"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"
by rcdaliva
CHAPTER I
INRODUCTION
According to Doris Kearns Goodwin, the past is not simply the past, but a prism which the subject filters his own changing self - image. In relation to this quote, the students like us should not forget the past because it was always perpendicular to ones life like a prism. Prism which means a polyhedron with two congruent parallel faces known as the bases, the other faces are called lateral faces are parallelograms and the height of a prism is the perpendicular distance between the planes of the bases (Soledad, Jose-Dilao Ed. D and Julieta G. Bernabe, 2009).There are formulas in finding the surface areas which means the sum of all areas faces of the prism. Perimeter is the outer boundary of a body or figure, or the sum of all the sides. Geometry is a branch of mathematics which investigates the relations, properties, and measurement of solids, surfaces, lines, and angles; the science which treats of the properties and relations of magnitudes; the science of the relations of space. This subject is being taught in the third year students.
When one of the researchers was playing footing with his friends, one of the third year students approached and asked him about their assignment on the surface area of hexagonal prism whose side and height were given. In the very start, the researcher thinks deeply and approached some of his classmates to solve the problem. By this instance, we as the fourth year researchers were challenged to find out the solution for the third year assignment.
The problem drove the researchers to investigate and that problem was: "What is the formula in finding the surface area of a regular hexagonal prism, with side s units and height h units?"
This investigation was challenging and likewise essential. It is important to the academe because the result of this investigation might be the bases of further discoveries pertaining to the formula in finding the surface area of a hexagonal prism. This is also beneficial to the Department of Education because it will give the administrators or the teachers the idea in formulating formulas for other kinds of prisms. And it is so very significant to the students and researchers like us because the conjectures discovered in this study will give them the simple, easy and practical formulas or approaches in solving the problems involving the surface area of prisms.
However, this investigation was limited only to the following objectives:
1. To answer the question of the third year students;
2. To derive the formula of the surface area of hexagonal prism; and
3. To enrich the students mathematical skills in discovering the formula.
In view of the researchers desire to share their discoveries, their conjectures, they wanted to invite the readers and the other students' researchers to read, comment and react if possible to this investigation.
CHAPTER II
STATEMENT OF THE PROBLEM
The main problem of this investigation was: "What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"
s
h
Specifically, the researchers would like to answer the following questions:
1. What are the formulas in finding the areas of the regular hexagon and of the rectangle as lateral faces?
2. What is the formula in finding the surface area of the regular hexagonal prism?
CHAPTER III
FORMULATING CONJECTURES
Based on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:
Table 1. Perimeter of a Regular Hexagon s
HEXAGON WITH SIDE (s) in cm
PERIMETER (P) in cm
1
6
2
12
3
18
4
24
5
30
6
36
7
42
8
48
9
54
10
60
s
6s
Table 1 showed the perimeter of a regular hexagon. It revealed that the perimeter of the said polygon was 6 times its side. Thus, the conjecture formed was:
CONJECTURE 1:
The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.
Table 2. The Apothem of the Base of the Hexagonal Prism
Hexagonal prism with side
(s) in cm
Measure of the apothem (a)
in cm
1
½ √3
2
√3
3
3√3
2
4
2√3
5
5√3
2
6
3√3
7
7√3
2
8
4√3
9
9√3
2
10
5√3
s
√3 s
2
s
Table 2 showed that the measure of the apothem is one-half the measure of its side times √3.
CONJECURE 2
The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.
In symbols: ½ √3 s or √3 s.
2
Table 3. The Area of the Bases of Regular Hexagonal Prism
SIDE (cm)
APOTHEM (cm)
PERIMETER (cm)
AREA OF THE BASES (cm²)
1
1√3
2
6
3√3
2
√3
12
12√3
3
3√3
2
18
27√3
4
2√3
24
48√3
5
5√3
2
30
75√3
6
3√3
36
108√3
7
7√3
2
42
147√3
8
4√3
48
192√3
9
9√3
2
54
243√3
10
5√3
60
300√3
s
√3s
2
6s
3√3 s²
Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.
CONJECTURE 3
The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².
Table 4. The Total Areas of the 6 Rectangular Faces of the Hexagonal Prism
SIDE (cm)
HEIGHT (cm)
TOTAL AREA (cm²)
1
1
6
2
2
24
3
3
54
4
4
96
5
5
150
6
6
216
7
7
294
8
8
384
9
9
486
10
10
600
s
h
6sh
Based on table 4, the total areas of the 6 rectangular faces of the regular hexagonal prism with side s units and height h units was 6 times the product of its side s and height h.
CONJECTRE 4
The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.
Table 5. The Surface Area of the Regular Hexagonal Prism
SIDE (cm)
HEIGHT (cm)
AREA OF THE BASES (cm²)
AREA OF THE 6 FACES (cm²)
SURFAE AREA (cm²)
1
1
3√3
6
3√3+6
2
2
12√3
24
12√3+24
3
3
27√3
54
27√3+54
4
4
48√3
96
48√3+96
5
5
75√3
150
75√3+150
6
6
108√3
216
108√3+216
7
7
147√3
294
147√3+294
8
8
192√3
384
192√3+384
9
9
243√3
486
243√3+486
10
10
300√3
600
300√3+600
s
h
3√3 s²
6sh
3√3s²+6sh
Table 5 showed the surface area of the regular hexagonal prism and based from the data, the surface area of a regular hexagonal prism with side s units and height h units was the sum of the areas of the bases and the areas of the 6 faces.
CONJECTURE 5
The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.
CHAPTER IV
TESTING AND VERIFYING CONJECTURES
A. Testing of Conjectures
CONJECTURE 1:
The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.
To test the conjecture 1, the investigators applied the said conjecture in finding the perimeter of the base of the following regular hexagonal prisms and regular hexagons. 5.5 cm
1. 10 cm 2. 3.
11 cm
4. 5.
12 cm
20m
Solutions:
1. P = 6s 2. P = 6s 3. P = 6s 4. P = 6s
= 6 (10cm) = 6 (5.5 cm) = 6 (11 cm) = 6 (12 cm)
= 60 cm = 33 cm = 66 cm = 72 cm
5. P = 6s
= 6 (20 cm)
= 120 cm
CONJECURE 2
The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.
In symbols: ½ √3 s or √3 s.
2
The investigators applied this conjecture to the problem below to test its accuracy and practicality.
Problem: Find the apothem of the base of each of the regular hexagonal prism in the figures under the conjecture 1.
Solutions:
1. a = √3 s 2. a = √3 s 3. a = √3 s 4. a = √3 s
2 2 2 2
= √3 (10 cm) = √3 (5.5 cm) = √3 (11 cm) = √3 (12 cm)
2 2 2 2
= √3 (5 cm) = √3 (2.75 cm) = √3 (5.5 cm) = √3 (6 cm)
= 5√3 cm = 2.75 √3 cm = 5.5√3 cm = 6√3 cm
5. a = √3 s
2
= √3 (12 cm)
2
= √3 (6 cm)
= 6√3 cm
CONJECTURE 3
The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².
To test this conjecture, the investigators applied its efficiency in the problem, "Find the total area of the bases of each regular hexagonal prism in figures 1, 2 and 3 under the testing of conjecture 1".
Solutions:
= 3√3 (10cm) ² = 3√3 (5.5 cm)² = 3√3 (11cm)²
= 3√3 (100cm²) = 3√3 (30.25) cm² = 3√3 (121 cm²)
= 300 √3 cm² = 90.75 √3 cm² = 363 √3 cm²
CONJECTRE 4
The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.
This conjecture can be applied in finding the total areas of the faces of regular hexagonal prism like the problems below.
a. Find the total areas of the faces of a regular hexagonal prism whose figure is
8 cm
Solution: A= 6sh
= 6 (8cm) (20cm) 20 cm
= 960 cm2
b. What is the total areas of the bases of the regular hexagonal prism whose side is 15 cm and height 15cm.
Solution: A = 6 sh
= 6 (15 cm) (15 cm)
= 1,350 cm 2
CONJECTURE 5
The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.
The investigators tested this conjecture by solving the following problems:
Solution:
SA= 3√3 s² + 6sh
= 3√3 (25cm) ² + 6 (25cm) (25cm)
= 3√3 (625cm²) + 3750 cm²
SA = 1,875 √3 + 3750 cm2
28 cm
Solution:
SA= 3√3 s²+ 6sh 18 cm
= 3√3 (18cm) 2 + 6 (18cm)(28cm)
= 3√3 (324cm²) + 3024 cm2
SA = 972 √3 cm² + 3024 cm²
B. Verifying Conjectures
CONJECTURE 1:
The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.
F E
A D
B C
s
Proof 1.
If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6S
Statements
Reasons
1. ABCDEF is a regular hexagon with BC=s.
2. AB=BC=CD=DE=FA
3.AB=s
CD=s
DE=s
FA=s
EF=s
4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s
5.AB+BC+CD+DE+EF+FA=6S
1. Given
2. Definition of regular hexagon
3.Transitive Property
4.APE
5. Combining like terms.
Proof 2.
Sides(s)
1
2
3
4
5
6
7
8
9
10
Perimeter f(s)
6
12
18
24
30
36
42
48
54
60
6 6 6 6 6 6 6 6 6
Since the first differences were equal, therefore the table showed linear function f(x) = mx+b. To derive the function, (1, 6) and (2, 12) will be used.
Solve for m:
m= y2-y1 Slope formula
x2-x1
= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.
2-1
= 6 Mathematical fact
1
m = 6 Mathematical fact
Solve for b:
f(x)=mx+b Slope-Intercept formula
6=6(1) + b Substituting y=6, x=1, and m=6.
6=6+b Identity
0=b APE
b=0 Symmetric
Thus, f(x) = 6x or f(s) = 6s or P = 6s.
CONJECURE 2
The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.
In symbols: ½ √3 s or √3 s. E D
2
Proof I.
Given: ABCDEF is a regular hexagon F C
AB=s
a
Prove: a= √3 s
2
A G B
s
Statements
Reasons
1. ABCDEF is a regular hexagon.
AB= s
1. Given
2.AG= ½ s
2. The side opposite to 30˚ is one half the hypotenuse.
3. a=(½ s)(√3)
3. The side opposite to 60˚ is equal to the side opposite to 30˚ times √3.
4. a= √3 s
2
4. Closure
Proof 2.
Side (s)
1
2
3
4
5
6
7
8
9
10
Apothem (a)
F(s)
√3
2
√3
3√3
2
2√3
5√3
2
3√3
7√3
2
4√3
9√3
2
5√3
√3 √3 √3 √3 √3 √3 √3 √3 √3
2 2 2 2 2 2 2 2 2
Since the first differences were equal, therefore the table showed a linear function in the form f(x) = mx+b.
Solving for m using (1, √3) and (2, √3).
2
m = y2-y1 Slope formula
x2-x1
m = √3 - √3 Substitution
2
2-1
m= √3 Mathematical fact/ Closure
2
1
m= √3
2
Solving for b: Use (1, √3)
2
f(x) = mx + b Slope - intercept form
√3 = (√3) (1) +b Substitution
2 2
√3 = √3+ b Identity
2 2
0=b APE
b=0 Symmetric
Thus, f(x) = √3 or f(s) = √3s or a = √3s
2 2 2
CONJECTURE 3
The total areas of the two bases of the regular hexagonal prism is 3√3 times
B C
the square of its side s. In symbols, A=3√3 s².
Proof 1 A D
Given: ABCDEF is a regular hexagonal prism.
FE = s units
Prove: AABCDEF = 3(√3)s² F s E
2
2AABCDEF = 3√3s²
Statements
Reasons
1. ABCDEF is a regular hexagon
FE =s
Given
2.a= 3√3s
The side opposite to 60 is the one half of the hypotenuse time's √3.
3.A = ½bh
The area of a triangle is ½ product of its side and height
4.A =(½)s(√3/2s)
Substituting the b=s and h=a=√3
2s.
5.A = (√3/4)s²
Mathematical fact
6.AABCDEF= 6A
In a regular hexagon, there are six congruent triangles formed
7.AABCDEF= 6(√3/4s²)
Substitution
8.AABCDEF= 3 (√3/2) s²
Mathematical fact
9.2AABCDEF= 2[3 (√3/2)]s²
MPE
10.2AABCDEF= 3 √3 s
²
Multiplicative inverse / identity
Proof 2
Based on the table, the data were as follows:
Side (s)
1
2
3
4
5
6
7
8
9
10
Area of the bases f(s)
3 √3
12√3
27√3
48√3
75√3
108√3
147√3
192√3
243√3
300√3
9√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3
First difference
6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3
Second difference
Since the second differences were equal, the function that the investigators could derive will be a quadratic function f(x) = ax²+bx+c.
Equations were:
Eq. 1 f(x) = ax²+bx+c for (1, √3)
6√3 = a (1)²+ b(1)+c Substitution
6√3 = a+b+c Mathematical fact / identity
a+b+c=6√3 Symmetric
Eq. 2 f(x) = ax²+bx+c for (2, 12√3)
24√3=a (2)²+b(2)+c Substitution
24√3=4a+2b+c Mathematical fact
4a+2b+c=24√3 Symmetric
Eq. 3 f(x) = ax²+bx+c for (3, 27√3)
54√3=a (3)²+b(3)+c Substitution
54√3=9a+3b+c Mathematical fact
9a+3b+c=54√3 Symmetric
To find the values of a, b, and c, elimination method was utilized.
Eliminating c
Eq. 2 4a+2b+c=24√3 Eq. 3 9a+3b+c=54√3
- Eq. 1 a+b+c=6√3 - Eq. 2 4a+2b+c=24√3
Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3
Eliminating b and solving a
Eq. 5 5a+b = 30√3
- Eq. 4 3a+b = 18√3
2a = 12√3
a = 6√3 MPE
Solving for b if a = 6√3
Eq. 5 5a+b = 30√3
5(6√3) +b= 30√3 Substitution
30√3+b=30√3 Closure
b = 0 APE
Solving for c if a = 6√3 and b=0
Eq. 1 a + b + c=6√3
6√3 + 0+c =6√3 Substitution
6√3 + c = 6√3 Identity
c = 0 APE
Therefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²
CONJECTURE 4
The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.
A
Proof
Given: ABCD is a rectangle. B
AB = s and BC = h
Prove:
AABCD = sh D
6AABCD= 6sh
C
Statements
Reasons
1. ABCD is a rectangle AB=s and BC=h
Given
2.AABCD=lw
The area of a rectangle is the product of its length and width
3. AABCD = sh
Substitution
4. 6AABCD = 6sh
MPE
CONJECTURE 5
The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s² + 6sh.
Proof
Given: The figure at the right
Prove: SA=3 √3s²+6sh
s h
Statements
Reasons
1. AHEXAGON= ½ aP
The area of a regular polygon is one -half the product of its apothem and its perimeter
2. a = √3/2s
The side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.
3. P = 6s
The perimeter of a regular polygon is the sum of all sides.
4. AHEXAGON = ½ (√3s)(6s)
2
Substitution
5. A HEXAGON = 3 √3s²
2
Mathematical fact
6. 2A HEXAGON= 3 √3s²
MPE
7. A RECTANGULAR FACES = sh
The area of a rectangle is equal to length (h) times the width (s).
8. 6ARECTANGULAR FACES = 6sh
MPE
9. SA = 2A HEXAGON + 6A RECTANGULAR FACES
Definition of surface area
10. SA = 3 √3s² + 6sh
Substitution
CHAPTER V
SUMMARY/CONCLUSIONS
After the investigation, the question of the third year student on "What is the surface area of the regular hexagonal prism whose side and height were given" was cleared and answered. Indeed, God is so good because of the benefits that the investigators gained like the discovery of various formulas and conjectures based on the patterns observed in the data gathered and most of all, the friendship that rooted between the hearts of the investigators and the third year students could not be bought by any gold.
The main problem of this investigation was: "What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"
s
h
Specifically, the researchers would like to answer the following questions:
1. What are the formulas in finding the areas of the regular hexagon and of the rectangle as lateral faces?
2. What is the formula in finding the surface area of the regular hexagonal prism?
Based on the results, the investigators found out the following conjectures:
CONJECTURE 1:
The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.
CONJECURE 2
The measure of the apothem of the base of the regular hexagonal prism is ½ the side s times √3.
In symbols: ½ √3 s or √3 s.
2
CONJECTURE 3
The total areas of the two bases of the regular hexagonal prism is 3√3 times the square of its side s. In symbols, A=3√3 s².
CONJECTRE 4
The total areas of the six rectangular faces of a regular hexagonal prism with side s units and height h units is equal to 6 times the product of its side s and height h. In symbols, A = 6sh.
CONJECTURE 5
The surface area of the regular hexagonal prism with side s units and height h units is equal to the sum of the areas of the two bases and the 6 faces. In symbols, SA =3√3s²+6sh.
These conjectures were proven based on the gathered data on different sources like books, practical applications, and internet. The formulas also followed the rules in finding the surface area of a prism.
CHAPTER VI
POSSIBLE EXTENSIONS
The investigators would like to elicit answers of the readers by applying the conjectures discovered and formulated through this study.
A. Find the surface area of the following regular hexagonal prism.
1. 8 cm 2. 7 cm 3.
9.8 cm
12 cm
10 cm 50 cm
4. .
a = 8 √3
22 cm
B. Derive a formula in finding the surface area of:
1. regular hexagonal prism whose side equals x cm and height equals y cm.
2. regular hexagonal prism whose side equals (x-1) cm and height equals (x2+4x+4) cm.
C. Derive the formula for the surface area of a regular octagonal prism. (Hint: Use Trigonometric Functions and Pythagorean Theorem)
Solution a.k.a. the coordinate point where the two lines cross/intersect. Example Solve system by elimination method~ Y= 5x+3 Y= 10x+17 Multiply top by -2. You now have~ Y=-10x+(-6) Y=10x+17 ********Y= 11(this is the "y" of your end solution) Now plug 11 into y on either equation (I chose the first one) 11=5x+3 -3 -3 8=5x ÷5 ÷5 ********1.6=x (this is the x of your end solution) So in conclusion, your SOLUTION (the thing you solved this whole system of equations for) is: (1.6, 11) *it is written as a coordinate graph point (x, y). Hope this helps, regards, Ryan S. :)
(a) y = -3x + 1
Usually expressed as,Y = 3X2============same thing as X * X, XX without multiplicative symbol, so yes it is exponential
11
The answer depends on whether it is 3x2 + 3y2 or 3x2 - 3y2. Unfortunately, limitations of the browser used by WA means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "equals" etc.
3x2-y=6 3x2-y=6
The question contains an expression, not an equation and so there is no solution.
-2
Let y = x3 - 8, then y' = 3x2 + 0 = 3x2.
It has one double solution.
(3x - y)(x - 3y) (6y - 11)(6y + 11)
3 * x * x * y
x = +/- sqrt(y/3)
3x2 - 5x - 22 (x + 2)(3x - 11) x = -2, 11/3 Neither is a multiple of 6.
72
You can find the x-coordinate of it's vertex by taking it's derivative and solving for zero: y = -3x2 + 12x - 5 y' = -6x + 12 0 = -6x + 12 6x = 12 x = 2 Now that we have it's x coordinate, we can plug it back into the original equation to find it's y coordinate: y = -3x2 + 12x - 5 y = -3(2)2 + 12(2) + 5 y = -12 + 24 + 5 y = 17 So the vertex of the parabola y = -3x2 + 12x - 5 occurs at the point (2, 17).
x = -6 and y = 1