13-year-old: Proof (Hippocrates' lune) — Lune formed from an isosceles right triangle in a circle has the same area as the triangle (ACARA v9 aligned)

What we have

Start with a circle of radius R and centre O. Draw an isosceles right triangle OAB with the right angle at O and the other two vertices A and B on the circle. So OA = OB = R and angle AOB = 90°.

What we must show

The semicircle built on AB (with AB as diameter) lies outside the original circle; the part of this semicircle that is outside the original circle is called the lune. Show that the area of that lune equals the area of triangle OAB.

Step-by-step proof

1. Area of triangle OAB.

   Triangle OAB has legs OA and OB of length R and the angle between them is 90°. So its area is
   Area(triangle) = 1/2 × OA × OB × sin(90°) = 1/2 × R × R × 1 = R²/2.

2. Length of AB and radius of the semicircle on AB.

   Since OA and OB are perpendicular, the chord AB has length AB = √(R² + R²) = R√2. The semicircle built on AB has radius r = AB/2 = R√2/2 = R/√2.

3. Area of the semicircle on AB.

   Area(semicircle) = (1/2)πr² = (1/2)π(R²/2) = πR²/4.

4. Area of sector AOB of the original circle.

   Sector AOB has central angle 90° (a quarter of the full circle), so
   Area(sector AOB) = (90/360) × πR² = πR²/4.

   Note: the area of this sector equals the area of the semicircle we just found.

5. Understand the circular segment cut off by chord AB.

   The chord AB divides the circle into two parts (two circular segments). The small segment bounded by the arc AB and chord AB (on the side away from O) has area
   Area(segment) = Area(sector AOB) − Area(triangle OAB).

   Geometrically, this segment is exactly the part of the original circle that lies on the same side of AB as the semicircle (that is, the part of the circle that overlaps the semicircle).

6. Relate the pieces to find the lune area.

   The semicircle area is made of two parts: the overlap with the circle (the circular segment) and the part outside the circle (the lune). So

   Area(semicircle) = Area(segment) + Area(lune).

   Therefore Area(lune) = Area(semicircle) − Area(segment). Substituting what we have:

   Area(lune) = Area(semicircle) − [Area(sector AOB) − Area(triangle OAB)].

   But Area(semicircle) = Area(sector AOB), so the two sector terms cancel and we get

   Area(lune) = Area(triangle OAB) = R²/2.

Conclusion

Hence the lune (the part of the semicircle on AB lying outside the original circle) has exactly the same area as the isosceles right triangle OAB.

Teacher comments (delivered in a Nigella Lawson cadence)

Oh, darling — what a delightfully neat little feast of geometry you’ve served. The reasoning is elegant and economical, like a well-baked tart: simple ingredients (radii, a chord, a semicircle) arranged so their flavours — I mean areas — perfectly balance. Be sure when you write it up to name each area clearly (triangle, sector, semicircle, segment, lune) so a reader can follow the recipe without guessing the measurements. A lovely result — quietly irresistible.

Rubric mapped to ACARA v9 (Geometry & Measurement — Year 8 level, age 13)

• Understanding and reasoning (A): Correctly identifies shapes and relationships (sector–triangle–segment) and clearly explains why the semicircle and sector areas are equal. (Excellent: clear, correct logical flow; Satisfactory: correct steps but some explanation missing; Needs improvement: unclear mapping between regions.)
• Mathematical procedures (B): Accurate computations for areas and lengths (R, AB = R√2, r = R/√2, areas πR²/4 and R²/2). (Excellent: all computations correct and shown; Satisfactory: correct final values but brief working; Needs improvement: arithmetic or formula errors.)
• Communication and notation (C): Uses labels (O, A, B), defines R and r, and states equalities clearly. (Excellent: neat diagram + labelled steps; Satisfactory: labels present but diagram minimal; Needs improvement: missing labels or unclear explanation.)
• Proof and justification (D): Demonstrates why area(segment) = sector − triangle and why the semicircle equals the sector, leading to Area(lune) = Area(triangle). (Excellent: logical and complete justification; Satisfactory: correct equalities but one step not explained fully; Needs improvement: gaps in the justification.)

Success Criteria (what I expect to see): a labelled diagram; expression of each area in terms of R; the equality Area(semicircle) = Area(sector AOB) shown; identification that the circle’s overlapping piece with the semicircle is the segment (sector − triangle); concluding Area(lune) = Area(triangle).

If you’d like, I can produce a clean diagram you can print or add to your write-up — a picture really makes this little geometric confection irresistible.