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Lesson 3 ~25 min Unit 3 · Measurement & Geometry +85 XP

Pythagoras Applications

Draw a diagram, spot the right angle, and solve any real-world problem.

Today's hook: A ship sails 12 km north then 9 km east. How far is it from the starting point? The path creates a right angle — Pythagoras has the answer.
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Think First
warm-up

A rectangular paddock is 60 m long and 80 m wide. A farmer wants to walk diagonally across it from one corner to the opposite corner. Without calculating, estimate the distance. Is it more or less than 100 m?

Record your answer in your workbook.
1
The Big Idea
+5 XP

Any real-world problem that contains a right angle can be solved with Pythagoras. The key skill is finding and drawing the right triangle hidden inside the problem.

Step 1: Draw a diagram. Step 2: Mark all known lengths. Step 3: Identify the right angle and the unknown side. Step 4: Apply $c^2 = a^2 + b^2$ or $a = \sqrt{c^2-b^2}$. Key clue words: vertical, horizontal, diagonal, direct distance, straight-line.

12 km N 9 km E d=?
Draw first → identify the right angle → apply Pythagoras
Always draw
Sketch the scenario before writing any numbers — it reveals the right triangle.
Label everything
Mark known sides and use "?" for the unknown. Makes substitution easy.
Units matter
State units in your final answer: cm, m, km. Don't leave them off.
2
What You'll Master
objectives

Know

  • Key clue words that signal a right triangle in word problems
  • The diagonal formula: $d = \sqrt{l^2 + w^2}$
  • How to find grid distances using horizontal and vertical legs

Understand

  • Why drawing a diagram is the most important first step
  • How to extract the right triangle from a real-world scenario

Can Do

  • Solve navigation, rectangle, grid and roof-rafter problems
  • Handle multi-step problems that require intermediate calculations
  • State answers with appropriate units and rounding
3
Words You Need
vocabulary
Direct distanceThe straight-line (shortest) distance between two points — often the hypotenuse.
DiagonalA line joining opposite corners of a rectangle or polygon — always the hypotenuse of the right triangle it creates.
Grid distanceDistance found using the horizontal and vertical gaps between two points on a coordinate grid.
RafterA sloping roof beam — its length is the hypotenuse of the triangle formed by the roof height and half-width.
Multi-step problemA problem requiring two or more separate calculations before reaching the final answer.
Compass directionNorth, South, East, West — perpendicular directions that form the legs of navigation triangles.
4
Spot the Trap
heads-up

Wrong: Adding the two travel distances: $12 + 9 = 21$ km direct distance.

Right: $d = \sqrt{12^2 + 9^2} = \sqrt{225} = 15$ km — the path, not the direct distance, adds to 21.

Wrong: Using the slant side as both a leg and a hypotenuse in roof problems.

Right: The rafter (slant) is always the hypotenuse; the height and half-width are the legs.

5
Extracting Right Triangles from Word Problems
+5 XP

Most applications hide a right triangle. Use these clue words to spot it: vertical, horizontal, diagonal, direct distance, shortest path, straight-line distance.

Common setups:

  • Navigation: N/S and E/W legs form a right angle at the turn.
  • Rectangle diagonal: Length and width are legs; diagonal is hypotenuse.
  • Grid points: Horizontal and vertical gaps are legs.
  • Roof rafters: Height above ridge and half-span are legs.
length (b) width (a) diagonal = c
$$d = \sqrt{l^2 + w^2}$$
6
Diagonal of a Rectangle
+5 XP

A diagonal splits a rectangle into two congruent right-angled triangles. The length and width become the legs; the diagonal is the hypotenuse.

Formula: $d = \sqrt{l^2 + w^2}$

Example: A 5 m × 12 m rectangle.

$d = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ m

Note: 5-12-13 is a Pythagorean triple — recognise it to save time.

l = 12 m w=5m d=13m
$$d = \sqrt{l^2 + w^2} = 13 \text{ m}$$
7
Distance Between Two Grid Points
+5 XP

On a coordinate grid, the horizontal and vertical gaps between two points are the legs of a right triangle. Their hypotenuse is the direct distance.

To find distance from $(x_1, y_1)$ to $(x_2, y_2)$:

  • Horizontal gap: $|x_2 - x_1|$
  • Vertical gap: $|y_2 - y_1|$
  • Distance: $d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

Example: $(0,0)$ to $(3,4)$: $d = \sqrt{9 + 16} = 5$

4 3 d=5
$$d = \sqrt{(\Delta x)^2 + (\Delta y)^2}$$
8
Multi-Step Problems
+5 XP

Some problems need an intermediate length before you can find the final answer. Complete each calculation fully before moving to the next.

Strategy:

  1. Draw a diagram and label everything.
  2. Identify which right triangle to solve first.
  3. Calculate the intermediate length (keep full precision).
  4. Use that result in the next calculation.
  5. Round only the final answer.

Never round intermediate values — rounding errors compound!

Step 1: find h h = √(c₁²−b₁²) Step 2: use h c₂ = √(h²+b₂²) Round at the end!
Keep full precision through intermediate steps.
9
Common Pitfalls
heads-up
Adding path lengths instead of using Pythagoras
Writing "direct distance = 12 + 9 = 21 km" for a north-then-east journey.
Fix: The two travel legs are perpendicular — use $d = \sqrt{12^2+9^2}$.
Using the full width instead of the half-width for roof problems
Using 7 m as the horizontal leg when the roof is 7 m wide and the ridge is at the centre.
Fix: The right triangle only covers half the roof — use $7 \div 2 = 3.5$ m.
Rounding too early in multi-step problems
Rounding an intermediate result to 1 decimal place and getting an inaccurate final answer.
Fix: Keep full calculator precision until the very last step, then round.
Copy Into Your Books

Application Strategy

  • Draw a diagram first
  • Label all known lengths
  • Mark the right angle
  • Identify the unknown side

Key Formulas

  • Rectangle diagonal: $d = \sqrt{l^2+w^2}$
  • Grid distance: $d = \sqrt{(\Delta x)^2+(\Delta y)^2}$
  • Navigation: legs are N/S and E/W distances

Clue Words

  • vertical / horizontal
  • diagonal / direct distance
  • straight-line / shortest
  • north/south + east/west

Multi-step Rule

  • Solve intermediate triangle first
  • Keep full precision
  • Round only the final answer
  • State units in the answer

How are you completing this lesson?

Watch Me Solve It · 3 examples

Watch Me Solve It · Navigation
+15 XP per step
Q1
PROBLEM
A ship sails 12 km north then 9 km east. How far is it from its starting point?
  1. 1
    Draw and label
    North leg $a = 12$ km, East leg $b = 9$ km, $c =$ direct distance
    N and E are perpendicular, forming the two legs of a right triangle.
  2. 2
    Apply Pythagoras
    $c^2 = 12^2 + 9^2 = 144 + 81 = 225$
  3. 3
    Square root
    $c = \sqrt{225} = 15$ km
    9-12-15 is a ×3 multiple of the 3-4-5 triple.
Answer15 km from start
Watch Me Solve It · Rectangle Diagonal
+15 XP per step
Q2
PROBLEM
A rectangular garden is 5 m wide and 12 m long. Find the diagonal.
  1. 1
    Identify legs
    $a = 5$ m, $b = 12$ m, $c =$ diagonal
    The diagonal splits the rectangle into a right triangle.
  2. 2
    Apply Pythagoras
    $c^2 = 5^2 + 12^2 = 25 + 144 = 169$
  3. 3
    Square root
    $c = \sqrt{169} = 13$ m
    5-12-13 is a Pythagorean triple — exact answer.
AnswerDiagonal $= 13$ m
Watch Me Solve It · Roof Rafter
+15 XP per step
Q3
PROBLEM
A roof is 7 m wide and the ridge is 2.4 m above the centre of the roof. Find the rafter length (slant side).
  1. 1
    Identify the right triangle
    Height $= 2.4$ m, half-width $= 7 \div 2 = 3.5$ m, rafter $= c$
    The rafter spans from ridge to eave — hypotenuse of the half-roof triangle.
  2. 2
    Apply Pythagoras
    $c^2 = 3.5^2 + 2.4^2 = 12.25 + 5.76 = 18.01$
  3. 3
    Square root
    $c = \sqrt{18.01} \approx 4.24$ m
    Each rafter is approximately 4.24 m long.
AnswerRafter $\approx 4.24$ m

Brain Trainer · 4 problems

D
Brain Trainer · Application Drills
4 problems

Apply Pythagoras to each scenario. Draw a diagram, then calculate.

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

    $c^2 = 25 + 144 = 169$ → $c = 13$ km

  2. 2 Rectangle 8 m × 15 m. Find the diagonal.

    $c^2 = 64 + 225 = 289$ → $c = 17$ m

  3. 3 Grid distance from $(0,0)$ to $(3,4)$.

    $d = \sqrt{9+16} = \sqrt{25}$ → $d = 5$ units

  4. 4 Square with side 6 cm. Find the diagonal (to 1 decimal place).

    $d = \sqrt{36+36} = \sqrt{72}$ → $d \approx 8.5$ cm
Complete in your workbook.

Quick Check · 5 questions

1
A ship sails 5 km north then 12 km east. Direct distance from start?
+10 XP
2
Diagonal of an 8 m × 15 m rectangle?
+10 XP
3
Distance from $(0, 0)$ to $(3, 4)$ on a grid?
+10 XP
4
Diagonal of a square with side 6 cm (to 1 d.p.)?
+10 XP
5
A ladder goes from the top of a 6 m wall to the top of a 2 m fence 4 m away. Ladder length?
+10 XP

Show Your Working · 3 questions

Show Your Working
9 marks total
ApplyMedium3 MARKS

Q6. A telephone pole is 6 m tall. A wire runs from its top to a point 4.5 m from its base. (a) Find the wire length to 2 decimal places. (b) If wire costs $3/m, find the total cost.

Answer in your workbook.
UnderstandEasy2 MARKS

Q7. A square has a diagonal of 10 cm. Find the side length. Give your answer to 2 decimal places.

Answer in your workbook.
ReasonHard4 MARKS

Q8. A rectangular park is 80 m long and 60 m wide. (a) Find the diagonal path length. (b) A person walks around two sides (length + width) vs the diagonal path. How much shorter is the diagonal route?

Answer in your workbook.
Comprehensive Answers

Quick Check

1. B — $\sqrt{5^2+12^2} = \sqrt{169} = 13$ km.

2. C — $\sqrt{8^2+15^2} = \sqrt{289} = 17$ m.

3. A — $\sqrt{3^2+4^2} = \sqrt{25} = 5$ units.

4. B — $\sqrt{6^2+6^2} = \sqrt{72} \approx 8.5$ cm.

5. B — Vertical diff $= 4$ m, horizontal $= 4$ m. $\sqrt{16+16} = \sqrt{32} \approx 5.66 \approx 5.7$ m.

Model Answers

Q6 (3 marks): (a) Wire $= \sqrt{6^2 + 4.5^2} = \sqrt{36+20.25} = \sqrt{56.25} = 7.5$ m. (b) Cost $= 7.5 \times \$3 = \$22.50$.

Q7 (2 marks): Each side satisfies $s^2 + s^2 = 10^2$, so $2s^2 = 100$, $s^2 = 50$, $s = \sqrt{50} \approx 7.07$ cm.

Q8 (4 marks): (a) Diagonal $= \sqrt{80^2+60^2} = \sqrt{6400+3600} = \sqrt{10000} = 100$ m. (b) Two-sides route $= 80+60 = 140$ m. Difference $= 140 - 100 = 40$ m shorter via diagonal.

Stretch Challenge · +25 XP, +10 coins

Multi-Leg Navigation

A helicopter flies 12 km north, then 16 km east. (a) How far is it from the start? (b) It then flies 9 km south. Find the new straight-line distance from the original start point. Show all working with diagrams.

Reveal solution

(a) $d = \sqrt{12^2+16^2} = \sqrt{144+256} = \sqrt{400} = 20$ km. (b) After flying 9 km south, the helicopter is $12-9=3$ km north of start and 16 km east. New distance $= \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265} \approx 16.28$ km.

R
Quick Review

Always draw first

A diagram reveals the right triangle hidden in every word problem.

Clue words

Vertical, horizontal, diagonal, direct distance, straight-line — these signal a right triangle.

Rectangle diagonal

$d = \sqrt{l^2+w^2}$ — length and width are the legs.

Grid distance

$d = \sqrt{(\Delta x)^2+(\Delta y)^2}$ — horizontal and vertical gaps are the legs.

Half-width for roofs

Use half the total width when the ridge is centred — the right triangle covers one side only.

Round last

Keep full calculator precision through all intermediate steps. Round only the final answer.

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