Mathematics Standard • Year 11 • Module 2 • Lesson 3
Pythagoras' Theorem
Apply Pythagoras to realistic 2D and 3D contexts — ladders, ramps, room diagonals, cones, and slant heights.
Problem 1 — Ladder safety check (worksite)
A worker leans a 6 m ladder against a wall. WorkSafe guidelines say the base of the ladder should be exactly 1.5 m from the wall for a 6 m ladder.
Set up: What are we solving for?
(i) Identify the right triangle: which side is the hypotenuse? 1 mark
(ii) Calculate the height the ladder reaches up the wall to 2 d.p. 2 marks
(iii) The worker needs to reach a window 5.8 m up the wall. Is this ladder configuration tall enough? Show the difference. 2 marks
Stuck? Revisit lesson § Practical Contexts — the ladder is the hypotenuse, the wall and ground are the shorter sides.Problem 2 — Wheelchair ramp (accessibility)
An accessibility ramp rises 0.5 m vertically over a horizontal run of 6 m. (The ramp itself is the hypotenuse.)
Set up: What are we solving for?
(i) Calculate the length of the ramp's surface to 2 d.p. 2 marks
(ii) The ramp must be covered with grip tape costing $14 per metre. The tape is sold in whole metres. Find the cost. 2 marks
(iii) The ramp width is 1.2 m. Find the total area of grip tape (width × length, in m²). 1 mark
Stuck on (ii)? Round UP to the next whole metre because tape only comes in whole-metre lengths and must cover the full ramp.Problem 3 — TV diagonal (reverse problem)
A television is advertised as "65-inch" — that refers to the diagonal of the screen. (1 inch = 2.54 cm.) The screen has an aspect ratio of 16:9, meaning if the width is 16x cm then the height is 9x cm.
Set up: What are we solving for?
(i) Convert the diagonal length from inches to cm. 1 mark
(ii) Using Pythagoras (width)² + (height)² = (diagonal)², and letting the width be 16x and height 9x, set up the equation and solve for x to 2 d.p. 3 marks
(iii) State the width and height of the screen in cm, each to 1 d.p. 2 marks
Stuck on (ii)? (16x)² + (9x)² = (diag)² gives 256x² + 81x² = 337x² = diag². Then x = √(diag²/337).Problem 4 — Conical tent (slant height for material)
A circus-style conical tent has a vertical height of 4.5 m and a circular base with diameter 7 m. The tent's slant height is needed so the makers can cut the curved fabric panel.
Set up: What are we solving for?
(i) Find the radius of the base. 1 mark
(ii) Identify the right triangle inside the cone (which sides are the legs and which is the hypotenuse), and find the slant height to 2 d.p. 3 marks
(iii) Compare: a maker estimates that "the slant height should equal the vertical height". By how many metres is this estimate wrong? 1 mark
Stuck? Revisit lesson § Worked Example 4 — Slant Height of a Cone. The slant is the hypotenuse, NOT the vertical height.Problem 5 — Will the table fit? (room diagonal)
A homeowner wants to know if a long dining table (3.6 m long) will fit diagonally across a rectangular room that measures 3.5 m by 2.8 m.
Set up: What are we solving for?
(i) Calculate the length of the room's floor diagonal to 2 d.p. 2 marks
(ii) Compare the diagonal to the table length and conclude with a one-sentence statement (fits / does not fit, and by how many cm). 2 marks
(iii) The homeowner adds 30 cm of clearance to each end of the table so chairs can be pulled out — effectively requiring 4.2 m of diagonal space. Does the room still fit the table? Justify with the numerical comparison. 2 marks
Stuck? Revisit lesson § Practical Contexts — Diagonal of a rectangle: d² = ℓ² + w².How did this worksheet feel?
What I'll revisit before next class:
Problem 1 — Ladder height
Set up. Find the wall-height the ladder reaches, then compare to a 5.8 m target.
(i) The ladder is the hypotenuse (6 m). The wall and ground are the shorter sides.
(ii) wall² = 6² − 1.5² = 36 − 2.25 = 33.75. Wall = √33.75 = 5.81 m (to 2 d.p.).
(iii) 5.81 m > 5.8 m, so the ladder reaches by 0.01 m (1 cm). It JUST reaches — but this leaves no safety margin and would not be acceptable in practice.
Problem 2 — Wheelchair ramp
Set up. Find the slope-length (hypotenuse), then tape cost and tape area.
(i) ramp² = 6² + 0.5² = 36 + 0.25 = 36.25. ramp = √36.25 = 6.02 m (to 2 d.p.).
(ii) Round UP to 7 m of tape. Cost = 7 × $14 = $98.
(iii) Area = 1.2 × 6.02 = 7.22 m² (to 2 d.p.).
Problem 3 — TV diagonal
Set up. Use Pythagoras with width:height = 16:9 to find x, then width and height.
(i) 65 × 2.54 = 165.1 cm.
(ii) (16x)² + (9x)² = 165.1². So 256x² + 81x² = 27 258.01 → 337x² = 27 258.01 → x² = 80.88. x = √80.88 = 8.99 cm (to 2 d.p.).
(iii) Width = 16 × 8.99 = 143.9 cm. Height = 9 × 8.99 = 80.9 cm (each to 1 d.p.).
Problem 4 — Conical tent
Set up. Find the radius, then apply Pythagoras to the (r, h, slant) right triangle.
(i) r = 7 ÷ 2 = 3.5 m.
(ii) Inside the cone, the right triangle has legs r = 3.5 and h = 4.5; the slant ℓ is the hypotenuse. ℓ² = 3.5² + 4.5² = 12.25 + 20.25 = 32.5. ℓ = √32.5 = 5.70 m (to 2 d.p.).
(iii) The estimate would give 4.5 m. Actual is 5.70 m. The estimate is 1.20 m too short. The slant is always longer than the vertical height (it is the hypotenuse).
Problem 5 — Dining table diagonal
Set up. Find the room's diagonal, then compare to the table length (and table + clearance).
(i) d² = 3.5² + 2.8² = 12.25 + 7.84 = 20.09. d = √20.09 = 4.48 m (to 2 d.p.).
(ii) 4.48 m > 3.6 m. The table FITS diagonally with 4.48 − 3.6 = 0.88 m (88 cm) to spare.
(iii) 4.48 m > 4.2 m by 28 cm. The room still fits the table plus 30 cm clearance at each end. Marking note: a bare "yes, it fits" without the numerical comparison loses one of two marks.