FINDING THE AREA
By Dario Gonzalez Martinez
We will start with the problem. To this end, we should observe figure 1 below:
Figure 1 
The problem is to find the area of the figure ABCD (yellow shape). The figure KIJL is a square of side a, and arcs KADJ, ICDL, KBCJ and LABI represent a quarter of a circumference each one.
I will show tow different ways to solve this problem. The first one involves integrals, and the second one just requires geometric concept. However, we should learn how to construct the yellow shape in Geometer’s Sketchpad before finding this area.
The steps are as follow:
1) Construct a line KI.
2) Construct the circles with center K and I and radius equal to segment KI.
3) Construct lines perpendicular to KI and through the points K and I.
4) Mark the intersections L and J between the perpendicular lines in step 3 and the circles in step 2.
5) Construct the line LJ.
6) Construct circles with centers L and J and radius equal to segment LJ.
7) Mark the intersections A, B, C and D among the circles in steps 2 and 6 inside the square KIJL.
8) Mark points P, Q, R and S on the arcs AB, BC, CD and DA, respectively.
9) Construct the arcs APB, BQC, CRD and DSA by using the tool Arc Through 3 Points in the menu Construct in GSP.
10) Select the arcs constructed in step 9 and use click on Construct, Arc Interiors and Arc Segments in GSP.
11) Select points ABCD and click on Construct and Quadrilateral Interior in GSP.
Then you can hide the unnecessary constructions, and you are done!!!!
Here is an animation that can help you to understand the steps above.

Animation 1 
SOLUTION 1
Suppose that we draw the figure 1 in a coordinate system as follow:
Figure 2 
I just considered the half green circle and the half blue circle since that is all that we need to solve the problem. First of all, I will explain how I obtained the values
and
The first value can be obtained if we
consider the equilateral triangle ADE.
Hence,
The second value is easy to see since it
is the half of the square’s side length. Since the square OEDF has side a, then
Now we need the equation for the green and blue curve. Since they are the half of circles, we will have
Green curve 

Blue curve 

Table 1 
If we compute the explicit equation, we will have
Green curve 

Blue curve 

Table 2 
Thus, if we calculate the yellow area shown in figure 2, then we will solve the problem since it is the half of the area required. Let’s do this, so we need to calculate the following double integral to find the yellow area:
The first integral is complicated to solve in orthogonal equation, so we should make a trigonometric substitution there. We should see figure 3 below:
Figure 3 
So, we can write
By replacing the above relation in the first integral, we will have:
Since this is the half of the area that
we are looking for, the desired area will be
Where a is the square’s side.
SOLUTION 2
To solve the problem by using only geometric concepts, we will consider the figure 4 below:
Figure 4 
I named each blue area as X, each green area as Y and the yellow area as Z. The latter, of course, is the area that we are looking for. Now consider the following figure:
Figure 5 
Consider the equilateral triangle KID. We will calculate the area closed between arcs KAD and DCI and segment KI. This area is equal to sum the area of the circular sectors KADI and DCIK minus the area of the equilateral triangle KID. In other words, if we remember the name that a put to each area in figure 4, we will have:
After we obtained the above area, you can
follow several strategies to find the desired area (area Z). I will consider again figure 4. The area X +Y is equal to the difference
between the area of a quarter of a circle with radius a and the above area (area 2X + Y + Z), that is,
Finally, the desired area (area Z) is
equal to the difference between the square KIJL’s area and four times X + Y (see
figure 4). So, we can write
We can see that this result coincides
with the result obtained by integration above.