Goal. Show that the lune (the part of the semicircle built on the hypotenuse that lies outside the original circle) has the same area as the isosceles right triangle whose right angle is at the circle's centre.
Setup and notation. Let O be the centre of the original circle and let its radius be r. Place two points A and B on the circle so that angle AOB = 90°. Then triangle OAB is an isosceles right triangle (OA = OB = r). The hypotenuse AB has length r√2. Let M be the midpoint of AB and draw the semicircle with diameter AB on the same side of AB as O (this semicircle has centre M and radius r/√2). The lune is the part of that semicircle lying outside the original circle.
Step 1 — area of the triangle OAB.
Because OA and OB meet at right angle, the area of triangle OAB is
Area_triangle = (1/2) * OA * OB = (1/2) * r * r = r^2/2.
Step 2 — area of the semicircle on AB.
The hypotenuse AB = r√2, so the semicircle built on AB has radius (r√2)/2 = r/√2. Its area is
Area_semicircle = (1/2) * π * (r/√2)^2 = (1/2) * π * (r^2/2) = π r^2/4.
Step 3 — area of the quarter of the original circle.
The sector AOB of the original circle has central angle 90°, so its area is
Area_sector = (90°/360°) * π r^2 = π r^2/4.
Note: numerically the semicircle area equals the quarter-circle sector area: π r^2/4.
Step 4 — understand the overlap.
The chord AB cuts the quarter-sector AOB into two pieces: (i) the triangle OAB and (ii) the circular segment between the chord AB and the arc AB of the big circle. That circular segment has area
Area_segment = Area_sector - Area_triangle = (π r^2/4) - (r^2/2).
Geometrically that circular segment is exactly the part of the quarter-sector which sits inside the semicircle (because the semicircle is the circumcircle of triangle OAB — by Thales' theorem the circle with diameter AB passes through O, A and B). Thus the intersection of the semicircle and the original circle is exactly the circular segment described above.
Step 5 — area of the lune (what we want).
Lune area = Area_semicircle - Area_of(intersection with big circle).
Using the facts above, the intersection area is Area_segment, so
Lune area = (π r^2/4) - [ (π r^2/4) - (r^2/2) ] = r^2/2.
Therefore Lune area = Area_triangle = r^2/2. That completes the proof.
Short summary (one-line).
The semicircle on AB and the quarter-sector AOB have equal area; the part they share is the sector minus triangle, so subtracting gives exactly the triangle's area as the lune.
Why this is true intuitively.
Because AB is the diameter of the circumcircle of triangle OAB (Thales' theorem), the semicircle built on AB is tightly related to the quarter of the original circle. The equal-area coincidence (both have area π r^2/4) makes the algebra tidy: the lune is what remains after removing the piece that the semicircle and the big circle share, and that leftover turns out to be precisely the triangular area.
Additional enrichment / extension problems.
- Generalize: choose angle θ at the centre (not necessarily 90°). Build the isosceles triangle with central angle θ and the semicircle on its base. Compute the lune-like region area and see when a similar simple relation holds.
- Explore Hippocrates' original collection of lunes: find other lune constructions (different angles and circles) whose areas are simple rational multiples of triangle or circular sectors.
- Coordinate challenge: put O at (0,0), A at (r,0), B at (0,r). Write equations of both circles and compute the lune area by integration to check the algebraic result directly.
- Construct the figure in dynamic geometry software (GeoGebra). Drag A and B while keeping angle AOB = 90° and watch how the semicircle, sector and lune move; measure areas numerically to see the equality persist.
- Investigate whether there are non-right isosceles triangles with a semicircle-on-hypotenuse lune equal to some simple polygon area — try experiment and conjecture.
Teacher comment (Nigella Lawson cadence — proficient/exemplary, ACARA mapped, ~100 words):
My dear, you have taken the dry bones of geometry and braised them into something sumptuous. With steady hands you arranged the circle, folded the right isosceles triangle into its center and slid the semicircle gently across the hypotenuse. The algebra and area work glazes cleanly; your reasoning is glossy, each step a ribbon of flavour. You show precision in measurement, clarity in argument, and creative appetite for extension. For ACARA proficiencies, your proof meets the criteria: accurate, logical and well-communicated. Keep stirring curiosity and serve each conjecture with confidence — mathematics, like food, rewards care and joyful discovery daily.
If you want, I can paste a GeoGebra link, draw a clear diagram for printing, or walk through the coordinate-integration version step by step.