Lesson 11ª

 

 

 

 

 

   

 

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:


bhj

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):

fvg

We join A and B points:

cgfbc

We find the middle point for m,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:

xdgder

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:

r

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:

sdz

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:

w5

The difference between the area of the circular sector and the area of the triangle will be the area of the circular segment:

er

4. The area of the lune will be gotten by subtracting the area of the semi-circle minus the area of the circular segment:

767

Another way for calculating the area of the lune is:

rer

area of the semi-circle: re46r5

We subtract the area of the circular sector:  9

8yh7u

We find the area of the triangle: ftruu

From this result, we add the difference between the area of the semi-circle and the area of the circular sector: tugu

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.

drt

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:

dryht

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:

fyghu

Thus, we have found the lunes which correspond to the catheti A and B:

huih

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:

uihi

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: hihuiui

uiiu

and the area of the semi-circle is: ouio

2. We find the area of the circular sector (in green) for the following shape, knowing that the radius equals: jik

jkl

If 360º corresponds to an area of jkljghy ,

103º will correspond to an area of    ytfy

drt

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:

rtyufu

6erf5y

4. The area of lune 1 will be:

fgtugy

 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.

dtry

and the area of the semi-circle is: rt57

2. We find the area of the circular sector (in green) for the following shape, knowing that the radius equals: 5ed46

r6ert

if 360º corresponds to an area of tfhy,

77º will correspond to an area of    frtju

fujvfg yj

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:

ftjuvh

cfg

4. The area of lune 2 will be:

c fhcg

The sum of the areas for the two lunes is: 

hcvhcvbh

The area of the right-angled triangle in the following image is:

cfgch

as we can see, it has a height of 5,22 cm. and a base of 11 cm., so the  area is:

chvbh

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:

vhn b

Here, you have 4 lunes turning 90º around a central point.

Horizontally:

fvghfgth

And finally, a sample:

dhdc bh

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:

cghf ghc