Year 8 (age 13) — Pythagoras and Geometry: rectangle diagonals and isosceles triangle side length

Step-by-step solutions and marking for two Year 8 geometry problems (rectangle corner distances and isosceles triangle) with ACARA v9 mapping, difficulty ranking, rubric and a 300-word teacher comment in a warm Nigella Lawson cadence.

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Order by difficulty (easiest → hardest) and why

  1. Isosceles triangle (base 24, area 60). Straight application: use area = (base × height)/2 to get the height, then Pythagoras. Mostly algebraic and arithmetic—routine.
  2. Rectangle corners (distances 3 and 5). Requires interpreting which distances are side and diagonal, then using Pythagoras. Slightly trickier because of the reading/interpretation step.

Question 1 (rectangle corners)

Problem: F and I, D and A, are standing exactly on the corners of a rectangular room. F is 3 m from D and 5 m from I. What is the minimum possible distance that F could be from A?

Step-by-step solution:

  1. Interpretation: the sentence means F and I are opposite corners and D and A are the other opposite corners of the rectangle. So from corner F the distances to the two adjacent corners are the two side lengths; the distance to the opposite corner is the diagonal.
  2. Let the two side lengths from F be x and y. If one side FD = 3, take x = 3. The diagonal FI = 5, so by Pythagoras: x^2 + y^2 = 5^2.
  3. So 3^2 + y^2 = 25 → 9 + y^2 = 25 → y^2 = 16 → y = 4 (positive length).
  4. The distance from F to A is the other side length y, so the minimum possible distance is 4 m.

Student answer: 4 — Correct.

Question 2 (isosceles triangle)

Problem: The base of an isosceles triangle is 24 and its area is 60. What is the length of one of the equal sides?

Step-by-step solution:

  1. Area = (base × height) / 2, so height = (2 × area) / base = (2 × 60) / 24 = 120 / 24 = 5.
  2. Drop the altitude from the apex to the base: it bisects the base in an isosceles triangle, so each half is 12.
  3. Each equal side is the hypotenuse of a right triangle with legs 12 and 5. By Pythagoras: side^2 = 12^2 + 5^2 = 144 + 25 = 169, so side = 13.

Student answer: 13 — Correct.

Marking rubric (per question, total 4 marks)

  • 1 mark — Correct interpretation of the geometric setup (labels, which is side vs diagonal, or that altitude bisects base).
  • 1 mark — Correct computation of intermediate quantity (height or other side).
  • 1 mark — Correct application of Pythagoras or formula to reach the final numerical answer.
  • 1 mark — Clear presentation and correct final answer.

Question 1: 4/4 — interpretation correct, computation and Pythagoras used correctly, answer 4.
Question 2: 4/4 — height computed correctly (5), used half-base (12) and Pythagoras to get 13.

ACARA v9 mapping (suitable Year level: Year 8 / age 13)

Relevant curriculum threads: solving problems involving the Pythagorean theorem to find lengths in right-angled triangles; using area formulas and relationships for triangles; interpreting geometric diagrams and relationships. These tasks map to learning that emphasises working with right-angled triangles, applying area = 1/2 × base × height, and reasoning about rectangle properties and diagonals.

Teacher comment (about 300 words, in a warm Nigella Lawson cadence)

My dear mathematician, what a lovely pair of problems you have offered — neat, modest, and unapologetically Pythagorean. There is a simple pleasure in watching numbers fit together the way flavours harmonise in a bowl: three and four, making a comfortable five; twelve paired with five to rise to thirteen. Your work is tidy; your instincts are good. For the rectangle, you read the room well and took the diagonal and side relationship exactly as required. For the triangle you coaxed the height out of the area with quiet efficiency and then let Pythagoras finish the dish with a flourish.

Where to move next? Practise the little act of labelling every diagram as if you were plating a dessert — labels and small notes keep your reasoning attractive and traceable. Try more problems where distances are not given as neat triples; that will encourage algebraic set-up and confidence with surds. And when you write solutions, a sentence or two that explains why you chose a step — for instance, “the altitude bisects the base in an isosceles triangle” — is like that pinch of salt that makes everything clearer.

Continue like this: curious, careful, and with a lightness in your method. Mathematics, like good cooking, rewards patience and small elegant choices. Bravo — four out of four on both counts. Let us savour the success and taste the next challenge.

Quick summary for the student

  • Question 1 answer: 4 m — correct.
  • Question 2 answer: 13 — correct.
  • Both solutions use the Pythagorean theorem and clear geometric interpretation. Well done!

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