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 12 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
The pole (12 m) is vertical and the ground distance (5 m) is horizontal, perpendicular legs of a right triangle. Guy-rope $= $ hypotenuse.
$c^2 = 12^2 + 5^2 = 144 + 25 = 169$, so $c = \sqrt{169} = \mathbf{13}$ m. (5-12-13 triple.)