Is (5 1) a solution of y-3x-2?

Answer 1

"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:

1. 
   
2. A= 3√3 s² 2. A= 3√3 s² 3.A= 3√3 s²

= 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:

1. How much material will be needed to make a regular hexagonal prism whose side equals 25cm and height 50cm?

Solution:

SA= 3√3 s² + 6sh

= 3√3 (25cm) ² + 6 (25cm) (25cm)

= 3√3 (625cm²) + 3750 cm²

SA = 1,875 √3 + 3750 cm2

1. Find the surface area of the solid at the right.

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)