Mathematics • Year 9 • Unit 3 • Lesson 3

Shorter Sides in the Real World

Apply $a = \sqrt{c^2 - b^2}$ to everyday situations where the slanted distance (hypotenuse) and one perpendicular distance (leg) are known: ladders against walls, TV screens, ramps, ropes, and ziplines. Then explain in your own words why the hypotenuse-squared must come first in the subtraction.

Apply · Real-World Maths

1. Word problems

Each problem gives you the HYPOTENUSE (the slanted distance) and one LEG, and asks you to find the missing perpendicular distance. Always sketch first, identify the hypotenuse, then subtract: $a^2 = c^2 - b^2$. Show your working — a single final answer with no working only earns half marks.

1.1 — Ladder height. A 5 m ladder leans against a wall. The foot of the ladder is 3 m from the wall.

(a) Sketch the situation, identifying which length is the hypotenuse.
(b) How high up the wall does the top of the ladder reach? (Use the 3-4-5 triple — no calculator.)    3 marks

Stuck? The ladder is the slanted side, so it's the hypotenuse. The base distance is one leg. The height up the wall is the missing leg.

1.2 — TV height from diagonal. A 65-inch TV (diagonal 65 inches) has a width of 56.7 inches.

(a) Find the height of the TV to 2 d.p.
(b) Briefly explain why we use SUBTRACTION here, not addition.    3 marks

Stuck on (b)? Recall the lesson's "Decision Question" — missing leg means SUBTRACT.

1.3 — Ramp horizontal run. A skate ramp's slope is 5 m long. The vertical rise at the high end is 3 m.

(a) Find the horizontal run (the base of the ramp).
(b) Which Pythagorean triple does this match?    3 marks

Stuck? The slope of 5 m is the hypotenuse. The rise of 3 m is one leg. Use 3-4-5.

1.4 — Tent guy-rope. A tent's guy-rope is 2.5 m long and is anchored 2 m from the base of the pole. The rope attaches to the top of the pole.

(a) How tall is the tent pole (the rope-attachment point above the ground)?
(b) Which Pythagorean triple does this match?    3 marks

Stuck? The rope is slanted — it's the hypotenuse (2.5 m). The 2 m on the ground is one leg. Multiply 3-4-5 by 0.5 to spot the triple.

1.5 — Zipline drop. An adventure park's zipline is 50 m long. The top platform is 14 m above the ground; the bottom platform is at ground level.

(a) Find the horizontal distance the zipline covers (the “run”) to 2 d.p.
(b) Comment on whether your answer is reasonable for an adventure park (rule of thumb: rise should be much smaller than the run).    3 marks

Stuck? The zipline cable is the slanted side — that's the hypotenuse (50 m). The 14 m rise is one leg. Subtract.

2. Explain your thinking

This question is about communication, not just answers. Use full sentences. 4 marks

2.1 A classmate is finding the missing leg of a right triangle with hypotenuse 13 cm and known leg 5 cm. They write $a^2 = 5^2 - 13^2 = 25 - 169 = -144$, then go quiet because they can't take the square root of a negative.

In your own words, explain (i) what mistake they have made, (ii) which rule from Lesson 3 they have forgotten, and (iii) what the correct value of the missing leg is.

Stuck? Revisit lesson § "Spot the Trap" and the tip "Sign test" — if the result under $\sqrt{}$ is negative, swap $b$ and $c$.

How did this worksheet feel?

What I'll revisit before next class:

Answers — Do not peek before attempting

1.1 — Ladder height

(a) Sketch: right triangle with ladder (5 m) as the slanted hypotenuse, base distance (3 m) as the horizontal leg, and height up the wall as the vertical leg.
(b) This is the 3-4-5 triple, so the height $= \mathbf{4}$ m. (Check: $a^2 = 5^2 - 3^2 = 25 - 9 = 16$, $a = 4$ ✓.)

1.2 — TV height

(a) $a^2 = 65^2 - 56.7^2 = 4225 - 3214.89 = 1010.11$, so $a = \sqrt{1010.11} \approx \mathbf{31.78}$ inches.
(b) The hypotenuse (the diagonal) is already known, and we are finding a leg. Adding $56.7^2 + 65^2$ would treat them BOTH as legs, which is wrong — the diagonal is already the longest side and squaring it goes on the OPPOSITE side of the equation from the legs. Subtracting cancels out the known leg's contribution to the diagonal-squared.

1.3 — Ramp horizontal run

(a) $a^2 = 5^2 - 3^2 = 25 - 9 = 16$, so $a = \sqrt{16} = \mathbf{4}$ m.
(b) $\mathbf{3\text{-}4\text{-}5}$ Pythagorean triple — the ramp's slope, rise and run match the classic 3-4-5.

1.4 — Tent pole height

(a) $a^2 = 2.5^2 - 2^2 = 6.25 - 4 = 2.25$, so $a = \sqrt{2.25} = \mathbf{1.5}$ m. The tent pole is 1.5 m tall (at the rope-attachment point).
(b) Multiply (3, 4, 5) by $0.5$ to get $(1.5, 2, 2.5)$ — so this matches the $\mathbf{3\text{-}4\text{-}5}$ triple scaled down by $0.5$.

1.5 — Zipline horizontal run

(a) $a^2 = 50^2 - 14^2 = 2500 - 196 = 2304$, so $a = \sqrt{2304} = \mathbf{48}$ m exactly (since $48^2 = 2304$). The horizontal run is 48 m.
(b) Reasonable: the rise (14 m) is much smaller than the run (48 m) — about a 1:3.4 slope — which is gentle enough for a beginner zipline. ✓

2.1 — Explain your thinking (sample response)

(i) The classmate has put the smaller number FIRST in the subtraction. They wrote $5^2 - 13^2$, but the hypotenuse-squared (the bigger one, $13^2$) must come first — otherwise you get a negative result, and $\sqrt{\text{negative}}$ isn't a real number. (ii) They have broken the rule "$c^2$ comes first" (this is exactly the trap flagged in the lesson's Spot the Trap card, and the Sign Test tip). (iii) Corrected working: $a^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $a = \sqrt{144} = \mathbf{12}$ cm. (This completes the 5-12-13 triple.)

Marking: 1 for spotting the swap; 1 for naming the rule ("hypotenuse-squared first"); 1 for the corrected answer of 12 cm; 1 for a clear, full-sentence explanation.