CALCULATING THE AREA OF A LUNE
First of all, we have to know what a lune is. On the nails on your fingers, specially thumbs, you can see a white semi-circle on the lower area. This area of the nail is called lunula (word that comes from the latin: lunula = small moon-lune). you can see this in the following image:
You might be wondering what this has to do with geometry.
We also study the lune in geometry.
1st. we draw a quadrant (we are talking about the first quadrant in gray):
We join A and B points:
We find the middle point for segment which measures 5,66 cm. With this measurement, we trace a semi-circle that goes through points A and B. The area in green is our lune:
To calculate the area in green (lune), we have to subtract the area of the circular segment (in red).
Lets calculate the area of the lune:
1. We calculate the area of the semi-circle, of which the lune is a part of (the areas in red and green), which measures 2,85 cm.in radius:
We divide it by 2 since we are working with a semi-circle.
2. To calculate the area of the circular segment, we find first the area of the circular sector corresponding to a 90º angle; we have the first quadrant in gray, which is to say, a fourth of the entire circle. This circle has as radius a segment of 4 cm., whose area is:
We have divided the area of the entire circle (with a radius of 4 cm.) by 4 because we want to find the area of one of the 4 quadrants.
3. Now we calculate the area of the triangle whose height is 2,85 cm., and its base 5,66:
The difference between the area of the circular sector and the area of the triangle will be the area of the circular segment:
4. The area of the lune will be gotten by subtracting the area of the semi-circle minus the area of the circular segment:
Another way for calculating the area of the lune is:
area of the semi-circle:
We subtract the area of the circular sector:
We find the area of the triangle:
From this result, we add the difference between the area of the semi-circle and the area of the circular sector:
15(2).33 Draw the corresponding lunes to the catheti (commonly known as legs or "the other two sides") of a right-angled triangle whose measurements in centimetres are shown in the following image. It shows that the sum of its areas correspond to the area of the triangle.
We find the centre of the circumference limited by the triangle, which means, the place where the bisectors cross (a bisector is the perpendicular line of a segment and which goes through its middle point) and we trace the circumference:
The bisectors, in black, are perpendicular to the sides of the triangle going through their middle points A, B and C respectively. Once the circumference has been traced, we can see the triangle inside.
Parting from the middle points of the catheti, we trace 2 circumferences (in magenta) with a radius equal to half the length of the leg:
Thus, we have found the lunes which correspond to the catheti A and B:
We will eliminate the curves and segments which wont be necessary for our calculation.
At the same time, using a protractor, a ruler, and a compass (drafting), we will calculate some measurements we will need:
Lets calculate the area of lune 1:
1. We find the area of the surface of the semi-circle (in red) in the following image, in which the radius equals:
and the area of the semi-circle is:
2. We find the area of the circular sector (in green) for the following shape, knowing that the radius equals:
If 360º corresponds to an area of ,
103º will correspond to an area of
3. We find the area of the triangle (in gray), of which we already know its height and base as we can see in the following image:
4. The area of lune 1 will be:
Lets calculate the area of lune 2:
1. We find the area of the surface of the semi-circle (in red) in the following image, in which the radius equals: 3,44 cm.
and the area of the semi-circle is:
2. We find the area of the circular sector (in green) for the following shape, knowing that the radius equals:
if 360º corresponds to an area of ,
77º will correspond to an area of
3. We find the area of the triangle (in gray), of which we already know its height and base as we can see in the following image:
4. The area of lune 2 will be:
The sum of the areas for the two lunes is:
The area of the right-angled triangle in the following image is:
as we can see, it has a height of 5,22 cm. and a base of 11 cm., so the area is:
You can test that the sum of the areas for both lunes (corresponding to the catheti of the right-angled triangle) is equal to the area of the triangle.
We can draw interesting geometrical shapes using lunes:
Here, you have 4 lunes turning 90º around a central point.
Horizontally:
And finally, a sample:
This shape has been turned to an angle of 6º around A vertex. This means that we have used the same drawing 60 times. This is the resulting image: