Proof for a 13-year-old: Isosceles right-triangle lune (circle centre at right angle) — ACARA v9 geometry task

Step-by-step, student-friendly proof (age 13) that the lune formed by a semicircle on the hypotenuse of an isosceles right triangle (with right angle at the circle centre) has the same area as the triangle; includes enrichment ideas.

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Here is a clear step-by-step proof you can follow.

  1. Set-up: Let O be the centre of the original circle and let OA and OB be radii of length R, with angle AOB = 90° so triangle AOB is an isosceles right triangle. Its area is
    Area_triangle = (1/2) * OA * OB = (1/2) * R * R = R^2/2.
  2. Hypotenuse AB has length AB = R*sqrt(2). The semicircle built on AB as diameter has radius r = AB/2 = R/√2. Its area is
    Area_semicircle = (1/2) * π * r^2 = (1/2) * π * (R^2/2) = π R^2 / 4.
  3. The part of the original circle bounded by chord AB is a sector with central angle 90°. Its area is
    Area_sector = (90/360) * π * R^2 = π R^2 / 4. The circular segment cut off by chord AB (the curved piece inside the circle and bounded by chord AB) equals Area_sector minus Area_triangle:
    Area_segment = (π R^2 / 4) - (R^2 / 2).
  4. The lune is the part of the semicircle that lies outside the original circle. So
    Area_lune = Area_semicircle - Area_segment
    = (π R^2 / 4) - [ (π R^2 / 4) - (R^2 / 2) ]
    = R^2 / 2.
  5. Conclusion: Area_lune = R^2 / 2 = Area_triangle. The lune has the same area as the isosceles right triangle.

Notes to help your diagram: label O at the centre, A on the x-axis at (R,0) and B on the y-axis at (0,R). The midpoint of AB is (R/2,R/2) — the centre of the semicircle on AB.

Teacher comment (proficient/exemplary, Nigella Lawson cadence — 100 words):

My darling, your construction is utterly crystalline: you began with a generous circle, placed the right-angled vertex at its heart, and coaxed out that elegant isosceles right triangle with its two faithful points perched on the rim. The semicircle on the hypotenuse appears and, like magic, the lune reveals itself. Your algebra and tidy area calculations are measured, confident and deliciously clear — the sector cancels the semicircle, leaving the triangle’s area in splendid isolation. Presentation, labelling and reasoning are exemplary. This solution invites applause and a gentle, satisfied sip of celebratory tea. Do keep exploring with curiosity and warmth.

Additional enrichment / extension ideas:

  • Investigate other lunes (Hippocrates' lunes): which configurations produce equal areas and why? Try changing the central angle from 90° and see what fails or holds.
  • Generalise: use coordinates to prove the result algebraically (we started that way with points (R,0) and (0,R)); try the same method for other right triangles to see patterns.
  • Use dynamic geometry software (GeoGebra) to drag points and measure areas numerically — this builds intuition and checks boundary cases.
  • Explore integration: compute the lune area by integrating the difference of the two circle functions and compare to the triangle area.
  • Research history: read about Hippocrates of Chios and the discovery of lunes; present a short report connecting history and math.

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