0
1 litre = 1000 ml
= 1000 cm3
1 m = 100 cm
→ 1 m x 1m x 1m = 100 cm x 100 cm x 100 cm
= 1000000 cm3
= 1000000 ÷ 1000 litres
= 1000 litres
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What is the volume of a cube whose side length is 1.5 meters?
The volume of water in a 1.5m x1m x1m cube is 396 US gallons.
What is the slope intercept form for the coordinates 5 2 -4 -7?
(5, 2) and (- 4, - 7)Find slope.m = Y2 - Y1/X2 - X1m = - 7 - 2/- 4 - 5= - 1======
What is the slope of the line that passes through the points -2 15 and -8 -5?
(- 2, 15) and (- 8, - 5)m = Y2 - Y1/X2 - X1m = - 5 - 15/- 8 - (- 2)m = 10/3=======
What is the equation of a line that includes points 5 5 and 2 5?
(5, 5) and (2, 5)Find slope first.m(slope) = Y2 - Y1/X2 - X1m = 5 - 5/2 - 5m = 0=======Now, point slope form. Use set of points; second set will do.Y - Y1 = m(X - X1)Y - 5 = 0(X - 2)Y - 5 = XY = X + 5=======
What equation represents a line containing points -3 and 10 and 6 and 7?
(- 3, 10) and (6, 7)Need slope.m = Y2 - Y1/X2 - X1m = 7 - 10/6 - (- 3)= - 1/3----------Now, take that nice set of points, (6, 7), and use the point slope form.Y - Y1 = m(X - X1)Y - 7 = -1/3(X - 6)Y - 7 = (- 1/3)X + 2Y = (- 1/3)X + 9==============The equation of the line
What is the equation of a line that passes through the poin -2-4 and -7-1?
(- 2, - 4) and (- 7, - 1)First the slope is needed.m = Y2 - Y1/X2 - X1m = - 1 - (- 4)/- 7 - (- 2)m = - 3/5==========now, point slope form, use these points (- 2, - 4)Y - Y1 = m(X - X1)Y - (- 4) = -3/5[X - (- 2)]Y + 4 = - 3/5(X + 2)Y + 4 = (- 3/5)X - 6/5Y = (- 3/5)X - 26/5===============equation of the line
How do you write the standard form of the equation for the line that passes through the points -8 and-7 and -5 and -5?
(- 8, - 7) and (- 5, - 5)Need slope first. m = Y2 - Y1/X2 - X1m = - 5 - ( - 7)/- 5 - (- 8)m = 2/3========Now, I use point slope form to get us half way thereY - Y1 =m(X - X1) ( - 8, - 7) points usedY - (- 7) = 2/3[X - (- 8)Y + 7 = 2/3(X + 8)Y + 7 = (2/3)X + 16/37 = 21/3, soY + 21/3 = (2/3)X + 16/3Y = (2/3)X - 5/3===========================standard form
What is cubic meter?
It depends on how you acquire it.If you put a barrel at the bottom end of a downspout on the corner of your house and let it fill with rain, its FREE.If you buy it from the city and it comes through a water meter at your house, its probably about $5.00 per HCF (hundred cubic feet). There are 35.314 cubic feet in a cubic meter, so if you buy it from a public utility, a cubic meter would cost about $0.35. (assuming your utility charges about what Akron, Ohio's does) Oh, and they include the cost of a sewer fee in this price, whether you use it or not.If you buy bottled water in 500ml single serving clear plastic containers at a dollar a bottle, a cubic meter of water would cost $2000.00. Now that they've made the bottles thinner and you always squish the bottle a bit trying to open it and often end up spilling some on your clothes, this method of buying water has lost some appeal, but its still popular. Go figure.
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)
