How do you beat Super Scribblenauts level S2-4?

What is the meaning of S24 in MS Excel?

S24 could refer to the cell in column S, row 24.

What are the two numbers that can be both the perimeter and the area of the same rectangle?

Area and perimeterA square is a special form of rectangle having four equal sides. If each side of the square measures four units, then the perimeter is 16 units and the area is 16 square units. Note that the magnitude -- 16 -- is the same but the units are different. The above can be shown algebraically.If P = 4s and A = s2, then setting P = A yields4s = s24 = s.Their is one other solution when the rectangle is not a square. If two sides are 6 and two sides are 3, the area and the perimeter are the same as well.2(6) + 2(3) = P12 + 6 = P18 = P3 * 6 = A18 = AP = A

How do you say numbers 1-70 in Spanish?

uno - ooo no dos - dose tres - trace cuatro - kwat-tro cinco- sink-o seis- siete - c-yet-tay otro - o-trow nueve - new-wev-ay diez - d-es once- own say dose - do-say trece - tray say catorce - ka-tour-say quince- king say dieziseis or diez y seis - d-es-e- for 17-19 you just say diez y, then the number...i.e. (diez y siete, diez y otro, diez y nueve) veinte- bayn-tay veinte y un- bay-n-tay-oo-n for 22-29 you just say veinte y, then the number.....i.e. (veinte y dos, veinte y tres, veinte y cuatro) treinta- train-tah treinta y un- train-tah-e-oo-n for 32-39 you just say treinta y, then the number....i.e. (treinta y dos, treinta y tres, treinta y cuatro) cuarenta - core-rain-tah cuarenta y un - core-rain-tah-e-oo-n for 42-49 you just say cuarenta y, then the number....i.e. (cuarenta y dos, cuarenta y tres, cuarenta y cuatro. if you haven't noticed, after 16, the numbers become extremely easy. you jus have to know how to say the first part, like 20, 30, 40, 50, 60, etc. the only thing that changes is 1. instead of saying uno, you just say un. so for the rest of this, I'm just going to give the first part you'll need and the pronunciation cincuenta - sin-kwayn-tah sesenta- say-saint-tah setenta- say-taint-tah ochenta- o-ch-aint-tah noventa - no-vein-tah cien - see-in

Does St Mary Abbotts Hospital Kensington still exist?

A former workhouse, St Mary Abbott's Hospital in Marloes Road, Kensington, London, W8 - where I was born 69 years ago next week - no longer exists, other than one listed building that has been retained in what is now a very exclusive, gated community of luxury apartments. I am in the process of trying to find out just when the main parts of the hospital were demolished. John Sedgmore Blundell - Harrow Mdx May 2009. New User. Adding to the above, my brother David (deceased) was born at St Mary Abbott's Hospital at 28 Marloes Road in November 1926, as was I myself in May 1929. It is nice to know that at least one listed building still exists. Will have a look next time I am in that part of town. ADDITION - James de Buitlear (now in Sweden) I was born there in November 1961. Later as a police officer on B District at Chelsea (BC Division) we were quite regularly called to the psychiatric unit at the hospital up to the late 80s to either pick people up or drop people off. It was in the early 90s that the psych unit closed down and demolition work for the flats commenced. In about 93-94 I remember attending the new luxury flats to dleiver domestic security/crime prevention advice to residents. So although I can't give an exact date for the closure of the various remaining hospital units nor an exact date for the commencement of demolition I'm 100% sure it was around the early 90s. ADDITION- Tom Flynn- Nottingham May 2010. I worked on the design of the flats in around 89-90. The architects were called Norman and Dawbarn, based in Percy Street off Tottenham Court Road.( famous for the design of the BBC TV centre in White City) We were able to reuse the Dutch gables from the original hospital building on the garden walls to some of the blocks. Sadly, I never got to really see the finished article because it's too exclusive for the likes of me! I heard that Jimi Hendrix died in that hospital, so I wanted to call one of the blocks "Hendrix Court". Hope this helps. Try looking at it on Google Earth, look for Marloes Road. Addition - Phillip Perry, Shepherd's Bush - London Further to earlier answers, I was born there in June 1966, and my younger brother in December 1971, he was one of the last to be born there as it closed shortly afterwards as far as maternity was concerned. I thought all of the buildings had gone, I was there yesterday (10/08/10), it's now a 'gated community' called Kensington Green. I wanted to have a look around and take some pictures, but decided against it. The only original building left along with the perimeter wall is the gatehouse, which is the main security lodge, that has a font and a plaque built into the wall in Marloes Road. Here's reference to an article saying it closed finally in 1992: http://digilondon.com/location/st-mary-abbots-hospital British Library article from 1986 http://www.british-history.ac.uk/report.aspx?compid=50331#s24 Addition: I was born there in 1954 and wondered if any old photos of the hospital exist as I haven't been able to find any online? Addition: I was born there in 1943 (well, actually, in the nearest tube station during an air raid where all ambulatory patients were moved). Photos of the hospital during its glory days can be found at: http://www.flickr.com/photos/backmanmal/3999092540/

Is it possible that the divine message of Bhagvad-Gita got distorted by mundane insertions?

Mundane distortions in the Divine discourseWilliam von Humboldt who wrote seven-hundred verses in praise of the Bhagvad-Gita averred that it is the most beautiful, perhaps the only true philosophical song existing in any known tongue. All the same, the boon of an oral tradition that kept the divine discourse of yore alive for millennia became the bane of the Gita going by the seemingly mundane distortions it had to endure. Strangely it was Sir Edwin Arnold the Englishman who sought to separate the divine wheat from the mundane chaff by branding s23-s27 of ch8 as the ranting of some vedanti in his century old 'Song Celestial'. While interpreting the Gita in English verse an attempt was made by the author to identify the interpolations in it and codify the same for the benefit of the modern reader. One way to scent the nature of these, if not zero in on every one of them, is to subject the text to the twin tests of sequential conformity and structural economy. Sequential conformity is all about uniformity of purpose sans digression and structural economy but represents the absence of repetitiveness.The pundits and the plebeians alike aver that the philosophy of the Gita is the practice of disinterested action. In this context it may be noted that while postulating nishkaama karma, the theory of disinterested action, Krishna was critical of the ritualistic aspects of and the mundane expectations from the Vedic ceremonies (s42 - 46 and s53 of ch.2.). Given that the pristine philosophy of the Gita is to tend man on the path of duty without attachment, the about turn in s9-s16 of ch.3 that formulate the procedural aspects of the rituals and the divine backing they enjoy cannot stand to either reason or logic. Such contradictory averments attributed to Krishna wherever occur can be taken as interpolations and the same are delved into in this article.Next on the agenda is the aspect of structural economy and one finds the similitude of a given content in many a sloka in the same or in a different context throughout the text. Obviously, some of them are interpolations but which were the originals and which are the imitations could be impossible to find out for they smugly fit into the overall structure. Whatever, save lengthening the discourse, they do not belittle the same and fortunately not even tire the reader, thanks to the exemplary charm of Sanskrit, which for the 18th Century British intellectual Sir William Jones 'is of wonderful structure, more perfect than Greek, more copious than Latin and more exquisitely refined than either.' Identified here are 110 slokas of deviant character or digressive nature that can be taken as interpolations with reasonable certainty. Readers may like to mark these verses in their Gita and then read it afresh by passing over them for a refreshing experience.Besides the interpolations s9-s16 of ch.3, s17, s18 and s24 of the same are clear digressions. Such others in the rest of the chapters wherever they crop up are dealt as follows:Chapter - 4: It should not be lost on one that s11's return of favour by the Lord is juxtaposing to the stated detachment of His as espoused in s14 of the same chapter. On the other hand, s12 that is akin to s20, ch.7, itself an interpolation, and s13 the contentious chaatur varnyam mayaa srustam - do not jell with the spirit of the philosophy. Why hasn't Krishna declared in s 29 ch.9, 'None I favour, slight I none / Devout Mine but gain Me true'. Slokas 24 to s32 that are of religious/ritualistic nature seem clearly out of context and character. Prior to this seemingly interpolated body of eleven slokas, the nature of the Supreme Spirit and the conduct of those who realize it are dealt with. Thus, the discontinuity in the text brought about by the body of these interpolative slokas would be self-evident. And s34 that advises Arjuna to seek wise counsel is irrelevant in the context of the discourse fashioned to set his doubts at rest in the battlefield itself.Chapter-5: S18 avers the Omnipresence of the Supreme in Brahmans, cows, elephants, dogs and dog eaters. This tasteless description could be but an interpolation as it ill behoves Krishna's eloquence and sophistication of expression seen throughout. Incidentally, the succeeding s19 makes it clear that whoever recognizes Him in all beings attains the Supreme State in life itself. S27-s28 that deal with yogic practices and s29, which asserts the Supreme as the beneficiary of sacrificial rituals, are but interpolation for reasons that bear no repetition.Chapter - 6: S10-s17 deal with aspects of ascetic practices which are but square pegs in the round philosophical hole the discourse is and so are interpolations, even going by what is stated in the very opening verse, 'Forego none if forsake chores / Eye not gain 'n thou be freed'. S41 and s42are clearly interpolations not only for affecting the continuity of the text but also for what they contain. S41 would have us that those who perform the asvamedha (ritualistic horse sacrifice) would reach heaven to be born again rich. Likewise, s 42 would have us that, 'or such would be born in learned homes'.Chapter-7: S20-s23 besides affecting the continuity in character of the discourse, would advocate worship of gods for boon seeking that Krishna chastises is s42-s44, ch.2 and that renders them interpolations.Chapter - 8: It can be seen that s5 places the cart before the horse. Besides, s9-s14 too are interpolations going by their content that's out of context. It is worth noting that s1-s4, s6-s8 and s15-s22, if read together would bear an unmistakable continuity of argument that the interpolations deprive. And s22 is a seemingly concluding statement of the Lord that only through un-swerved devotion the Supreme could be reached from which there is no return (s21). Then appear s23 to s27 which if literally taken would imply that if one dies when the moon is on the ascent he would go to heaven and, to hell if it's other way round. Needless to say, these slokas spelling superstition in an otherwise thought-elevating treatise are but interpolations which Sir Edwin Arnold dismissed as the work of some vedanti and thought it fit, justifiably at that, not to include them in his 'Song Celestial'. In this connection it may be noted that the relationship between the state in which a person dies and his imminent rebirth is covered in s14 - s15 of c14, which seem to be authentic.Chapter -9: S7, that contravenes s15-s16 of ch.8, and which echoes interpolative s18-s19 of the current chapter, is an interpolation. Also s15 of is but a digression to facilitate the interpolations in s16-s21 and s23-s25. What is more, there could be some omissions from the original, given the seemingly incomplete exposition of the promised dharma in s2. Further, in s 30 and s 31, it is said that even a reformed sinner is dear and valuable to Him. Then in s 32 it is stated that women, Vaisyas and Sudras could win His favour through devotion, sounding as if they are all in an inferior league. Leave aside the Lord's averment in many a context in this text that the Supreme Spirit lies in all beings, it is specifically stated in s34 of ch.10 that He symbolizes all that is glorious in woman. Given this, and the background of the interpolations, s32 surely is a case of trespass. S33 of this chapter is but a jointing medium of the said obnoxious verse and in itself is patronizing in nature towards the virtuous Brahmans and thus is an interpolation.Chapter - 11: Owing to the improbability of their being, s9-s14, make an amusing reading. S3 states that Krishna grants Arjuna the divine sight required to espy His Universal Form. Of course, the ESP that Vyasa granted Sanjaya (s75 ch.18) might have enabled him to monitor the goings on at the battleground in order to appraise the blind king Dhrutarashtra about the same. Thus, only from Arjuna's averments could have Sanjaya gathered what he was divining of the Universal Form, which obviously was beyond his (Sanjaya) own comprehension. But s10-s14 would have him describe the Universal Form as though he himself was witnessing the same, even before Arjuna utters a word about it. In this context it is worth noting that the Lord made it clear in s52, 'Ever craved gods 'n angels too / Just to behold what thee beheld'. Thus, the Universal Form that was seen by Arjuna surely was beyond the scope of Sanjaya's ESP and hence, s9-s14 that picture beforehand what Arjuna would witness later on are clear interpolations. Contrast this with the parallel situation in s50-s51, when the Lord reassumes His human form, but handled differently by Sanjaya. The s29 which seeks to emphasize what was already pictured in s28, albeit with not so appropriate a simile, could be but an interpolation.Chapter -13: One might notice that s10, advocating asceticism to which Krishna is opposed, doesn't jell with the rest, either contextually or philosophically, and thus should be seen as an interpolation. S22, which states that the Supreme Soul lay in beings as a sustainer, consenter, enjoyer and overseer, contravenes its very nature expostulated in s16-s18, ch.15. Besides, as can be seen, it affects the continuity between s21 and s23 of this chapter. S30, akin to s15 is an irrelevant interpolation.Chapter- 14: In this chapter that details the three human proclivities - virtue, passion and delusion- s3, s4 and s19 that deal with the Nature and the Spirit are digressive interpolations.Chapter - 15: S9, s12, s13, s14 and s15 being digressions are clearly interpolations.Chapter - 16: S19 which implies that the Supreme Spirit condemns to hell those who hate Him is an obvious interpolation that contravenes Krishna's affirmative statement in s29 ch.9 and other such averred in many a context in this text.Chapter-17: S11-s13 that deal with the virtuous, the passionate and the deluded in ritualistic sense and s 23 -28 concerning Om, Tat, Sat and Asat of the Vedic hymns are clear interpolations for reasons the reader is familiar with. However, s7-s10 that deal with the food habits of the virtuous, the passionate and the deluded would pose a problem in determining whether or not they are interpolations. Can eating habits be linked to the innate nature of man in an infallible manner? Perhaps, some future research and analysis might resolve the universality or otherwise of this averment, and till then, it is appropriate to reserve the judgment on these.Chapter -18: One can note that s12 breaks the continuity between s11 and s13 with hyperbolic averments and s56 combines what is stated in the preceding and the succeeding slokas, and thus both are seemingly interpolations. S41- s48 that describe the allotted duties of man on the basis of his caste are clearly interpolations. In essence, the discourse till s40 is about the human nature and how it affects man. As can be seen, the duties on caste lines detailed in the said interpolations have no continuity of argument. As in earlier chapters, the text acquires continuity if only these verses are bypassed. S61 avers that the Supreme dwells in humans and deludes them all by his Maya. This is contrary to what is stated in s14, c5, 'It's his nature but not Spirit / Makes man act by wants induced'. Thus, s61 clearly is an interpolation as it contravenes the neutrality of the Supreme Spirit in the affairs of man affirmed throughout by Lord Krishna.For those who may like to see how the Gita reads if the above cited 110 slokas are bypassed, the same are summarized as under.Ch. 3: s9 -s18, s24 and s35 (12 slokas); Ch.4: s11 - s 13, s24- s32 and s34 (13 slokas); Ch.5: s18 and s27 -29 (4 slokas) ; Ch. 6: s10-s17 and s41 -s42 (10 slokas) ; ch.7: s20 -s23 (4 slokas) ; ch.8: s5, s9- s14 and s23-s28 ( 13 slokas) ; ch.9: s7,s15-s21, s23-s25, and s32-s34 (14 slokas) ; ch.11: s9- s14 and s29 (7 slokas) ; ch.13: s10, s22 and s30 (3 slokas) ;ch.14: s3 -s4 and s19(3 slokas) ; ch.15: s9 and s12- s15 (5 slokas );ch.16: s19 (1 sloka) ;ch.17: s11- s14 and s23- 28 (10 slokas) and ch.18: s12, s41-48, s56 and s61(11 slokas).One may like to read the 591-verse 'Bhagvad-Gita: treatise of self-help' in verse sans the above at Vedanta Spiritual Library - (related link 1)or hear the audio rendition of the same(related link 2)

Answers of intelegence tests parts 1to5?

024 H in a D126 L of the A26 LETTERS OF THE ALPHABET27 D of the W7 DAYS OF THE WEEK37 W of the W7 WONDERS OF THE WORLD412 S of the Z12 SIGNS OF THE ZODIAC566 B of the B66 BOOKS OF THE BIBLE652 C in a P (W J)52 CARDS IN A PACK (WITHOUT JOKERS)713 S in the U S F13 STRIPES IN THE UNITED STATES FLAG818 H on a G C18 HOLES ON A GOLF COURSE939 B of the O T39 BOOKS OF THE OLD TESTAMENT105 T on a F5 TOES ON A FOOT1190 D in a R A90 DEGREES IN A RIGHT ANGLE123 B M (S H T R)3 BLIND MICE (SEE HOW THEY RUN)1332 is the T in D F at which W F32 IS THE TEMPERATURE IN DEGREES FAHRENHEIT AT WHICH WATER FREEZES1415 P in a R T15 PLAYERS IN A RUGBY TEAM153 W on a T3 WHEELS ON A TRICYCLE16100 C in a D100 CENTS IN A DOLLAR1711 P in a F (S) T11 PLAYERS IN A FOOTBALL (SOCCER) TEAM1812 M in a Y12 MONTHS IN A YEAR1913 is U F S13 IS UNLUCKY FOR SOME208 T on an O8 TENTACLES ON AN OCTOPUS2129 D in F in a L Y29 DAYS IN FEBRUARY IN A LEAP YEAR2227 B in the N T27 BOOKS IN THE NEW TESTAMENT23365 D in a Y365 DAYS IN A YEAR2413 L in a B D13 LOAVES IN A BAKER'S DOZEN2552 W in a Y52 WEEKS IN A YEAR269 L of a C9 LIVES OF A CAT2760 M in an H60 MINUTES IN AN HOUR2823 P of C in the H B23 PAIRS OF CHROMOSOMES IN THE HUMAN BODY2964 S on a C B64 SQUARES ON A CHESS BOARD309 P in S A9 PROVINCES IN SOUTH AFRICA316 B to an O in C6 BALLS TO AN OVER IN CRICKET321000 Y in a M1000 YEARS IN A MILLENNIUM3315 M on a D M C15 MEN ON A DEAD MAN'S CHEST24 H in a D15 to 7 C of the W5 to 7 continents of the world21 O G E 4 Y1 Olympic games every 4 years3208 to 214 B in a H B208 to 214 bones in a human body42 M a P2 makes a pair56 S of a C6 sides of a cube68 B in a B8 bits in a byte71.6 K in a M1.6 kilometers in a mile84 W of a C4 wheels of a car98 L on a S8 legs on a spider10191 U N M S191 united nations member states112 D in a W2 days in a weekend12168 H in a W168 hours in a week13101 D101 dalmatians147 D S7 deadly sins159 P in the S S9 planets in the solar system1617 is a P N17 is a prime number179 M of P9 months of pregnancy1817 S in a H P17 syllables in a haiku poem191 7 S H in the W1 7 star hotel in the world2052 W K on a P52 white keys on a piano2121 Q S F21 questions so far224 H of the A4 horsemen of the apocalypse23115 K C E115 known chemical elements246 S W M (S F)6 Star Wars Movies (So Far)5 S of G5 Stages of Grief24 S of the Y4 seasons of the year33 P C3 primary colors42 H of a C2 hands of a clock54 (b) P of a C4 (basic) points of a compass68 N in an O8 notes in an octave732 P (A) T32 permanent (adult) teeth86 is H a D6 is half a dozen97 C in a R7 colors in a rainbow1024 F P S24 frames per second114 S and C in a S4 sides and corners in a square12101 K on a S K101 keys on a standard keyboard132 P of a M2 poles of a magnet1425 P is 1 Q25 percent is 1 quarter156 is the S P N6 is the smallest perfect number165 A in a S P5 acts in a shakespearean play17N C be D to 0nothing can be divided to 01830 S is H a M30 seconds is half a minute195 R on the O F5 rings on the olympic flag203 D3 dimensionsIntelligence Test Part 4 Solutions1.1024 B in a K1024 Bytes in a Kilobyte2.The S R of -1 is an I NThe Square Root of -1 is an Imaginary Number3.666 is the N of the B666 is the Number of the Beast4.12 N on a C12 Numbers on a Clock5.4 C in the H H4 Chambers in the Human Heart6.40 D of L40 Days of Lent7.99 B of B on the W99 Bottles of Beer on the Wall8.2 L in a S P of G2 Lenses in a Standard Pair of Glasses9.12 D of J12 Disciples of Jesus10.2 P in a D2 Players in a Duet11.S W and the 7 DSnow White and the 7 Dwarfs12.2 S of a C2 Sides of a Coin13.4 S in a D of C4 suits in a Deck of Cards14.8 H in a R W D8 Hours in a Regular Work Day15.150 P in the B150 Psalms in the Bible16.E S H M P N S with 555Every Single Hollywood Movie Phone Number Starts with 55517.42 is the A to L, the U, and E42 is the Answer to Life, the Universe, and Everything18.3 M T S - C, F and K3 Major Temperature Scales - Celsius, Fahrenheit and KelvinSolutions to Intelligence Test Part 515 P of I T5 parts of intelligence tests210 is 2 in B10 is 2 in binary3X 360xbox 3604800×600 S R800×600 screen resolution5A B and the 40 Tali baba and the 40 thieves66 Z in a M6 zeros in a million722 P on a F F22 players on a football field854 S on a R C54 squares on a rubik's cube921 D on a S D21 dots on a standard die104 W in a M4 weeks in a month11The 9 MThe 9 Muses1212 D of J12 disciples of jesus13The D to A C is A 4 L Ythe distance to alpha centauri is approximately 4 light years14122 S in a C C B122 Spaces in a Chinese Checkers Board159/11 A9/11 attacks16J has 62 Mjupiter has 62 moons174 L on a T4 legs on a table182 B on a P of S2 blades on a pair of scissors199 C of H in D I9 circles of hell in dante's inferno207 L in the W L7 letters in the word letters

Numbers 50-100 in Spanish?

1-10: Uno, dos, tres, cuatro, cinco, seis, siete, ocho, nueve, diez. 11-20: once, doce, trece, catorce, quince, dieciséis, diecisiete, dieciocho, diecienueve, veinte. or veinite 21-30: veinte-uno, veinte-dós, veinte-tres, veinte-cuatro, veinte-cinco, veinte-séis, veinte-siete, veinte-ocho, veinte-nueve, treinta. 31-40: treinta - uno, treinta - dos, treinta - tres, treinta - cuatro, treinta - cinco, treinta - seis, treinta - siete, treinta - ocho, treinta - nueve, cuarenta. 41-50: cuarenta y uno, cuarenta y dos, cuarenta y tres, cuarenta y cuatro, cuarenta y cinco, cuarenta y seis, cuarenta y siete, cuarenta y ocho, cuarenta y nueve, cincuenta. 51-60: cincuenta y uno, cincuenta y dos, cincuenta y tres, cincuenta y cuatro, cincuenta y cinco, cincuenta y seis, cincuenta y siete, cincuenta y ocho, cincuenta y nueve, sesenta. 61-70: sesenta y uno, sesenta y dos, sesenta y tres, sesenta y cuatro, sesenta y cinco, sesenta y seis, sesenta y siete, sesenta y ocho, sesenta y nueve, setenta. 71-80: setenta y uno, setenta y dos, setenta y tres, setenta y cuatro, setenta y cinco, setenta y seis, setenta y siete, setenta y ocho, setenta y nueve, ochenta. 81-90: ochenta y uno, ochenta y dos, ochenta y tres, ochenta y cuatro, ochenta y cinco, ochenta y seis, ochenta y siete, ochenta y ocho, ochenta y nueve, noventa, 91-100: noventa y uno, noventa y dos, noventa y tres, noventa y cuatro, noventa y cinco, noventa y seis, noventa y siete, noventa y ocho, noventa y nueve, cien.

What weapons did china use in World War 2?

japan was n the war china stayed out of it and the japaneese usedRiflesType 38 RifleType 38 Cavalry RifleType 44 Cavalry RifleType 97 Sniper RifleType 99 RifleType 99 Sniper RifleType I RifleTERA RiflesExperimental Automatic Rifles (Type 5 rifle)PistolsType 26 9 mm PistolType 14 8 mm Nambu PistolType 94 8 mm PistolSubmachine gunType 100Machine gunsType 11 Light Machine GunType 96 Light Machine GunType 99 Light Machine GunType 3 Heavy Machine GunType 92 Heavy Machine GunType 1 Heavy Machine GunGrenadesType 10 GrenadeType 91 GrenadeType 97 GrenadeType 99 GrenadeCeramic GrenadeGrenade launchersType 10 Grenade DischargerType 89 Grenade DischargerType 2 rifle grenade launcherInfantry mortarsType 11 70 mm Infantry MortarType 94 90 mm Infantry MortarType 96 150 mm Infantry MortarType 97 81 mm Infantry MortarType 97 90 mm Infantry MortarType 97 150 mm Infantry MortarType 99 81 mm Infantry MortarType 2 120 mm Infantry MortarType 98 50 mm MortarHeavy mortars & rocket launchers15cm MortarType 14 27 cm Heavy MortarType 98 32 cm Spigot MortarType 4 20 cm Rocket LauncherType 4 40 cm Rocket LauncherType 5 Mortar Launcher "Tok"FlamethrowersType 100 FlamethrowerMilitary swordsType 98 Military SwordHeavy weaponsField artillery7cm Mountain GunType 31 75 mm Mountain GunType 41 75 mm Mountain GunType 94 75 mm Mountain GunType 99 10 cm Mountain GunType 41 75 mm Cavalry Gun7 cm Field GunType 31 75 mm Field GunType 38 75 mm Field GunType 90 75 mm Field GunType 95 75 mm Field GunKrupp 12 cm HowitzerKrupp 15 cm HowitzerType 38 12 cm HowitzerType 38 15 cm HowitzerType 4 15 cm HowitzerType 91 10 cm HowitzerType 96 15 cm HowitzerKrupp 10.5 cm CannonType 38 10 cm CannonType 14 10 cm CannonType 92 10 cm CannonFortress guns28cm HowitzerType 45 24 cm HowitzerType 45 15 cm CannonType 7 30 cm HowitzerType 7 10 cm CannonType 7 15 cm CannonType 11 75 mm CannonType 89 15 cm CannonType 96 24 cm HowitzerType 96 15cm CannonExperimental 41 cm HowitzerType 90 24 cm Railway GunInfantry gunsType 11 37 mm Infantry GunType 92 70 mm Infantry GunAnti-tank gunsType 94 37 mm Anti-Tank GunType 1 37 mm Anti-Tank GunType 1 47 mm Anti-Tank GunType Ra 37 mm AT GunAnti-tank weaponsInfantry ArmorType 97 20 mm AT RifleType 99 AT MineArmor Piercing Anti-Tank GrenadeType 93 Pressure Anti-Tank/Personnel MineType 2 AT Rifle GrenadeType 3 AT GrenadeLunge AT Mine57 mm Tank Cannon37 mm Tank CannonExperimental AT Gun (Japanese equivalent to Bazooka, PIAT, Panzerfaust or Panzershreck)Type 5 45 mm Recoilless GunType 4 70 mm AT Rocket LauncherAnti-aircraft weaponsOccasional anti-aircraft gunsType 97 20 mm AT/AA RifleType 11 Light Machine GunType 96 Light Machine GunType 99 Light Machine GunType 3 Heavy Machine GunType 92 Heavy Machine GunType 1 Heavy Machine GunLight anti-aircraft gunsType 3 Heavy Machine GunType 4 Heavy Machine GunType 97 20 mm AT/AA RifleType 98 20 mm AA Machine CannonType 2 20 mm AA Machine Cannon20 mm Twin AA Machine CannonType 4 20 mm Twin AA Machine CannonModel 96 25 mm AT/AA GunAA Mine DischargerMedium & heavy anti-aircraft gunsModel 96 25 mm AT/AA Gun (triple Muzzle)Vickers Type 40 mm AT/AA GunType 11 75 mm AA GunType 88 75 mm AA GunType 4 75 mm AA GunType 3 80 mm AA GunType 99 88 mm AA GunType 10 120 mm AA GunType 14 10 cm AA GunType 3 12 cm AA GunType 5 15 cm AA GunMobile anti-aircraft vehiclesType 98 20 mm AAG Tank "Ho-Ki"20 mm AA Machine Cannon Carrier Truck20 mm Anti-Aircraft Tank "Ta-Se"Type 96 AA Gun Prime MoverType 98 20 mm AA Half-Track Vehicle "Ko-Hi"TanksBritish Mk IV - World War I vintageBritish Medium A "Whippet" - World War I vintageRenault FT17 "Ko" Light Tank - World War I vintageType 89 Medium Yi-GoRenault NC27 "Otsu" Light TankCarden Loyd Mk.VIType 92 Combat CarType 94 Tankette "TK"Type 95 Heavy TankType 95 Ha-GoType 97 Te-KeType 97 Chi-HaType 97-improved Medium Tank "Shinhoto Chi-Ha"M3 Light Tank - CapturedType 98 Light Tank "Ke-Ni"Type 1 Medium Tank "Chi-He"Type 2 Light Tank "Ke-To"Type 2 Amphibious Tank "Ka-Mi"Type 3 Amphibious Tank "Ka-Chi"Type 3 Medium Tank "Chi-Nu"Type 4 Medium Tank "Chi-To"Type 4 Light Tank "Ke-Nu"Type 5 Medium Tank "Chi-Ri"Type 92 TanketteType 95 Ke-Go Light TankType 93 Light TankType 94 Medium TankType 95 Kyo-Go Light TankType 95 Ke-Ri Light TankType 95 Keni Light TankType 95 So-Ki Rail TankType 95 Ke-Nu Rail TankType 95 Heavy TankType 97 Chi-NiType 1 Ti-Ho Medium TankType 2 Medium TankType 2 Ho-I Infantry Support TankType 4 Ke-Hy Light TankType 5 Ke-Xo Light TankExperimental Tank - Number 1Type 91 Heavy Tank - ExperimentalExperimental Medium Tank Chi-NiExperimental Medium TankExperimental Medium Tank Chi-HoExperimental O-I Super Heavy TankExperimental Ultra Heavy TankExperimental Type 98 Medium TankExperimental Flying Tank Ku-Ro or So-RaExperimental Flying Tank Maeda Ku-6Model 94 3/4 Ton Tracked TrailerSRII Amphibious TankType 1 "Mi-Sha" Amphibious TankType 3 "Ka-Chi" Amphibious TankType 2 "Ka-Mi" Amphibious TankType 4 "Ka-Tsu" Amphibious LaunchType 4 "Ka-Sha" Amphibious TankType 5 "To-Ku" AmphibiousNote: Amphibious Tanks were used by the IJN.Self-propelled gunsType 1 75 mm SPH "Ho-Ni I"Type 1 105 mm SPH "Ho-Ni II"Type 2 Gun Tank "Ho-I"Type 3 Gun Tank "Ho-Ni III"Type 4 150 mm SPH "Ho-Ro"Short Barrel 120 mm Gun Tank75 mm SP AT Gun "Na-To"Type 4 120 mm Ho-ToType 4 155 mm Ho-Ro75 mm SPG "Kusae"120 mm Short Barrel Gun TankExperimental Type 2 Ho-Ri Tank DestroyerExperimental Jiro-Sha Self Propelled GunExperimental Type 2 105 mm Ka-To Tank DestroyerExperimental Type 2 75 mm Ku-Se SPGExperimental Type 5 47 mm Ho-Ru SPATGExperimental 105 mm Tank DestroyerExperimental Type 5 155 mm Ho-Chi SPGExperimental 120 mm SPGExperimental Type 2 105 mm Ka-To Tank DestroyerExperimental Type 2 75 mm Ku-Se SPGExperimental Type 5 47 mm Ho-Ru SPATGArmored vehiclesArmored carsAustin Motor Company Armoured CarVickers Crossley Armored CarWolseley Motor Company Armoured CarModel 91 Broad-gauge Railroad Tractor Sumi-DaType 91 Armored Railroad Car "So-Mo"Type 95 Armored Railroad Car "So-Ki"Aikoku Armored CarHokoku Armored CarType 2592 Osaka HoKoKu-Go Armored CarType 2592 "Kokusan" Armored CarType 2592 "Chokei Sensha" Armored Car*Type 2592 Chiyoda Armored CarType 2593 "Sumida" Armored CarModel 92 Naval Armored CarModel 92 "Chiysda" Armored CarFord Armored CarType 2598 Railroad CarArmored carriersType 98 Armored Ammunition Carrier "So-Da"Type 1 APC "Ho-Ki"Type 1 Ho-Ha "Half-Track"Type 4 Amphibious Vehicle "Ka-Tsu"Experimental Light Armored ATG Carrier "So-To"Type-4 Ka-Tsu "Toku 4 Shiki Naikatei" Amphibious Personnel CarrierExperimental armored vehiclesAnti-Aircraft Tank "Ta-Se"Type 4 30 cm SP Heavy Mortar Carrier "Ha-To"105 mm SP Gun Tank "Ho-Ri"Type 5 47 mm SP Gun "Ho-Ru"Super-Heavy Tank "O-I"Armored trainsType 94 Armored TrainImprovised Armored TrainExperimental Armored TrainWagonsWagon-1 Reconnaissance WagonWagon-1 Protective WagonWagon-2 Heavy Canone WagonWagon-3 Light Canone WagonWagon-4 Infantry WagonWagon-5 Command WagonLocomotivesWagon-6 Auxiliary TenderWagon-7 Materials WagonWagon-7 Power Supply WagonWagon-8 Infantry WagonWagon-9 Light Canone WagonWagon-10 Howitzer WagonWagon-11 Protective WagonOther armored vehiclesArmored Engineer Vehicle "SS"Armored Lumberjack "Ho-K"Type 95 Crane Vehicle "Ri-Ki"Armored Recovery Vehicle "Se-Ri"Swamp Vehicle "FB"Command Tank "Shi-Ki"Type 100 Observation Vehicle "Te-Re"Type 97 Mini Engineer Vehicle "Yi-Go"High-Voltage Dynamo Vehicle "Ka-Ha"Type 97 Pole Planter and Type 97 Cable LayerType 94 Disinfecting VehicleType 94 Gas Scattering VehicleOther land vehiclesTrucksType 94 6-Wheeled TruckType 95 Mini-truckType 97 4-Wheeled TruckType 1 6-Wheeled TruckType 2 Heavy TruckToyota KB/KC TruckNissan 80 TruckNissan 180 TruckAmphibious Truck "Su-Ki"Type 1 4x2 Toyota GB2 Wheel Towed CartIsuzu Hucks StarterPassenger carsType 92 Armored CarType 93 6/4-Wheeled Passenger CarType 95 Passenger Car "Kurogane"Type 98 Passenger CarModel 97 Nissan Staff Car, Nissan 70MotorcyclesType 97 MotorcycleJapanese Military "Trike"Railroad carsType 98 Railroad CarType 100 Railroad CarTractors & prime moversType 92 5 t Prime Mover "I-Ke"Type 98 6 t Prime Mover "Ro-Ke"Type 92 8 t Prime Mover "Ni-Ku"Type 95 13 t Prime Mover "Ho-Fu"Type 94 4 t Prime Mover "Yo-Ke"Type 98 4 t Prime Mover "Shi-Ke"Type 98 Half-tracked Prime Mover "Ko-Hi"Type 98 Half-tracked Prime Mover Large ModelType 96 AA Gun Prime Mover NEWExperimental Heavy Gun Tractor Chi-KeExperimental Crawler TruckT G Experimental Crawler TruckFordson Prime MoverThe Pavessi Gun TractorThe 50 hp Gun TractorKomatsu 3 ton TractorLight Prime MoverClarton Prime MoverThe Holt 30Miscellaneous vehiclesType 94 AmbulanceType 94 Repair VehicleArmy vesselsRiver-crossing craftsType 95 Collapsible BoatType 99 Pontoon BridgeRubber RaftsLanding craftPersonnel Landing Craft "Shohatsu"Personnel Landing Craft "Chuhatsu"Vehicle Landing Craft "Daihatsu"Vehicle Landing Craft "Toku-Daihatsu"Vehicle Landing Craft "Mokusei-Daihatsu"MotorboatsSpeedboat Model KoSpeedboat Model OtsuSuicide-Attack Motorboat "Maru-Re"Gun boatsArmored Boat "AB-Tei"Submarine-chaser "Karo-Tei"Landing craft/aircraft carriersLanding Craft Carrier "Shinshu Maru"Landing Craft Carrier Model Ko, Otsu, HeiTransport vesselsTank Landing Ship "SS-Tei"Fast Transport Vessel "Yi-Go"Transport Submarine "Maru-Yu"Engineering vehicles & weaponsArmored Engineer Vehicle Soukou Sagyou Ki "SS-Ki"Armored Cable LayerType 97 Mini Engineering Vehicle "Yigo"Tank Recovery Vehicle "Ri-Ki"Tank Recovery Vehicle "Seri"T G Armored Bridge LayerJapanese Armored Lumberjack "Ho-K"Japanese Armored Lumberjack "Basso Ki"Armored Cable LayerExperimental Trench ExcavatorExperimental ExcavatorNavy ships and warvesselsMain article: List of Japanese Navy ships and warvessels in World War II Secret weaponsMain article: Secret and special weapons in Showa Japan Army secret weaponsRemote control special vehicle "I-Go"Unmanned miniature special vehicle "Ya-I"Remote control special working cable car with 90 mmMortar cannon "Ite-Go"Remote control boat "Isu-Go"Rocket cannon "Ro-Go"Nuclear project "Mishina"Engine stop gas "Ha-Go"Radio jammer "Ho-Go"Electric cannon "To-Go"Ultrashort waves application research "Chi-Go"High voltage weapon "Ka-Go"High voltage obstacle destruction device "Kaha-Go"High voltage wire obstacles "Kake-Go"High voltage net launching rocket "Kate-Go"Infrared rays detection device "Ne-Go"Mine detection sonar for the landing operations "Ra-Go"Radio control device "Mu-Go"Radio controlled boat with sonar and a depth bomb "Musu-Go"Plan to make artificial thunderclouds "U-Go"Noctovision "No-Go"Death ray "Ku-Go"Infrared ray homing bomb "Ke-Go"Balloon bomb "Fu-Go"Optical communication device "Ko-Go"Rope launching rocket "Te-Go"Blind ray "Ki-Go"Propaganda electronic ray device "Se-Go"Sonar "Su-Go"Experimental armour for MG-gunnerNavy secret weaponsI-Go 14 Type (Type "Ko-Kai 2":Modified A Type2) I-Go 14 Aircraft SubmarineI-Go 15 Type (Type "Otsu":Type B) I-Go 26 Aircraft SubmarineI-Go 54 Type (Type "Otsu-Kai 2":Modified B Type2) I-Go 54 Aircraft SubmarineI-Go 400 Type (Special Submarine) I-Go 402 Aircraft SubmarineAichi M6A1 Seiran Torpedo-Bomber carried in subs.Suicide Attack Frogman "Fukuryuu""Kaiten" Type1 Suicide Attack Midget Submarine"Kairyu" Midget SubmarineNuclear Project "F-Go"Aircraft Battleship "Ise" Class "Ise"Imperial Japanese Army Electronic Warfare SystemsGround-based radarTa-Chi 1 Ground-Based Target Tracking Radar Model 1Ta-Chi 2 Ground-Based Target Tracking Radar Model 2Ta-Chi 3 Ground-Based Target Tracking Radar Model 3Ta-Chi 4 Ground-Based Target Tracking Radar Model 4TypeA Bi-static Doppler Interface Detector (High Flequency Warning Device "Ko")Ta-Chi 6 TypeB Fixed Early Warning Device (Fixed Early Warning Device "Otsu")Ta-Chi 7 TypeB Mobile Early Warning Device (Mobile Early Warning Device "Otsu")Ta-Chi 13 Aircraft Guidance SystemTa-Chi 18 TypeB Potable Early Warning Device (Portable Early Warning Device "Otsu")Ta-Chi 20 Fixed Early Warning Device Receiver (for Ta-Chi 6)Ta-Chi 24 Mobil Anti-Aircraft Radar (Japanese Würzburg)Ta-Chi 28 Aircraft Guidance DeviceTa-Chi 31 Ground-Based Target Tracking Radar Model 4 modifiedAirborne radarTa-Ki 1 Model 1 Airborne Surveillance RadarTa-Ki 1 Model 2 Airborne Surveillance RadarTa-Ki 1 Model 3 Airborne Surveillance RadarTa-Ki 11 ECM DeviceTa-Ki 15 Aircraft Guidance Device Receiver (for Tachi 13)Shipborne radarTa-Se 1 Anti-Surface RadarTa-Se 2 Anti-Surface RadarImperial Japanese Navy Electronic Warfare SystemsLand-based radarType 2 Mark 1 Model 1 Early Warning Radar ("11-Go" Early Warning Radar)Type 2 Mark 1 Model 1 Modify 1 Early Warning Radar ("11-Go" Model 1 Early Warning Radar)Type 2 Mark 1 Model 1 Modify 2 Early Warning Radar ("11-Go" Model 2 Early Warning Radar)Type 2 Mark 1 Model 1 Modify 3 Early Warning Radar ("11-Go" Model 3 Early Warning Radar)Type 2 Mark 1 Model 2 Mobil Early Warning Radar ("12-Go" Mobil Early Warning Radar)Type 2 Mark 1 Model 2 Modify 2 Mobil Early Warning Radar ("12-Go" Modify 2 Mobile Early Warning Radar)Type 2 Mark 1 Model 2 Modify 3 Mobil Early Warning Radar ("12-Go" Modify 3 Mobile Early Warning Radar)Type 3 Mark 1 Model 1 Early Warning Radar ("11-Go" Modified Early Warning Radar)Type 3 Mark 1 Model 3 Small Size Early Warning Radar ("13-Go" Small Size Early Warning Radar) *Type 3 Mark 1 Model 4 Long-Range Air Search Radar ("14-Go" Long-Range Air Search Radar)Type 2 Mark 4 Model 1 Anti-aircraft Fire-Control Radar (Japanese SCR-268)Type 2 Mark 4 Model 2 Anti-aircraft Fire-Control Radar (Japanese SCR-268) (S24 Anti-aircraft Fire-Control Radar)Airborne radarType 3 Air Mark 6 Model 4 Airborne Ship-Search Radar (H6 Airborne Ship-Search Radar) (N6 Airborne Ship-Search Radar)Type 5 Model 1 Radio Location Night Vision DeviceShipborne radarType 2 Mark 2 Model 1 Air Search Radar ("21-Go" Air Search Radar)Type 2 Mark 2 Model 2 Modify 3 Anti-Surface, Fire assisting Radar for Submarine ("21-Go" Modify 3 Anti-Surface, Fire-assisting Radar)Type 2 Mark 2 Model 2 Modify 4 Anti-Surface, Fire-assisting Radar for Ship ("21-Go" Modify 4 Anti-Surface, Fire-assisting Radar)Type 2 Mark 3 Model 1 Anti-Surface Fire-Control Radar ("31-Go" Anti-Surface Fire-Control Radar)Type 2 Mark 3 Model 2 Anti-Surface Fire-Control Radar ("32-Go" Anti-Surface Fire-Control Radar)Type 2 Mark 3 Model 3 Anti-Surface Fire-Control Radar ("33-Go" Anti-Surface Fire-Control Radar)Missiles & bombsJapanese Army guided missilesMitsubishi Ki-67 I Go-IA (Carrier Missile Bomber)Kawasaki Ki-147 I-Go Type1 - Ko Air to Surface Radio Guidance MissileMitsubishi Ki-148 I-Go Type1 - Otsu Air to Surface Radio Guidance MissileI-Go Type 1 - Hei"Ke-Go" IR Guidance Air to Surface MissileJapanese Navy guided missilesFunryu Type1Funryu Type2 Surface to Air Radio Guidance MissileFunryu Type3Funryu Type4 Surface to Air Radio Guidance MissileImperial Japanese Army bombsType Ro-3Type Ro-5Type Ro-7Imperial Japanese Navy bombsType 3 No.1 28-Go Bomb Type 2Type 3 No.1 28-Go Bomb Type 2 Modify 1Type 3 No.1 28-Go Bomb Type 2 Modify 2Type 3 No.1 28-Go Bomb "Maru-Sen"No.6 27-Go BombType 3 No.25 4-Go Bomb Type 1Type 3 No.50 4-Go Bomb

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)

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)