What we are given (simple picture to imagine)
Start with a circle of radius r and centre O. Pick two points A and B on the circle so that triangle OAB is an isosceles right triangle with the right angle at O. That means OA = OB = r and angle AOB = 90°.
Draw the semicircle that has AB as its diameter (the semicircle sits on the side of AB opposite O). The part of this semicircle that lies outside the original circle is called the lune. We must show the area of that lune equals the area of triangle OAB.
Step 1 — area of the triangle OAB
Triangle OAB is right-angled at O and the two legs OA and OB both have length r. So the triangle's area is
Area(triangle) = (1/2) × OA × OB = (1/2) × r × r = r²/2.
Step 2 — area of the semicircle on AB
Find the length of the hypotenuse AB. By Pythagoras, AB = r√2. The radius of the semicircle whose diameter is AB is
R = (AB)/2 = (r√2)/2 = r/√2.
So the area of that semicircle is
Area(semicircle) = (1/2) × πR² = (1/2) × π × (r²/2) = π r²/4.
Step 3 — area of the 90° sector of the original circle
The arc AB of the original circle subtends a central angle of 90° (because angle AOB = 90°). The area of the corresponding sector of the original circle is 90/360 = 1/4 of the whole circle, so
Area(sector AOB) = (1/4) × π r² = π r²/4.
Notice something important: Area(semicircle on AB) = π r²/4 = Area(sector AOB). So the semicircle and the 90° sector have the same area.
Step 4 — identifying the overlap inside the circle
Inside the original circle the chord AB cuts off a circular segment. The area of that segment is exactly the area of the sector AOB minus the area of triangle OAB (because the sector is formed by the triangle plus that segment). So
Area(segment cut off by chord AB) = Area(sector AOB) − Area(triangle OAB) = π r²/4 − r²/2.
That circular segment is precisely the part of the original circle that lies on the same side of AB as O. If you look at the semicircle on AB, the part of that semicircle that falls inside the original circle is exactly this same circular segment (you can check this by drawing a diagram: the semicircle lies on the opposite side of AB from O except for where it meets the original circle, and the overlap with the original circle matches the segment cut off by AB).
Step 5 — area of the lune
By definition, the lune is the part of the semicircle that lies outside the original circle. So
Area(lune) = Area(semicircle) − Area(overlap inside original circle).
We already saw Area(semicircle) = π r²/4, and Area(overlap inside circle) = Area(segment) = π r²/4 − r²/2. Therefore
Area(lune) = π r²/4 − (π r²/4 − r²/2) = r²/2.
But r²/2 is exactly the area of triangle OAB (from Step 1). So
Area(lune) = Area(triangle OAB).
Quick numeric check (r = 1)
- Triangle area = 1/2 = 0.5
- Semicircle area = π/4 ≈ 0.7854
- Sector area = π/4 ≈ 0.7854
- Segment (overlap) = sector − triangle ≈ 0.7854 − 0.5 = 0.2854
- Lune = semicircle − overlap ≈ 0.7854 − 0.2854 = 0.5, which equals the triangle area.
Conclusion
Using simple area formulas and Pythagoras, we proved that the lune formed in this construction has the same area as the isosceles right triangle OAB.
Teacher comments (Amy Chua "Tiger Mother" cadence — strict, clear feedback)
Good. If you drew the picture carefully and labelled lengths, you would not make silly mistakes — do that every time. You must always state what you know (OA = OB = r, angle AOB = 90°), compute lengths (AB = r√2), and write area formulas cleanly. Your algebra must be tidy: cancel and substitute correctly. If your working is messy, I will mark it down. Reproduce this proof by hand three times until you can explain each step without pausing.
ACARA v9 – curriculum mapping and learning goals (plain-language)
- Geometry and measurement: use properties of circles (sectors and segments) and Pythagoras to solve area problems.
- Reasoning: construct a logical area argument, justify each step, and use algebra to simplify expressions.
- Modelling: draw and label diagrams accurately, substitute measured quantities, and check with a numerical example.
Rubric (clear, strict — what I expect)
Assessment criteria: (1) Correct diagram and labelling; (2) Correct calculation of lengths (AB) and areas (triangle, sector, semicircle); (3) Clear logical argument linking the overlap to sector − triangle; (4) Correct final conclusion and neat presentation.
| Level | Descriptor (what you must show) |
|---|---|
| Excellent (A) | Perfect diagram and labelling. Correct use of Pythagoras. Accurate areas for triangle, semicircle and sector. Clear justification that overlap = sector − triangle. Correct algebra leading to Area(lune)=Area(triangle). Work is neat and explained in sentences. |
| Satisfactory (C) | Diagram present but may be slightly untidy. Most calculations correct; minor arithmetic or algebra slips not affecting final reasoning. Argument given but with brief or slightly unclear justification for the overlap step. |
| Needs improvement (E) | Missing or incorrect diagram. Errors in computing AB or areas. The connection between sector, triangle and overlap is missing or incorrect. Conclusion not justified. Rework required. |
Suggested next steps
- Redraw the figure carefully, label all lengths, and rewrite the proof neatly — no skipping steps.
- Try the same idea with a different right triangle configuration (compare results) to develop deeper understanding.
- Explain the overlap = sector − triangle step in words to a friend — if you can teach it, you truly understand it.
Remember: neat diagram, correct substitution, clean algebra. Do not be sloppy.