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What is the total area of a regular hexagonal prism that has a base edge of 8 and height of 8?
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How many total edges would 35 hexagonal prisms have?
1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!
What shape has 8faces total?
There are many possible shapes. For example, a heptagonal pyramid, a hexagonal prism, a rectangular dipyramid.
What is an example of math investigatory project?
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording 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; and3. 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 IISTATEMENT OF THE PROBLEMThe 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?"shSpecifically, 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 IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 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 PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The 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.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The 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 PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased 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 4The 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 PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 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 5The 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 IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 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 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions: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 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The 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.2The 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 s2 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 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The 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: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 cm²CONJECTRE 4The 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 is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. 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 2CONJECTURE 5The 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: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 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since 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 formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The 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 D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. 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 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since 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).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince 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 Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 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√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The 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.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The 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.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter 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?"shSpecifically, 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 2The 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.2CONJECTURE 3The 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 4The 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 5The 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 VIPOSSIBLE EXTENSIONSThe 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 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. 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)
Is (5 1) a solution of y-3x-2?
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording 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; and3. 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 IISTATEMENT OF THE PROBLEMThe 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?"shSpecifically, 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 IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 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 PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The 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.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The 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 PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased 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 4The 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 PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 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 5The 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 IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 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 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions: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 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The 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.2The 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 s2 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 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The 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: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 cm²CONJECTRE 4The 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 is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. 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 2CONJECTURE 5The 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: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 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since 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 formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The 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 D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. 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 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since 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).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince 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 Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 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√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The 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.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The 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.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter 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?"shSpecifically, 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 2The 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.2CONJECTURE 3The 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 4The 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 5The 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 VIPOSSIBLE EXTENSIONSThe 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 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. 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)
What is the formula for finding the total surface area of a triangular prism?
Assume that a = apothem length of the triangular prism, b = base length of the triangular prism, and h = height of the triangular prism. The formulas to find the surface area is SA = ab + 3bh.
How many faces does a hexagonal prism have?
A hexagonal prism has 6 rectangular faces.
How many faces on a hexagonal prism?
The prefix hex- denotes a six sided polygon, so a hexagonal prism will have a six-sided base and, in total, will have eight faces: it is an octohedron.Provided the faces are all regular the hexagonal prism is referred to as a semiregular polyhedron, but could be defined in other terms if the faces are irregular.
What shape has 6 rectangular faces and 2 hexagonal faces?
A hexagonal prism has 6 rectangular faces and 2 hexagonal faces. Since it has 8 faces in total, it is a type of irregular octahedron, however it is correctly classified as a prismatic uniform polyhedron to avoid ambiguity with the regular octahedron with triangular faces. The hexagonal prism forms the basis of the hexagonal prismatic honeycomb.
How many vertices does a prism hexagon have?
It has 12 vertices
How many total edges would 35 hexagonal prisms have?
1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!1 hexagonal prism = 18 edges.35 hexagonal prisms = 18*35 = 630 edges. Simple!
What are the total number of diagonals in a hexagonal prism?
From each vertex, N-4 diagonals can be drawn, where N is the number of vertices. The total number of diagonals in a prism will be N(N-4)/2, where N is the number of vertices. For N=12 (a hexagonal prism) The total number of diagonals is 12*8/2=48 Answer:48
How many faces vertices edges does a hexagonal prism have?
8 faces, 18 edges, and 12 vertices------------------------In geometry, the hexagonal prism is a prism with hexagonal base.It is an octahedron. However, the term octahedron is mainly used with "regular" in front or implied, hence not meaning a hexagonal prism; in the general meaning the term octahedronit is not much used because there are different types which have not much in common except having the same number of faces.-----------------An n-gonal prism has 3n edges, 2n vertices and n+2 faces.
What is the lateral area of a hexagonal prism where all edges are 10cm and the height is 16cm?
Each lateral side is 10*16 = 160 cm2 So the total lateral area is 6*160 = 960 cm2
What shape has 8faces total?
There are many possible shapes. For example, a heptagonal pyramid, a hexagonal prism, a rectangular dipyramid.
Is (1-1) a solution of y -3x-2?
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording 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; and3. 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 IISTATEMENT OF THE PROBLEMThe 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?"shSpecifically, 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 IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 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 PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The 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.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The 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 PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased 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 4The 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 PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 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 5The 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 IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 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 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions: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 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The 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.2The 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 s2 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 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The 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: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 cm²CONJECTRE 4The 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 is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. 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 2CONJECTURE 5The 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: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 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since 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 formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The 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 D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. 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 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since 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).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince 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 Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 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√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The 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.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The 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.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter 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?"shSpecifically, 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 2The 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.2CONJECTURE 3The 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 4The 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 5The 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 VIPOSSIBLE EXTENSIONSThe 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 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. 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)
What is an example of math investigatory project?
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording 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; and3. 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 IISTATEMENT OF THE PROBLEMThe 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?"shSpecifically, 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 IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 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 PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The 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.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The 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 PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased 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 4The 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 PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 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 5The 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 IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 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 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions: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 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The 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.2The 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 s2 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 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The 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: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 cm²CONJECTRE 4The 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 is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. 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 2CONJECTURE 5The 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: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 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since 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 formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The 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 D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. 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 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since 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).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince 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 Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 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√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The 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.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The 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.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter 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?"shSpecifically, 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 2The 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.2CONJECTURE 3The 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 4The 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 5The 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 VIPOSSIBLE EXTENSIONSThe 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 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. 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)
Is (5 1) a solution of y-3x-2?
"What is the formula in finding the surface area of a regular hexagonal prism with side s units and height h units?"by rcdalivaCHAPTER IINRODUCTIONAccording 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; and3. 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 IISTATEMENT OF THE PROBLEMThe 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?"shSpecifically, 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 IIIFORMULATING CONJECTURESBased on the thorough investigation of the researchers, the tables and conjectures discovered and formulated were as follows:Table 1. Perimeter of a Regular Hexagon sHEXAGON WITH SIDE (s) in cmPERIMETER (P) in cm162123184245306367428489541060s6sTable 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 PrismHexagonal prism with side(s) in cmMeasure of the apothem (a)in cm1½ √32√333√3242√355√3263√377√3284√399√32105√3s√3 s2sTable 2 showed that the measure of the apothem is one-half the measure of its side times √3.CONJECURE 2The 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.2Table 3. The Area of the Bases of Regular Hexagonal PrismSIDE (cm)APOTHEM (cm)PERIMETER (cm)AREA OF THE BASES (cm²)11√3263√32√31212√333√321827√342√32448√355√323075√363√336108√377√3242147√384√348192√399√3254243√3105√360300√3s√3s26s3√3 s²Table 3 revealed that the area of the base of regular hexagonal prism was 3√3 times the square of its side.CONJECTURE 3The 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 PrismSIDE (cm)HEIGHT (cm)TOTAL AREA (cm²)11622243354449655150662167729488384994861010600sh6shBased 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 4The 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 PrismSIDE (cm)HEIGHT (cm)AREA OF THE BASES (cm²)AREA OF THE 6 FACES (cm²)SURFAE AREA (cm²)113√363√3+62212√32412√3+243327√35427√3+544448√39648√3+965575√315075√3+15066108√3216108√3+21677147√3294147√3+29488192√3384192√3+38499243√3486243√3+4861010300√3600300√3+600sh3√3 s²6sh3√3s²+6shTable 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 5The 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 IVTESTING AND VERIFYING CONJECTURESA. Testing of ConjecturesCONJECTURE 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 cm1. 10 cm 2. 3.11 cm4. 5.12 cm20mSolutions: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 cm5. P = 6s= 6 (20 cm)= 120 cmCONJECURE 2The 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.2The 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 s2 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 cm5. a = √3 s2= √3 (12 cm)2= √3 (6 cm)= 6√3 cmCONJECTURE 3The 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: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 cm²CONJECTRE 4The 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 is8 cmSolution: A= 6sh= 6 (8cm) (20cm) 20 cm= 960 cm2b. 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 2CONJECTURE 5The 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: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 cm²SA = 1,875 √3 + 3750 cm2Find the surface area of the solid at the right.28 cmSolution:SA= 3√3 s²+ 6sh 18 cm= 3√3 (18cm) 2 + 6 (18cm)(28cm)= 3√3 (324cm²) + 3024 cm2SA = 972 √3 cm² + 3024 cm²B. Verifying ConjecturesCONJECTURE 1:The perimeter of a regular hexagon with side s is equal to 6 times the side s. In symbols P = 6s.F EA DB CsProof 1.If ABCDEF is a regular hexagon with BC=s, then AB+BC+CD+DE+EF+FA= 6SStatementsReasons1. ABCDEF is a regular hexagon with BC=s.2. AB=BC=CD=DE=FA3.AB=sCD=sDE=sFA=sEF=s4.AB+BC+CD+DE+EF+FA=s+s+s+s+s+s5.AB+BC+CD+DE+EF+FA=6S1. Given2. Definition of regular hexagon3.Transitive Property4.APE5. Combining like terms.Proof 2.Sides(s)12345678910Perimeter f(s)61218243036424854606 6 6 6 6 6 6 6 6Since 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 formulax2-x1= 12-6 Substituting y2= 12, y1=6, x2=2 and x1=1.2-1= 6 Mathematical fact1m = 6 Mathematical factSolve for b:f(x)=mx+b Slope-Intercept formula6=6(1) + b Substituting y=6, x=1, and m=6.6=6+b Identity0=b APEb=0 SymmetricThus, f(x) = 6x or f(s) = 6s or P = 6s.CONJECURE 2The 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 D2Proof I.Given: ABCDEF is a regular hexagon F CAB=saProve: a= √3 s2A G BsStatementsReasons1. ABCDEF is a regular hexagon.AB= s1. Given2.AG= ½ s2. 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 s24. ClosureProof 2.Side (s)12345678910Apothem (a)F(s)√32√33√322√35√323√37√324√39√325√3√3 √3 √3 √3 √3 √3 √3 √3 √32 2 2 2 2 2 2 2 2Since 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).2m = y2-y1 Slope formulax2-x1m = √3 - √3 Substitution22-1m= √3 Mathematical fact/ Closure21m= √32Solving for b: Use (1, √3)2f(x) = mx + b Slope - intercept form√3 = (√3) (1) +b Substitution2 2√3 = √3+ b Identity2 20=b APEb=0 SymmetricThus, f(x) = √3 or f(s) = √3s or a = √3s2 2 2CONJECTURE 3The total areas of the two bases of the regular hexagonal prism is 3√3 timesB Cthe square of its side s. In symbols, A=3√3 s².Proof 1 A DGiven: ABCDEF is a regular hexagonal prism.FE = s unitsProve: AABCDEF = 3(√3)s² F s E22AABCDEF = 3√3s²StatementsReasons1. ABCDEF is a regular hexagonFE =sGiven2.a= 3√3sThe side opposite to 60 is the one half of the hypotenuse time's √3.3.A = ½bhThe area of a triangle is ½ product of its side and height4.A =(½)s(√3/2s)Substituting the b=s and h=a=√32s.5.A = (√3/4)s²Mathematical fact6.AABCDEF= 6AIn a regular hexagon, there are six congruent triangles formed7.AABCDEF= 6(√3/4s²)Substitution8.AABCDEF= 3 (√3/2) s²Mathematical fact9.2AABCDEF= 2[3 (√3/2)]s²MPE10.2AABCDEF= 3 √3 s²Multiplicative inverse / identityProof 2Based on the table, the data were as follows:Side (s)12345678910Area of the bases f(s)3 √312√327√348√375√3108√3147√3192√3243√3300√39√3 15√3 21√3 27√3 33√3 39√3 45√3 51√3 57√3First difference6√3 6√3 6√3 6√3 6√3 6√3 6√3 6√3Second differenceSince 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 Substitution6√3 = a+b+c Mathematical fact / identitya+b+c=6√3 SymmetricEq. 2 f(x) = ax²+bx+c for (2, 12√3)24√3=a (2)²+b(2)+c Substitution24√3=4a+2b+c Mathematical fact4a+2b+c=24√3 SymmetricEq. 3 f(x) = ax²+bx+c for (3, 27√3)54√3=a (3)²+b(3)+c Substitution54√3=9a+3b+c Mathematical fact9a+3b+c=54√3 SymmetricTo find the values of a, b, and c, elimination method was utilized.Eliminating cEq. 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√3Eq. 4 3a+b = 18√3 Eq. 5 5a+b = 30√3Eliminating b and solving aEq. 5 5a+b = 30√3- Eq. 4 3a+b = 18√32a = 12√3a = 6√3 MPESolving for b if a = 6√3Eq. 5 5a+b = 30√35(6√3) +b= 30√3 Substitution30√3+b=30√3 Closureb = 0 APESolving for c if a = 6√3 and b=0Eq. 1 a + b + c=6√36√3 + 0+c =6√3 Substitution6√3 + c = 6√3 Identityc = 0 APETherefore, f(x) = 6√3x² or f(s) = 6√3s² or A= 6√3s²CONJECTURE 4The 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.AProofGiven: ABCD is a rectangle. BAB = s and BC = hProve:AABCD = sh D6AABCD= 6shCStatementsReasons1. ABCD is a rectangle AB=s and BC=hGiven2.AABCD=lwThe area of a rectangle is the product of its length and width3. AABCD = shSubstitution4. 6AABCD = 6shMPECONJECTURE 5The 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.ProofGiven: The figure at the rightProve: SA=3 √3s²+6shs hStatementsReasons1. AHEXAGON= ½ aPThe area of a regular polygon is one -half the product of its apothem and its perimeter2. a = √3/2sThe side opposite to 60˚ is a 30˚-60˚-90˚ triangle is one-half the hypotenuse times √3.3. P = 6sThe perimeter of a regular polygon is the sum of all sides.4. AHEXAGON = ½ (√3s)(6s)2Substitution5. A HEXAGON = 3 √3s²2Mathematical fact6. 2A HEXAGON= 3 √3s²MPE7. A RECTANGULAR FACES = shThe area of a rectangle is equal to length (h) times the width (s).8. 6ARECTANGULAR FACES = 6shMPE9. SA = 2A HEXAGON + 6A RECTANGULAR FACESDefinition of surface area10. SA = 3 √3s² + 6shSubstitutionCHAPTER VSUMMARY/CONCLUSIONSAfter 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?"shSpecifically, 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 2The 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.2CONJECTURE 3The 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 4The 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 5The 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 VIPOSSIBLE EXTENSIONSThe 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 cm12 cm10 cm 50 cm4. .a = 8 √322 cmB. 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)
