Why the lune equals the isosceles right triangle (age 13) — clear step-by-step proof and teacher feedback

Step-by-step proof (for a 13-year-old AoPS student) that the lune formed by drawing an isosceles right triangle inside a circle and a semicircle on its hypotenuse has the same area as the triangle. Includes a simple geometric calculation, a numeric check, and teacher comments + rubric mapped to ACARA v9.

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Problem set-up (what you draw)

Let O be the centre of a circle of radius r. Draw an isosceles right triangle OAB with the right angle at O, so OA = OB = r and angle AOB = 90°; AB is the hypotenuse. Now draw the semicircle that has AB as its diameter on the side opposite O. The lune is the part of that semicircle lying outside the original circle. Show that the area of this lune equals the area of triangle OAB.

Step 1 — area of the triangle OAB

  • Triangle OAB is right with legs OA and OB each of length r, so area(triangle OAB) = (1/2)·OA·OB = (1/2)·r·r = r²/2.

Step 2 — area of the semicircle on AB

  • Length of the hypotenuse AB = r·√2 (because legs are r and r).
  • Radius of the semicircle on AB is (AB)/2 = (r·√2)/2 = r/√2.
  • Area(semicircle on AB) = (1/2)·π·(radius)² = (1/2)·π·(r/√2)² = (1/2)·π·(r²/2) = (π·r²)/4.

Step 3 — area of the part of the original circle that lies inside that semicircle

  • Look at the chord AB on the original circle. The portion of the original circle cut off by chord AB on the side toward O is the sector AOB (central angle 90°) minus triangle OAB.
  • Area(sector AOB) = (central angle/360°)·π·r² = (90°/360°)·π·r² = (π·r²)/4.
  • So the circular segment (the piece of the original circle bounded by chord AB and arc AB) that lies toward O has area area(segment) = area(sector AOB) − area(triangle OAB) = (π·r²)/4 − r²/2.
  • That circular segment is exactly the portion of the original circle that lies inside the semicircle on AB (visual: the two circles meet at A and B and that segment is the overlap on the original circle's side).

Step 4 — compute the lune area

  • By definition, lune area = area(semicircle on AB) − area(of original circle inside that semicircle). Substitute the values we found: lune = (π·r²)/4 − [ (π·r²)/4 − r²/2 ] = r²/2.
  • But r²/2 is exactly area(triangle OAB). Therefore the area of the lune equals the area of triangle OAB. QED.

Quick numeric check (sanity)

  • Take r = 2. Triangle area = (1/2)·2² = 2. Semicircle area = (π·4)/4 = π ≈ 3.1416. Sector area = π also. The overlap is sector − triangle = π − 2 ≈ 1.1416. So lune = semicircle − overlap = π − (π − 2) = 2, which matches the triangle area.

Why this idea is clean and powerful

The two π·r²/4 terms (the semicircle area and the sector area) cancel; what remains is the triangle area. The geometry behind this cancellation is that AB is a diameter of the circumcircle of triangle OAB, so the semicircle on AB has the same sector-area number (π·r²/4) that appears in the sector at the centre O of the main circle. Recognising the right-angle and the chord/sector relations leads immediately to a neat algebraic cancellation.

Teacher comments (Amy Chua cadence — firm, precise, encouraging exactness)

Good. You drew a correct diagram and you can compute areas. But do not be sloppy: label the centre O, points A and B, write down r once, and show each algebraic step. I expect a tidy diagram and justification for why the sector area is (π·r²)/4 — do not skip that. If you present this on an exam, write the cancellation clearly: semicircle minus segment = semicircle − (sector − triangle) = triangle. Neatness and logic matter.

ACARA v9 mapping (task alignment)

  • Mathematical reasoning and problem solving: geometric proof and area calculation using properties of circles and right triangles.
  • Measurement and geometry: use of area formulae for triangles, sectors and semicircles; using Pythagorean relation (for AB = r√2) and chord/sector relations.
  • Suggested year band: Years 7–9 (suits a capable 13-year-old, especially with AoPS background).

Assessment rubric (strict, clear — what I expect)

Criteria (score each): diagram & labelling, correct formulas, correct algebra, logical justification, communication & neatness.

  • Excellent (A) — All parts correct. Clear labeled diagram. Correct areas for triangle, sector and semicircle. Correct identification of overlap as sector − triangle and clean cancellation to get triangle area. Work is neat and fully justified.
  • Good (B) — Minor arithmetic or wording slips but reasoning is correct. Diagram present but maybe one label missing. Final answer correct after small fixes.
  • Satisfactory (C) — Shows main idea (triangle area, semicircle area) but misses justification that the overlap equals sector − triangle or has an algebra error. Needs clearer explanation or labels.
  • Needs Improvement (D) — Missing diagram, incorrect formulas (for example wrong radius for semicircle), or no clear chain of logic linking geometry to areas. Rework step-by-step with teacher guidance.

Quick marking notes I will write on your work

  • "Label the diagram properly and write r once — do not repeat hidden assumptions."
  • "Show why the overlap equals sector − triangle. That is the heart of the proof."
  • "Excellent cancellation — you recognised the equal π·r²/4 terms. Clear thinking."

Final word: You have a beautiful and classical result. The algebra is short, so make sure your diagram and verbal justification are precise: mathematics rewards clarity.

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