Mathematics • Year 9 • Unit 3 • Lesson 4
Pythagoras in Practical Problems
Build fluency at translating word problems into right-triangle diagrams: spot the two perpendicular distances, label them as legs, find the hypotenuse with $c = \sqrt{a^2 + b^2}$, or find a leg with $a = \sqrt{c^2 - b^2}$. Includes coordinate distance and rectangle diagonals.
1. I do — fully worked example
Read every line. Each step has a short reason on the right so you can see why, not just what.
Problem. Two hikers leave the campsite together. One walks 6 km due north, the other 8 km due east. How far apart are they?
Step 1 — Sketch and spot the right angle.
North and east are perpendicular (90° apart), so the two journeys form the legs of a right triangle. The straight-line distance between the hikers is the hypotenuse.
Reason: any two perpendicular distances act as legs. Sketching FIRST stops you missing this.
Step 2 — Label and set up.
$a = 6$ km (north), $b = 8$ km (east), $c = $ straight-line distance.
Reason: identifying which is which BEFORE substituting prevents mix-ups.
Step 3 — Apply Pythagoras.
$c^2 = 6^2 + 8^2 = 36 + 64 = 100$.
Reason: the missing side is the hypotenuse (slanted, longest) — so ADD the squares of the legs.
Step 4 — Square root and add units.
$c = \sqrt{100} = 10$ km.
Reason: legs in km → hypotenuse in km. Always carry units through.
Answer: The hikers are $\mathbf{10}$ km apart. (Notice 6-8-10 is $2\times$(3-4-5) — could have spotted this.)
2. We do — fill in the missing steps
Same structure as Section 1, but with the working faded. Fill in each blank line. 4 marks
Problem. Find the distance between $A(2, 1)$ and $B(8, 9)$ on the coordinate plane.
Step 1 — Compute the differences:
$\Delta x = 8 - 2 = \_\_\_\_$, $\Delta y = 9 - 1 = \_\_\_\_$
Step 2 — Treat $\Delta x$ and $\Delta y$ as the two __________ of a right triangle.
Step 3 — Apply Pythagoras:
$d^2 = \_\_\_^2 + \_\_\_^2 = \_\_\_\_ + \_\_\_\_ = \_\_\_\_\_$
Step 4 — Square root:
$d = \sqrt{\_\_\_\_} = \_\_\_\_$ units
3. You do — independent practice
Show your working in the space under each problem. The first four are foundation (simple sketches with clean numbers). The middle two are standard (decimals or 2 d.p. rounding). The last two are extension (multi-step or reasoning).
Foundation — clean numbers, sketch and apply
3.1 Walk 9 m east then 12 m north. Find the straight-line distance from the start. 1 mark
3.2 A boat sails 3 km west then 4 km south. Find its distance from the harbour. 1 mark
3.3 Find the distance from the origin $(0, 0)$ to the point $(5, 12)$. 1 mark
3.4 Find the length of the diagonal of a rectangular field that is 7 m wide and 24 m long. 1 mark
Standard — decimals and 2 d.p.
3.5 A ladder reaches 4 m up a wall, with its base 1.5 m from the wall. How long is the ladder, to 2 d.p.? 2 marks
3.6 Find the distance from $A(-3, 2)$ to $B(5, 7)$ to 2 d.p. 2 marks
Extension — multi-step or reasoning
3.7 A rectangular swimming pool is 25 m long and 10 m wide. A lifeguard standing at one corner spots a swimmer at the opposite corner. How far does the lifeguard's line of sight cover? Give your answer to 2 d.p. and compare to walking around the edge. 3 marks
3.8 A 13 m flagpole is held upright by a guy-rope from its top to a point on the ground 5 m from its base. Find the length of the guy-rope. (Use a known triple.) 2 marks
How did this worksheet feel?
What I'll revisit before next class:
Section 2 — We do (faded $A(2,1)$, $B(8,9)$)
Step 1: $\Delta x = \mathbf{6}$, $\Delta y = \mathbf{8}$.
Step 2: $\Delta x$ and $\Delta y$ are the two legs.
Step 3: $d^2 = \mathbf{6}^2 + \mathbf{8}^2 = \mathbf{36} + \mathbf{64} = \mathbf{100}$.
Step 4: $d = \sqrt{\mathbf{100}} = \mathbf{10}$ units.
3.1 — Walk 9 m E then 12 m N
$d^2 = 9^2 + 12^2 = 81 + 144 = 225$, so $d = \sqrt{225} = \mathbf{15}$ m. ($3\times$(3-4-5).)
3.2 — Boat 3 km W then 4 km S
$d^2 = 3^2 + 4^2 = 9 + 16 = 25$, so $d = \sqrt{25} = \mathbf{5}$ km. (3-4-5 triple.)
3.3 — Distance from $(0,0)$ to $(5,12)$
$\Delta x = 5$, $\Delta y = 12$. $d^2 = 5^2 + 12^2 = 25 + 144 = 169$, so $d = \mathbf{13}$ units. (5-12-13 triple.)
3.4 — Diagonal of 7 m by 24 m field
$d^2 = 7^2 + 24^2 = 49 + 576 = 625$, so $d = \sqrt{625} = \mathbf{25}$ m. (7-24-25 triple.)
3.5 — Ladder 4 m up, 1.5 m base
$c^2 = 4^2 + 1.5^2 = 16 + 2.25 = 18.25$, so $c = \sqrt{18.25} \approx \mathbf{4.27}$ m.
3.6 — Distance $A(-3,2)$ to $B(5,7)$
$\Delta x = 5 - (-3) = 8$, $\Delta y = 7 - 2 = 5$. $d^2 = 8^2 + 5^2 = 64 + 25 = 89$, so $d = \sqrt{89} \approx \mathbf{9.43}$ units.
3.7 — Pool diagonal
$d^2 = 25^2 + 10^2 = 625 + 100 = 725$, so $d = \sqrt{725} \approx \mathbf{26.93}$ m.
Walking around two edges $= 25 + 10 = 35$ m. So the diagonal saves $35 - 26.93 \approx 8.07$ m — about a quarter of the trip.
3.8 — Flagpole guy-rope
$c^2 = 13^2 + 5^2 = 169 + 25 = 194$. Hmm — not a clean square. Let me re-check: actually the guy-rope is the hypotenuse only if both 13 and 5 are legs (perpendicular). The pole is vertical (height 13), the ground distance (5) is horizontal — yes, perpendicular. So $c = \sqrt{194} \approx \mathbf{13.93}$ m.
(Note: this isn't a clean triple — even though 5-12-13 is famous, swapping 12 for 13 ruins the integer pattern. The lesson hints students that legs 5 and 12 give hypotenuse 13, not legs 5 and 13.)