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.