw the diagonal. The diagonal splits it into two right ______ triangles.

Step 2 — Label:

a = ______ m (width), b = ______ m (length), c = ? (diagonal)

Step 3 — Apply Pythagoras:

c² = ______² + ______² = ______ + ______ = ______

Step 4 — Square root and units:

c = √______ = ______ m (recognise the triple: ______-______-______ )

Stuck? Revisit lesson § Card 6 — d = √(l² + w²). 5-12-13 is a core Pythagorean triple.

3. You do — independent practice

Show all working — always draw a quick diagram first. The first four are foundation (one-step). The middle two are standard (grid distances and rafters). The last two are extension (multi-step and a longer scenario).

Foundation — direct application

3.1 A ship sails 5 km north then 12 km east. Find the direct distance from start.    1 mark

3.2 A rectangular pool is 8 m by 15 m. Find the length of the diagonal.    1 mark

3.3 A square has side length 6 cm. Find the diagonal to 1 decimal place.    1 mark

3.4 Find the grid distance from (0, 0) to (3, 4).    1 mark

Standard — grid points and roof rafters

3.5 Find the grid distance from (1, 2) to (7, 10). (Hint: Δx = 6, Δy = 8.)    2 marks

3.6 A roof is 7 m wide and the ridge is 2.4 m above the centre. Find the rafter length (the slant side) to 2 decimal places. (Hint: half-width = 3.5 m.)    2 marks

Extension — multi-step

3.7 A walker goes 3 km east, then 4 km north, then 6 km east. How far (in a straight line) is she from her starting point? (Hint: the total east is 3 + 6 = 9 km; the total north is 4 km.)    2 marks

3.8 A rectangular paddock is 60 m wide and 80 m long. A drone flies straight from one corner to the opposite corner, then back along two sides (one full length plus one full width). What is the total distance flown?    2 marks

Stuck on 3.8? Diagonal: 60² + 80² = 10 000, so diagonal = 100 m. Then add the two side distances on top.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

Section 2 — We do (faded 5×12 rectangle)

Step 1: right angled triangles.
Step 2: a = 5 m, b = 12 m.
Step 3: c² = 5² + 12² = 25 + 144 = 169.
Step 4: c = √169 = 13 m. Triple: 5-12-13.

3.1 — Ship 5 km N then 12 km E

c² = 5² + 12² = 25 + 144 = 169, so direct distance = 13 km. (5-12-13 triple.)

3.2 — Pool 8 × 15

c² = 8² + 15² = 64 + 225 = 289, so diagonal = 17 m. (8-15-17 triple.)

3.3 — Square side 6

c² = 6² + 6² = 36 + 36 = 72, so c = √72 ≈ 8.5 cm (to 1 d.p.).

3.4 — (0,0) to (3,4)

d² = 3² + 4² = 9 + 16 = 25, so d = 5 units. (3-4-5 triple.)

3.5 — (1,2) to (7,10)

Δx = 7 − 1 = 6; Δy = 10 − 2 = 8. d² = 6² + 8² = 36 + 64 = 100, so d = 10 units. (6-8-10 = 3-4-5 × 2.)

3.6 — Roof rafter

c² = 3.5² + 2.4² = 12.25 + 5.76 = 18.01, so rafter ≈ 4.24 m (to 2 d.p.).

3.7 — Walker (3 E, 4 N, 6 E)

Total east = 3 + 6 = 9 km; total north = 4 km. d² = 9² + 4² = 81 + 16 = 97, so d = √97 ≈ 9.85 km.

3.8 — Drone over 60 × 80 paddock

Diagonal: c² = 60² + 80² = 3600 + 6400 = 10 000, so diagonal = 100 m. Return along two sides = 60 + 80 = 140 m. Total distance flown = 100 + 140 = 240 m.