Skip to content
π
LearnMath
TutorialsCalculatorsPractice
DE
Calculator

Lune of Hippocrates Calculator

Find the area of the lunes of Hippocrates — the two lunes over the legs together equal the right triangle itself: a·b/2. Step by step with examples and FAQ.

Quick answer
What is the area of the lunes of Hippocrates?
For a right triangle with legs a and b, the two lunes over the legs together equal exactly the area of the triangle itself: A = a·b/2. Example a = 3, b = 4 → A = 6.
The tool

Enter values — get full working

Comma or dot as decimal separator, negative values allowed.
Step-by-step
Press Calculate to see every step.
HowTo

Lune area — 3 steps

Using a = 3, b = 4
  1. 1
    Step 1 of 3

    Note the legs

    Write the two legs a and b of the right triangle, e.g. a = 3 and b = 4.

  2. 2
    Step 2 of 3

    Form the triangle area

    A△ = a·b/2 = 3·4/2 = 6. The semicircles are defined over the hypotenuse c = √(a²+b²) = 5.

  3. 3
    Step 3 of 3

    Apply Hippocrates’ theorem

    The two lunes together equal the triangle: A = a·b/2 = 6.

Examples

Lune of Hippocrates — worked examples

Legs and lune area
a=3, b=4
c=√(9+16)=5
3·4/2
6
a=1, b=1
c=√2
1·1/2
0.5
a=5, b=12
c=13
5·12/2
30
a=6, b=8
c=10
6·8/2
24
a=2, b=5
c=√29
2·5/2
5
Theory

Hippocrates’ theorem

The lunes of Hippocrates are among the most beautiful results of ancient geometry. Take a right triangle with legs a and b and hypotenuse c = √(a²+b²), and draw a semicircle outward over each side. Over the hypotenuse sits a large semicircle, over each leg a smaller one. The crescent-shaped regions between the small semicircles and the large one are called lunes (Latin lunulae). Around 440 BC Hippocrates of Chios showed that the two lunes over the legs have a combined area exactly equal to the triangle, that is a·b/2. The proof uses the fact that semicircle areas are proportional to the square of their diameter, so that when you subtract, the Pythagorean theorem makes all the circular parts (and hence π) cancel. The striking point is that a region bounded by circular arcs is exactly squared by a straight-sided one — long taken as hope that the circle itself could be squared, which later proved impossible.

Pitfalls

Common mistakes

Adding the semicircle areas separately

You don’t need to work through the circle areas — the theorem gives a·b/2 directly. The circular parts cancel.

Using the hypotenuse as a leg

a and b are the two short sides at the right angle; c = √(a²+b²) is the hypotenuse, not an input.

Counting only one lune

The theorem is about the sum of both lunes over the legs, not one alone.

Triangle not right-angled

Hippocrates’ theorem assumes a right angle between a and b.
FAQ

Frequently asked questions

Glossary

Glossary — key terms explained simply

Lune
Crescent-shaped region between two circular arcs.
Leg
One of the two sides at the right angle.
Hypotenuse
The longest side, opposite the right angle, c = √(a²+b²).
Hippocrates’ theorem
The lunes together equal the area of the triangle.
Semicircle
Half a circle’s area drawn over a side of the triangle.
Quadrature
Representing a curved-edged region by a straight-edged one.