The Relationship of Areas That Heron's Formula and Brahmagupta's Formula Produces

Abstract 

The present paper seeks to discover the connection in the application of Heron’s and Brahmagupta’s formulae in geometry. Heron’s formula is named after the Hero of Alexandria, and its primary purpose is to help calculate the area of a triangle when the lengths of its three sides are given. When using Heron’s formula, there is no need to use the angles of the given triangle for calculation and other distances in the triangle since the calculation relies on the lengths of the sides. Brahmagupta’s formula, on the other hand, is applied in Euclidean geometry to find the area of a cyclic quadrilateral, especially those that are usually inscribed within circles given the length of the sides. In the paper, error detection will be used between the two formulae to help in understanding and improving the process of calculating the area of a quadrilateral triangle. The study will also use six different quadrilateral triangles in the GeoGebra to demonstrate the application of the two formulas. 

Introduction 

This writing works on how Brahmagupta’s Formula only works on the quadrilateral that lies on the edge of the circle. This seems not essential, but it is possible to find a better way to find the area by finding the relationship between Brahmagupta and Heron’s Formulas. Also, if we can find a better way of finding an area, then it will help math educators show students different ways of finding the area of the quadrilateral. It will also improve student’s mathematical skills by giving more opportunities by providing several ways of solving problems. 

First, we must find the value of S to find a general equation of the Brahmagupta’s Formula: . After finding S, it will be plugged into the A= . This equation will give the area of the quadrilateral. When our group was trying this formula in the classroom, it gave us a few questions that triggered our curiosity. What will happen if one of the points gets farther away from the edge of the circle? What happens to the area if the point gets closer to the center of the circle? 

To answer questions, we will have to know the difference between Brahmagupta’s Formula and Heron’s Formula and how their answer is different. The first step is drawing a circle since we want to place three points on the edge of the circle. Next, we will set a total of six points that lie outside, on edge, and inside of the circle so we can have enough data to find the relationship of each quadrilateral. Then, we will find the area of each quadrilateral by using Brahmagupta’s Formula and Heron’s Formula. Heron’s Formula will give the correct value of the area, while Brahmagupta’s Formula doesn’t provide the right amount when they are not on the edge of the circle. Lastly, we will compare the areas that Heron and Brahmagupta’s formula supplied to us. 

Background 

Heron’s Formula  If a, b, and c are the sides of a triangle,  and if the semi-perimeter,  ,  then the area of a triangle is  square units.

Problem-solving is a critical skill that any mathematician is required to have. The way mathematicians look at any problem, which is already solved or unsolved, is always different. To solved problems, they always ask themselves whether there are any other solutions, the better and shorter ones. Because of that view, mathematicians create a lot of new things with more applications. 

Figure 1 

Formula for finding the area of quadrilateral (Heron)

Figure 1 

Formula for finding the area of quadrilateral (Heron)

Heron was a famous mathematician and engineer who was born in Egypt during 10 AD and died 70 AD. His achievements were inventing the first vending machine, a wind-wheel, the force pump, etc. As a mathematician, he created the formula for finding the area of the quadrilateral (see Figure 1). 

Figure 2 

Formula for finding the area of quadrilateral (Brahmagupta)

Figure 2 

Formula for finding the area of quadrilateral (Brahmagupta)

Brahmagupta’s Formula  If a, b, c, and d are the sides of a cyclic quadrilateral,  and if the semi-perimeter,  ,  then the area of the cyclic quadrilateral is  square units.

Brahmagupta is an Indian mathematician who was born during 598 CE and died 668 CE. He is well-known for his contributions to algebra, arithmetic, Diophantine analysis, geometry, and trigonometry. One of his most significant contributions to geometry is Brahmagupta's formulas. Unlike Heron's Formula, Brahmagupta's Formula does not require us to find areas of two triangles. If four points of the quadrilateral are on the circle and distance of four sides are given, then we can calculate the area (see Figure 2). 

Our group’s topic can be like Babylonian’s square root problem, which is finding the approximation of the distance. Their primary goal is to find an approximation of the range, and our group is concentrating on finding the area of quadrilateral by using Brahmagupta’s Formula. Brahmagupta’s Formula won’t give accurate value, but it will be able to find an approximation of the area. 

Problem Description and Method 

This section deals with finding the area of quadrilateral by using Brahmagupta’s Formula, which is shown below: 

Brahmagupta’s Formula is one of the methods that find the area of the quadrilateral that lies only on the edge of the circle. Our group wanted to see the relation of the formula to the circle and what happens to the result as one of the points get farther away from the circle and get close to the circle by using Brahmagupta equation. By using GeoGebra, one of the computer systems that find the result of math problems, our group was able to find the six pairs of areas by using Brahmagupta’s Formula and Heron’s Formula. If GeoGebra or any computer system is not available, we can also use a calculator since we know the formulas that we are using and the length of the quadrilateral. Since we know that Heron’s formula can be used on any quadrilateral, we compared both results to find the error. First, we observed the area when one of the points lies farther away from outside of the circle. Next, we observed the area when the point lies on the edge of the circle. Then, we found the area of the quadrilateral when the point lies inside the circle. Lastly, our group compared Brahmagupta’s Formula and Heron’s Formula to find the error and find the relation of them when the point gets close to the circle. 

Findings 

Figure 3 

Picture of six quadrilaterals

Here we will look at the picture of the six quadrilaterals; the first thing we will do is finding the area of the quadrilateral that has the farthest point from the circle. After finding the farthest one, we will find the area that gets close to the edge of the circle. Then, we will find the area where the point gets inside of the circle. (See Figure 3) 

BCDF
Segment	Distance
BC	5.5
CD	4.5
DF	11.9
BF	11
S	16.45
AREA	54.1
(B’s FORM)	56.96328
	Chart 1

Observation 1 

First area that we are going to find is the farthest quadrilateral shown in Figure 3. When 

we used the Brahmagupta’s Formula, first step was finding S (see Chart 1). which was 16.45. Then, we will plug in 16.45 to the Formula. 

and the result would be 56.96 . Heron’s Formula will be finding the area of the two triangles and add them together which gives 54.1 . By using these two numbers, error was -5.29%. By looking at the error, we know that Brahmagupta’s Formula did not give us the correct measurement of the area. Then, what will happen to the area if the point gets closer to the edge of the circle? 

Observation 2 

BCDG
Segment	Distance
BC	5.5
CD	4.5
DG	10
GB	9.1
S	14.55
AREA	46.1
(B’s FORM)	47.491
	Chart 2

This time, we will find the area of second farthest (Figure 3) quadrilateral (Chart 2) which is second farthest from the edge of the circle. First step is finding S. which gives 14.55. Then we will plug the S to the formula. 

A=

This will give the area of 47.491 . Next, we will have to find the area by using Heron’s Formula, and it will give the area of 46.1 . By using these two results, we know that the error is about -3.02%. Surprisingly, the percentage of error decreased as point got closer to the edge of the circle. Next, we will try to find the area of the quadrilateral that has the closest point from the circle. 

Observation 3 

BCDH
Segment	Distance
BC	5.5
CD	4.5
DH	7.9
HB	7.1
S	12.5
AREA	37.1
(B’s FORM)	37.29665
	Chart 3

Lastly, we are going to find the area of quadrilateral that is close to the edge of the circle (Figure 3). In the date that is shown at Chart 3, we will see how error of the area decreased. First step is finding S. which gives about 12.5. Then we will plug the S to the formula. 

A= The result was about 37.3 when it was rounded up. Next step is finding actual area which is 37.1 . So, the error of the area is -0.53%. What would be the relationship of the area based on the location of the point? 

Overall 

Figure 4 

A graph that shows the change of error (Used figure 3) 

By doing similar work on the area of the last two quadrilaterals in the inside of the circle shown in Figure 3. The quadrilateral that was close to the edge had about -1.08%, and another quadrilateral that was close to the center of the circle had about -10.5% error. The relationship of the point would be as the point of the quadrilateral gets close to the edge of the circle, the error of the area decreases (Figure 4). If the point of the quadrilateral gets farther from the circle, the error increases. This was also applied inside of the circle. When the point got closer to the center of the circle, the error got greater than the area of the quadrilateral that was outside of the circle. 

Conclusion 

As has been discussed in this paper, we can generalize one mathematical concept to a new or complex problem. Heron’s formula calculates the area of triangles based on the given lengths while Brahmagupta’s formula calculates the area of quadrilaterals inscribed in a cycle (Shell-Gellasch & Thoo, 2015). The two formulae are similar since it is possible to set the length of the fourth side of a quadrilateral to zero, which makes it a triangle (Alexander & Koeberlein, 2011). In turn, this represents a compelling application concept that can be used in different areas and topics in mathematics. Mathematicians can learn from this study about the importance of being creative in finding pairs of concepts that can be applied in their areas of specialization and applying them in pairs that bear the same or closely linked connections. 

Mathematics today is a dynamic subject that requires a lot of creativity and critical thinking in problem solving. These critical skills can be exploited when finding solutions to everyday problems and challenges. To effectively come up with solutions to a problem, a mathematician should look beyond one solution and discover other potential solutions to the same problem in order to achieve progress.