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What kind of triangle have a 10 cm10 cm12 cm?
The dimensions given relate to an isosceles triangle
How much is 10.5 feet in centimeters?
1 ft = 30.48 cm10 ft = 304.80 cm0.5 ft = 15.24 cm +10.5 ft = 320.04 cm
How do you convert 1250 mm to feet?
Use this formula: mm x 0.00328 = feet1250 mm x 0.00328 = about 4.1 feet.Algebraic Steps / Dimensional Analysis Formula1,250 mm*1 cm10 mm*1 in2.54 cm*1 ft12 in=4.101049869 ftDirect Conversion Formula 1,250 mm*1 ft304.8 mm=4.101049869 ft
How do you convert millimeters into inches?
To convert millimeters to inches you divide the number of millimeters by 25.4Example: 8 mm / 25.4 = 0.31496 inches25.4 mm = 1 inch. To convert mm to inches. divide mm by 25.4. To convert inches to mm, multiply inches by 25.4Direct Conversion Formula 1 mm*1 in25.4 mm=0.03937007874 inTo convert mm to inches, use this formula - mm x 0.03937 = inchesAlgebraic Steps / Dimensional Analysis Formula ____ mm*1 cm10 mm*1 in2.54 cm=? inDirect Conversion Formula ____ mm*1 in25.4 mm=? in1 inch are 25.4 millimeters.1 millimeter are 0.0393700787 inches.Scroll down to related links and look at "Inches - feet - yards - millimeters - centimeters - meters".Direct Conversion Formula____ mm*1 in25.4 mm=? in
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)
What kind of triangle have a 10 cm10 cm12 cm?
The dimensions given relate to an isosceles triangle
How many square millimeters in a square centimetere?
10010 mm = 1 cm10 x 10 = 100 sq mm =1 sq cm
How much is 10.5 feet in centimeters?
1 ft = 30.48 cm10 ft = 304.80 cm0.5 ft = 15.24 cm +10.5 ft = 320.04 cm
What is the formula to convert millimeters to inches?
Algebraic Steps / Dimensional Analysis Formula____mm*1 cm10 mm*1 in2.54 cm=?inDirect Conversion Formula____mm*1 in25.4 mm=?in
How big is 140 mm in inches?
140 millimeters = about 5.5 inches. (5.51181102 inches)Algebraic Steps / Dimensional Analysis 140 mm*1 cm10 mm*1 in2.54 cm=5.511811024 in
How much in inches is 12 mm?
12 millimeters = 0.47 inches.Algebraic Steps / Dimensional Analysis Formula 12 mm*1 cm10 mm*1 in2.54 cm=0.4724 in
How many cm are in 568mm and show work?
8 mm = 0.8 cm10 mm = 1 cm20 mm = 2 cm50 mm = 5 cm60 mm = 6 cm500 mm = 50 cm560 mm = 56 cm568 mm = 56.8 cm
Which is larger 1 m or 10 mm?
1 m is larger than 10 mm.Here are the details:1 m = 1,000 mm = 100 cm10 mm = 1 cm100 cm > 1 cm
How many mm in .125 inches?
125 millimeters = 4.92 inches.Algebraic Steps / Dimensional Analysis Formula 125 mm*1 cm10 mm*1 in2.54 cm=4.921259843 inDirect Conversion Formula 125 mm*1 in25.4 mm=4.921259843 in
How many centimeter are in ten inches?
25 Algebraic Steps / Dimensional Analysis Formula10 in*2.54 cm 1 in=25.4 cm
How do you convert 1250 mm to feet?
Use this formula: mm x 0.00328 = feet1250 mm x 0.00328 = about 4.1 feet.Algebraic Steps / Dimensional Analysis Formula1,250 mm*1 cm10 mm*1 in2.54 cm*1 ft12 in=4.101049869 ftDirect Conversion Formula 1,250 mm*1 ft304.8 mm=4.101049869 ft
How do you calculate mm into inchs?
1 mm = about 0.04 inches. So, to convert from mm to inches, multiply inches by 0.04.Algebraic Steps / Dimensional Analysis Formula ____ mm*1 cm10 mm*1 in2.54 cm=? inDirect Conversion Formula ____ mm*1 in25.4 mm=? in
