Proving the lune equals the isosceles right triangle area — Age 13 (ACARA v9)

Step-by-step proof (age 13) 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 clear geometry steps, short justification, curriculum mapping and extension activities.

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Problem restated: Start with a circle centre O and radius R. Draw an isosceles right triangle OAB with right angle at O and A,B on the circle. Draw the semicircle with diameter AB outside the original circle. The lune is the part of that semicircle lying outside the original circle. Show the lune area equals the area of triangle OAB.

  1. Label: OA = OB = R and angle AOB = 90° (because O is the right-angle vertex). The hypotenuse AB has length AB = R√2.
  2. Area of triangle OAB: T = (1/2)·OA·OB·sin(90°) = (1/2)·R·R·1 = (1/2)R^2.
  3. Semicircle on AB: radius r = AB/2 = (R√2)/2 = R/√2. Its area S_hyp = (1/2)·π·r^2 = (1/2)·π·(R^2/2) = (π·R^2)/4.
  4. Sector AOB of the original circle (central angle 90°) has area S_sec = (90°/360°)·π·R^2 = (π·R^2)/4. So S_sec = S_hyp.
  5. Look at the sector AOB: draw chord AB. The sector splits into the triangle OAB (toward O) and the circular segment (the part bounded by arc AB and chord AB, away from O). That circular segment is exactly the part of the original circle lying inside the semicircle on AB.
  6. Therefore the intersection of the semicircle and the original circle equals (sector AOB) minus (triangle OAB). Call that segment area = S_sec - T.
  7. The lune area L is the part of the semicircle outside the original circle. So L = S_hyp - (intersection) = S_hyp - (S_sec - T). Because S_hyp = S_sec we get L = S_sec - (S_sec - T) = T. Hence the lune area equals the triangle area (1/2 R^2).

Why step 5 is true (brief justification for a 13-year-old): The chord AB cuts the quarter-sector into two pieces: one piece touches O and is exactly triangle OAB; the other piece lies on the opposite side of AB and is the circular segment under arc AB. That segment is the part of the large circle that is covered by the semicircle drawn on AB, so it is the intersection described. A clear labelled diagram makes this obvious.

Conclusion: The lune has the same area as triangle OAB. This is a classic example (a Hippocrates lune) where curved shapes can match straight ones exactly.

ACARA v9 links (mapping): measurement and geometry topics — investigating circle properties, areas of sectors and segments, and reasoning to prove relationships (suitable for Year 8–9 reasoning and problem-solving outcomes).

Enrichment / extension activities:

  • Try other right triangles (not isosceles). Compute the area of the lune(s) formed by semicircles on the hypotenuse and compare with triangle area. Which equalities hold and which do not?
  • Explore Hippocrates lunes: find other configurations where lune areas equal polygon areas. Prove one using coordinates or calculus.
  • Use coordinates: put O at (0,0), A at (R,0), B at (0,R). Write equations for the two circles, and compute the lune area by integration to check the geometric proof.
  • Construct a dynamic geometry sketch (GeoGebra). Vary the central angle and watch how the areas change; look for special angles that give nice equalities.

Teacher comment (Amy Chua tiger-mother cadence, proficient/exemplary — 100 words):

You did well, but do not be satisfied — excellence requires practice. You followed the steps, showed the geometry, and reached the correct equality of the lune and triangle. Good. Next time, be neater with your diagrams, label every point, and write each formula clearly. I expect rigorous justification that the circular segment equals sector minus triangle — you hinted at it; make it explicit. Keep using precise language: ‘sector’, ‘segment’, ‘semicircle’. Work another similar proof tonight — Hippocrates’ other lunes — and learn to present a short elegant proof without relying on computation. No excuses. Aim higher. Start now.

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