Lune — Definition, Formula & Examples

A lune is a crescent-shaped region bounded by two circular arcs, where one arc curves inward and the other curves outward. It appears when two circles overlap and you consider the area inside one circle but outside the other.

A lune is the region enclosed between two circular arcs of unequal radii that share two endpoints, specifically the area belonging to one circle but not to the other at their intersection. In classical geometry, the lune of Hippocrates is the most famous example, formed between a semicircle and an arc of a larger circle.

Key Formula

A_{\text{lune}} = A_{\text{segment}_1} - A_{\text{segment}_2}

Where:

$A_{\text{lune}}$ = Area of the lune
$A_{\text{segment}_1}$ = Area of the larger circular segment (the outer boundary)
$A_{\text{segment}_2}$ = Area of the smaller circular segment (the inner boundary)

How It Works

To find the area of a lune, you subtract one circular segment from another. Each circular segment's area depends on the radius and central angle of its circle. When two circles intersect, the overlapping region consists of two circular segments. The lune is what remains of one circle's segment after removing the overlap. Hippocrates of Chios (c. 440 BC) proved that certain lunes have areas exactly equal to those of triangles or other rectilinear figures — a remarkable result in the history of squaring curved regions.

Worked Example

Problem: A semicircle with diameter 4 is drawn on the hypotenuse of a right isosceles triangle with legs of length 2√2. A semicircle with diameter 2√2 is drawn on one leg. Find the area of the lune formed outside the larger semicircle but inside the smaller semicircle.

Step 1: Find the area of the smaller semicircle (on the leg of length 2√2). Its radius is √2.

A_{\text{small}} = \frac{1}{2}\pi (\sqrt{2})^2 = \pi

Step 2: Find the area of the larger semicircle (on the hypotenuse of length 4). Its radius is 2.

A_{\text{large}} = \frac{1}{2}\pi (2)^2 = 2\pi

Step 3: The area of the right isosceles triangle with legs 2√2 is 4. By the classical result of Hippocrates, the lune's area equals the triangle's area. This follows because the lune area equals the small semicircle minus the circular segment of the large semicircle that overlaps it, which simplifies to exactly the triangle's area.

A_{\text{lune}} = 4

Answer: The area of the lune is exactly 4 square units — equal to the area of the triangle, with no π involved.

Why It Matters

The Lune of Hippocrates was one of the first curved regions in history to be "squared" — shown equal in area to a polygon. This result is a milestone in the ancient Greek quest to understand the relationship between curved and straight-edged figures, and it connects directly to the deeper (and ultimately impossible) problem of squaring the circle.

Common Mistakes

Mistake: Assuming a lune is the same as the overlapping region (lens shape) of two circles.

Correction: The lens or vesica piscis is the intersection of two circles. A lune is the opposite: the part of one circle that excludes the overlap. They are complementary regions.