A geometrical practise, named Pantometria diuided into three bookes, longimetra, planimetra, and stereometria, containing rules manifolde for mensuration of all lines, superficies and solides: with sundry straunge conclusions both by instrument and without, and also by perspectiue glasses, to set forth the true description or exact plat of an whole region: framed by Leonard Digges gentleman, lately finished by Thomas Digges his sonne. Who hathe also thereunto adioyned a mathematicall treatise of the fiue regulare Platonicall bodies, and their Metamorphosis or transformation into fiue other equilater vniforme solides Geometricall, of his owne inuention, hitherto not mentioned of by any geometricians.

The second kynde of Geome∣trie called Planimetra. (Book 2)

HAVING accomplished the first part called Longimetra, con∣tayning sundrie rules to measure lēgthes, breadthes, heigths, depthes, and distances: I thinke it meete novv to proceede to the seconde kynde named Planimetra, vvherein ye shall haue rules for the mensuration of all manner playne Figures, to knovv their contente superficiall. And forasmuch as there is no Superficies, but is enuironed vvith lines either streight or curue. And all Figures com∣prehended vvith streight lines may be resolued into Triangles: It seemeth most meete, first to teache the measuring of Triangles as follovveth.

The fyrst Chapter. Of Triangles.

AS there are thrée kind of Angles, so is there also thrée kind of Triangles: the first is called Orthogonium, hauing one of his angles a right: the other Ambligonium, contayning one obtuse angle: the third kinde is called Oxigonium, whose thrée angles are all acute. Of right angled Triangles also there are two kindes, for eyther it hath two equall sides, and then is it called Isoscheles, or thrée vnequall, and that is Scalenum.

The right angled Isoscheles is thus measured.

MUltiplie one of the equall sides in it selfe, the halfe of the producte is the Area or superficiall contente.

Example.

ADmit ABC an Isoscheles right angled triangle, ha∣uing

[illustration]

the sides AB and BC that contayne the righte angle B equall, eyther of them being 10, I multiplie 10 in it selfe, thereof ariseth 100, the halfe 50, is the Area of that Isoscheles right angled triangle.

A right angled Scalenum you shall thus mete.

MEasure the longitude of two sides contayning the right angle, aug∣menting the one by the other, the halfe thereof is that triangular su∣perficiall contente.

Example.

Suppose the right angled Scalenum ABC, his

[illustration]

two sides that contayne the right angle AB 10, BC 16, the one augmented by the other, yeeldeth 160, whose halfe being 80, is that triangulare Area.

In euery of these right angled Triangles you shall not neede to measure more than two of their sides, for any two knowen by supputation the third may be found, whereof I shall giue you rules with examples.

The second Chapter. Any tvvo sides of right angled triangles knovven by calcula∣tion to finde the thirde.

EVCLIDE in the 47 propositiō of his first booke of Elemēts, hath proued by demonstration, that y^e squares of the two cō∣tayning sides ioyned togither, are equall to the square of y^e Hypothenusa, or third side subtending y^e right angle: wherby we may redely (any two sides being giuē) find y^e third thus:

〈◊〉〈◊〉 the two sides comprehending the right Angle be knowne, adde their ••••uares together, and the Radix Quadrate of the Product, is the Hypothe∣••••sa, but if you knowe the length of that Hypothenusa, and one other side, 〈◊〉〈◊〉 shall subtract from the square of that Hypothenusa, the square of that ••••her geuen side, and the roote Quadrate of the remainder is the third side ••••sired.

Example.

In the Triangle ABC, AB is 6, and

[illustration]

C ••, their squares ioyned make 100, ••hose roote is 10, and that is the Hypothe∣••usa. In the Triangle DEF, the Hypothe∣••usa is ••3 DE, 5. I desire the lengthe of EF, 〈◊〉〈◊〉 the square of DE, subtracted from 169 〈◊〉〈◊〉 square of the Hypothenusa, leaueth 144, ••hose Quadrate roote 12, is the side EF.

Of Obtuse angled Triangles, there are also two sortes, Isoscheles ••nd Scalenum.

The .3. Chapter. Ambligonium Isoscheles is thus measured.

YOu must first finde out the Perpendicular line, from the Ob∣tuse Angle to the contrary side, by deducting the square of halfe the side subtending the Obtuse angle, from the Square of one whole side containing the same angle, the Roote Quadrate of ••he Product is the Perpendicular, which multiplied in the halfe of the for∣••ayd subtending side, produceth the Area.

Example.

ABC the Isoscheles Ambligonium, AB

[illustration]

••0, BC 16, 10 squared is 100, 8 squared yel∣deth 64, which deducted from 100, leaueth ••6, whose Quadrate roote is 6, the Per∣pendicular AD: which multiplied by 8, pro∣duceth 48, and that is the Area of this Obtusiangle Isoscheles.

Ambligonium Scalenum, you shall thus measure.

FIrst you must search out his Perpendicular line in this manner, square euery side, then adde the square of the side subtending the Obtuse angle,

to one of the other squares, from the offcome abating the thirde square, the half of the remainder diuide by the forsaid subtending side, the square of the Quotient you shal deduct from that square which before you did adde to the square of the subtending side, and from the remaine, extract the roote Qua∣drate, for that is the line Perpendicular, which multiplied by half the fore∣named subtending side, will produce the content Superficiall of y^e triangle▪

Example.

ABD the Triangle, AB 20, AD 34, DB 42, the square of AB 400, of AD 1156, of BD 1764, 1764 added to 400 maketh 2164 from which subtracting 1156, there re∣maines 1008, the halfe beeing

[illustration]

504, diuided by 42, the side subtending the Obtuse angle, yeldeth in the Quotient 12, so much is the line BC, which squa¦red maketh 144, that subtra∣cted frō 400, leaueth 256, whose roote Quadrate being 16, is the Perpendicular AC. Now mul¦tiply 16 in 21, the halfe of the forsaid subtending side BD, so haue you 336 the super∣ficies of that Ambligonium Scalenum.

The 4. Chapter. Of Acutiangle Triangles called Oxigonia, there are three kindes.

FOr either the thrée sides are equall, and then is it an Equilate•• Triangle, or two sides only equal, which is Isoscheles, or al thrée vnequall, and that is a Scalenum.

Of Triangles Equilater.

You shall multiply the square of the side in it selfe, and the offcome in 3, the Product diuide by 16 the roote Quadrate of the Quotient is the Area, or multiply the Cube of halfe the Equiangle Triangles side in the Semi∣perimetrie of the Triangle, the Quadrate roote of the Product is the fore∣said Area also.

Example.

Admit the Triangles side 6, the squared square is 1296, which augmented by

•• yeldeth 3888, and that diuided by 16, bringeth in the Quotient▪ 243, whose

[illustration]

roote Quadrate beeing 15, and betweene 1/•• and 1••/31, is the superficiall content of that E∣quilater Triangle. The same number is founde by the seconde rule, for the Cube of 3 is 27, which multiplied in 9 the Semiperime∣try yeldeth 243, whose roote Quadrate is the Area of that Triangle agreeing with the former supputation.

Of Isoschele Acutiangle Triangles.

••Rom the square of one of the equall sides subtract the square of halfe the base or vnequall side, the Quadrate roote of the remainder multiplied in ••••lfe the base, produceth the content superficiall.

Example.

AB and AC the equall sides of the Isoschele Trian∣••••••,

[illustration]

either of them 6, the base BC 4, the halfe thereof 2, ••••ose square deducted from the square of 6, leaueth 32, ••••ose roote multiplied in 2, bringeth the roote of 128, which ••••ery nigh 11 7/2•• the Area of that Triangle.

Of Oxigonium Scalenum.

To finde the Area of Acutiangle triangles that haue side equall to other, it behoueth you to searche out 〈◊〉〈◊〉 line Perpendicular falling from one of the angles the contrary side, and multiply the same in halfe the base or side wheron ••alleth, the Producte is your desire. But to get the line Perpendiculare, 〈◊〉〈◊〉 shall thus worke: square euery side, then adde the square of the base, to 〈◊〉〈◊〉 square of one side, deducting fro the Producte the third square, halfe of 〈◊〉〈◊〉 Remainder diuided by the base, the Quotient you shall againe square 〈◊〉〈◊〉 deduct this number from that square which you adioyned to the square your base, the root Quadrate of the remainder is the line perpendicular.

Example.

〈◊〉〈◊〉 Acutiangle triangle Scalenū ABC, AB 13, BC 14, CA 15, the square of AB 169,

of BC 196, of CA 225, 169 added to 196, pr••∣duceth

[illustration]

365, from which 225 detracted, leaueth 140, whose medietie being 70, diuided by 14, yel∣deth 5 the line BD, the square thereof abated from the square of AB, leaueth 144, his root•• Quadrate is 12, the Perpendiculare AD▪ which augmented by 7 the medietie of the base, bringeth 84, the Area or Superficiall content of that Triangle.

Finally I thincke it not amisse to geue you one Rule vniuersal to mea∣sure all manner Triangles of what sorte so euer they be, and that without any regarde of the Perpendicular.

The .5. Chapter. A rule generall to measure all manner Triangles according to their plaine.

ADde all the sides of that Triangle together, taking halfe of the number which surmounteth. Now out of this halfe, you must pull by Subtraction euery side by it selfe, noting diligently the differences, that is, how much euery side dif∣fereth from that halfe, multiply the differences the one in the other, and the Product in the thirde, with that which riseth augment the halfe aboue mentioned, then séeke of that summe the Quadrate roote, so haue you the true content Superficial of that Triangle, the example shal ensue, wherby al that I haue said, shal the better appéere.

Admit the Triangle whose sides you haue measured ABC, of whome the

[illustration]

left side AB i•• 12, BC 16, AC 20, these sides ioy∣ned together, make 48, whose halfe is 24, from whome BC 16, differeth 8, AB 12, AC 4, so the differences are 8, 12, 4. Nowe multiply 8 with 12, riseth 96, the which augmented with 4, commeth 384, that multiplied in 24, the halfe if 48, sur∣mounteth 9216, of this summe the Radix is 96, which is the verye content of this Triangle.

Of Quadrangles (that is to saye plaine figures, hauing foure Angles and foure sides,) there are fiue sortes, as appéereth in the Diffinitions, the Square, the Rectangle, Rhombus, Rhomboides, and Trapezia.

The .6. Chapter. Of Squares.

FOr the square you shall onely measure

[illustration]

one side, multiplying the same in it self, so haue you the Area or content super∣ficiall.

Example.

I finde AB the one side of the square ABCD 10, whiche augmented by it selfe riseth 100 the Area of that square.

Of rectangles or right angled Paralelogrammes.

In right angled paralelogrammes ye muste measure the two vnequal sides, multiplying the one in the other, the product is the content superfi∣ciall of the figure.

Example.

I suppose the two sides AB, AC of that righte

[illustration]

angled Paralelogramme. ABCD the one 45, the other 110, these two multiplyed togither yelde 4950 the Area.

Of Rhombus and Rhomboides.

THese two figures haue one rule, it behoueth you to measure one side, and the perpendicular fallyng from one of the opposite angles to the same side, these multiplyed the one in the other produceth the Area.

Example.

Admitte ABCD the Rhombus, whose Area I desire: I measure the ••••de BD finding it 20, then muste I measure also the lengthe of a lyne perpen∣dicular

fallyng from A vppon BD,

[illustration]

which I suppose here 16, these two mul¦tiplied the one in the other bring 320, the superficiall content of the Rhom∣bus. Likewise in the Rhomboides sup∣pose I finde by mensuration the side BD 42, the lyne perpendicularly fal∣lyng from A vpon BD 16, multiplie these nūbers, I produce 672, the Area of that Romboides ABCD.

But bicause it may séeme somewhat difficult to get the length of those perpendiculars, bicause it is vncertain on what point of BD the perpen∣dicular line shall fall, I thinke good to prescribe you a rule how you may exactly and redily get all such lines perpendicularly fallyng, and that not only in these figures, but also in triangles, whiche shall be no small ease and discharge of laborsom trauayling, when you shall measure great fiel∣des, or champion playnes, &c.

The .7. Chapter. For measuring of lines perpendicular.

YOu shall prepare a right angle by conioyning thrée staues proportioned according to these nūbers, 3 4 and 5, as you were taught in the former boke, the one of the staues con∣teyning the right angle you must place directly vpon that side of your playn figure that subtendeth the angle from whence youre perpendiculare shoulde fall, marking ther∣withall whither the other conteining side directeth, if it lye euen with the foresayd opposite angle, then is the right line betwéene your station, & that opposite angle, the perpendicular. But if it directe not iustely to the an∣gle, you shall moue to and fro in that subtendent syde, till you fynde it so, then by the preceptes giuen in the fornier booke, you may sundry wayes measure the distaunce of the angle opposite from youre station, whereby you are brought in knowledge of this perpendiculars length.

Admit ABCD the paralelogramme, or ABC the triangle, whose per∣••••ndiculares falling from A to the contrary side I desire to measure, but because 〈◊〉〈◊〉 know not to what point in BC, or BD, these perpendicular lynes from A will 〈◊〉〈◊〉, I take my right angle made of the three staues GEF, placing EF in the

[illustration]

〈◊〉〈◊〉 BC, and BD, passing to and 〈◊〉〈◊〉 in those lynes till I find EG 〈◊〉〈◊〉 directly with the angle A, then 〈◊〉〈◊〉 I that a straight line from A, to 〈◊〉〈◊〉 station or angle E, is the perpen∣••••cular, whose length I may mea∣••••re either by Quadratum geome∣••••icū, or otherwise, without instru∣••ent, as in the first booke is decla∣••••d. Or thus deducting the square 〈◊〉〈◊〉 EB, the distance of my station ••o the ende or angle of eyther field 〈◊〉〈◊〉 B, from the square of the line A••, rising from that subtendent side 〈◊〉〈◊〉 the opposite angle, the roote qua∣••rate of the remaynder shall be the 〈◊〉〈◊〉 perpendicular, whiche multi∣••lyed in BD, yeldeth the Area of that paralelogramme: or in halfe BC, so ••aue ye the superficiall content of the triangle ABC.

The .8. Chapter. To measure Trapezia.

YOu shall measure the length of a diagonall or crosse line exten∣ded to opposite angles through the Superficies of the Trapezium and likewise the length of the two perpendicular lines, falling from the other angles vppon the sayd crosse lyne as you were ••aught in the last chapter, then adde those two perpendiculars together, ••ultipling the halfe of the product in that diagonall line, so haue yée the Area of that Trapezium.

Example.

ABCD the Trapezium, CB the diagonall or crosse lyne extnded fro

the angle C, to BE and F, the two poyntes where the perpendiculars shall 〈◊〉〈◊〉

[illustration]

from the other angles A, D, vppon the crosse line CB, these points I find with my right angled triangle, as was taughte in the laste Chapiter. Now to attein the Area, I measure the length of BC 20, AE 8, FD 6: 6 and 8 ioyned make 14, the half is 7, multiplied in 20, the line diagonal produceth 140, and that is the con∣tent superficiall of that Trapezium.

The .9. Chapter. Rules to measure all equiangle superficies hovve many sides soeuer they haue.

FIrste you must get the centre of youre figure, then from it pull a perpendicular line to the middes of some syde, sée how many perches or other measures it conteyneth, adde all the sides toge∣ther, multiplying halfe the summe in the perpendicular or han∣••ing line, so haue ye your purpose.

Example.

[illustration]

Imagine this figure BCDEFG and euery side of length 12, the centre A founde, drawe a lyne perpen∣dicular from it to the middle of the side BC, this line being 10 ⅖ multiplyed in 36 the halfe number of the sydes, bryngeth 374 ⅖ the superficiall contente of that figure.

An other Example.

FGH is here a Superficies beeing 5 square,

[illustration]

euery syde 10, Nowe from youre centre to the myddes of the syde pull the perpendicular lyne whych is 6 22/25 thys multiplyed in 25, the semipe∣rimetrie of that Fygure, yeldeth 172, and that is the Area of this pentagonall superficies.

A note to finde the centres of those equiangle Figures.

THe centre is found drawing lines from one angle to the contrarie, or from the middle of some side being odde to the opposite angle, the cut∣ting or concourse of those streight lines sheweth the place of the Centre: and thus as is declared, you maye readely measure all equiangle figures, what capacitie or number of sides soeuer they bée of.

The .10. Chapter. To measure the Superficiall content of any rightlined Figure of vvhat forme so euer it bee,

THe best rule I can perscibe, is to resolue it into Triangles, by drawing or ymagening lines from Angle to Angle, and so measure euery Triangle seuerally: Finally, adding all the productes togyther, ye shall haue the Area or contente Superficiall of that whole Figure, which althoughe it be of it selfe (the premisses well vnderstande) playne ynoughe, yet to auoyde all doubtes, I shall adioyne one example.

Admit ABCDE an irregular

[illustration]

pentagonum, whose side AB is 20, BC 30, CD 16, DE 24, EA 18 this figure you maye diuide in three triāgles, by drawing the two lines BE BD, thus by the rules tofore giuen, I measure first the Area of the tri∣angle ABE, whiche I finde 144, likewise BED, 312, the thirde tri∣angle BDC 240: these three nūbers ioyned togither, produce 696 which is the true Area or contente Superficiall of that irregular Superficies.

The .11. Chapter. A readie mean to find the content superficial of any great field, or cham∣pion playne, hovv irregular of forme or fashion soeuer it bee, vvithoute painfull trauayling about it, onely by measuring one side.

YE shall, as I haue taught in Longimetra (either with Theodelitus or your Topographicall instrument) draw vpon some leuell, smothe playne superficies, the exacte platte and symetry of that field, then either by your in∣strument, or otherwise: measure howe many rodde or perche, is contayned in some one side of that playne, so many equall diuisions shall you make, opening youre compasse accordingly in his correspondent side of your platte. This done, note well the fashion of the field, and drawing lines from angle to angle, or otherwise, diuide it into triāgles, or other regulare figures, as you shall sée cause: then opening your compasse to one of those diuisions, you maye therewith measure euery side, and line, in that figure as exactely as with corde, or pole, ye should paynfully pase it ouer, whereby with the ayde of these former preceptes, you shall seuerally measure euery triangle or o∣ther regular figure, and ioyning togither their contentes the resulting summe is your desire.

Example.

Suppose GHIKL that irregu∣gular

[illustration]

pentagonum, the true platte of some great field or playne made by one of the instruments Geometrical, as is before in the first booke declared and that by mensuration I finde the side GL 120 perches in length, ope∣ning therfore my cōpasse according∣ly, I diuide GL into 120 equall por∣tions, and proceding on to euery side I finde GH 90, HI 100, IK 150, KL 80, then beholding the forme thereof, I see it maye aptely bee parted in three Triangles by lines drawen from I, 60 G and L. In like manner I searche howe many such Diuisions there are in them, finding either of them 170 perches in length. Finally, by the rule generall of Triangles, I searche the Area of euery

Triangle, finding the fyrst that is IHG 360, IGL 780, IKL 600, all three ••••yned togyther, maketh 1740, so manye perches you maye conclude the Area of ••hat Figure, which reduced 10 Acres diuiding by 160 bringeth 10 Acres 3 ½ R••odes.

A note for Wooddes.

THIS way shall you moste spéedely and exactly measure all manner Woddelande whiche otherwise is very harde and tediouse preciselye ••o doe, forasmuche as ye can not at one viewe beholde euery parte there∣of, nor measure such Diagonall lines passing thoroughe it in sundrie ••laces, as were requisite in irregular Figures, but with the ayde of your ••nstrument Topographicall encompassing it rounde aboute, alwaye no∣••ing the Angles of Position at euery station, you maye firste make an exacte platte thereof, and after measure as you were instructed in the for∣mer Chapter, the Superficiall capacitie.

The .12. Chapter. Hovve you maye from an highe Hil, or Cliffe, measure hovv manye Acres, Roodes, or Perches, is contayned in any Fielde, Parke, VVood, or other playne Superficies, in the countrie rounde aboute you, not approching nighe them.

CALl to remembraunce howe you were taught in the first Booke by the Instrument Topographicall to set foorth the true platte of an whole Countrie, and euery parte there∣of, whiche, for asmuche as it is there at large sette out, it were héere superfluous to recite agayne, admitting therefore, (by the Arte there taughte) an exacte platte forme of the Fielde, Woodde, or other playne made, ye shall with your•• Compasse diuide some one side thereof into 40, 60, or 100 equall partes, as

you liste, and kéeping youre Compasse immouable, measure all suche o∣ther lines Perpendiculares, &c. as shall séeme requisite to attayne the Area thereof, and by the former preceptes, diuiding it into Triangles, Rightangled Parallelogrammes, or other regular figures, ye shall mea∣sure the contentes superficiall thereof, that is to say, how manye of those small squares, whereof euery little diuision was a side, is contayned in that superficies or platforme. This done, ye must also with youre square geometricall or other instrument from the hill or cliffe, measure y^e length of that side in the fielde, that the first diuided side in your platte did repre∣sente, I meane how many rodde or perche it is long then square aswell the number of perches in the side of the fielde, as also the number of diui∣sions in his corresponding side of youre platte, the number procéeding of the perches squared, ye shall multiplie in the Superficiall contente of youre platte tofore founde, and the Producte diuide by the square of the Diuisions in the syde of youre platte, the Quotient will be the number of perches, whiche diuided by 160, and the remayne by 40, the Quoti∣entes will shewe the Acres and roodes contayned in that fielde, parke, or woodde, you measure.

Example.

ADmitte ABCDEF the platte or proportionall patterne of a parke which from some Hill or Cliffe a farre off I haue drawen by the ayde of myne Instrument Topographicall, as was declared in the firste Booke, and for that it is a Figure of many sides, I searche howe it maye beste be resolued into regular Figures, whiche (as you maye perceyue) is readely done withdrawing the two lines, AC and FD, which partes the whole Figure in two Triangles and one right angled Parallelogramme, then opening my Compasse to some small di∣stance, I diuide some one side (for Example the side AF) into 40 partes, and keeping my Compasse immouable, I measure howe manye of those Diuisions are contayned in AC and FD, finding eyther of them 60. Now, to measure the Triangles ABC, DEF, I drawe Perpendiculares from B too AC, and from E to FD, finding BG 20, and HE 16, Nowe 40 multiplied in 60, bringeth 2400, the right angled Parallelogramme AD, 20 the Per∣pendicular in 30 halfe the Base yeeldeth 600, the Triangle ABC, 8 halfe the other perpendiculare, in 60, the base ariseth 480, the triangle DEF, these three

ioyned togither, produce 3480, the whole contente of ABCDEF, now as was

[illustration]

taught in Longimetra, with my square geo∣metricall, I must measure from the hill or cliffe the length of that side of the parke represented here by AF, which I admitte founde as is before sayde 356 rodde or perche, whose square beeing 126736 aug∣mēted in 2400, the content of my plat brin∣geth 304166400, this diuided by 1600, the square of the side AF yeldeth 190104, and this diuided by 160 produceth in the quoti∣ente 1188, and the remayne is 24. I con∣clude therefore that there is in that parke 1188 acres, and 24 perches.

In this sorte with small exercise vsing industrie, in making your plattes you may moste exactely and speedely surueye an whole countrey, with all his pastours, medowes, marshes, woddes, and euery perticulare inclosure, whereof although I might propound an infinite number of examples, and filt many leaues with varietie of rules, yet considering the premis∣ses to the ingeniouse will suffise, I thinke good to passe them ouer, referring the rest to the practisioner, who shal by his experience (well vnderstanding the pre∣misses) inuent manyfold mo, euen as occasion shall be offred in viewing the ground with situation of places.

The 13. Chapter. A note hovv to suruey an vvhole Region or playne champion Coun∣trey by the ayde of a playne pullished glasse.

IT behoueth you to resorte vnto the firste booke where you were taught by the ayde of a playne pullished glasse of stéele) vpon an high hill or leueled platfourme) to set foorth the syme∣trie or proportion of an whole countrey with al his partes. Ye shall therfore by the arte there shewed, get the proportion or plat of that fielde, wood, marshe, or other inclosure, whiche you desire to suruey, and therewith worke euen in like maner as you were taught in the last••

chapter, to do with the platte drawen with your instrument Topogra∣phicall, one ground, one reason, and like operation serueth them both, the former well vnderstande maketh this so manyfest as it néedeth no o∣ther example, onely this I muste giue you warning of, that you attempt not this kinde of measuring, but onely in champion and playne leuell Countreys, for in vnleuell and hilly places, without moste exquisite ob∣seruation and learned handeling great errour may ensue.

Thus hauing declared seuerall rules for euery kinde of rightlyned Superficies, it seemeth méete somewhat to say of suche playne Super∣ficies as are enuironed with curue lynes or mixte of bothe, and firste of the circle and his partes.

The .14. Chapter. Of Circles.

BEfore I entreate of the mensuration of circles, it shall be requisite to declare Archimedes rules, concerning the proportion of the circumfe∣rence to his diameter, and of the Superficies to his dimetientes square. The rules ensue.

The fyrst Theoreme.

〈 in non-Latin alphabet 〉〈 in non-Latin alphabet 〉.

EVery circle is equall to that rightangled triangle, of vvhose conteyning sides the one is equall to the semidiameters, the other to the perimetrie or circumference.

The seconde Theoreme.

〈 in non-Latin alphabet 〉〈 in non-Latin alphabet 〉.

THe proportion of euerye circle to the square of his dimetiente is as 11 to 14.

The thirde Theoreme.

〈 in non-Latin alphabet 〉〈 in non-Latin alphabet 〉.

THe circumference of a circle is more than triple his diameter by suche a parte as is lesse than 1/7, and more than 12/7•• thereof.

Hovv the Area of a circle is founde.

BY the former Theoremes ye may collecte these rules. Multiplye the circumference of any circle by 7, and diuide by 22, your quotient is the circles diameter, whose medietie multiplied in halfe the circumference yeldeth the Area, or multiplying the square of the circles dimetient by 11, and diuiding the ofcome by 14, your quotient will declare the same.

Example.

Admit the circumference of the circle 44, which en∣creased

[illustration]

by 7 yeldeth 308, this diuided by 22 bringeth in the quotiēt 14, the diameter, whose medietie being 7 aug∣mented with 22 the halfe of the circumference, produ∣ceth 154. Likewise the square of 14 multiplyed in 11, maketh 2156, whiche diuided by 14, bringeth 154, the superficiall capacitie of that circle.

Of the halfe circle.

EUen as I haue declared that in multipleing the halfe diameter of any circle in his halfe circumference, the producte to vtter his contente, so by augmenting the halfe diameter in the fourth parte of the circumfe∣rence, that is in halfe the arke of the semicircle, ye haue the whol summe of that halfe circle.

Example.

ADmit the semicircle DEF were to be measured, whose diameter DF dra∣wen

by the centre C is 14, as afore, the Arcke

[illustration]

DEF 22. Nowe multiplie the semidiameter 7 in 11 riseth 77, the playne of that semicircle, euen so of all portions or partes of a circle. Althoughe to the wittie this may suffise, yet to satisfie also the meane witted, I shall not thinke it tediouse to pro∣pounde an other example.

Of the fragments or partes of a circle.

SVppose GHIA were a portion of a circle to be moten, the whole circumfe∣rence

[illustration]

of the circle wherof this is a por∣tion is as afore 44. Now you shall multi∣plye 7 the semidiameter HA in the halfe arke 15, so haue ye 105, whiche is the su∣perficial content of that portion GHIA, Yf you desire to know the Area of GIB, conteyned of the corde GI, and arcke GIB, ye muste adde to the number before found, the Area of the triangle GIA which I suppose (found by the rules tofore giuen) 22, that maketh 127 the Area of the segment GHI to the corde GI, which subtracted from 154, the whole circle leueth 27 the area of the segment GIB.

I thinke none will doubt how these figures folowing are measured, bicause they are made of parts or portions of a circle, whose plaine is ga∣thered as I haue alredy declared. In the left figure there is at eche end a semicircle, euery of them conteyning (found by the art afore mentioned) 77, which added make 154, the quadrangle hath 84, as the rules of qua∣drangles declareth, that ioyned with 154, bringeth 238, the whole summe of that figure.

[illustration]

The other are but two suche segmentes as GIB, which measured as is before shewed and ioyned together bringeth 54, the Area of that figure. Likewise y^e last figure called a Lunula, ABCD is mesured by deducting the segment ADC (found by the former rules) 27 from ABC known in lyke maner 56 4/•• there remayneth 29 4/7 the Area or Superficies of the Lu∣nula ABCD. Forasmuche as it is necessarye in measuryng of these por∣tions or segmentes of circles to knowe the diameter of the whole circle, wherof they are the fragmentes. I thynke it not amisse herevnto to ad∣ioyne a rule for the same purpose.

Admit ABCD a portion or fragment of a circle, I desire to know the lon∣gitude of his circles diameter, first I measure AC whiche I fynde 12, lykewyse the length of the lyne DB from D the middle of the

[illustration]

streyght lyne to B the middle or highest poincte of the arke whiche I suppose 4, Now diuide the square of AD that is 36 by 4▪ so haue ye 9, which added to 4 bringeth 13, and that is the length of the circles dimetient, whose parte or fragment this figure ABCD is.

These preceptes well noted there can no playne Superficies offer it self whether it be regularor irregular, framed of right lines enuironned with circular or mixte of both, but ye shal readily by diuiding it into triangles & cular fragmentes, finde out the Area and content superficiall thereof. I will nowe therfore only, adioyne certayne questions for the partition and diuision of grounde. And so passe on to the thirde kynde of Geometrie, where you shall haue rules to measure, not onely the solide, but also the superficiall contentes of all maner bodies, but to returne to these parti∣tions my first question shall be of triangles, and so on to figures of more difficultie.

The .15. Chapter. There is a trianguler field hauing on the one side a vvell or foun∣tayne, this fielde must be equally diuided betvvene tvvo partie••, and that in such sort that either of them may haue cōmoditie of that fountayn, not cōming on the others lande. I demaunde hovv that partition shall be made.

LEtte ABC represente the triangular fielde, hauing on the syde BC a well or fountayne at D, from whence I woulde directe a hedge or ditthe in suche sorte that it parteth the Triangle in two euen partes, first therefore I searche the myddle of the syde BC, whiche I suppose E, then my instrument Topographicall at the fountayne situate as he oughte to bee, I turne the Ashidada till I can espye thorough the sightes the op∣posite

[illustration]

angle here represented by A, then re∣mouinge myne instrument to the foresayde middle at E, the Ashidada remayning im∣mouable on the degrees tofore cutte, the in∣strument agayne duely situate, I view tho∣roughe the sightes what parte of the syde AC I can espye, Admitte it F. Now say I that a hedge dyke or other partition runnyng from D the fountayne to F the marke espyed in the syde AC, wyll diuide that triangular figure exactely into two equall portions, and thus maye you withoute supputation onely by the ayde of youre instrument Topographical diuide any triangle in two equal parts, and that from any parte or poincte in any of hir sydes assigned.

The .16. Chapter. To cut off from any triangular fielde as many acres, rodde or other measures, as shall be required, and that by a lyne dravvne from any angle assigned.

YOu shall fyrste measure the syde subtendyng the Angle assigded whiche here I wyll call the base. Secondely, you shall measure the Area of the whole triangle: Then multiplye the number of Acres or Roddes whiche you woulde cutte off from the triangle, by the lengthe of the base, and the producte diuide by the Area of that Trian∣gle, the Quotiente sheweth howe many pearches you shall measure in the base from one of the Angles to cutte of youre desyred number of roddes▪

Example.

Suppose ABC a triangular peece of ground, hauyng his three sides AB 30▪ AC 40, CB 50, my desire is with a right lyne from the angle A to cut of one ••cre of grounde, to performe this by the

[illustration]

••rte before taught, I searche the Area of that figure which is 600 roddes. Now bycause I would cut off from that figure one acre, and an acre cōteineth 160 rods: I multiply 160 in 50, the whole side subtē∣ding the assigned angle, the product being 8000, I diuide by 600 the triangles Area, my quotient is 13 1/•• so many perches I recken in the base BC from B to D, finally extending a lyne from A to D, I conclude the Superficies ABD one exacte acre, and this is there from that tri∣angular field one acre cut of with the lyne AD.

The .17. Chapter. To cut off from any triangular piece of grounde vvhat quantitie of perches ye lyste vvith a lyne equidi∣stant to one of the sides.

TO perfoorme this conclusion it shall be requisite to serch out what parts of the other two sides this equidistant line shal cut, wherof I will giue you two rules, forasmuche as this partition may two seuerall ways be made: for the portion to be cut of eyther is adioyning to the parallele syde, and then is it a quadrangular figure, or else to his subtēding an∣gle, and then is it a triangle. For the triangle ye shal thus doe, square the sides that the parallele line shal cut, and the products seuerally multiplie in the number of perches to bée taken away, the surmountyng summes di∣uide by the Area of the whole triangle, the rootes quadrate of the quotien∣tes is the number of pearches to be accompted in the correspondent sides,

from the angle that subtendeth the Parallele side. But if ye would haue the portion cut off, lye adioyning to the paralele syde, then shall you first deduct that portion from the whole triangular Area, and with the remain∣der woorke in like maner as ye were taught in the former rule, so will your quadrate rootes (accompted from the paralele syde his subtendente angle) shewe to what part of eyther syde the lyne shall be drawne, to cut of the forenamed portion.

Example.

Suppose ABC the triangle from which I would cut off one acre, that is to say 160 rodde by a paralele lyne to the side AB, first therefore I measure the Area of that whole triangle as was before taught in this boke, fynding AB 50 perch, AC 120, BC 130, and so consequentely the Area of that triangle 3000 rodde, the square of AC is 14400, the square of BC is 16900, these augmented by 160, bring 2304000, and 2704000, and these diuided by 3000, produce in the quotientes 768 and 901 ⅓, the roote of 768 perches) being 27 perches 12 foote) I mesure out in the side AC,

[illustration]

beginning from C, admit it ende at D, likewise the roote of 901 ⅓ perches is 30 pear∣ches, and betwene 4 and 9 in∣ches, measuring therefore 30 perches from C in the other side CB I sette vp a marke at E. Nowe if you drawe a streyght lyne from D to E it shall be a paralele to AB, and the peece of grounde, represented by DEC, an exacte acre. But if you desire to laye oute this acre at one of the sides as ye may see in the figure signified by the quadrilater Superficies ABGF, Then must ye deducte the aforesayde 160 roddes fro the Area of that triangle, the re∣mayne is 2840, whiche I multiplye as before in the square of AC, so haue I 40896000, Lykewise the same 2840 augmented by the square of BC produ∣ceth 48016000, these products seuerally deuided by 3000, the Area of the whole triangle will yelde in the quotientes 13632 and 16005 ⅓ theyr quadrate rootes are 116 perches 12 ½ foote, the length of the lyne CF, and 126 perches 8 ½ foote, the lyne GC, Or if yee deducte those rootes from the whole sides AC, and BC there wil remayn 3 rodde 4 foote from A to F, and 3 rodde 8 foote from B to G And thus may you in all Triangular peeces of ground, exactly lay forth an acre,

••r any other quantitie of grounde ye will require, and that either against the side 〈◊〉〈◊〉 Angle, euen as you will desire.

The .18. Chapter. For partition of Paralelogrammes vvhat kinde so euer they be of, note these Rules ensuing.

BIcause partition may sundry wayes be made according to the seuerall situation of the line wherewith it is deuided, I wil first entreat of that diuision that is made by a Pa∣rallel vnto two of the Paralellogrammes sides.

Admit therfore ABCD the peece of ground which by men∣suration I finde to containe 65. Acres 16 Perches, and that I would cut of with a Parallel to AC 5 Acres, as ye may see done by the line EF, whiche were easily brought to passe if I knewe the quantitie of the lines AE or CF, for knowledge of

[illustration]

them therfore I worke thus, first with my square Geometricall or otherwise as hath ben declared in Longimetra, I finde the lēgth of AB 308 rodde, I multiplie ther∣fore 800 for so many rodde are there in 5 Acres, by 308, there amounteth 246400 which diuided by 10416 the number of Perches contained in the whole quadran∣gle, yeldeth in the Quotient 23 Perches 11 fote, the length of the two lines AE & CF. Thus by making the partition EF, the portion AECF, shall conteyne ex∣actly 5 acres, this rule is generall for all quadrangular peeces of grounde, whose sides be parallele, whether it be square rectangle, Rhombus or Rhomboides.

But if you desire to make like partition by a righte line issuyng out of one angle you shall thus woork: first consider whether the portion ye wold cut of be greater or lesse than halfe the parallelogramme: if lesse multiply as before the numbre of perches, that ye wold separate from the quadran∣gle in one of the opposite sides to that angle whence your diuiding lyne is∣sueth, and the product diuide by the medietie of the Area, the quotient will shewe on what parte of that opposite side your diuiding lyne will fal. But if the aforesayd portion to be separate be greater than the medietie of the Parallelogramme ye shal deduct it from the whole Area therof, and with the residue procéede in like maner as was before taught.

Example.

Admit ABCD a Quadrangular peece of ground, which being measured, is founde to containe 50 Acres, and that my desire is to cut of 10 acres from that figure, with a right line passing foor••he from the Angle C, first therefore I mea∣sure the length of the side AB, finding it 200 rodde, 200 therefore I augment by 1600 the number of Perches in 10 acres, there ariseth 320000, whiche diuided by 4000 the number of rodde contained in halfe that figure, your Quotient wil be 80, the number of Perches from A to E, so that

[illustration]

by the Partition CE, ye shall seperate the Triangle ACE containing exactly 10 a∣cres, but if the summe of acres had ben grea∣ter than halfe, ye should haue deducted it from the whole Area, and with the re∣sidue woorke in all respectes as ye haue done with this, the only difference is, that at the ende of your operation, whereas heere the Triangle ACE is the portion seperate, there it shoulde be the residue or Quadrangle ECDB, for that there is no difference in the working, it were superfluous to vse mo Examples.

The .19. Chapter. To cut of from any Trapezium or Quadrangular peece of ground, vvhat part therof ye list.

Example.

Admit ABCD a Trapezium, or Irregular Quadrangle, and that I woulde cut of a quarter of his Area, with a line issuing from the Angle A, first

[illustration]

by the Rules tofore geuen, I measure his content superficiall, which I sup∣pose 72 acres, secondlye I meete the Area of the Triangle ADC, fin∣ding it 50 acres. Nowe bicause I would cut of a quarter of that Tra∣pezium, I diuide his content that is 72 by 4, my Quotient is 18, and

that multiplied in DC, (which heere I suppose founde by mensuration 146 Per∣ches,) produceth 2520, and this diuided by 50 the number of acres in the Triangle ADC, yeldeth in the Quotient 50 Perches 6 ½ foote, which measured out from D in the line DC, sheweth where the line of partition issuing forthe of the angle A shall fall: and thus may you conclude that ADE, is the foresaide quarter of the whole Trapezium, and containeth 18 Acres.

The .20. Chapter. To diuide the superficies of any irregular Pollygonium, vvith a straight line proceeding from any one of the Angles assigned in suche sorte, that the partes shal retaine any proportion appointed.

WHen anye proportion is geuen, there are two Numbers wherewithall it is expressed, and they are called Termini, those two you shall adde together, reseruing the Producte for a Diuisor, then measure the Area of that whole Irre¦gulare Polligonium, which you shall multiply in the lesser of those Termini, the Producte shall you diuide by your re∣serued Diuisor, the Quotient is the content of the lesser Portion, whiche Deducted from the whole, leaueth the Superficies of the greater parte. But to drawe the line or Partition that shall diuide this Pollygonium, it were necessary first to learne on what side and part therof this line should fall, which you shall thus doe: Drawe or imagine right lines extended from the assigned Angle to euery Angle on either side, so shall ye make se∣uerall Triangles, whereof by the Rules tofore geuen, you must search the content Superficiall. This done, ye shall compare the Area of the first tri∣angle, with the Superficial content of the lesser portion found as is before declared. If the Triangle be greater, then may you by the preceptes before geuen, cut of from that Triangle so many Acres or Perches as that les∣ser Portion shoulde containe, but if the Triangle be not greater, ye shall deducte it from the foresayde lesser portion, and the Remaine compare wyth the nexte Triangle, whiche if it be greater than that Tri∣angle

also, subtract from it the Superficies of the triangle, & so procéede on til ye finde a triangle greater than your remayn, then may you say, that your partition or diuiding right line proceding from the assigned angle shall fall on the base or side subtendent of the assigned angle in that last triangle, but to know on what part therof, you must worke with your last remayn and the Area of your last triangle, as you were before taught in the diuision of triangles. And in like manner draw your diuiding line which shal exactly, seperate that Irregular Polligonum, in two Superficies retaining the pre∣scribed proportion: for more plainnesse peruse the example ensuing.

Admit ABCDEF, an irregular Hexagonum which I would diuide with a right lyne issuing foorth of the angle A in suche sort that the greater part might be triple to the lesser; this proportion triple may be expressed with these two num∣bers 3 and 1, and they are called Termini or boundes of that proportion, these two added make 4, wherewith diuide the whole Superficies of that Pollygonum, which I suppose by mensuration founde 64 Acres, my Quotient will be 16. Ad∣mit

[illustration]

also I finde the Area of my first triangle ABC 12 acres, & the Area of the second ACD, 18 acres. Nowe comparing 16 with the Triangle ABC, I finde it greater, deducting ther∣fore the one from the other, ther resteth 4 Acres, whiche foras∣much as it is lesse than the Area of the Triangle ACD, I con∣clude the partition or diuiding line shal fall on the side CD, but to learne on what part thereof, I worke as I was taught in the partition of Triangles thus, first with my square Geometticall or otherwise, I measure the side CD. Admit it 36 Perches, whiche multiplied in 4 the laste Remaine bryngeth 144, and that diuided by 18 the acres contained in the triangle ACD, produceth in the quotient 8, so many rod length shall you meete from C to G, in the side CD, finally extending a right line from the assigned angle A, to the poynt G, you may conclude AFEDG Triple to AGCB.

The 21 Chapter. To diuide any irregular Pollygonium into as many equall partes as ye vvill desire, vvith right lines dravven from any poynt vvithin the superficies therof assigned.

FIrste get the Area of the whole irregulare Pollygonium, which you shall diuide by the number of partes, whervnto ye would disseuer it, and the quotient reserue, then from your poynt assigned deducte or imagine lines to euery an∣gle of that figure, so shall ye parte it into sundry triangles, then compare your reserued quotient with some one trian∣gle beginning where ye li••te, and yf ye finde the triangle greater, then cut of from it a portion equall to your quotient, with a streight line pro∣céeding from the angle adiacent to the assigned poynt, but if your quoti∣ent be greater than the firste triangle, deducte the one from the other, and compare the remayne with the seconde triangle, which triangle yf ye finde greater, cut of from it a portion equall to your remayne, with a line issuing out of the angle ioyning to your assigned poynte, and then compare ageyne your reserued quotient with the remayning triangle, yf your quotient be greater, seperate from the triangle next ensuing a portion equall to the excesse or ouerplus, and that alwayes with a lyn•• issuing oute of the angle adiacent to the poynte firste giuen. Thus pro∣céeding till ye haue circulate the figure, alwayes drawing streyght lines from the poynt assigned, ye shall in the ende departe the whole figure in∣to as many equall portions as ye determined. This one thing is to be noted, that so ofte as your remayne or reserued quotient falleth out e∣quall to any of the triangles, then shal you not draw any line from the asigned poynt, for the side of that triangle serueth for the partition or diuiding lyne.

Example.

LEt ABCDEFG represent an irregular Pollygonium, or seuen sided peece of ground, which I would diuide with streight lines issuing out of the poynt H into three equall partes, first therefore I imagine streight lines from H to euery

angle, so is it diuided into 7 seueral triangles: now by the rules tofore giuen I mete the area of euery triangle, finding HAB 5 acres, HBC 5 acres two roodes, HCD 4 acres, HDE 6 acres 1 rode, HEF 5 acres, HFG 5 acres 12 perches, HAG 4 acres 2 roodes. Thus by adding all these triangles togither you shal find the area of that whole figure 35 acres, 1 roode, 12 perches, which resolued into perches yel∣deth this number 5652, and that diuided by 3 bringeth 1884 perches, and that is the contente of euery portion that this figure should be parted into. First there∣fore

[illustration]

comparing it with the triangle ABH, I finde the triangle the lesser, the one deducted frō the other, there remayneth 1084 perches, and this re∣maine I fynde likewise greater than the triangle CHB, detracting there∣fore the one from the other, there re∣mayneth 204, which compared with the third triangle CHD, forasmuch as it is lesse than the area thereof, I searche by the rules tofore giuen in diuision of triangles, on what parte of the base, CD the diuiding line shal fall that proceedeth from the angle H: admit it cut CD in the poynt I, from I to H extende a right line, and that shalbe the first partition. Now seeing HID ioy∣ned to DEH the fourth triāgle, is lesse than 1884 perches my reserued quotient, I deduct the one frō the other, so there remayneth 448 perches, I serch therfore by the rules tofore giuen on what parte of EF the line shall fall, proceeding from H to cut of 448 perches from HEF: but heere you must consider that this portion of 448 roddes or perches must be adiacent to the side HE, bicause the quotient doth excede the triangles by somuch. This considred admit the partition HK. now bicause my desire was to diuide it onely into three partes, I nede proceede no fur∣ther, for those two lines do exactly departe that whole irregular Pollygonium into three equall partes, wherof the first is that rightlined figure IHABC, the seconde is IHKED, the third is KHAGF, and this partition is made with right lines issuing from the poynt H assigned, whiche was required. In this maner you may diuide any Superficies into as many equal or vnequall partes as you liste, or into what proportion ye will desire, as by these few examples the ingeniouse will redily conceaue of any other: manyfolde rules might I giue you in these and like partitions, but knowing the premisses to the wittie are sufficient; I wil onely

adioyne certayne questions of some more difficultie, with the resolution of them, and so finish the seconde kinde of mensuration.

The .22. Chapter. To diuide any circle vvhose semidiameter is knovven, vvith an other circumference concentricall, in tvvo suche partes that the one por∣tion to the other shal retayne any proportion assigned.

WHen any proportion is giuen, it consisteth of two num∣bers, as I haue before saide, that are called Termini ra∣tionis, those numbers ye shall adde togither, reseruing the producte for a diuisor, then multiplie the square of these midiameter knowē by the lesser of those Termini, and the ofcome diuide by the reserued diuisor, from the quotient thereof, resulting, extracte the quadrate roote, for that is the se∣midiameter of the concentricall circle, whose circumference shall diuide the former giuen Circle in two partes, retayning the proportion assigned.

Example.

ADmit 120 the semidiameter of the circle, BCDE, whose area I would diuide with a concentrirall circumference in

[illustration]

suche sorte that one parte might be triple to the other, this proportion consisteth of 3 and 1, which added together make 4, now the square of the semidiameter beeing 14400, augmented by 1 the lesser of those Termini, produceth the same summe agayn whiche diuided by 4 tofore reserued for that purpose yeldeth 3600, whose quadrate roote is 60, so muche is AF the semi∣diameter of that inwarde Circle, whose circumference hath diuided the Circle

BCDE in two partes, the lesser is the circle FGH, and the greater is the anular Superficies conteined betweene those circumferences, the one beeing triple to the other. In like maner may you diuide that anular Superficies into three other, eue∣ry one of them equall to the same inwarde circle, whereof it were superfluous to adioyne any farther example, forasmuche as the proportion of the partes once knowen, the operation is in all poyntes agreable with the former.

The .23. Chapter. The three sydes of a triangle knovven, by supputation to get the greatest circles semidimetiente that may be described vvithin that circle.

YE shall firste adde the two shorter sides together, reseruing the producte, then substract those sides the one from the other, the re∣mayne multiply in the former product, and the amounting summe diuide by the thirde or longest side, the quotient detracte from that lon∣gest side that was your last diuisor, the medietie of this remayne you shall multiplye in it selfe, and deducte the ofcome from the square of the shortest side, the roote quadrate of the remaynder is the perpendiculare falling from the greatest angle to the greatest side, whiche retayneth the same proportion to the semidiameter of the inscribed circle, that the pe∣rimetrie of the triangle doth to the base or greatest side, working there∣fore by the rule of proportion by thrée knowen, ye may redily finde the fourth.

Example.

LEt ABC be the triangle, AB 36,

[illustration]

BC 48, AC, 60, nowe to get the greatest circles semidiameter that may be described within that triangle, I woorke thus, fyrst 36 added to 48, bringeth 84, then 36 detracted from 48 leueth 12, whiche multiplyed in 84 bringeth 1008, whiche diuided

by 60, the longest side yeldeth in the Quotient 16 ⅘. This deducted from 60 the Diuisor, the remayne will be 43 ⅕, the medietie thereof is 21 ⅖: whose square de∣••ucted frō the square of 36 leueth 829 11/25, Therfore do I conclude √{powerof2}829 11/25, ••he perpendicular BG, for the semidiameter EF, I vse the rule of proportion ••aying 144 the permetrie, giueth 60 the base, what shall √{powerof2}829 11/25 by multiplica∣••ion of the latter and diuision with the first, ye shall finde the fourth proportional number 12, and that is the line EF or semidimetiēt of the greatest circle that may ••e described within that triangle.

This is also to be noted, that the medietie (whose square ye subtracte from the square of the left side to get the perpendiculare) is the line AG: which deducted from the whole base, leueth the other line GC. and those two partes of the bases are called Casus.

The .24. Chapter. To finde the greatest squares side that may be described vithin any triangle vvhose sides are knovven.

SUndrie rules might be giuen to resolue this question, but to auoyde confusion, I will only declare the easiest: ye shal therfore (as is before in this booke taught) searche the per∣pendicular line falling from the greatest angle to the lon∣gest side: this perpēdicular it behoueth you to diuide in such sorte, that the partes retaine such proportion as doth the base or longest side to the sayd perpendiculare, so doing the greater portion is y^e squares side. But Arithmetically to attayne the quantitie of this longer portion, ye shall thus worke: Multiplie the perpendicular in it selfe▪ and diuide the producte by the Base or longest side of the triangle and the perpendi∣cular ioyned togither, the quotient detracted from the whole perpendicu∣lare leueth your desire. Or multiplie the base in the perpendicular, and di∣uide by them bothe ioyned togither as before, so shall your Quotient pro∣duce the squares side also.

Example.

ABC is the triangle whose sides are knowen, AB 24, BC 40, AC 32, the perpendicular founde, a•• is tofore declared, is 19 ⅕, which diuided at I, in suche

sorte that DI to IA retayne the

[illustration]

proportion of BC to DA, DI shall bee the side of the greatest square that may be made within that Triangle: but too finde the length of DI, you muste multi∣plie 40 the base BC in 19 ⅕ the perpendiculare, so haue you 3840/5 which diuided by 206/5 the perpē∣diculare and Base ioyned togi∣ther, yeldeth 12 36/37, the line ID. Likewise, if ye square the perpendicular, the mounting summe will be 9216/25 which diuided by 196/5 the former diuisor bringeth in the quotient 6 ••2/185, which deducted fro 19 ⅕ the perpendicular, there remayneth 12 180/185, the side of the greatest square agreeing with the former operation. In like manner if ye diuide BD, whose longitude you were taught by the laste Chapter to finde, and also AB, the left side of the Triangle in such sorte, that the greater sections to the lesser retayne the same proportion, that the base doth to the perpen∣dicular, the squares of the two greater sections one deducted from the other le∣ueth the Area of the greatest square that maye be described within that Tri∣angle. Also, if ye diuide any of the two lesser sides in two partes retayning the fo∣resayd proportion of the perpendiculare to the Base, the lesser of those Portions augmented by the base and the product diuided by the side bringeth in the quoti∣ent that greatest squares side. Thus also an other way, you maye attayne the same: Diuide both the Casus, that is ••o say, BD, and DC the distance of ey∣ther Angle from the perpendicular, in like manner as hath been said of the Per∣pendicular, then adde both the smaller sections togither, the resulting summe is the squares side: Euery of these wayes working, ye may resolue this question, great pleasure shall the Arithmetrician (especially if he be seene in surde and radi∣call numbers) receyue, when he shall perceyue so diuerse intricate and different o∣perations alwaye in fine to produce the selfe same certayntie.

The ende of the second Booke.