Developing Formulas for Quick Calculation of Polyhedron Volume in Spatial Geometry: Application to Vietnam

: In the age of globalization, an effective leadership skill is the ability for quick calculation of work-related problems. From an economic perspective, fast computation often provides a competitive advantage in business, where speed, efficiency and accuracy are required. Quick calculation techniques are a central problem in modern mathematics because it shortens the time for solving technical problems. The purpose of the paper is to provide an explanation that will lead to a quick solution to a volume problem. Specifically, some convenient formulas are provided for quick calculation of the volume of the common polyhedron, together with a number of multiple-choice questions with IATA software to practice. Based on the evaluation results, reliable multiple-choice questions are used for an empirical study in Can Tho City, Vietnam on the effectiveness of the formulas for quick calculation of the polyhedron volume in spatial geometry. Statistical analysis shows that quick formulas help students to complete lessons at a higher rate, thereby contributing to improvements in the effectiveness of teaching geometry, especially the volume of the Polyhedron.


INTRODUCTION
In the age of globalization, an effective leadership skill is the ability for quick calculation of work-related problems. From an economic perspective, fast computation often provides a competitive advantage in business, where speed, efficiency and accuracy are not only useful but also essential. Quick calculation techniques are a central problem in modern mathematics because it shortens the time for solving technical problems.
As an example of an interesting problem in spatial geometry, consider a company that produces a monolithic wooden souvenir in the shape of a pyramid, as shown below. Knowing the wood block with the bottom side is 6cm square, the side is 9cm long. How many cm 3  In order to address this problem, one needs to build a mathematical model, then use geometric formulas to calculate the solution. This work is quite simple for a mathematician, but it may not be quite as simple for others. The question to be addressed can be presented as: "How can a system of formulas be constructed for a quick calculation of polyhedron volume in spatial geometry in order to solve problems in real life?" Returning to the initial problem, call the bottom and side edges of the wooden pyramid a and b, respectively. Then it is possible to formulate the volume of wood blocks as: V = a 2 4b 2 ! 2a 2 6 With a = 6cm and b = 9cm, using the formula leads to a quick calculation of the volume of wood.
From the illustration above, it can be seen that the construction of a quick calculation system of the pyramid volume has two benefits:

1.
The time to calculate the volume of the pyramid problem will be faster; 2.
The quick calculation formula can be programmed easily.
The procedure for establishing the formula can be considered as a type of knowledge production. According to Bribena (2019), the production of such knowledge is closely related to the process of globalization, with both offering important elements in the global commodity and commercial markets, where speedy computation can provide a competitive advantage. The process of producing knowledge also plays an essential role in categorizing global organizations in different systems.
From a different perspective, according to Sousa et al. (2019), to be a good leader in the industrial age 4.0 requires at least two qualities, namely good work and computational thinking. With the construction of quick calculation formulas, its use by skillful leaders will save time and improve work efficiency. Creating such a recipe also helps to train people to learn computational skills develop a specific mindset, thereby connecting the real world to its virtual counterpart.
From an educational perspective, mathematics is a very important and influential subject for admission to a university or college. Therefore, in order to obtain a high score in this subject, we need to have knowledge and understanding of mathematical concepts and essences. In particular, spatial geometry has the most obvious difference for algebra which is the theory of space, thus learners should study theories as well as theorems to be able to apply effectively.
In order to address the problems of spatial geometry, it is important to draw the right shape. Imagine how you can see the surface, not see the surface, from which one can know the shapes that we need to calculate it. But to be able to solve quickly to spatial geometry, it is crucial to practice this issue regularly, because only practice will increase the ability to think and apply theorems quickly. About spatial geometry, this issue is currently researched theoretically and practically by several scientists and has gained many attentions in the last two decades.
For instance, Bertoline (1991) presented to use 3D geometric models to teach spatial geometry concepts. Bako (2003) researched about different projecting methods in teaching spatial geometry. Laborde et al. (2006) introduced about teaching and learning geometry with technology. Kosa and Karakus, F. (2010) researched to dynamic geometry software Cabri 3D for teaching analytic geometry. Kurtulus and Uygan (2010) presented to the effects of Google Sketchup based geometry activities and projects on spatial visualization ability of student mathematics teachers. Moss et al. (2015) introduced about adapting Japanese Lesson Study to enhance the teaching and learning of geometry and spatial reasoning in early year classrooms: a case study. Peter (2012) introduced to critical thinking: Essence for teaching mathematics and mathematics problem solving skills.
In Vietnam, the renewal of the form of multiplechoice assessment of mathematics from National high school exams from 2017 shows progress in education reform. Since then, the renewal of teaching and learning methods to suit new exam forms has also been continuously set and achieved significant results. Nevertheless, to meet that requirement, students need a tremendous amount of knowledge to solve 50 multiple-choice questions within 90 minutes. In which questions about spatial geometry still occupy relatively large time of students. About multiple-choice questions, there are many scientists have studied and utilized it, for example, Lindberg et al. (2010) researched to new trends in gender and mathematics performance: a meta-analysis. Torres et al. (2011) presented about improving Multiple-Choice Questions. Little et al. (2012) introduced to multiple-choice tests exonerated, at least of some charges: Fostering test-induced learning and avoiding test-induced forgetting. Lesage et al. (2013) researched about scoring methods for multiple choice assessment in higher education-Is it still a matter of number right scoring or negative marking? Lindner et al. (2014) presented about tracking the decision--making process in multiple--choice assessment: Evidence from eye movements. Azevedo (2015) introduced to e-assessment in mathematics courses with multiple-choice questions tests.
Regarding the use of formulas in teaching mathematics, Chazan (2000) proved effective when applying the Algebraic formula system in High Schools. For quick manipulations, Wheatley (2016) has developed rapid drawing techniques applied in mathematics; Boslaugh (2012) provides a statistical view of fast citation tools on the desktop. In Geometry, Grabinski et al. (2002) also studied simple formulas related to straight lines. From studies and context of teaching mathematics in Vietnam, a question arises: "How to build formulas for quick calculation of polyhedron volumes in geometry for the teaching of mathematics?" In order to provide an answer to this question, this paper first introduce to formulas to quickly calculate of polyhedron volume in spatial geometry based on the knowledge learned to help students solve multiplechoice questions quickly and effectively. This is the primary motivation for us to study this article. The rest of the paper is organized as follows. The literature review is presented in Section 2. We develop formulas to quickly calculate of polyhedron volume in spatial geometry in Section 3. In Section 4, we investigate to empirical analysis at Can Tho City to evaluate efficiently of the proposed formulas. Concluding remarks and inference will be provided in the last section.

Pyramid
Given convex polygons A 1 A 2 ...A n and a point S is outside the plane containing that polygon. Connecting S with vertices A 1 , A 2 ,..., A n to obtain n triangles:

Shapes consists of n triangles and polygons
A 1 A 2 ...A n is called a pyramid and it denoted by S.A 1 A 2 ...A n .

Equilateral Pyramid
A pyramid is called an equilateral pyramid if its base is a regular polygon and its sides are equal. Sometimes we also define the following: A pyramid is called an equilateral pyramid if its base is a regular polygon and has the ending point of perpendicular lines that coincides with the center of the base polygon.
In an equilateral pyramid, it can be seen that, side edges and side surfaces create the base polygons of equal angles.

Prisms
Shapes is created by parallelograms In a prism, it has been seen that, the side edges are equal, the side surfaces are parallelograms and two bottoms are equal polygons.

Box Shape
A box shape is a prism has the base is a parallelogram.
In a box shape, it can be seen that the side surfaces are parallelograms and the diagonal of the box shape intersects at the center of each line. We now introduce to formulas to quickly calculate of polyhedron volume in spatial geometry in the next section.

DEVELOPING FORMULAS FOR QUICK CALCULATION OF POLYHEDRON VOLUME IN SPATIAL GEOMETRY
According to Jones (2002), the demonstration of Mathematical properties is conducted in the following order: Explanation → Discovery → Intellectual challenge → Verification → Systematization. This process has two major problems for learners: proof time and intelligence. To shorten the process, we demonstrate the general formulas of the volume of common polyhedron. In order to shorten the time, we have developed a recipe table for learners. The following presentation relates to the proof and formulation of the general formula.

Equilateral Triangular Pyramid
Formula 1: Given an equilateral triangular pyramid S.ABC has a base ABC is an equilateral triangle with its edges equal to a and the side edge equals to b, then one has: Solution Let G is a center of !ABC " SG # ( ABC) It can be observed that !ABC is an equilateral triangular In addition, we have !SGA is a right triangle at G thus: In case of a = b ! V = a 3 2 12 .
Formula 2: Given an equilateral triangular pyramid S.ABC has a base edge with length equal to a and the side surface created with the base plane is an angle is ! , then one has: Thus the angle of between SBC ( ) and ABC ( ) is SG = GM.tan! = 1 3 AM.tan! = a 3.tan! 6 and obtain V S . ABC = 1 3 .SG.S !ABC = 1 3 . a 3.tan" 6 . a 2 3 4 = a 3 tan" 24 .
Formula 3: Given an equilateral triangular pyramid S.ABC has a side edge with length equal to b and the side edge created with the base plane is an angle is ! , then one has: !SGA is a right triangle at G , one has: On the other hand, !ABC is an equilateral triangular, one has Therefore, V S . ABC = 1 3 .SG.S !ABC = 3b 3 .sin " cos 2 " 4 .

Formula 4:
Given an equilateral triangular pyramid S.ABC has a base edge with length equal to a and the side edge created with the base plane is an angle is ! , then one has : It has been seen that, !ABC is an equilateral triangular !SGA is a right triangle at G , thereafter we get: Formula 5: Given an equilateral triangular pyramid S.ABC has a base edge with length equal to a. Let (P) is the plane passing through A and parallel to BC, and perpendicular to the plane (SBC) , the angle between the plane (P) and the base plane is ! , then one has: It can be observed that, !ABC is an equilateral triangular On the other hand, !SGA is a right triangle at M , thereafter we obtain: and get, S S . ABC = 1 3 .SG.S !ABC = 1 3 . a 3.cot " 6 . a 2 3 4 = a 3 .cot " 24 .

Equilateral Quadrilateral Pyramid
Formula 6: Given an equilateral quadrilateral pyramid S.ABCD with its base ABCD is a square with the length of side is a and its side edge with the length equal to b: If an equilateral quadrilateral pyramid has all of sides with the length equal to a, one has: In the other side, !SOM is a right triangular at O, one has: Formula 7: Given an equilateral quadrilateral pyramid S.ABCD has its base edge with the length is a, the angle between the side surface and the base plane is ! :

and M M is a midpoint of CD
Thus the angle of between (SCD) and ( ABCD) is Formula 8: Given an equilateral quadrilateral pyramid S.ABCD has its base edge with the length is a, the angle between the side edge and the base plane is: Formula 9: Given an equilateral quadrilateral pyramid S.ABCD has its base edge with the length is !SOM is a right triangular at O , thereafter we get: Formula 10: Given an equilateral quadrilateral pyramid S.ABCD has its side edge with the length is a, the angle between the side surface and the base plane ) . Then one has: M is a midpoint of CD Thus we have the angle of (SCD) and ( ABCD) is It can be seen that !SMC is a right triangular at M , one has: In addition, one has !SOM is a right triangular at O thus:

Normal Pyramids
Formula 11: Given a pyramid S.ABC has SA perpendicular with ( ABC) , and (SAB) perpendicular Solution !SAB is a right triangular at A , one has: It can be seen that, !SBC is a right triangular at B In addition, we have:

Regular Tetrahedron
Formula 12: Given a regular tetrahedron S.ABC with the length of its edge is a:

Nearly Regular Tetrahedron
Formula 13: Give a nearly regular tetrahedron:

Square Tetrahedron
Formula 14: Given a tetrahedron SABC has (SAB), (SBC), (SAC) perpendicular to each other in pairs, the triangle area of SAB, SBC, SAC are S 1 , S 2 , S 3 , respectively: Formula 15: Given a tetrahedron SABC has SA, SB, SC perpendicular to each other in pairs and the length of them are a, b, c , respectively:

Normal Tetrahedron
Formula 16: Given the length of 3 edges and 3 degrees at the top: 1! cos 2 " ! cos 2 # ! cos 2 $ + 2 cos" cos # cos $ Formula 17: Given the length of the opposite sides, the distance and the angle of that two sides: abd sin! Formula 18: Given the length of an edge, area and angle between two adjacent surfaces: Formula 19: Given the length of 3 edges, 2 vertices and 1 dihedral angle:

Volume of Other Polyhedral
Formula 20: The octahedron has vertices that centers of the sides of the cube with the length of its edge is a: Thereafter, we get: Formula 21: Given a regular octahedral with the length of its edge is a. Connect the center of side surfaces to the cube: Solution: It can be seen that GP = So the volume of the cube V = a 2 3

Formula 22:
The regular dodecahedron with the length of its edge is a. Then one has : V = a 3 (15 + 7 5) 4 In order to evaluate efficiently of the proposed formulas, we next turn on investigate to empirical analysis at Can Tho City in the next section.

Objectives Experiment
In this paper, we evaluate 40 objective multiple choice questions with 4 options on the subject volume of polyhedron by IATA software. We aim to compare traditional work with the work has the support of quick calculation formula.

Tasks Experiment
We first build test questions and a summary table of recipes for experiment, and then we obtain the results of student work to data analysis. Finally, we evaluate experimental results.

Experimental Objects
Test questions are experimented on subjects who have learned through chapter 2, geometry 12. We have surveyed 175 students at Nguyen Viet Hong High School, Chau Van Liem High School, Phan Van Tri High School, Ly Tu Trong High School for the Gifted, High School Pedagogical Practice, Phan Ngoc Hien High School, An Khanh High School, Nga Sau High School, Bui Huu Nghia High School, Luu Huu Phuoc High School.

Experimental Subject
Question 1: Given an equilateral triangular pyramid S.ABC has its base is an equilateral triangle ABC with the length of its edge is a , the length of the side edge is 3a . Volume of pyramid S.ABC will equal to: Question 2: Given an equilateral triangular pyramid S.ABC has its base is an equilateral triangle ABC with the length of its edge is a , the side edge with the length is 2a . Volume of pyramid S.ABC is:
Question 3: Given an equilateral triangular pyramid S.ABC has the length of its base edge is a and the side surface creates a base plane at an angle of 60°. Volume of pyramid S.ABC is given by: Question 5: Given an equilateral triangular pyramid S.ABC has the length of its base edge is a 2 and the side edge creates a base plane at an angle of 60°.
Volume of pyramid is illustrated by: a 3 3 6 16 .
Question 6: Given an equilateral triangular pyramid S.ABC has its base edge with the length is a and the side edge creates a base plane at an angle of 30°.
Volume of pyramid will equal to: 3a 3 32 .
Question 7: Given an equilateral triangular pyramid S.ABC has the length of its base edge is 2a and the side edge creates a base plane at an angle of 30°.
Volume of pyramid S.ABC is:
Question 8: Given an equilateral triangular pyramid S.ABC has the length of its base edge is a and the side edge creates a base plane at an angle of 60°.
Volume of pyramid S.ABC is given by:
Question 9: Given an equilateral triangular pyramid S.ABC has its base edge with the length is a . Let (P) is the plane passing through A parallel to BC and perpendicular to (SBC) , the angle between the plane (P) and the base plane is 30°. Volume of pyramid S.ABC is provided by: a 3 3 24 .
Question 10: Given an equilateral triangular pyramid S.ABC has the length of its base edge is 2a . Let (P) is the plane passing through A parallel to BC and perpendicular to (SBC) , the angle between the plane (P) and the base plane is 60°. Volume of pyramid S.ABC is illustrated by:
Question 11: Given equilateral quadrilateral pyramid S.ABCD has base plane is a square ABCD with the length of its edge is a and the length of the side edge is 2a . Volume of pyramid S.ABCD will equal to:
Question 12: Given equilateral quadrilateral pyramid S.ABCD have all edges equal to 2a . Volume of pyramid S.ABCD is:
Question 13: Given equilateral quadrilateral pyramid S.ABCD has its base edge with the length is a 3 2 , the side surface creates a base plane at an angle of 45°. Volume of pyramid is given by: Question 14: Given equilateral quadrilateral pyramid S.ABCD has the length of its base edge is 2a , the side surface creates a base plane at an angle of 60°. Volume of pyramid is provided by:
Question 15: Given equilateral quadrilateral pyramid S.ABCD has its base edge with the length is a 3 , the side edge creates a base plane at an angle of 30°. Volume of pyramid S.ABCD is illustrated by: a 3 2 2 .
Question 16: Given equilateral quadrilateral pyramid S.ABCD has the length of its base edge is 2a , the side edge creates a base plane at an angle of 45°. Volume of pyramid S.ABCD will equal to: Question 17: Given equilateral quadrilateral pyramid S.ABCD has its base edge with the length is a , SAB ! = 60°. Volume of pyramid S.ABCD is:
Question 18: Given equilateral quadrilateral pyramid S.ABCD has the length of its side edge is 2a , the side surface creates a base plane at an angle of 45°. Volume of pyramid S.ABCD is given by: Question 19: Given equilateral quadrilateral pyramid S.ABCD has its side edge with the length is a , the side surface creates a base plane at an angle of 60°. Volume of pyramid S.ABCD is provided by:  Question 24: Given a tetrahedron S.ABC has SA = BC = 4a, SB = AC = 5a, SC = AB = 6a . Volume of pyramid S.ABC is provided by: Question 26: Given a tetrahedron S.ABC has planes (SAB), (SBC), (SAC) perpendicular to each other in pairs, area of triangles SAB, SBC, SAC are 15cm 2 , 20cm 2 ,18cm 2 , respectively. Volume of pyramid S.ABC will equal to: A. 60 3cm 3 . B. 10 6cm 3 .
Question 28: Given a tetrahedron S.ABC has SA, SB, SC perpendicular to each other in pairs and its length are 5cm, 10cm , 2dm , respectively. Volume of pyramid S.ABC is given by: Volume of pyramid S.ABC will equal to: A. a 3 6 6 . B. 0. C. a 3 2 6 . D. a 3 6 .
Question 32: Given a tetrahedron S.ABC has SA = a, BC = a 3 , the distance between two straight lines SA and BC has the length of 8a and the angle between them equal to 60°. Volume of pyramid S.ABC is:

D. a 3 4 .
Question 33: Given a tetrahedron S.ABC has SA = BC = 3a , two straight lines SA and BC perpendicular to each other and the distance between them equal to 2a . Volume of pyramid S.ABC is given by: A. a 3 .

B.
3 2 a 3 . C. 18a 3 . D. 3a 3 . Question 38: Given a regular octahedron has the vertex is the center of the sides of the cube with its length is 2a . Volume of the regular dodecahedron is given by: a 3 6 .
Question 39: Given a regular octahedron with the length of its edge a . Connecting the center of its sides will be a cube with volume V . The score of a 3 V is provided by :

2 4 .
Question 40: The regular dodecahedron with the length of its edge is 2a , its volume is illustrated by:

Experimental Process
Experiments are designed in 2 approaches:

Approach 1
The main aim is to assess 40 multiple choice questions in the experiment using IATA software. With this objective we conduct to 120 students to solve multiple choice tests within 90 minutes.

Approach 2
The primary objective is to evaluate the effectiveness of formulas for quick calculation of volume in math problems. With this target we choose a control group (group A) and experimental group (group B) and jointly conduct a multiple choice test, with the require that group B use the formula table, while group A does not use it.
Before conducting the experiments, we have studied the learning results of the groups. The results of the first semester study of the experimental group and the control group are listed in the following table.
The first step shows that the control group has a high mathematics score: 100% good. At that time, the experimental group was lower than the control group: 45% good and 55% rather good. In order to evaluate comprehensively a student's test, we need prior and posterior analysis. We first discuss the prior analysis in the next sub-section.

Question 1
The regular tetrahedral volume is So choosing Answer B.
• Answer A is wrong because students forget to minus a 2 in the formula.
• Answer D is wrong because students neglect to divide 12.

Question 2
The regular tetrahedral volume is Hence selecting Answer D.
• Answer A is wrong because students forget to replace 2a 2 .
• Answer B is wrong because students neglect to divide 12.
• Answer C is wrong because students mistake Question 3 Thus choosing Answer C.
• Answer A is wrong because students used the wrong formula should only be divided by 12.
• Answer B is wrong because students forget to multiply tan ! .
• Answer D is wrong because students mistake cos 60°. Therefore, selecting Answer A.
• Answer B is wrong because of only replace 2a 3 .
• Answer C is wrong because students used the wrong formula should only be divided by 12.
• Answer D is wrong because students forget to multiply tan ! .
• Answer B is wrong because students neglect to divide 4. • Answer C is wrong because students forget to square cos 60°.
• Answer D is wrong because students use to square sin 60° and students forget to square cos 60°.
• Answer A is wrong because students neglect to square cos 30°.
• Answer B is wrong because students forget to divide 4 in the formula.
• Answer D is wrong because of only replace edge length with a .
Question 7 Thus, choosing Answer D.
• Answer A is wrong because of only replace edge length with a .
• Answer B is wrong because of only substituting 2a 3 .
• Answer C is wrong because students mistake the formula, thus they divided to 24 in the formula.
Question 8 Therefore selecting Answer B.
• Answer A is wrong because students mistake the formula, thus they divided to 24 in the formula.
• Answer C is wrong because students mistake the formula: 3.a 3 .sin 60°.cos 2 60°4 • Answer D is wrong because students forget to divide 12 in the formula. So choosing Answer D.
• Answer A is wrong because of only substituting tan 30°.
• Answer B is wrong because of only replacing cos 30°.
• Answer C is wrong because students forget to divide 12 in the formula.
Question 10 Hence selecting Answer C.
• Answer A is wrong because of only substituting 2a 3 .
• Answer B is wrong because students mistake cos 60°.
• Answer D is wrong because students mistake tan 60°. Thus choosing Answer A.
• Answer C is wrong because students substitute edge lengths with a .
• Answer D is wrong because of only replacing Question 12 Volume of equilateral quadrilateral pyramid is Therefore selecting Answer B.
• Answer A is wrong because of only the original formula, has not replaced the length of the sides to the formula.
• Answer C is wrong because students only substituting 2a 3 .
• Answer D is wrong because of only squared the length of the edge.
Question 13 So choosing Answer C.
• Answer A is wrong because wrong formula of equilateral triangular pyramid.
• Answer B is wrong because students used the wrong formula: 2a 3 tan ! 6 .
• Answer D is wrong because students neglect to cube the length of edge.
Question 14 Hence selecting Answer D.
• Answer A is wrong because of only substituting 2a 3 .
• Answer B is wrong because of the wrong formula: 2a 3 tan ! 6 .
• Answer C is wrong because students only replace the length of edge is a .
Question 15 Thus choosing Answer D.
• Answer A is wrong because students only replace the length of edge is a .
• Answer B is wrong because students utilized of the wrong formula.
• Answer C is wrong because of the wrong formula: a 3 tan! 6 .
Question 16 Therefore selecting Answer B.
• Answer A is wrong because of not cube to the length of edge 2a .
• Answer C is wrong because students only replace the length of edge is a .
• Answer D is wrong because students mistake cos 45° or the wrong formula.
• Answer A is wrong because students neglect to square tan 60°.
• Answer B is wrong because students forget to divide 6 in the formula.
• Answer B is wrong because students forget to cube 2a .
• Answer C is wrong because of only replace the length is a .
• Answer D is wrong because students neglect to cube to expression in the root.
• Answer A is wrong because students forget to multiply 4 in the formula.
• Answer C is wrong because students neglect to cube to expression in the root.
• Answer D is wrong because students forget to divide 3 in the formula.
Therefore selecting Answer A.
• Answer B is wrong because of only multiply sin 30°.
• Answer D is wrong because students forget to divide 12 in the formula.
So choosing Answer C.
• Answer D is wrong because students forget to divide 12 in the formula.
Question 22 Hence selecting Answer D.
• Answer A is wrong because students mistake in a 3 .
• Answer B is wrong because of only square in a 2 or forget to multiply 2 in the formula. • Answer A is wrong because of only substitute the length of edge is a .
• Answer C is wrong because students forget to cube 2a .
• Answer D is wrong because students mistake in options.
• Answer C is wrong because students forget to multiply 2 12 in the formula.
• Answer D is wrong because students mistake in the addition operation.
• Answer A is wrong because students forget to multiply 2 12 in the formula.
• Answer D is wrong because students neglect to take the square root.
Hence selecting Answer D.
• Answer A is wrong because students neglect to divide 3 in the formula.
• Answer B is wrong because students forget to multiply 2 in the root.
• Answer C is wrong because students mistake in abc 6 .
Question 27 Thus choosing Answer A.
• Answer B is wrong because students neglect to divide 3 in the formula.
• Answer C is wrong because students forget to multiply 2 in the root.
• Answer D is wrong because mistake in abc 6 .
• Answer A is wrong because students forget to change cm .
• Answer C is wrong because of mistake in 2.S 1 S 2 S 3 3 .
Question 29 So choosing Answer C.
• Answer A is wrong because students forget to divide 6 in the formula.
• Answer D is wrong because of mistaking in 2.S 1 S 2 S 3 3 . Hence selecting Answer B.
• Answer A is wrong because students neglect to cube cos! and cos ! .
• Answer D is wrong because students forget to take the square root. Thus choosing Answer C.
• Answer B is wrong because students neglect to cube cos! and cos ! .
• Answer D is wrong because students forget to take the square root.
• Answer B is wrong because of mistaking sin 60° to cos 60°.
• Answer C is wrong because students neglect to divide 6 in the formula.
• Answer D is wrong because students forget to multiply to d . So choosing Answer D.
• Answer A is wrong because students forget to mutltiply lack of an edge 3a .
• Answer B is wrong because students neglect to multiply to d .
• Answer C is wrong because students forget to divide 6 in the formula.
Question 34 Hence selecting Answer B.
• Answer A is wrong because students neglect to divide 3 in the formula.
• Answer C is wrong because students forget to multiply 2 in the formula.
• Answer D is wrong because of missing in SA = a . Thus choosing Answer A.
• Answer B is wrong students neglect to divide 3 in the formula.
• Answer C is wrong because of mistaking the length of edge is a .
• Answer D is wrong because students forget to multiply 2 in the formula. Therefore selecting Answer B.
Answer A is wrong because students forget to divide 6 in the formula.
Answer C is wrong because students mistake the product of 3 edges to the sum of 3 edges.
Answer D is wrong because students mistake sin to cos in the formula. • Answer A is wrong because students forget to divide 6 in the formula.
• Answer C is wrong because of mistaking the length of edge is a .
• Answer D is wrong because of missing in options.
Question 38 Hence selecting Answer A.
• Answer B is wrong because students forget to cube the length of edge.
• Answer C is wrong because students only square the length of edge.
• Answer D is wrong because students only replace the length of edge is a .
• Answer A is wrong because of missing in the formula.
• Answer B is wrong because of mistaking in the formula and the score of a 3 V .
• Answer C is wrong because students neglect to take the score of a 3 V , only calculating the volume. Therefore selecting Answer C.
• Answer A is wrong because students forget to cube the length of edge.
• Answer B is wrong because of only square the length of edge.
• Answer D is wrong because students only replace the length of edge is a .
We now turn on the posteriori analysis in the next sub-section.

Introducing IATA Question Analysis Software
We execute IATA software to analyze the above results. This software can be downloaded via the link: https://polymetrika.com/Downloads/IATA. Discr is discriminant; Pval is difficulty; PBis is correlation coefficient according to classical test theory. The coefficients a and b are discriminants, the difficulty according to the test theory answers the IRT question. The results of the IATA show that the questions that are well used are 14 questions, which can be used as 23 questions and 3 questions have issues are questions 20, 21, and 27. Therefore, 3 questions have issues are excluded to proceed to approach 2.

Experimental Results of Approach 2
From 40 exercises of group A students drawn from approach 1 and 37 exercises of group B students from Ly Tu Trong High School for the Gifted, High School Pedagogical Practice, Chau Van Liem High School, Bui Huu Nghia High School, Nguyen Viet Hong High School are analyzed as follows:

Evaluation of Experimental Results
Based on Tables 2 and 3, the control group has an average score of 100% in good mathematics, but the correct rate of sentences is low, for example: questions 34 and 35 have a rate of 21.62%; questions 37 and 39 have a rate of 27.03%; and question 40 has a rate of 35.14%.
Meanwhile, the experimental group had a lower mathematics score than the control group, but the rate of correct answers was higher, most of them were over 80%, the lowest rate was 28 with 75%. This shows the effectiveness of using quick calculation formulas.
Through the experimental process, the results showed that: the experimental purpose has been completed, and the feasibility and effectiveness of using the formulas for calculating the volume of the polyhedron in spatial geometry has been affirmed. Implementing geometric exercises with quick calculation formulas is a tool to support students to receive, understand and apply in appropriate situations, thereby helping students to achieve good results in exams.

CONCLUDING REMARKS AND INFERENCE
As stated in the Introduction, in the age of globalization, an effective leadership skill is the ability for quick calculation of work-related problems. From an economic perspective, fast computation often gives a competitive advantage in business, where speed, efficiency and accuracy are essential. Quick calculation techniques are a central problem in modern mathematics because it shortens the time for solving technical problems. This paper has developed convenient and straightforward formulas to calculate quickly the polyhedron volume in spatial geometry. The empirical analysis was conducted at Can Tho City to evaluate efficiently the proposed formulas. This paper provides a foundation for teachers and students to approach the method of quick solving spatial geometry exercises using simple formulas. The experimental results show that students can solve exercises faster with the help of quick calculation formulas. This is a useful tool that can help students be more confident in the face of problems of calculating volumes in space geometry, which can be useful for commercial purposes.  Moreover, the formulas for quick calculation of the polyhedron volume in spatial geometry contributed importantly and significantly to the solution of multiple choice tests. This is also a premise to develop simple and faster calculation formulas in solving problems of spatial geometry and a tool for research in developing cognate topics. As an extension of the paper, we can directly use software such as R, GeoGebra for programming in learning and teaching. This is a topic of ongoing research.