Statement of the problem

We start with a circle of radius R and centre O. On that circle we pick two points A and B so that angle AOB = 90° (so OA = OB = R). The triangle OAB is an isosceles right triangle (right angle at O). Let AB be the hypotenuse. Draw a semicircle whose diameter is AB on the side of AB away from O. The "lune" is the part of that semicircle that lies outside the original circle. Show that the area of the lune equals the area of triangle OAB.

Key idea (one-sentence summary)

The semicircle built on AB has exactly the same area as the 90° sector of the big circle (sector AOB). The lune is the semicircle minus the part of that semicircle that lies inside the big circle; that inside part is exactly the circular segment cut off by chord AB, so when you subtract it from the equal-area sector you get precisely the triangle OAB. Hence lune area = triangle area.

Step-by-step proof

1. Label and compute lengths. Let the big circle have radius R and centre O. Because OA = OB = R and angle AOB = 90°, the chord AB (the hypotenuse of triangle OAB) has length AB = R√2 (right triangle with legs R and R).
2. Radius of the semicircle on AB. The diameter of the semicircle is AB, so its radius is r = AB/2 = (R√2)/2 = R/√2.
3. Area of the semicircle on AB.

   Area = (1/2)·π·r^2 = (1/2)·π·(R^2/2) = πR^2/4.

4. Area of the 90° sector AOB of the big circle.

   The whole circle has area πR^2. The sector with central angle 90° is one quarter of the circle, so its area is also πR^2/4.

5. Conclude semicircle area = sector area. Both areas equal πR^2/4, so they are equal.
6. Break the sector and the semicircle up by the chord AB.

   The sector AOB (bounded by OA, OB and the arc AB of the big circle) consists of two pieces: the triangle OAB plus the circular segment S that lies between arc AB and the chord AB (the part of the sector outside the triangle).

   The semicircle on AB consists of the circular segment T bounded by the semicircular arc and the chord AB (on the side opposite O). Its area is the semicircle area we computed.

7. Compare the two decompositions.

   Sector AOB area = triangle OAB + segment S.

   Semicircle area = segment T.

   We already found sector area = semicircle area, so

   triangle OAB + S = T.

8. Identify the lune and finish.

   The lune is the part of the semicircle that lies outside the original circle. That is exactly T minus S. So

   lune area = T − S = (triangle OAB + S) − S = triangle OAB.

   Therefore the area of the lune equals the area of the isosceles right triangle OAB. QED.

Numeric check (example)

Choose R = 2. Then AB = 2√2, semicircle radius r = √2, semicircle area = (1/2)π(2) = π. The 90° sector area = (1/4)πR^2 = (1/4)π(4) = π. Triangle area = (1/2)·R·R = (1/2)·4 = 2. The lune area equals triangle area (2) by the argument above; checking the decompositions numerically confirms the algebraic proof.

Teacher comments and rubric (Amy Chua "Tiger Mother" cadence) — ACARA v9 mapped task

Note to student (direct, strict, encouraging): You must show every step with clarity. No hand-waving. Geometry asks for exact reasons: labeling, computing, and partitioning areas are not optional. If you skip a reason you will not earn full credit. Now do this neatly, and do it correctly.

ACARA v9 learning focus (mapped): geometric reasoning with circles and areas; using properties of central angles, sectors and circle-area formulas; decomposing regions. This task develops problem solving and proof skills expected in middle-secondary mathematics.

Rubric (total 12 marks)

• Correct diagram and labels (A, B, O, chord AB, semicircle) — 2 marks. Be precise: indicate which side of AB the semicircle is on.
• Correct computation of AB and semicircle radius and their areas — 3 marks. Show formulas and simplify (no mysterious shortcuts).
• Correct computation of sector area and statement that sector area = semicircle area — 2 marks. Must show numeric/algebraic equality.
• Correct decomposition argument (triangle + segment S = sector; semicircle = segment T; lune = T − S) and conclusion lune = triangle — 3 marks. Give names to the regions S and T and explain what they are.
• Neatness, clarity, and written explanation enough that someone else can follow each step — 2 marks.

Teacher feedback style (Amy Chua cadence): "You can do better. Your diagram must be neat. If you omitted why the semicircle area equals the sector area, you have not done the work. Label everything; compute precisely; write sentences explaining each step. If you follow the rubric and write every step, you will get a perfect score. No excuses."

Additional enrichment / extension

• Explore Hippocrates’ other lunes: Hippocrates found several configurations where lunes have areas equal to certain polygons — research and present one other example and give a proof.
• Generalize with coordinates: place O at (0,0), A at (R,0), B at (0,R). Use circle equations to compute the exact area of the region of intersection and confirm algebraically the area equality. This practices analytic geometry and integration if desired.
• Ask: for which central angles θ (with OA = OB = R and angle AOB = θ) does the semicircle built on chord AB have the same area as the sector AOB? (Compute AB = 2R sin(θ/2), semicircle radius r = R sin(θ/2), equate (1/2)πr^2 to (θ/2π)πR^2, and solve for θ.)
• Challenge: find all angles θ for which the corresponding lune-area equals the triangle OAB area for that θ (this leads to interesting trigonometric equations).
• Try a numerical exploration: fix R = 1, vary θ, and plot semicircle area minus sector area; see where they cross zero.

Work each extension with the same discipline. Label everything. Write each step. If you do that, you will understand the beauty of these classical results.

Finish strong: practice this proof until you can reproduce it without notes. Geometry rewards precision and persistence.