Mathematics Standard • Year 11 • Module 2 • Lesson 3
Pythagoras' Theorem
Practise HSC Mathematics Standard 2-style writing on Pythagoras — multi-mark short answers and one structured extended response.
1. Short-answer questions
1.1 A right-angled triangle has shorter sides of 11 cm and 5 cm. Find the value of the hypotenuse correct to 2 d.p. 2 marks Band 3
1.2 A cone has vertical height 10 cm and base diameter 12 cm. 3 marks Band 3-4
(a) Write down the radius of the base.
(b) Find the slant height correct to 2 d.p.
1.3 A builder is installing a diagonal brace across a rectangular gate that is 2.4 m wide and 1.8 m tall. 4 marks Band 4
(a) Find the length of the diagonal brace correct to 2 d.p.
(b) Timber costs $12.50 per metre. Find the cost of the brace, rounding the timber length UP to the nearest 10 cm to ensure it is long enough.
2. Extended response
2.1 A landscape architect is designing a triangular garden bed. The garden has a right angle at one corner. Two sides — the legs — measure 7.5 m and 4 m.
The brief:
• The two short sides are made of timber edging.
• The diagonal (hypotenuse) is a curved sandstone wall that costs more per metre.
• Timber edging: $42 per metre. Sandstone wall: $185 per metre.
(a) Find the length of the hypotenuse correct to 2 d.p.
(b) Find the total perimeter of the garden bed.
(c) Calculate the total construction cost for the timber edging plus the sandstone wall, stated in dollars and cents in a clear conclusion sentence. 6 marks Band 5-6
Explicit marking criteria
Part (a) — 2 marks
• 1 mark — correct substitution into c² = a² + b² using 7.5 and 4.
• 1 mark — correct hypotenuse value rounded to 2 d.p.
Part (b) — 1 mark
• 1 mark — correct total perimeter (7.5 + 4 + hypotenuse), rounded appropriately.
Part (c) — 3 marks
• 1 mark — applies the timber rate ($42/m) ONLY to the two legs (11.5 m total), not to the whole perimeter.
• 1 mark — applies the sandstone rate ($185/m) ONLY to the hypotenuse.
• 1 mark — explicit conclusion sentence stating the total cost in dollars and cents.
Your response:
Stuck on (c)? Cost = (timber rate × timber length) + (sandstone rate × sandstone length). The two rates apply to DIFFERENT sides.How did this worksheet feel?
What I'll revisit before next class:
1.1 — Hypotenuse from 11 and 5 (2 marks)
Sample response.
c² = 11² + 5² = 121 + 25 = 146. c = √146 = 12.08 cm (to 2 d.p.).
Marking notes. 1 mark — correct setup and substitution. 1 mark — correct rounded answer with unit. Common error: stopping at "c² = 146" without taking the square root scores 1/2.
1.2 — Cone with diameter (3 marks)
Sample response.
(a) r = 12 ÷ 2 = 6 cm.
(b) ℓ² = r² + h² = 6² + 10² = 36 + 100 = 136. ℓ = √136 = 11.66 cm (to 2 d.p.).
Marking notes. (a) 1 mark — correct radius shown as a SEPARATE step. (b) 1 mark — correct substitution using r (not d). 1 mark — correct slant height to 2 d.p. Common error: using d = 12 directly: ℓ² = 144 + 100 = 244 → 15.62 cm (4 cm too long).
1.3 — Gate brace (4 marks)
Sample response.
(a) d² = 2.4² + 1.8² = 5.76 + 3.24 = 9. d = √9 = 3.00 m.
(b) The brace is exactly 3.0 m, so no rounding up is needed. Cost = 3 × $12.50 = $37.50.
Marking notes. (a) 1 mark — correct substitution. 1 mark — correct diagonal value. (b) 1 mark — recognises that exact length needs no rounding. 1 mark — correct cost. Common error: rounding 3.00 to 3.10 (unnecessary) gives $38.75 — wrong because the brace is already exactly 3 m.
2.1 — Triangular garden bed (6 marks): sample Band-6 response with annotations
Sample Band-6 response.
(a) Hypotenuse of the garden.
c² = 7.5² + 4² = 56.25 + 16 = 72.25. [1 mark — substitution.]
c = √72.25 = 8.50 m (to 2 d.p.). [1 mark — hypotenuse value.]
(b) Total perimeter.
P = 7.5 + 4 + 8.50 = 20.00 m. [1 mark — perimeter.]
(c) Construction cost.
Timber edging (two legs): (7.5 + 4) × $42 = 11.5 × $42 = $483.00. [1 mark — timber rate applied only to the two legs.]
Sandstone wall (hypotenuse): 8.50 × $185 = $1,572.50. [1 mark — sandstone rate applied only to the hypotenuse.]
Total = $483.00 + $1,572.50 = $2,055.50.
Conclusion: the total construction cost of the garden bed is $2,055.50 (timber edging $483.00 plus sandstone wall $1,572.50). [1 mark — explicit conclusion in dollars and cents with units.]
Total: 6/6.
Band descriptors for marker.
Band 3: Hypotenuse found but only one rate applied (e.g. $42/m to entire 20 m perimeter = $840). ≈ 2-3 marks.
Band 4: Hypotenuse and perimeter correct; two different rates attempted but mixed up (timber rate on hypotenuse, sandstone rate on legs). ≈ 4 marks.
Band 5: Rates correctly assigned, totals correct, but conclusion is a bare number without units or naming what was calculated. ≈ 5 marks.
Band 6: Complete, rates correctly assigned, conclusion sentence in dollars and cents with a breakdown of the two components. 6/6.