Mathesis enucleata, or, The elements of the mathematicks by J. Christ. Sturmius ; made English by J.R. and R.S.S.

SOME EXAMPLES OF SPECIOUS ANALYSIS, In each kind of Equations.

I. In Simple Equations.

PROBLEM I.

HAving the sum of any two sides given for forming a Triangle ABC, to find each of the sides, and form the Triangle.

Supppose e. g. three Lines given in Fig. 14. the first =AB+AC in the Triangle sought, the second =AB+BC, the third =BC+AC, to find each of the sides e. g. to find AB, which being known, the rest will be so also.

SOLƲTION.

1. Denomination. Make AC+AB=a; AB+BC=b; BC+AC=c; AB=x; then will AC=a−x, and BC=b−x, and so the Denomination be compleat.

2. Equation. Now if the values of the two last Lines BC and AC be added into one Sum, which we had before given; you'l have this Equation a+b−2x=c.

3. Reduction. By adding on both sides 2x, you'l have 〈 math 〉〈 math 〉; and substracting from both sides 〈 math 〉〈 math 〉; and dividing both sides by 〈 math 〉〈 math 〉.

4. The Effection or Geometrical Construction, which the Equation thus reduced will help us to

Join AE=a and ED=b in one Line AD, and from this backwards cut off DF=c; and divide AF which remains into two equal parts in B, and you'l have AB the first side of the Triangle to be formed; and BE will give the other side AC, which substracted from ED, will leave GD= to the third side BC; of which you may now form the Triangle ABC.

5. A general Rule for Arithmetical Cases. Add the two former Sums, and from the Aggregate substract the third Sum; half the Remainder will give the side AB common to the two former Sums. For an Example take this Question: There are three Towns of ancient Hetruria, viz. Forum Cassii (which the Letter A denotes in ▵ ABC) Sudertum (B) and Volsinii (C) which are at this distance one from another; if you go from Volsinii to Forum Cassii and thence to Sudertum, you mst go 330 Furlongs; from Forum Cassii to Sudertum and thence to Volsinii there are 306 Furlongs; lastly, from Sudertum to Volsinii and thence to Forum Cassii 272 Furlongs. How far is each Town distant from each other.

PROBLEM II.

IN a right-angled Triangle ABC, having given the Base AB, and the difference of the Perpendicular AC and the Hypothenusa BC to find the Perpendicular and Hypothenusa, and form the Triangle.

Make e. g. the Base AB (Fig. 15.) and the difference of the Perpendicular and Hypothenusa BD, to find the Perpen∣dicular AC; which being known, the Hypothenusa AC will be known also, if the given difference be added to the found Perpendicular.

SOLƲTION.

1. Denomination. Make AB=a, BD=b, AC=x; then will BC=x+b.

2. Equation by the Pythagorick Theorem, 〈 math 〉〈 math 〉, viz. the two Squares of the Sides to the Square of the Hypothenusa.

3. Reduction. Substracting from both sides xx, you'l have 〈 math 〉〈 math 〉; and moreover by substracting also 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

4. Effection or Geometrical Construction. Having described upon the given Base AB a Semi-circle, apply therein the given difference BD, and draw AD, whose Square is =aa−bb. Since this must be divided by 2b, make, as AE=2b to AD = 〈 math 〉〈 math 〉, so AD = 〈 math 〉〈 math 〉 to AC the Per∣pendicular sought. To which if you add CF=BD, you will have AF= to the Hypothenusa sought BC; which will come of course together with the whole Triangle sought, if the found Perpendicular AC be erected at right Angles on the given Base AB.

5. The Rule for Arithmetical Cases. From the square of the given Base substract the square of the given difference, and divide the Remainder by the double difference; and you'l have the Perpendicular sought. E. g. suppose the Base = 20 foot, and the difference between the Perpendicular and Hypothenu∣sa 10.

PROBLEM III.

IN the right angled Triangle ABC, having given the side AC and the sum of the other side AB and the Hypothenusa BC, to find the other side and the Hypothenusa separately, and form the Triangle. Suppose the given side (that is to be) AC (Fig. 16.) and the sum of the other sides AD, to find the side AB, which being known the Hypothenusa BC will be known also.

SOLƲTION.

1. Denomination. Make AC=a, AD=b, AB=x, then will BC=b−x.

2. Equation. 〈 math 〉〈 math 〉 and substract∣ing xx.

3. Reduction. 〈 math 〉〈 math 〉; and adding 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

4. The Effection or Construction is like the former, and so will be manifest only by inspecting that Scheme.

5. The Arithmetical Rule. From the square of the given sum substract the square of the given side, and divide the Re∣mainder by double the given sum; and youll have the other side, and substracting that from the given sum, you have the Hypothenusa also. E. g. let one side be 15, and the sum of the other two 45.

PROBLEM IV.

HAving given the Perpendiculars and sum of the Bases of two right-angled Triangles having equal Hypothenuses, to find the Bases separately, and form the Triangles. Sup∣pose e. g. to form the Triangle ABC (see Fig. 17.) you have given the Perpendicular AB, and for the other ▵ ADC, the Perpendicular CD, and the given sum of the Bases BE, to find the Bases singly, viz. the less for the greatest Perpendicular, and the greater AD for the less Perpendicular.

SOLƲTION.

1. Denomination. Make AB=a, CD=b, the sum BE=c; make the lesser Base BC=x; the greater AD will =c−x.

2. Equation. Since the Hypothenuses of the two Trian∣gles are supposed equal, the two □ □AB+BC, i. e. xx+aa will be = to the two □□AD+CD, i. e. 〈 math 〉〈 math 〉.

3. Reduction. By taking away therefore xx and adding 2cx, aa+2cx will = bb+cc; and further taking away from both sides 〈 math 〉〈 math 〉; and dividing both sides by 〈 math 〉〈 math 〉.

4. Geometrical Construction. Join the Lines b and c CD and BE at right Angles, (n. 2.) and the square of Hypothenusa DE will = bb+cc. Upon this Hypothen•• having described a Semi-circle, apply therein the Line AD, •• the square of AE will = bb+cc−aa. Which, since it m•• be further divided by 2c, make (n. 3.) as BF=2c to •• = 〈 math 〉〈 math 〉, so BG to BC, the lesser Base sought.

5. The Arithmetical Rule. From the sum of the squa•• of the lesser Perpendicular and the sum of the Bases substract 〈◊〉〈◊〉 square of the greater Perpendicular, and the Remainder di••¦ded by the double sum of the Bases, will give the lesser Ba•• E. g. make AB 76, CD 57, and BE 114.

PROBLEM V.

HAving given the Perpendiculars of two right-angled T••¦angles standing on the same given Base, to find the Se••¦ments of the Hypothenuses. E. g. suppose the common give Base be AB (Fig. 18.) and the Perpendicular of one Triang•• AD of the other BC; to find the segments of the Hypothen••¦ses, cutting one another geometrically.

SOLƲTION.

If a Geometrical Solution be required there is no need 〈◊〉〈◊〉 any Analysis; for having erected perpendicularly on the co••¦mon given Base AB the given Perpendiculars AD and B•• the Hypothenuses AC, BD being drawn immediately, exhib•• their Segments EA, EB, EC, ED. But if it be to be do arithmetically by a general Rule, then an Analysis will be n••¦cessary.

1. Denomination. Make the common Basis AB=a, B•• =b, AD=c; and, having found AF, all the rest may had (for as AB to BC so AF to FE; which being given, y•• have also GD and HC, and consequently also DE, CE, &•• make AF=x, then will BF or HE=a−x.

2. The Equation from FE found twice. 1. As AB to BC so AF to FE. 〈 math 〉〈 math 〉

[illustration]

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As BA to AD so BF to FE. 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Reduction. Multiplying both sides by a you'l have bx, c−cx; and adding on both sides cx, bx+cx=ac dividing both sides by 〈 math 〉〈 math 〉.

The Arithmetical Rule. Multiply the commom base ••••e least Perpendicular, and divide the Product by the sum ••e Perpendiculars; and you'l have the lesser segment of the 〈◊〉〈◊〉 which being given you'l have all the rest. E. g. sup∣•• AB=10, BC=9, AD=6.

PROBLEM VI.

O inscribe a Rhombus in a given Oblong, i. e. having the sides of the Oblong AB and BC given, (Fig. 19.) to find Segment BF or DE, which being cut off, the remainder FC ••E will be the side of the Rhomb sought.

SOLƲTION.

Make AB=a, BC=b, BF=x: FC or FA will be =x (so far the Denomination.) Therefore the square of FA, ••ch is bb−2bx+xx will be =aa+xx, viz. to the two ••res of AB and BF (so far the Equation;) and substracting both sides 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉. (so far the Reduction.)

The Geometrical Construction. Having described a semi∣••le on BC (n 2.) apply CD or AB, and the □ BD will be bb−aa. Which since it must be divided by 2b, make, ••E=2b to BD = 〈 math 〉〈 math 〉, so BD to BF sought, and ••e cut off the side of the Oblong BC (n. 1.)

The Arithmetical Rule. From the square of the greater side substract the square of the lesser, and divide the remainder by double the greater side, and the quotient will give the Seg∣ment BF sought. E. g. suppose AB=4, and BC=8.

PROBLEM VII.

TO inscribe the greatest square possible in a given Triangle, i. e. having given the heighth of the Triangle CD (Fig. 20) and the Base AB, to find a portion of the altitude CE, which being cut off there shall remain ED=FG.

SOLƲTION.

Make the base AB=a; the altitude CD=b, CE=x; then will ED or FG=b−x.

By reason of the similitude of the Triangles ABC and FGC you'l have as AB to CD so FG to CE. 〈 math 〉〈 math 〉

Therefore the Rectangles of the means and extremes will be equal, i. e. ax=bb−bx; and adding on both sides bx, ax+bx=bb, and dividing by 〈 math 〉〈 math 〉.

Construction. Upon the side of the Triangle CB produced, make CH=b, and HI=a, so that the whole Line shall be a+b. And having joined ID and parallel to it HE drawn from H, the part CE will be cut off, which is that sought.

For as CI to CD so CH to CE, 〈 math 〉〈 math 〉 according to the second case of simple Effections.

Arithmetical Rule. Square the given heighth of the Tri∣angle, and divide the Product by the sum of the base and al∣titude; and the quote is the part to be cut off CE. E. g. sup∣pose CD=10, and AB=15.

PROBLEM VIII.

IN an acute-angled Triangle having all the sides given, to find the Perpendicular that shall fall from the Vertex on the Base, i. e. having given AB, AC, BC (Fig. 21.) to find AD

or BD (for having found the one you may easily find the o∣ther) Coroll. Prop. 13. Lib. 2. Eucl.

SOLƲTION.

If there be only required a Geometrical Construction of this Problem, there will be no need of any Analysis; for having formed a Triangle ABC of the three given sides, you need only let fall the perpendicular AD from the Vertex A, which would determine the Segment BD. To find the general A∣rithmetical Rule, which is the general Corollary of Euclid, or if any one for exercise sake had rather determine the Per∣pendicular DA by the segment of the Base BD, than the latter by the former, the Analysis will proceed thus:

Make AB=a, BC=b, AC=c, BD=x; then will CD be = b−x. Wherefore by the Pythagorick Theorem □AD=aa−xx, and by the same reason the same □ AD = 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; and by adding on both sides 〈 math 〉〈 math 〉; and by transferring 〈 math 〉〈 math 〉, and dividing by 〈 math 〉〈 math 〉.

The Arithmetical Rule. Substract the square of the lesser side from the Sum of the □□ of the base and greater side, and the remainder divided by double the base will give its greater segment: If the □ of the greatest side be substracted from the sum of the other squares, &c. you will have the less segment CD.

Geometrical Construction Having described a semi circle upon AB (n. 2.) apply therein AC, and the □ BC will = aa−ccx, and continuing AC to B, till CB be = CB (n. 1.) the □ of BB will = aa−cc+bb; which since it is to be divided by 2b make as BE=2b to BB = 〈 math 〉〈 math 〉, so BF= BB to BD the segment sought = BD n 1.

PROBLEM IX.

IN an obtuse angled Triangle having the thr••e sides given to find the Perpendicular let fall from the Vertex to the Base being continued: i. e. Having given AB, BC, AC (Fig. 22. n. 1.) to find AD or CD (for the one being found the other will readily be so also) Coroll Prop. 12. Lib. 2. Eucl.

SOLƲTION.

What we premonished about the former Problem, we un∣derstand to be premonish'd here also. For the rest make here also AB=a, BC=b, AC=c, CD=x; then will BD= b+x: Wherefore by the Pythagorick Theorem the □ AD will = cc−xx, and by the same Theorem the same □ AD = 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉; and adding to both sides 〈 math 〉〈 math 〉; and transposing cc and 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

The Arithmetical Rule. Substract from the square of the greater side the sum of the squares of the base and lesser side; and the Remainder divided by double the base will give its continuation to the Perpendicular.

Geometrical Construction from the Equation reduc'd: Having described a semi-circle upon AB (n. 2.) apply therein AC, and the □ of CB drawn will 〈 math 〉〈 math 〉; which since it must be divided by 2b make, as CF=2b to CE 〈 math 〉〈 math 〉, so CE to CD the segment sought, n 1.

PROBLEM X. Commonly ascribed to Archimedes.

THE Diameter AB of a given semi-circle (Fig. 22 n. 1.) being any how divided in L, and from L erecting a Per∣pendicular LX, and upon the segments LA and LB having de∣scribed two other semi-circles, whose semi-diameters are also gi∣ven as well as that of the great Circle CB; to find the Radii FM and Vy of the little Circles that are to be so described, that they shall touch the Perpendicular LX, the Cavity of the greater semi-circle, and the Convexities of the less.

SOLƲTION.

I. For the Radius FM.

1. Denomination. Make CB=a, EB=b then will CE =a−b; for which for brevities sake put c. And let FM or FN or FK = x: Therefore EF will be = b+x, and CF (sub∣stracting FK from CK) = a−x. Wherefore now you'l have at least the names of the three sides in the ▵ CFE, so that according to Poblem 8. the Segment of the base GE may be determined (which indeed is determined already, as being =LE−LG or MF i. e. b−x) for which in the mean time we will put y; and now will CG=c−y.

2. For the Equation. If the □ GE=y be substracted from the □ EF = 〈 math 〉〈 math 〉, you'l have the square of the Perpendicular FG = 〈 math 〉〈 math 〉; and, if □ CG = 〈 math 〉〈 math 〉 be substracted from the □ CF = 〈 math 〉〈 math 〉, you'l have the same □ of the Perpendicular FG = 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

3. Reduction. And taking from both sides the quantities xx and yy, 〈 math 〉〈 math 〉; and adding 2ax and cc, but taking away aa from both sides, 〈 math 〉〈 math 〉; and adding 2ax and cc, and taking away from each side 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉. but the same y or EG is = EL−MF i. e. b−x. There∣fore 〈 math 〉〈 math 〉; which is a new and more principal Equation: And multiply∣ing both sides by 2c (you have a new Reduction) 〈 math 〉〈 math 〉; and adding 2cx, and transpo∣sing the others, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

The Geometrical Construction of this first Case. Add the de∣terminate (n 2.) quantity 2bc into one sum with the quantity aa, as in the beginning, n. 3. Then from this sum substract successively the quantities bb and cc, and there will come out (the same n 3) FH, whose □ 〈 math 〉〈 math 〉: Which since it must be divided by 〈 math 〉〈 math 〉, make (the same n. 3) as FI = 〈 math 〉〈 math 〉 to FH = 〈 math 〉〈 math 〉, so FH to FM the Radius sought of the little Circle to be de∣scribed. This quantity FM being thus found, place it from L to G (n. 1.) and from G erect a Perpendicular, which be∣ing cut off at the interval CF (which may be had, if from CB or CK you cut off FK=FM) or from E at the interval EF (which is composed of the Radii EN and FN) gives the Center of the little Circle to be described.

The Arithmetical Rule. Add twice the □ CEB to the square of the greatest semi-diameter CB, and from the sum substract the Aggregate of the □□ CE and EB; divide the remainder by the sum of all the three Diameters, (AB, AL and LB) i. e. by double the greatest AB; and you'l have the Radius FM, &c. For Example sake let a be = 12, b=4; c will be = 8, and x will be produced = 2 ⅔.

II. For the Radius Vy by help of the obtuse-angled ▵ DVC.

1. Denomination. CA=a as above, DA or DL=b, and putting x again for the sought Vy or VK, CV will be = a −x, DL or DR=b, and consequently DV=b+x, and DC=a−b, for which for brevity's sake we will put c. Now you'l have at least in Denomination in the ▵ CVD the three sides, so that aecording to Problem 9. the segment CW may be determined, for which in the mean while we will put y; then will DW=c+y, which is the same as DL−WL or Vy i. e. b−x.

2. For the Equation. If the □ CW=yy substract it from the □ CV = 〈 math 〉〈 math 〉, and you'l have the □ of the Perpendicular VW, 〈 math 〉〈 math 〉; and if the □ DW = 〈 math 〉〈 math 〉, substract it from the □ DV = 〈 math 〉〈 math 〉, and you'l have the same □ of the Perpendicular VW = 〈 math 〉〈 math 〉 Therefore 〈 math 〉〈 math 〉.

3. Reduction. Therefore taking from both sides xx and yy, 〈 math 〉〈 math 〉; and adding 2cy and 2ax, 〈 math 〉〈 math 〉; and substract∣ing 〈 math 〉〈 math 〉, and dividing by 2c, 〈 math 〉〈 math 〉

But if you add to the same y or CW DC=c, you'l have DW = 〈 math 〉〈 math 〉 i. e. reducing this c to the same Denomination, 〈 math 〉〈 math 〉 = DW. But the same DW =DL−WL=b−x. Therefore 〈 math 〉〈 math 〉, and multiplying 〈 math 〉〈 math 〉, and adding 2cx, and transposing the rest, 〈 math 〉〈 math 〉, and dividing by 〈 math 〉〈 math 〉 just as above in the first Case.

4. The Geometrical Construction therefore will be the same as there. See Fig. 23. n. 4. and 5.

5. The Arithmetical Rule is also the same, but the given quantities in this Example, which the figure of the Problem will shew, thus vary, while a remains 12, b will be 8, and c 4, from which data (or given quantities) there will not∣withstanding come out again, for x or the Radius Vy 2 ⅔.

II. Some Examples of simple or pure Quadratick Equations.

PROBLEM I.

TO make a Square equal to a given Rectangle; i. e. having given the sides of the Rectangle, to find the side of an equal Square, Eucl. Prop. 14. Lib. 2. Suppose e. g. the gi∣ven sides of the Oblong to be AB and BC (Fig. 24.) to find the Line BD whose square shall be equal to that Rectangle.

SOLƲTION.

Make AB=a and BC=b, and the side of the square sought = x, and the Equation will be ab=xx; and extra∣cting the root on both sides 〈 math 〉〈 math 〉.

Geometrical Construction. Join AB and BC in one right line, and describing a semi-circle upon the whole AC, from the common juncture B erect the Perpendicular BD which will be the side of the square sought, according to Case 1. of the Effection of pure quadraticks.

Arithmetical Rule. Multiply the given sides of the Oblong by one another, and the square root extracted out of the Pro∣duct will be the side of the square sought.

PROBLEM II.

THE square of the Hypothenusa in a right-angled ▵ being given, as also the difference of the other two squares to find the sides. E. g. If the Hypothenusa be BC (Fig. 25.) and the difference of the squares of both the legs, and conse∣quently its Leg also BE given (for the squares being given the sides are also given geometrically) to find the sides of the right-angled ▵ which shall have these conditions; or more plainly, to find one side e. g. the lesser which being found, the other, or the greater, will be found also.

SOLƲTION.

Let the □ of the given Hypothenusa = aa, and the square by which the two other differ = bb. Let the less side = x, and its □=xx. Wherefore the greater will be xx+bb. And since the sum of these is = to the □ of the Hypothenusa, you'l have 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. Having described a semi-circle on BC, and applyed therein BE, the □ EC will = aa−bb; and having described another semi circle upon EC divided into two Quadrants the □ DC will be 〈 math 〉〈 math 〉, and so DC = 〈 math 〉〈 math 〉 or the side sought; which being also transferr'd upon the other semi-circle describ'd on BC, viz. from C to A gives the other side AB and the whole ▵ sought.

The Arithmetical Rule. From the square of the Hypothe∣nusa substract the given difference, and the square root extra∣cted out of half the remainder gives the lesser side of the ▵ sought.

PROBLEM III.

HAving an equilateral ▵ ABC given (Fig. 26. n. 1.) to find the Center and Semi diameter of a Circle that shall circumscribe it. i. e. Find the BD the side of an Hexagon that may be inscribed in it. For if we consider the thing as already done; it will be manifest that the side of the Hexagon BD will fall perpendicularly on the side of the ▵ AB, as making an angle in a semi-circle, so having bisected the Hypothenusa DA you'l have E the Center sought.

SOLƲTION.

Make the side of the Triangle AB=a, BD=x, then will AD=2x. Since therefore the square BD i. e. xx being sub∣stracted out of the square AD i. e. 4xx, there remains the square AB 3xx, you'l have the Equation 〈 math 〉〈 math 〉; and dividing by 3 〈 math 〉〈 math 〉; therefore 〈 math 〉〈 math 〉

The Geometrical Construction. Having produced AB (n. 2.) to F a third part of it, the square of a mean proportional BD between BF and BA will be ⅓ aa or 〈 math 〉〈 math 〉, and so the Line BD = 〈 math 〉〈 math 〉. Therefore the Hypothenusa DA being divided in two in E, or at the interval BD, making the intersection from B and A, you'l have the Centre sought.

The Arithmetical Rule. Divide the square of the given side into three equal parts, and the square Root of a third part will give the semi-diameter AE or BE sought, by the intersection of two of which you'l have the Centre.

PROBLEM IV.

HAving given, in a right-angled Parallelogram, the Dia∣gonal, or for a right-angled ▵, the Hypothenusa and the proportion of the sides, to find the sides separately and construct the Parallelogram or ▵. Suppose e. g. the given Diagonal to be AB (Fig. 27. n 1.) and the given reason of the sides as AD to DE, to find the sides.

SOLƲTION.

Make AB=a, the reason of AD to DE as b to c; make the lesser side = x, then will the greater be 〈 math 〉〈 math 〉.

For the Equation, the □□ of the sides are 〈 math 〉〈 math 〉, □ AB; and multiplying both sides by bb, 〈 math 〉〈 math 〉; and dividing by bb + cc, 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 i. e. Extracting the roots as far as possible 〈 math 〉〈 math 〉

Another Solution.

Call the name of the given reason e, so that assuming any line for unity, the value of e may also be expressed by a right line, which shall be equal e. g. to DE above. Wherefore be∣cause we make the less side x, the greater will be ex, and so 〈 math 〉〈 math 〉 i. e. dividing by 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉

The Geometrical Construction. The last Equation above being reduc'd to this proportion as the 〈 math 〉〈 math 〉 to b so a to x, make (n. 2) AD and DE at right angles, and AE will be = 〈 math 〉〈 math 〉, and continuing AE and AD make as AE to AD so AB to AC the lesser side sought. Having therefore drawn BC which determines the lesser side AC, the greater side and so the ▵ ABC will be already formed, and may be easily compleated into a Rectangle. In the other Solution the last Equation agrees with the precedent (for it gives us this pro∣portion as 〈 math 〉〈 math 〉 to 1 so a to x in which 1 is = b, and ee =cc by what we have supposed) and so the Construction will be the same.

The Arithmetical Rule may be more commodiously expres∣ed by this last Equation under the last form but one, after this way, divide the □ of the Diagonal by the □ of the name of the Reason lessen'd by unity, and the root extracted out of the Remainder is the lesser side sought.

PROBLEM V. (Which is in Pappus Alexandrinus, and in Cartes's Geometry, p. 83. in a Biquadratick affected E∣quation, and p. 84. he gives us thereon a very remarkable Note.)

HAving given the Square AD (Fig. 28) and a right line BN, you are to produce the side AC to E, so that EF drawn from E towards B shall be equal to BN.

It will be evident, if you imagine a semi-circle to pass thro' the points B and E, that the most commodious way will be to find the line DG, that you may have the Diameter BG; upon which having afterwards described a semi circle, there will be need of no other operation to satisfie the question, than to produce the side AC 'till it occur to the prescrib'd Peri∣phery.

SOLƲTION. (As found by Van Schooten, p. 316. in his Com∣ment on Cartes's Geometry, which we will here give somewhat more distinct.)

1. Denomination. Make BD or DC=a, BN or FE=c, BF=y, and DG=x; the Perpendicular EH will be = a, and EG=BF, viz y (because the ▵ EHG is similar to ▵ BDF, by n. 3 Schol 2 Prop. 34. Lib. 1. Math••s. Enucl. and BD in the one = to EH in the other) and BG=a+x, BE=y+c; and BH will have its Denomination, if you m••ke (by reason of the similarity of the ▵ ▵ BFD and ••EH)

as BF to BD so BE to BH 〈 math 〉〈 math 〉. and you'l have also HG = 〈 math 〉〈 math 〉, i. e. having reduc'd them all to the same Denomination, 〈 math 〉〈 math 〉, i. e. 〈 math 〉〈 math 〉. Having therefore na∣med all the lines you have occasion for, you must find two E∣quations, because there are assumed two unknown quantities, viz. x and y.

2. For the first Equation and its Reduction. By reason of the similiarity of the ▵ ▵ BGE and BEH, as BG to GE so BE to EH 〈 math 〉〈 math 〉: Therefore the Rectan∣gle of the Extremes will be = to the Rectangle of the means, i. e. 〈 math 〉〈 math 〉; and taking from both sides 〈 math 〉〈 math 〉.

3. For the second Equation and its Reduction. Since BH, HE and HG 〈 math 〉〈 math 〉 are continual proportionals, the Rectangle of the extreams are equal to the square of the mean, i. e. 〈 math 〉〈 math 〉; and multiplying both sides by yy, and dividing by 〈 math 〉〈 math 〉, and taking away ayy and transposing the rest, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; i. e. dividing actually as far as may be by 〈 math 〉〈 math 〉.

4. The comparison of these two Equations thus reduc'd▪ gives a third new one, in which there will be only one un∣known quantity, viz. 〈 math 〉〈 math 〉; and adding to both sides cy,

〈 math 〉〈 math 〉; and multiplying by 〈 math 〉〈 math 〉; i. e. 〈 math 〉〈 math 〉 and dividing both sides by 〈 math 〉〈 math 〉; and adding 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉.

5. The Geometrical Construction, which is the same Pap•• prescribes in Cartes, viz. having prolong'd the side of 〈◊〉〈◊〉 square BA to N, so that BN shall be = to a given right li•• since BA is = a, and BN=c, the Hypothenusa DN will 〈 math 〉〈 math 〉. Having therefore made DG=DN, a•• describ'd a semi-circle upon the whole line BG, if AC be pr••¦longed until it occur to the Periphery in E, you'l have do that which was requir'd.

PROBLEM VI. (Which Van Schooten has in his Comment, p. m. 150, and following.)

HAving given a right line AB, from the ends of it A an•• B (Fig. 29.) to inflect two right lines AC and B•• which shall contain an angle ACB = to the given one D, an•• whose squares shall be in a given proportion to the Triang•• ACB, viz. as 4d to a.

Viz. You must determine the point C, which the two rig•• lines AH and HC or EH and HC will do, assuming the mi••¦dle point E in the line AB. Wherefore here will be two u••¦known quantities HE and HC, and consequently two Equ••¦tions to be found in the Solution; one whereof the giv•• proportion in the Question supplies us with, and the other 〈◊〉〈◊〉 have from the similar Triangles AIC and GFD, which repr••¦sent equal angles.

SOLƲTION.

Denomination. Make AE, half AB=a, HE=x a•• HC=y; therefore AH will be = a−x and HB=a+•• whence the Denomination of the squares AC and BC is ea•• had; viz. the one 〈 math 〉〈 math 〉, and the other, a••

[illustration]

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xx+yy, so that the sum of the squares is 2aa+2xx+

And the ▵ ACB will be = ay: And since the ▵ ▵ •• and AIC are similar, and the sides of the former FD and ••rbitrary, so that for FD we may put b and for FC, c; ••he sides of the latter are determined by the similitude of ▵▵ ABI and HDB, as being right-angled ones, and ha∣•• the common angle B; they will be obtain'd by making the Hypothen. BC to the Hypoth. AB so the base HB to ••ase BI, i. e.

〈 math 〉〈 math 〉, whence substracting BC=e, there remains CI = 〈 math 〉〈 math 〉.

For the first Equation, by virtue of the Problem as 〈 math 〉〈 math 〉 to ay. And the Rectangle of the ••mes is = to the Rectangle of the means, i. e. 〈 math 〉〈 math 〉.

For the other Equation, since as DF to FG so CI to AI 〈 math 〉〈 math 〉 the Rectangle of the extreams will again be = to the Re∣••gle of the means, i. e. 〈 math 〉〈 math 〉; multiplying both sides by 〈 math 〉〈 math 〉; ••h is the second Equation.

The Reduction of both Equations. 〈◊〉〈◊〉 first was 〈 math 〉〈 math 〉. ••refore dividing by 〈 math 〉〈 math 〉. substracting 〈 math 〉〈 math 〉. ••he latter Equation was 〈 math 〉〈 math 〉, i. e. ••ituting again the value ee, which was 〈 math 〉〈 math 〉; and by transposition, 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉, or 〈 math 〉〈 math 〉,

or (putting 2 f for 〈 math 〉〈 math 〉) 〈 math 〉〈 math 〉.

Wherefore we have the value of xx twice expressed, but by quantities partly unknown, because y is found on both sides. Wherefore now we must make a new comparison of their values, whence you'l have this new

5. Third Equation, in which there is only one of the un∣known quantities: 〈 math 〉〈 math 〉; and adding on both sides both yy and aa, 〈 math 〉〈 math 〉; or dividing by 2, 〈 math 〉〈 math 〉; and transposing 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; which is the value of the quantity y in known terms.

But this value in one of the precedent Equations, viz. in this 〈 math 〉〈 math 〉, being substituted for y and its square for yy, will give 〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉.

6. The Geometrical Construction, which Schooten gives us p. 153. Having made the angle KAB (n. 2. Fig. 29.) e∣qual to the given one D, erect from A, AL perpendicular to KA, meeting the Perpendicular EM in L; and from the Centre L, at the interval of the given right line d, describe a Circle that shall cut KA and EL produced to K and M. Then assuming EN=KA, join MA, and from N draw NH pa∣rallel to it, which shall meet AB in H. Afterwards, having described from L, at the interval LA, the segment of a Circle ACB, draw from H, HC perpendicular to AB meeting the circumference in C, and join AC, CB.

NB. The reason of this elegant Construction, which the Author conceal'd, for the sake of Learners we will here shew. 1. Therefore, he reduc'd the last Equation (extracting the root, as well as it could bear, both of Numerator and Deno∣minator) to this: 〈 math 〉〈 math 〉 multipl. by 〈 math 〉〈 math 〉, so that after this way the Construction would be reduc'd to this proportion, as d+f to a so 〈 math 〉〈 math 〉. 2. He made the angle KAE = to the given one D, and the angle KAL a right one, so that having described the segment of a Circle from L the inscribed angle will also be made equal to the gi∣ven one, according to the 33. Lib. 3. Eucl. 3. By doing this, EL expresses the quantity f, since by reason of the simi∣larity of the ▵ ▵ KOA s. GFD, n. 1. and AEL (for the angles LAE and AKO are equal, because each makes a right one with the same third Angle KAO) you have as KO to OA so AE to EL i. e. 〈 math 〉〈 math 〉

4. Making now LM and LK = d you had EM=d+f, and AK = 〈 math 〉〈 math 〉 (for the □ AL is = to aa+ff, which being substracted from □ LK = 〈 math 〉〈 math 〉.)

5. Wherefore there now remains nothing to construct the last Equation above, but to make EN=AK, and to draw HN parallel to AM; for thus was the whole proportion as EM to EA so EN to EH 〈 math 〉〈 math 〉 to x. Q. e. f.

For the point H being determined, a perpendicular HC thence erected in the segment already described defines the Point C, which answers the Question.

PROBLEM VII.

HAving given the four sides of a Quadrangle to be inscribed in a Circle, to find the Diagonals and their Segments, and so to construct the Quadrangle, and inscricbe it in the Circle. As e. g. suppose the given sides are AB, BC, CD, DA (Fig. 30 n. 1.) which now we suppose to be joined in••o a quadr••n∣gle

inscrib'd in the Circle, the Diagonals also AC and BD being drawn (n. 3.) to find first the segments of the diagonals Ae, Be, &c. which being had, the Construction is ready.

SOLƲTION.

Denomination. Make AB=a, AC=b, CD=c, DA =d, Ae=x [for this segment alone being found, the rest will be found also, as will be evident from the process.] Since therefore the vertical angles at e are equal, and likewise the angles in the same segment BCA, BDA, also DAC, DBC, &c. are equal, the Triangles AeD and BeC, also AeB and CeD are similar: wherefore it will follow that,

• 1. As DA to Ae so CB to Be 〈 math 〉〈 math 〉
• 2. As AB to Be so CD to Ce 〈 math 〉〈 math 〉
• 3. As AB to Ae so CD to De 〈 math 〉〈 math 〉

Therefore the whole Diagonal AC will be = 〈 math 〉〈 math 〉 and BD = 〈 math 〉〈 math 〉.

2. The Equation. But now by Prop. 48. Lib. 1. Math. Enucl. the Rectangle of the Diagonals is equal to the two Rectangles of the opposite sides.

Diag. AC, 〈 math 〉〈 math 〉

Diag. BD, 〈 math 〉〈 math 〉

Therefore the □ of the Diagonals 〈 math 〉〈 math 〉.

3. Reduction, i. e. taking (a) for unity 〈 math 〉〈 math 〉; i. e. the quantities on the left hand being reduc'd to the same denomination.

〈 math 〉〈 math 〉; and multiplying both sides by dd, 〈 math 〉〈 math 〉; and dividing both sides by d, 〈 math 〉〈 math 〉; and then dividing both sides by 〈 math 〉〈 math 〉; i. e. in the present case, where b by chance happens to be = a, 〈 math 〉〈 math 〉

Therefore 〈 math 〉〈 math 〉 or in our case 〈 math 〉〈 math 〉

4. The Geometrical Construction, which, by supposing a (and in the present case also b) to be unity, ought to deter∣mine,

1. The quantities cd, 〈 math 〉〈 math 〉, cc, and their aggregate with unity. 2. The aggregate of cd and dd. 3. To divide the one by the other. And, 4. To extract the root out of the quotient, or also to extract the roots first out of each quantity, and divide them by one another; which may all of them be separately done in so many separate Diagrams, but more ele∣gantly connected together after the following or some such like way. 1. Join AD and DC (n. 2.) into one line, and having described a semi-circle thereupon, erect the Perpendicular DE; and the line AE drawn will = 〈 math 〉〈 math 〉. 2. Making the angle CAG at pleasure, make AF=AB, and draw CG pa∣rallel to the line DF; so FG will be =〈 math 〉〈 math 〉. Now if, 3. in the vertical angle you make AH=CD, the line HI drawn parallel to DF will cut off AI=cd. 4. In AK erected = to AB, if you take AL=AH or CD, and draw LM paral∣lel to KH, you'l have AM=cc. 5. Having prolonged AG to N and AH to O, so that GN shall be =AI+AM and AO=AB or AK, and having described a semi-circle upon the whole line NO, a perpendicular erected AP will be = 〈 math 〉〈 math 〉 and so, 6. if AQ be made =AE and AR=AF or AB, and you draw a line RS from R pa∣rallel to PQ; AS will be = x, i. e. the segment sought Ae of the Diagonal AC; which being given, by force of the first Inference premis'd in the Denomination above, by drawing DS and, having made DT=BC, TV parallel to it; you'l have also the other segment Be=SV and by their Intersection on the line AB (n. 3.) the point e, thro' which the Diagonals must be drawn which will be terminated by the other given sides, and thence you'l have the quadrilateral figure ABCD sought, to be circumscribed about the Circle, according to Con∣sect. 6 Defin. 8. Mathes. Enucl.

NB. Unless we had here consulted the Learner's ease, the artifice of this Construction might be proposed after a more short and occult way, thus: Make DE a mean proportional between AD and DC, and draw AE. Then having made any angle CAG, make AF=AB, and at this Interval de∣scribe

the circle FROK, and draw CG parallel to DF. More∣over in the opposite vertical angle, having made AH=CD, draw HI parallel to DF, and having erected the perpendicular AK, and thence the abscissa AL=AH, make LM parallel to HK, and thence having prolonged AH to O, and GN being made equal to AI+AM, make AP a meam proportional be∣tween AO and AN, cutting the hidden circle in R; and last∣ly having made AQ=AE, if RS be drawn parallel to QP, you'l have AS the value of x sought, &c.

III. Some Examples of Affected Quadratick Equations.

PROBLEM I.

HAving given, to make a right angled Triangle ABC, the differences of the lesser and greater side, and of the greater, and the Hypothenusa, to find the sides separately and form the Triangle. E. g. Having given the right line DB (Fig. 31.) for the difference of the perpendicular and base, and CE for the difference of the base and Hypothenusa, to find the perpendicular AC, which being found, you'l have al∣so, by what we have supposed, the base AB, and the hypothenu∣sa BC.

SOLƲTION.

Make the difference DB=a, CE=b; put x for the perpendicular; the base, which is greater than that will be x+a and the Hypothenusa x+a+b. Therefore by vertue of the Pythagorick Theorem,

〈 math 〉〈 math 〉; and substracting from both sides 〈 math 〉〈 math 〉.

Wherefore by the first case of affected quadratick Equati∣ons 〈 math 〉〈 math 〉.

Construction. Find a mean proportional AK between AH=2b and AI=a (n. 2.) (Fig. 31.) and having made both AF and AG=b, place the Hypothenusa KF from AL, and

cut off GC equal to the hypothenusa GL; thus you'l have AC the perpendicular of the Triangle sought, and adding DB you'l also have the base AB, and from thence having drawn the hy∣pothenusa BC, it Will be found to differ by the excess required CE.

The Arithmetical Rule. Join twice the product of the differences multiplyed by one another, to twice the square of the difference of the base and the hypothenusa; and if the square root of this sum being extracted be added to the afore∣said difference, you'l have the perpendicular sought. Suppose e. g. both the differences of CE and DB=10.

PROBLEM II.

IN a right-angled ▵ having given the Hypothenusa and sum of the sides, to find the sides. E. g. If the Hypothenusa BC be given (Fig. 32.) and the sum of the sides CAB, to find the sides AB and AC separately, to form the Triangle.

SOLƲTION.

Make the Hypothenusa BC=a, the sum of the sides = b. Make one side e. g. AB=x, then will the other side AC be =b−x. Therefore

〈 math 〉〈 math 〉; and adding 2bx, and taking a∣way 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

Therefore according to Case 1. of affected Quadraticks. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉.

The Geometrical Construction. Having described a semi-circle upon BD=BC so a apply therein the equal lines BE and DE, and having described another semi-circle on BE apply therein BF=½b, to be prolonged farther out. Lastly, if another little semi circle be described at the ••nterval EF, the whole line AB

will be the true root or the side sought, and GB the false root, &c.

The Arithmetical Rule. From half the square of the Hy∣pothenusa substract the fourth part of the square of the given sum, and the root extracted out of the remainder, if it be ad∣ded to half the sum, will give one side of the Triangle; and substracted from the given sum, will give also the other Suppose e. g. BC to be 20, and the sum of the sides 28.

PROBLEM III.

HAving given again in the same ▵ the Hypothenusa, as above, and the difference of the sides DB (Fig. 33.) to find the sides.

SOLƲTION.

Make the less side x, the difference of the sides = b; the greater side will be x+b. Let the Hypothenusa be = a. Therefore, 〈 math 〉〈 math 〉; and taking away 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉.

Therefore by case 2, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉.

The Geometrical Construction. Having described a semi-circle upon BD=BC or a, apply therein the equal lines BC and DC, and having described another semi-circle on DC apply in it DF =½b, and if at the same interval you cut off FA from FC, the remainder AC will be the lesser side sought, &c.

The Arithmetical Rule. From half the square of the Hy∣pothenusa substract the square of half the difference, and if you take half the difference from the root extracted out of the re∣mainder, you'l have the lesser side of the Triangle required, and by adding to it the given difference you'l have also the greater. E. g. Let the Hypothenusa be 20, and the difference of the sides 4.

PROBLEM IV.

HAving given the Area of a right-angled Parallelogram, and the difference of the sides to find the sides. E. g. If the Area is = to the square of the given line DF, and the difference of the sides ED (Fig. 34.) to find the sides of the rectangle.

SOLƲTION.

Make the given Area = aa, the difference of the sides = b, the lesser side x; then the greater will be x+b. Therefore the Area xx+bx=aa; and substracting bx xx=−bx+aa.

Therefore according to case 2, 〈 math 〉〈 math 〉.

The Geometrical Construction. Join at right angles AG=a, and GH=½b, and having drawn AH and prolong'd it, describe the little Circle at the interval GH: so you'l have AE the lesser side, and AD the greater of the Rectangle sought, &c.

The Arithmetical Rule. Add the given Area and the square of half the difference, and having the sum, substract and add the difference from or to the root extracted, and so you'l have the greater and less sides of the rectangle.

PROBLEM V.

HAving given for a right-angled Triangle the difference of both the Legs from the Hypothenusa, to find the sides and so the whole Triangle. E. g. Suppose the difference of the less side to be BD and of the greater DE (n. 1. Fig. 35.) to find the sides themselves, and so make the Triangle.

SOLƲTION.

For BD put a, for DE, b. Let the greater side be x; the Hypothenusa will be x+b; therefore the lesser side will be x+b−a. Now the □□ of the sides are = to the □ of the Hypothenusa, i. e. 2xx−2ax+2bx+bb−2ab+aa

=xx+2bx+bb; and taking away aa, 〈 math 〉〈 math 〉; and adding 2ax and 2ab, and taking away 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉.

The Geometrical Construction. Between the given diffe∣rences BD and DE▪ find (n 2.) a mean proportional DF▪ and join to it at right angles the equal line FG, and cut off DH equal to DG; and so you'l have BH the greater side of the triangle sought. This being prolong'd to C, so that HC shall be = b, having described a semi-circle upon the whole line BC apply therein BA=BH; and having drawn AC, the Triangle sought ABC, will be formed.

The Arithmetical Rule. If the square root extracted from the double rectangle of the differences be added to the greater difference, you'l have the greater side sought, &c.

PROBLEM VI.

HAving given, to make two unequal Rectangles, but of equal heighth, the sum of their Bases with the Area of ……her (viz. the greater,) and the proportion of the sides of the ……her (viz the least,) to find the sides separately. E. g. Let the sum of the bases be AB (n. 1. Fig. 36) and the square of the line BC= to the Area of the greater rectangle; and let the sides of the lesser rectangle be to one another as CD to DE: To find the sides of both the rectangles; i. e. to find the com∣mon altitude, which being found the other sides will be easily obtain'd from the Data; or to find the base of the greater which, with the same ease, will discover the rest.

SOLƲTION.

Make AB=a, and the Area of the greater rectangle = bb; ••••d the proportion of the altitude to the base in the lesser, as c ••d; to find e. g. the greater base which call x. Therefore 〈◊〉〈◊〉 common altitude will be =〈 math 〉〈 math 〉, and the base of the lesser ••••ctangle=a−x.

Wherefore you'l have for the Equation, as c to d so 〈 math 〉〈 math 〉 to a−x.

Therefore ac−cx=〈 math 〉〈 math 〉; and multipl. by x, acx−cxx=bbd; and adding cxx and taking away bbd, acx−bbd=cxx. Now that you may conveniently divide both sides by c, make first as c to b so d to a fourth which call f, and then put cf for bd, and you'l have 〈 math 〉〈 math 〉; and dividing by 〈 math 〉〈 math 〉; and so according to case 3. 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉.

The Geometrical Construction. Find first the quantity f•• (num. 2. Fig. 36.) according to the following proportion, as c to b so d to f; and a mean proportional between b and f will be = 〈 math 〉〈 math 〉. Then having at the interval ½a described a semi circle (n. 3.) upon the given line AB, and erected BD= 〈 math 〉〈 math 〉, and having made EF= to it, CF will be 〈 math 〉〈 math 〉▪ To which AC being added will give x for one value and FB for the other. And for the common altitude, which we called 〈 math 〉〈 math 〉, make as x to b, so b to a fourth, i. e. as AF to FH so FH to FG; which will be the altitude of both rectangles Ag and Bg which may now easily be constructed.

The Arithmetical Rule might easily be had from this Equa∣tion reduced; but you may have it more commodiously from this other

SOLƲTION.

Let the Denomination remain the same as above, only her•• put x for the common altitude, and express the reason of th•• lesser base of the rectangle to this altitude by e, and that bas•• will be = ex: Therefore the base of the greater Rectangl•• will be = a−ex. Having now multiplyed the com••mon altitude by each base, the area of the greater rectangl•• will be ax−exx, and hence you'l have the Equation

ax−exx=bb; and adding exx, and taking away bb, ax−bb=exx; and dividing by 〈 math 〉〈 math 〉. Therefore by case 3.

〈 math 〉〈 math 〉.

Wherefore now this will be the Arithmetical Rule. If from the fourth part of the square of the sum of the bases divided by the □ of the name of the reason you substract the given area divided by the same name of the reason, and if the root extracted ••ut of the remainder be added to or substracted from half the sum of the bases divided by the same name of the reason; this sum or ••emainder will give the altitude of the given Rectangles, and ••hat multiplyed by the name of the reason one of the bases: And that being substracted from the given sum of the bases ••ill give the other base. For Example, let the sum of the ba∣•• be 16, the area of one of the rectangles 30, the ••ame of the reason which the common altitude has to the base ••f the other rectangle = 2. There will come out the com∣••on altitude, on the one side 5, on the other 3, &c.

PROBLEM VII.

HAving given the Perpendicular of a right-angled Trian∣gle let fall from the right angle, and its Base, to find the ••ments of the Base, and so to form the Triangle.

E g. If the base of the right-angled Triangle you are to ••m be AB (Fig. 37.) and the length of the perpendicular •••• or BF; to find the segments of the base, and so the point 〈◊〉〈◊〉, from which you are to make the perpendicular CD, to form ••e Tiangle ABG.

SOLƲTION.

Let the given base be = a; and the given perpendicular ••b: Then will one of the segments of the base be =x; ••d the other =a−x, and b a mean proportional between ••e said segments, i. e.

x to b as b to a−x;

••herefore ax−xx=bb; and by adding xx and taking a∣••y bb,

ax−bb=xx. Therefore by case 3. 〈 math 〉〈 math 〉.

The Geometrical Construction. Having described a semi-circle upon the given line AB, if you erect the perpendicular BE, and from the point G (which is determined by EG pa∣rallel to AB) let fall GD equal to it, you will have the two segments sought, AD = 〈 math 〉〈 math 〉 and DB = 〈 math 〉〈 math 〉, which Construction, it cannot be denyed, but it may be evident to any attentive person even without the A∣nalysis.

But that case may by the by be taken notice of wherein the given perpendicular would not be BE but BF. For in this case the perpendicular BF being erected upon AB, the paral∣lel FG would not cut the semi-circle; which is an infallible sign that the Problem in this case is impossible, where the perpendicular is supposed to be greater than half the base; which is inconsistent with a right angle.

The Arithmetical Rule. From the square of half the base take the square of the given perpendicular, and add or sub∣stract the square root extracted out of the remainder, to or from half the base; and on the one hand the sum will give the greater segment, and on the other the difference will give the less.

PROBLEM VIII.

HAving given the perpendicular of a right-angled Triangle that is to be let fall from the right angle, and the diffe∣rence of the segments of the base, to find the segments, and de∣scribe the Triangle.

E. g. If the perpendicular is, as above, BE, and the diffe∣rence of the segments AH (n. 1. Fig. 38.) to find the seg∣ments AD and DB, from whose common term you are to e∣rect a perpendicular DG or DC to form the Triangle.

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SOLƲTION.

Make the lesser segment =x, and the difference of the seg∣••ts =a, the greater segment will be x+a Make the gi∣•• perpendicular as before =b: Therefore you'l have as x+a to b so b to x; and consequently, xx+ax=bb; and substracting ax, xx=bb−ax. Wherefore according to case 2. 〈 math 〉〈 math 〉.

The Geometrical Construction. Make HD=½a, DG equal perpendicular to b; HG will be = 〈 math 〉〈 math 〉 = HB or ••, viz having drawn a semi-circle from H thro' G. ••erefore DB is the less segment, and AD the greater; and ••ing drawn AG and BG, or on the other side, (making the ••pendicular DC=DG) having drawn AC and BC, the ••iangle will be constructed. Or with Cartes, make (n. 2.) ••=½a and EB=b, and having described a Circle from ••hro' E draw BHA; and so you'l have the two segments ••ght AB the greater and DB the lesser.

The Arithmetical Rule. Join the squares of the half diffe∣••ce and perpendicular into one sum, and then having extra∣•• the root substract half the difference from it; and the re∣••nder will be the lesser segment sought; and having added •• difference you'l have also the greater.

PROBLEM IX.

HAving given for a rigbt-angled Triangle one segment of the base and the side adjacent to the other segment, to find rest and construct the Triangle.

As if the lesser segment of the base DB be given (Fig. 39.1.) and the side AC adjacent to the other segment; to find greater segment of the base, which being found the rest easily obtain'd, and consequently the whole Triangle.

SOLƲTION.

Make the greater segment = b, the given side = c, the segment sought = x. Now if we suppose the triangle ABC to be already found, it is evident, 1. If from the square of AC you substract the □ AD, you'l have the □CD=cc−xx. 2. The same □ CD may also be otherwise hence obtain'd, be∣cause, the angle at C being a right one, CD is a mean pro∣portional between BD and DA, i. e. between b and x; whence the rectangle of the extremes bx is = □ of the mean CD. Wherefore now it follows, 3. that cc−xx=bx; and adding xx, cc=bx+xx; and substracting bx,−bx+cc=xx. Therefore according to case 2. x=−½b+√¼bb+cc.

Geometrical Construction. Join EF=½b (n. 2. Fig. 39.) and FA=c at right angles, and having described a Circle from E thro' F draw AEB; so you'l have DA the greater segment and DB the less; having erected therefore a perpendicular from D, and described a semi-circle upon AB, you'l have C the vertex of the triangle sought, whence you are to draw the sides AC and BC.

The Arithmetical Rule. Join the □ of half the given seg∣ment, and the □ of the given side into one sum; and having extracted the root of it, if you thence take half the given seg∣ment, you'l have the segment sought.

PROBLEM X.

HAving given in an oblique angled Triangle the perpendi∣cular height, and the difference of the segments of the base, and the difference of the other sides, to find the sides and form the triangle.

As, if the altitude CD be given (n. 1. Fig. 40.) and also the difference of the segments of the base EB, and the diffe∣rence of the sides FB (as is evident from the triangle ABC (n. 2.) conceived to be so formed beforehand) and you are to determine the base it self and both sides, &c.

SOLƲTION.

Make the given perpendicular CD=a (see n. 2. Fig. 40.) EB=b, FB=c: For the lesser segment of the base AD put x, and the greater will be x+b. It is now evident that you may obtain the □ CB by the addition of the □ □ DC and BD, 〈 math 〉〈 math 〉, and the □ AC by the addition of the □ □ AD and DC, viz. aa+xx: So that the side AC will be = 〈 math 〉〈 math 〉, and the side BC = 〈 math 〉〈 math 〉. But since also this same side BC may be obtain'd by adding the difference c to the side AC, so that it shall be = 〈 math 〉〈 math 〉: you'l have this Equation, 〈 math 〉〈 math 〉; and squaring both sides, 〈 math 〉〈 math 〉; and substracting from both sides 〈 math 〉〈 math 〉; and again squaring it, 〈 math 〉〈 math 〉; and substracting from both sides 4ccxx (because c is less than b) and transposing the rest, 〈 math 〉〈 math 〉 and dividing by 〈 math 〉〈 math 〉 i. e. dividing the affected quantities by 4 both above and un∣derneath, 〈 math 〉〈 math 〉 and actually dividing the former part by 〈 math 〉〈 math 〉.

Therefore according to the second case, 〈 math 〉〈 math 〉; or reducing ¼bb to the same denomination with the rest 〈 math 〉〈 math 〉; and leaving out those quantities that destroy one another 〈 math 〉〈 math 〉.

The Arithmetical Rule. Multiply four times the square of FB by the square of the perpendicular CD, and add to it the product of the square of EB into the □ FB, and from the sum substract the biquadrate of FB; and the remainder will be the first thing found. Then substract four times the square of FB from four times the square of EB; and the remainder will be the second thing found. Lastly, divide the first thing found by the second, and from the quotient take half after having extracted the root: Thus you'l have the lesser segment of the base sought, &c. E. g. In numbers you may put 2 for FB, 4 for EB, 12 for CD.

As for the Geometrical Construction, the quantity of the last Equation contain'd under the radical sign will help us to this proportion,

as 〈 math 〉〈 math 〉 so cc to a fourth; or divi∣ding all by 4,

as 〈 math 〉〈 math 〉 so ¼cc to a fourth, which is ¼ of the quantity under the radical sign. Assuming there∣fore the quantity c for unity, make (n. 3.) IK=c IN and KL=b; NO will be = bb, and substracting OP=cc (i. e. to unity) there will remain NP=bb−cc. In like manner IS and KM=a, ST will = aa; to which if you add SX=¼NO, and take thence XV=¼ unity; TV will be = 〈 math 〉〈 math 〉. Wherefore if you make NR equal to this TV, and PQ=¼ of unity or cc, since NP is = bb−cc; by the rule of proportion there will come out DR¼ of that quantity, which is under the radical sign. Therefore this being taken four times will give DZ for the whole quantity; to which if

you join Dy= to unity, and, having described a semi-circle upon the whole line YZ, erect the perpendicular DE; this will be the root of the said quantity, and taking hence more∣over EF=½b, you'l have DE or DA the less segment of the base sought. Therefore adding GB=b to DG, DB will be the greater segment, and, having let fall the perpendicular DC=a, BC and AC will be the sides sought. Q. E. F.

IV. Some Examples of Affected Biquadratick Equa∣tions, but like Affected Quadratick ones.

PROBLEM I.

TO find a square ABCD (such as in the mean while we'll suppose n. 1. to be in Fig. 41.) from which having taken away another square AEFG, which shall be half the former, ••here will be left the Rectangle GC whose Area is given. E. g. Suppose the given area equal to the square of the given line LM, to find the true sides of the squares AB and AE, answer∣••ng to these supposed ones, n. 1.

SOLƲTION.

Make the area of the rectangle that is to remain = bb, and GB=x; BC or AB will be = 〈 math 〉〈 math 〉, and substracting hence GB, the remaining side of the lesser square AG=〈 math 〉〈 math 〉−x, 〈◊〉〈◊〉. 〈 math 〉〈 math 〉. Since therefore the square of this is supposed ••o be half of the square of AB, this will be the Equation: 〈 math 〉〈 math 〉; ••nd multiplying by xx, 〈 math 〉〈 math 〉 ••nd multiplying by 2, 〈 math 〉〈 math 〉;

and substracting 2b{powerof4}, and adding 〈 math 〉〈 math 〉; and dividing by 2, 〈 math 〉〈 math 〉.

NB. The same Equation may be obtain'd, if, putting x for GB or FH, and having found the □ of AG or GF as a∣bove, you infer 〈 math 〉〈 math 〉.

This last Equation, tho' it be a biquadratick, yet may be rightly esteem'd only a quadratick one, because there is neither x{powerof3} nor single x in it, and so you may substitute this for it, 〈 math 〉〈 math 〉, viz. by supposing y=xx. Whence according to the third case of affected quadraticks, y will = 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 or = 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. Now if the given line b be assu∣med for unity, bb and b{powerof4} will be = = to the same line Therefore, if between LM as unity, and MN=½b viz 〈 math 〉〈 math 〉 you find a mean proportional MO (n. 2. Fig. 41.) that wil•• be = 〈 math 〉〈 math 〉, which being substracted from LM, and added to it, will give the two values of the quantity y. Moreover there∣fore by extracting its roots, i. e. by finding other mean pro∣portionals LR and LS between the quantities found LP and LQ and unity (n. 3.) they will be the two values of the quantity x sought; the first whereof LR will satisfie the que∣stion, and the other LS be impossible. Wherefore to form

the square it self, since its side will be = 〈 math 〉〈 math 〉; by making (n 4.) as x to b so b to a fourth, it will be obtain'd: And this may be further prov'd, if finding a mean proportional BK between BI=LR and the side of the □ BC, it be equal to the given quantity LM.

Arithmetical Rule. From the given area or the square of the given line LM substract the root of half the biquadrate of the same line; thus you will have the value of the □ FC, viz. xx: Therefore extracting further the square root of ••••is, it will be the value of x sought.

PROBLEM II.

TO find another square ABCD (Fig. 42. n. 1.) out of the middle whereof if you take another square EFGH, which ••all be a fourth part of the former, the area of the rectangle ••K intercepted between BC and FG prolonged, shall be equal 〈◊〉〈◊〉 the square of a given line LM; i. e. having these given to ••nd the segment BI, and consequently also the side BC or ••B.

SOLƲTION.

Make the area of the given rectangle, or the square of LM 〈◊〉〈◊〉 to bb, and the side sought of the rectangle BI=x; the o∣••er side BC will be = 〈 math 〉〈 math 〉, and having substracted out of it 〈◊〉〈◊〉 and GK (i. e. 2x) the side of the lesser square FG will ••〈 math 〉〈 math 〉−2x, i. e. 〈 math 〉〈 math 〉; whose square since it is the ••orth part of the greater square by the Hypothesis, you'l have 〈 math 〉〈 math 〉; 〈◊〉〈◊〉 multiplying both sides by 〈 math 〉〈 math 〉; 〈◊〉〈◊〉 taking away 4b{powerof4}, and adding 〈 math 〉〈 math 〉;

and dividing by 4, 〈 math 〉〈 math 〉.

Therefore according to the third case of affected quadratick ••¦quations, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉.

Geometrical Construction. If the given line b be taken 〈◊〉〈◊〉 unity, b{powerof4} and bb will be equal to it. Therefore if betwe•• LM as unity, and MN=3 ⅓b, you find a mean proporti••¦nal MO (n. 2. Fig. 42.) 'twill be 〈 math 〉〈 math 〉; which subst••¦cted from MQ=2b, or added to it, will give two values the quantity xx, viz. PQ and IQ; the first whereof will 〈◊〉〈◊〉 only a true one, and of use here. Now therefore a mea•• proportional QR found between PQ and unity will expre•• the quantity sought x.

Therefore for forming the square it self, since its side AB=〈 math 〉〈 math 〉, you may proceed as in the former Constructon, (vi•• n. 3.)

PROBLEM III.

HAving given the base of a right-angled Triangle, and mean proportional between the Hypothenusa and Perpe••¦dicular, to find the Triangle. As if the given base be A (Fig 43.) and the mean proportional between AC and B•• be CD; to find the perpendicular BC, and Hypothenu•• AC.

SOLƲTION.

Make the given base = a, and the mean proportional = the perpendicular BC=x, then will the Hypothenusa be 〈◊〉〈◊〉 the Hypoth, −〈 math 〉〈 math 〉.

[illustration]

Pag. 56.

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••erefore 〈 math 〉〈 math 〉; ••d multiplying both sides by 〈 math 〉〈 math 〉; ••d substracting 〈 math 〉〈 math 〉. ••erefore by the second case of affected quadraticks 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉.

Or thus.

Make the Hypothenusa AC=x, then will the perpendi∣••ar be BC=〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; ••d multiplying by 〈 math 〉〈 math 〉. ••••erefore by case 1. 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction; the first for the latter Equation. 〈◊〉〈◊〉 be put for unity, the line AB will be also = aa, and king, as a to b so b to a third, i. e. as LM to MN so LO OP, and you'l have bb. Having erected the perpendicu∣•• AQ=OP upon AM, and drawn MQ or Mn equal to 〈 math 〉〈 math 〉, and consequently An will be = 〈 math 〉〈 math 〉, 〈◊〉〈◊〉 the value of xx. Moreover a mean proportional AC ••nd between An and AR unity will be the value of x, i. e. Hypothenusa sought; which being found, you may easily ••mplete the Triangle ABC.

2. In the case of the former Equation, making every thing before AK would be the value of the quantity xx, i. e. 〈 math 〉〈 math 〉. Therefore a mean proportional RT ••nd between RS=AK and AR unity will be the value of x,

i. e. the perpendicular sought, and so AT the Hypothenusa of the Triangle sought.

Arithmetical Rule. In the first Solution add the biquadrate of the given mean proportional to the biquadrate of half the given base; and having extracted the square root of the sum, take from it half the square of the given base; the root of the remainder will give the perpendicular of the triangle sought, and the root of the sum will give the hypothenusa of it.

PROBLEM IV.

HAving the Hypothenusa of a right-angled Triangle given, and a mean proportional between the sides to find the Tri∣angle. As if the hypothenusa be AC (Fig. 44.) and a mean proportional between the sides BD, to find the sides AB and BC.

SOLƲTION.

Make the given Hypothenusa = a, and the mean propor∣tional =b, and the perpendicular BC=x; the basis AB by the hypoth. will be 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉; and multiplying by xx, 〈 math 〉〈 math 〉; and substracting b{powerof4}, 〈 math 〉〈 math 〉.

Therefore by the third case, 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction. If a be put for unity, AC will be also = aa, and by making as AC to CG (a to b) so AF to GH (b to a third) this third will be GH=bb. Assuming therefore OC=½aa=OB the radius of a semi-circle, and ha∣ving erected CD=bb=BE parallel to it, EO will be

〈 math 〉〈 math 〉, and consquently EC = 〈 math 〉〈 math 〉, and EA 〈 math 〉〈 math 〉, viz. the double value of the quantity xx. Therefore for the double value of x, you must extract the roots out of them, i. e. you must find the mean proporti∣onals AL and AM between unity AC and AI=EC on the one side, and AK=AE on the other; Altho' these last may be more compendiously had, and the triangle it self immedi∣ately constructed, if having found EC and EA, you draw CB and AB: For these will be those two last mean proportionals = = AL and AM; for by reason of the ▵ ▵ ABC, AEB, and BEC, BC is a mean proportional between AC and CE, and AB a mean proportional between the same AC and AE by the 8. Lib. 6. Eucl. which is Consect. 3. Schol. 2. Prop. 34. Lib. 1. Math. Enucl.

PROBLEM V.

HAving given the Area and Diagonal of a right-angled Parallelogram, to find the sides and so the Parallelogram. As if the given Area be = to the square of a given line BD (Fig. 45.) and the Diagonal AC, to find the sides AB and BC.

SOLƲTION.

If for the given Area, or square of the line BD you put bb, and make the Diagonal AC=a, and put for the lesser side BC, x; the other side will be 〈 math 〉〈 math 〉.

Therefore 〈 math 〉〈 math 〉; and multiplying by xx, 〈 math 〉〈 math 〉; and substracting b{powerof4} 〈 math 〉〈 math 〉. Which Equation, since it is the same with that of the preceding Problem (which is no wonder, since this fifth perfectly coincides with the fourth; for the Di∣agonal AC is the hypothenusa, and BD, whose square is = to the given area of the rectangle, is a mean proportional be∣tween the sides AB and BC) and so will have the same Con∣struction,

(see Fig. 45.) and the same Arithmetical Rule, which may be easily formed from the last Equation of the preceding Problem.

PROBLEM VI.

HAving given the first of three proportional lines, and ano∣ther whose square is equal to both the squares of the other two, to find those two proportionals. As if AC (Fig. 46.) be the first of the three proportionals, and another line ED gi∣ven, whose square equals the two squares of the others taken together; to find those two as second and third proportio∣nals.

SOLƲTION.

If for AC you put a, and make the given line ED=c, and the second proportional = x, the third will be 〈 math 〉〈 math 〉. Where∣fore the squares of the two last will be 〈 math 〉〈 math 〉+xx=cc, □ ED by the hypoth. and multiplying both sides by 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉.

Geometrical Construction. If a be put for unity, AC will also = aa and a{powerof4}, and making as AC to CD (a to c) so AF to DE (c to a third) DE will be = cc. Now having made AK=DE, i. e. cc, a mean proportional AI found between AC and AK will be 〈 math 〉〈 math 〉. Therefore taking AO=½AC, viz. ½aa, the hypothenusa OI will be = 〈 math 〉〈 math 〉. And OA=½a being substracted from OI or OH equal to it will leave AH the value of xx; and the root of that being extra∣cted, i. e. finding another mean proportional AG between AC and AH, it will be the value of x, i. e. the second of the proportionals sought, and since AC is the first given, AH will be the third. Q. E. F.

NB. This Construction may be abbreviated, and the first Operation, by which you find DE, to which afterwards AK is made equal, may be omitted. For since you make use of a mean proportional between CA and DE sought, which is = to the given line ED, and afterwards AI a mean pro∣portional between AC and AK is sought; it is evident that AI will be equal to ED given, and consequently that they may be immediately joined at right angles at the beginning of the given line AC, and the rest may then be done as before.

Some Examples of Cubick and Biquadratick Equations, both simple and affected, whether reducible or not.

PROBLEM I.

BEtween two given right angles to find two mean proportio∣nals. E. g. Suppose given AB the first and CD the fourth, (Fig. 47. n. 1.) between these to find two mean pro∣portionals.

SOLƲTION.

Make the first of the given quantities AB=a, the other CD=q, the first mean = x, then will the latter be 〈 math 〉〈 math 〉, and consequently 〈 math 〉〈 math 〉=q; and multiplying by aa, x{powerof3}=**aaq.

The Central Rule will b 〈 math 〉〈 math 〉=AD 〈 math 〉〈 math 〉=DH. i e. according to a supposition we shall by and by make, ½a=AD, and ½q= DH.

Geometrical Construction. If AB or a be made unity, and also the Latus Rectum of your Parabola, and you describe, by means of this Latus Rectum, the Parabola, according to Schol. 1. Prop. 1. Lib. 2. Math. Enucl. [see n. 2. and 3. Fig 47.] ••n which AB is the Latus Rectum; A1, A2, &c. the Abscis∣••a's; AI, AH, &c. the semiordinates; make moreover (n. 4) AD=½a, and having from D erected a perpendicular =½q, describe a circle at the interval AH, cutting the para∣bola

in N: Which being done, a perpendicular to the Ax will be the root sought or the value of x, i. e. the first of the means, and consequently OA the other; since NO by the first property of the Parabola (see Prop. 1. Lib. 2. Mathes. Enucl.) is a mean proportional between the Latus Rectum AB, and the abscissa AO. And by this means there will come out, by Baker's Central Rule the very construction of Des Cartes, Ge∣om. p. m. 91.

The Arithmetical Rule. Multiply the square of the first by the fourth given, and the cube root extracted out of the pro∣duct, will express the first of the means sought.

PROBLEM II.

HAving given the solid Contents of a solid or an hollow Pa∣rallelepiped, and the proportion of the sides, to find the sides. As, if the given capacity or solid contents be = to the cube of a certain given line IK (Fig. 48. n. 1.) and the pro∣portion of the heighth to the length be as AB to BC, and to the latitude as the same AB to BD; to find first the altitude, which being had, the other Dimensions will also be known, by the given proportions.

SOLƲTION.

Make IK=a, AB=b, BC=c, and BD=d; and lastly the heighth sought = x, then as b to c, so x to the length required 〈 math 〉〈 math 〉; and as b to d, so x to the latitude sought 〈 math 〉〈 math 〉.

Multiplying therefore these three dimensions of the Parallelepi∣ped together, you'l have its capacity or solid contents 〈 math 〉〈 math 〉 =a{powerof3}; and multiplying by bb, cdx{powerof3}=a{powerof3}bb; and dividing by cd, 〈 math 〉〈 math 〉 i e. 〈 math 〉〈 math 〉.

Therefore the Central Rule will be the samc as above, 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH, i. e. according to the supposi∣tion which will by and by follow, 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH.

Geometrical Construction. If IK or a be made unity, and at the same time Latus Rectum, and by means of it you de∣scribe a Parabola, after the way we have shewn, Fig. 47. n. 2. and 3. and shall always hereafter make use of; and then to prepare the quantity 〈 math 〉〈 math 〉 (which in the Central Rule is the the quantity 〈 math 〉〈 math 〉) make (n. 2.) IK−IM=BC−KL=BD−MN as a to c so d to e; so that for cd you may put ae, and afterwards divide by a both above and underneath; you'l have the quantity 〈 math 〉〈 math 〉.

Therefore by further inferring as 2e to bb, so aa to a fourth IO−IP=IT−IK−IQ, and you'l have the quantity DH, which will determine the centre, after AD is made equal 〈 math 〉〈 math 〉. Having therefore described from that centre a circle through the vertex of the Parabola A (n. 3.) a semi∣ordinate NO drawn from the intersection will be the altitude sought, which will easily give you the length and breadth by the reasons above shewn.

Another Solution.

Which will be more accommodated to the Arithmetical Rule.

Let the rest of our Positions or Data remain as above, but the name of the proportion which the altitude has to the length be = e, and of that which it has to the latitude = i, the

length will be = to ex, and the latitude to ix. Wheref•• multiplying the sides together, you'l have the whole solidity 〈 math 〉〈 math 〉; and dividing by ei, 〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉 Hence

The Arithmetical Rule. Multiply together the given na•• of the reasons, and divide the given cube by the produ•• which done, the cubick root extracted out of the quotient 〈◊〉〈◊〉 be the altitude of the solid sought.

Another Geometrical Construction. Now if we would a•• construct this Equation x{powerof3}=〈 math 〉〈 math 〉 geometrically, putting AB=•• for unity, BC and BD will be the names of the reasons = •• and i. Making therefore first IK−IM−KL−MN as a to e so i to a fourth f (Fig. 49. n 1.) af •• be = ei, and the proposed Equation will have this form: 〈 math 〉〈 math 〉 i. e. aa / f.

Therefore 2. making as f to a so a to a third IQ, that will be the value of 〈◊〉〈◊〉 But hence 3 by extracting the cube root, i. e. by finding t•• mean proportionals between unity b, viz. AB and the 〈◊〉〈◊〉 found ••Q; the first of them will give the root sought.

NB. The same Central Rule would come out according Baker••s Central Rule, which would have the same form of Equation, as the precedent Example x{powerof3}**−aa=o,

• taking b for the Lat.
• Rect. and unity,
  • 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH,
  • 〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉 i. e. ½IQ=DH

(see Fig. 49. n. 2.)

[illustration]

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PROBLEM III.

HAving given the solid contents of a solid or hollow Paral∣lelepiped, and the difference of the sides, to find the sides. 〈◊〉〈◊〉 if the given capacity be equal to the cube of any given ••e LM (Fig. 50. n. 1.) and the difference whereby the ••gth exceeds the breadth = NO, and the difference by which 〈◊〉〈◊〉 breadth exceeds the altitude or depth = PQ, to find the ••gth, breadth and depth.

SOLƲTION.

Make the side of the given cube = a, the excess of the ••gth above the breadth NO=b, and of the breadth above 〈◊〉〈◊〉 depth PQ=c, make the depth = x, the breadth = x+c, ••d the length = x+b+c. Multiplying therefore these ••ree dimensions together.

〈 math 〉〈 math 〉

•• according to the forms of Baker and Cartes 〈 math 〉〈 math 〉

Wherefore the Central Rule contracted by the supposition ••hich will hereafter follow will be this, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH. •• e. by virtue of the supposition just now mentioned (which ••kes LM viz. a for unity and also for Lat. Rectum)

〈 math 〉〈 math 〉 = AD. And 〈 math 〉〈 math 〉 = DH.

Or more short; 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 (∽ 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Geometrical Construction. If LM or a be taken for unity or Latus Rectum of the Parabola to be described, that being described (Fig. 50. n. 3.) you are first of all to determine two quantities AD and DH; which may be done two ways; either by Baker's form of his Central Rule, or by ours immediately divided by the quantities of the last Equation.

1. For AD by our form, 〈 math 〉〈 math 〉 = AD, you must make (Fig. 50. n. 1.) as a to b so b to a third (AB to AC so BE to CF) which will be bb, and D{powerof2} the eighth part of this CF must be added to A{powerof2} the half of AB. And by Baker's form you must make, 1. as AB (=a, n. 2.) to AC (=½p i. e. ½b+c) so BE (=¼p i. e. ¼b+½c) to a fourth CF (which will be = 〈 math 〉〈 math 〉▪) 2. Make moreover as AB to AG (a to b) so BH to GI (c to bc) and, as AB to AK (a to c) so BH to KL (c to cc) and the two quantities found GI and KL (bc and cc) being added into one sum will give the quantity q=MN, the half whereof MO will express the quantity 〈 math 〉〈 math 〉 in the

Rule, and to be substracted from the former 〈 math 〉〈 math 〉. Actually therefore to determine the quantity AD not on the Ax, but on another Diameter of the described parabola n. 3. (because the quantity p is in the Equation) having made a perpendi∣cular to the Ax aE=〈 math 〉〈 math 〉 i. e. to the line BE n. 2. and from E having drawn EA parallel to the ax, according to our form AD n. 1. transfer it only on the diameter of the parabola n. 3. from A to D, either by parts 〈 math 〉〈 math 〉 i. e. ½LM from A to c, and 〈 math 〉〈 math 〉 i. e. ⅛CF n. 1. from c to D: But according to Baker's form, first you must put 〈 math 〉〈 math 〉=½LM n. 1. from A to 1. Se∣condly you must put from 1 to 2 the quantity 〈 math 〉〈 math 〉=CF n. 2. Thirdly you must put from 2 to 3 backwards the quantity to be substracted 〈 math 〉〈 math 〉=MO n. 2. which being done, the point D will be determined.

[It is evident by comparing these two ways of Construction, that we may join our forms not incommodiously to Baker's; because by ours the quantity AD was obtained more compendiously than by Baker's, which will also often happen hereafter. And where this Compendium cannot be had, there is another not inconside∣rable one, that, if the quantities AD and DH determined ac∣cording to both ways shall coincide, (which happens in the pre∣sent case) we may be so much the more sure of the truth]

2. As for DH by our form, you must put it from D to e on a perpendicular erected from D on the left hand, the quantity 〈 math 〉〈 math 〉 = BE n. 2. falling here upon the Ax. Then for the quantity 〈 math 〉〈 math 〉 make n. 4. as a to ½bb (LM to LN) so ⅛b to 〈 math 〉〈 math 〉 (MO to NP) and this quantity must be put from e to f in a perpendicular to the Diam. Thirdly, for the quantity 〈 math 〉〈 math 〉

you must farther make n. 4. as a to ½bb (DM to LN) so ¼c to a fourth (MQ to NR) which must be put from f to g. Lastly the quantity 〈 math 〉〈 math 〉 (which is = MO n. 2.) must be put backwards (because to be substracted) from g to H, which is the centre sought. In like manner by Baker's form, first 〈 math 〉〈 math 〉=BE is put from D to 1 even to the Ax. Secondly, for the quantity 〈 math 〉〈 math 〉 make, n. 4. as a to 〈 math 〉〈 math 〉 (LM to LS=CF n. 2.) so 〈 math 〉〈 math 〉 to a fourth (MT=AC n. 2. to SV) and this SV is further put (n. 3.) from 1 to 2. Thirdly, make in the same Fig. n. 4. for the quantity 〈 math 〉〈 math 〉, as a to 〈 math 〉〈 math 〉 (LM to MT) so 〈 math 〉〈 math 〉 to a fourth (LX to XZ) and this XZ is put back∣wads (n. 3.) from 2 to 3, which precisely coincides with the point g. Lastly the remaining quantity 〈 math 〉〈 math 〉 (=MO n. 2. and so by what we have said above, precisely coinciding with the interval g H) is put backwards from g to H the Centre sought.

[Hence it appears again that Baker's form is more laborious than ours; tho' both accurately agree, and hereafter, for the most part, we shall use them both together, tho in' the work it self, rather in Figures, than in that tedious process of words, which we have here for once made use of, that it might be as an Example for the following Constructions.]

Having therefore found by one or both ways the Center H, and thence described a circle thro' the vertex of the parabola A, the intersection N will give the perpendicular to the dia∣meter NO, the value of the quantity x sought.

PROBLEM IV.

TO divide a given angle NOP, or a given arch NQTP (Fig. 51. n. 1.) into three equal parts; i. e. having gi∣ven or assumed at pleasure the Radius NO, and consequently also the Chord of the arch NP, to find NQ the subtense of the third part of the given arch.

SOLƲTION.

If NO be made unity, NP=q, and NQ be supposed =z; having drawn QS parallel to TO, you'l have three similar triangles NOQ, QNR and RQS. For since the an∣gle QOP is double of the angle QON, and the same (as be∣ing at the Center) double also of the angle at the Periphery QNR; it will be equal to the angle QON. But the angle at Q is common to both triangles: Therefore the whole are equi-angular, and consequently the legs NQ and NR equal, as also NO and QO; and by the like reason also PY and PT. Wherefore if RS should be added to RY, the line NP by this addition would be triple of the line NQ; and so would give the Equation, if RS was determined; which may be done by means of the ▵ QRS, similar to the two former NOQ and QNR; for the angle RQS is equal to the alternate one QOP=QNR, and the angle at R common to the tri∣angles QNR and RQS, &c. Wherefore as NO to NQ so NQ to QR 1−z−z−zz and as NQ to QR so QR to RS z−zz−zz−〈 math 〉〈 math 〉

Therefore according to what we have above said 〈 math 〉〈 math 〉; and substracting 〈 math 〉〈 math 〉; or 〈 math 〉〈 math 〉

Therefore the Central Rule will be (supposing also unity NO for the Latus Rectum,)

〈 math 〉〈 math 〉 = AD i. e. by our ½+〈 math 〉〈 math 〉 i. e. 2NO=AD form, and 〈 math 〉〈 math 〉=DH. and 〈 math 〉〈 math 〉=DH.

Geometrical Construction. Having described your parabola (Fig. 51. n. 2.) take on its Ax (because the quantity p is want∣ing in the Equation) AD=2NO, and from D having e∣rected a perpendicular = 〈 math 〉〈 math 〉NP to H; that will be the Center, from which a circle described thro' A, by cutting the parabola in three places, will give the three roots of the Equation, viz. NO and no true ones, the first whereof will express the quantity sought NQ (n. 1.) the latter the line NV, being the subtense of the third part of the compl. of the arch; and MO will express the false root, which is equal to the former taken together: All the same as in Cartes p. 91. but here some∣what plainer and easier.

PROBLEM V.

HAving three sides given of a quadilateral Figure to be in∣scribed in a circle, AB, BC, CD, (Fig. 52. n. 1.) to find the fourth side, which shall be the Diameter of the Cir∣cle.

SOLƲTION.

If we consider the business as already done, and make AB=a, BC=b, CD=c, and AD=y; we shall have first in the right-angled ▵ ABD, □BD=yy−aa, and (since in the obtuse angled ▵ BCD the □ BD is = □□BC+CD+2▭ of BC into CE) if those two □□BC+CD (i. e. bb+cc) be substracted from □BD (yy−aa) you'l have 2. yy−aa−bb−cc= to the two said rectangles of BC into CE. But these two rectangles may also be otherwise ob∣tain'd, 3, if the segment CE be otherwise determined; which may be done by help of the similar ▵ ▵ ABD and CED (for the angles at B and E are right ones, and ECD and BAD equal, because each with the same third BCD makes two right

ones; the one ECD by reason of its contiguity, the other at A by the 22. Eucl. Lib. 3.) viz. by inferring as DA to AB so DC to CE y−a−c−〈 math 〉〈 math 〉 for now multiplying CE=〈 math 〉〈 math 〉 by BC=b, the ▭ of BC into CE will be = 〈 math 〉〈 math 〉 and two such 〈 math 〉〈 math 〉.

Now therefore yy−aa−bb−cc=〈 math 〉〈 math 〉; and multipl. by y,

• y{powerof3}*−aa y=2abc; i. e. by Baker's and Cartes's forms.
• −bb y=2abc; i. e. by Baker's and Cartes's forms.
• −cc y=2abc; i. e. by Baker's and Cartes's forms.
• y{powerof3}*−aa y−2abc=o.
• −bb y−2abc=o.
• −cc y−2abc=o.

Therefore the Central Rule will be (supposing the same quantity e. g. a for unity and Lat. Rect.) 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉=DH, i. e. according to our form, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD, and

Geometrical Construction, which, without any circumlocuti∣on, from our form is founded on the foll. in Fig. 52. n. 2.

LM=a n. 3. A 1.=LM from n. 2.

MN and LP=b 1, 2=½PQ

PQ=bb 2, 2=½ST

LS and MO=c DH=PR

PR=bc MO and mo two false roots

ST=cc NO the true root; upon which having de∣scribed a semi-circle the quadrilateral will be easily made. Ac∣cording to Baker's form, for AD there would be n. 3. first

Ac=½LM, then cD=〈 math 〉〈 math 〉=VX, half the line VZ, which is compounded of LM, PQ and St; but DH is = PR as above.

PROBLEM VI.

HAving given to form a right-angled Triangle the least side BA (Fig. 53. n. 1.) and the difference of the seg∣ments of the base, to find the difference of the sides, and so form the Triangle. If we represent the business as already done, ha∣ving given AB and EC to find FC.

SOLƲTION.

Make AB=a, and EC=b, and FC=x; then will BC=a+x: Therefore the □AC=2aa+2ax+xx and the line AC 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉. Now therefore ACE i. e. 〈 math 〉〈 math 〉 multiplyed by b or √bb i. e. 〈 math 〉〈 math 〉 is = GCF [but GC is = 2a+x] i. e. 2ax+xx by Cons. 1. Prop. 47. Lib. 1. Mathes. E∣nucl. and squaring both sides

〈 math 〉〈 math 〉; and transposing all, 〈 math 〉〈 math 〉. −bb (p) (q) (r) (s)

Therefore (taking a for 1 and the Latus Rectum of the parabola) the Central Rule will be, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH i. e. according to our form and Reduction, 〈 math 〉〈 math 〉

Geometrical Construction. First from our form, the com∣pendiousness whereof will here appear, for it requires only one preparation n. 2. in which LM=a, MN and LO=b, OP=bb, which being premis'd, and the diameter Ay (be∣cause the quantity p is in the Equation) being drawn (n. 3.) make AI=½LM, and 1, 2 or ••D=½OP, and DH=LM. The rest therefore being also perform'd, which the quantity S occurring in the present Equation requires, according to the last precepts of our introduction, you'l have NO the va∣lue of x sought; whence (n. 4.) at the interval AB having describ'd a Circle, and made a right angle at B, if FC be made = to the found NO, you'l have the ▵ ABC required, and EC will be found at the beginning of the prescribed mag∣nitude.

Now if you were to find the center H by Baker's form with∣out our Reduction, 1. you must put of (n. 3.) ½AB from A to c. 2. For the quantity 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to a fourth, which would be = 2AB, viz. 2DM, and to be set from c to d. 3. The quantity 〈 math 〉〈 math 〉 (to obtain which you must substract OP (n. 2.) from the quadruple of LM, and divide the re∣mainder by 2) must be set off backwards from d to e, by thus determining the point D. 4. The quantity p, which here is precisely = LM, must be transferr'd from D to f on a perpen∣dicular erected from D. 5. For the quantity 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 (= 2AB) so 〈 math 〉〈 math 〉 (also = 〈 math 〉〈 math 〉) to a fourth quadruple of LM; and this must further be produc'd from f to the point g (which here the paper wil not permit) 6. For the quanti∣ty 〈 math 〉〈 math 〉 make as 1 to 〈 math 〉〈 math 〉 so 〈 math 〉〈 math 〉 to a fourth, (which would be = to the quadruple of LM, but taking away OP) which must be set backwards from g to h. 7. Lastly the quantity 〈 math 〉〈 math 〉 (=OP) set off from h backwards or towards the right hand to i will at length give the point H required.

Another Solution of the same Problem.

This Problem may be more easily solved, and will give a far more simple Equation, if you are to find not FD but AE. Make therefore (n. 1. Fig. 53.) = x, and the rest as above; AC will be = x+b, and its □xx+2bx+bb; therefore the □BC=xx+2bx+bb−aa; therefore the □ of the tan∣gent HC=xx+2bbx+bb−2aa. But the rectangle ACE will be bb+bx. Therefore by Prop. 47. Lib. 1. Math. E∣nucl.

〈 math 〉〈 math 〉; and turning all over to the right hand, 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉. Therefore by case 2 of affected quadraticks, 〈 math 〉〈 math 〉.

The Geometrical Construction may be performed according to the rules of quadraticks Fig. 54. n. 1. as will be evident to any attentive Reader. Having therefore described a circle at the interval BA, whether it be done from any arbitrary cen∣ter (see n. 4. Fig. 53.) or upon AE found in the pres. Fig. making an intersection at the said interval in B; and apply∣ing AE, and producing it until EC become equal to the gi∣ven quantity q, and at length having drawn BC you'l have the triangle right-angled at B, and also the difference of the sides FC. But to make it more short and elegant; having determined AE by a little circle (Fig 54 n. 3.) prolong it to the opposite part of the circumference in C, and draw AB and CB; for as the radius of the little circle is ½b, so EC is =b.

Now if you would construct the Equation above by Baker's rule (that its universality may also be confirm'd by an Exam∣ple in quadraticks) viz.

〈 math 〉〈 math 〉;

The centtal rule will be (taking a for 1 and the L. R.) 〈 math 〉〈 math 〉 = AD

[illustration]

LIV

LV

LVI

LVII

〈 math 〉〈 math 〉 = DH. i. e. by our Reduction, 〈 math 〉〈 math 〉 = AD 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

The Construction therefore will consist in these: LM (n. 2. ••g. 54)=a, MO=¼b, LP=½b, therefore PR=〈 math 〉〈 math 〉: ••N=PR viz. 〈 math 〉〈 math 〉, therefore Q=〈 math 〉〈 math 〉. In Fig. 53. n. 3. ••=¼b, A1=½LM, 1, 2, or 1D=PR; D1=LP MO, 1, 2, or 1H=PQ Having drawn a circle from 〈◊〉〈◊〉 thro' A you'l have the true root RO= to AE sought; ••d MO the false root.

NB. Hence it is evident that one Problem may have seve∣••l Solutions and Constructions, some more easy and simple, ••hers more compound and laborious; viz. according as the ••••known quantity is assumed more or less commodious to the ••urpose: which may not be amiss here to note for the sake 〈◊〉〈◊〉 Learners.

PROBLEM VII.

SUppose a right line BD (Fig. 55.) any how divided in A, to divide it again in C, so that the square of BA shall ••e to the square of AC as AC to CD.

SOLƲTION.

Since the first segments BA and AD are given, call the first 〈◊〉〈◊〉 and the other b, and call AC the quantity AC x; and CD ••ill be = b−x. Now therefore since we suppose as the □ of AB to the □ AC so AC to CD 〈 math 〉〈 math 〉

aab−aax will = x{powerof3}, and transferring the qu n••ities on ••he left hand to the right, x{powerof3}*aax−aab=o.

Wherefore the Central Rule (taking a for 1 and the L. R.) will be ½−〈 math 〉〈 math 〉=AD and 〈 math 〉〈 math 〉=DH. i. e. by our form 〈 math 〉〈 math 〉 i. e. o=AD, that D may fall on the vertex of the parabola; and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉=DH. The Construction therefore will be very simple, as is evident from the fifty fifth Figure.

PROBLEM VIII.

THere is given AB the capital line of a horn work (which we represent (tho' rudely) n. 1. Fig. 56) and the Gorge AD, also part of the line of defence EF, to find the face BE, the flank DE, the curtain (or the chord) DF, also the angle of the Bastion ABE, &c. and so the whole delineation of the horn work. It is evident if you have the flank DE or the curtain DF, the rest will be had also. Suppose therefore, the capital line AB and the gorge AD, and part of the line of defence EF to be of the magnitudes denoted by the Letters a, b, c, on the right hand.

SOLƲTION.

Make AB=a, AD=b, and EF=c, and DF=x; then will AF=x+b, and by reason of the similitude of the ▵▵BAF and EDE and ECB, as FA to AB so FD to DE 〈 math 〉〈 math 〉

But now □ □ DF+DE are = = □EF i. e. 〈 math 〉〈 math 〉; or giving the same denomination to all the quan∣tities on the left hand, 〈 math 〉〈 math 〉

and multiplying both sides by 〈 math 〉〈 math 〉 and translating all the quantities on the right hand by the contrary signs to the left, 〈 math 〉〈 math 〉

Wherefore (putting a for 1 and L. Rectum) the Central Rule will be, 〈 math 〉〈 math 〉 (because the quantity q is negative in the E∣quation, for cc is greater than aa+bb)=AD; and 〈 math 〉〈 math 〉 = DH. or according to our Redu∣ction, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Wherefore the Geometrical Construction requires no other preparatory determination by our form than of the quantities 〈 math 〉〈 math 〉 for AD, 〈 math 〉〈 math 〉 for DH at the center, and bbcc to determine the radius of the circle; which are exhibited by n. 2. Fig. 56. viz NP is = cc, RS=bcc, RV=bbcc, which are found by means of LM=a, LR=b, LN and MO=c, MQ=NP and MT=RS. Having therefore described a para∣bola, n. 3. and drawn its diameter, transfer AD=½NP upon it (because the quantity p is in the Equation) and also ½RS from D upon H perpendicularly, and on the right hand, (because DH=−〈 math 〉〈 math 〉;) and so you'l have the center H;

thro' which having drawn KAI so that AK shall be = to the quantity bbcc or S, i. e. RV, &c. a circle described at the interval HL will cut the parabola in M and N, and applying the magnitude NO it will be that of the Curtain sought; up∣on which, n. 4. having laid down the circumference of the horned work by help of the given lines AB and AD, you'l have the line EF, of the magnitude which was above supposed. Now if any one has a mind to do the same thing by Baker's way; by laying down first the interval AB =〈 math 〉〈 math 〉 and then making bc=〈 math 〉〈 math 〉, and lastly, putting cd for the quantity 〈 math 〉〈 math 〉; he will fall upon the same point D, and in like manner may express the other parts of the Central Rule by the intervals De, ef, fg, and setting back the last gh, he will fall upon the same center H: But this is done with a great deal more trou∣ble and labour to determine so many quantities, and also is in more danger of erring by cutting off so many parts separately, as experience will shew; and thus we have by a new argument shewn the advantage of our Reduction.

Another Solution of the same Problem.

Things remaining as before (only assuming the given lines AB, AD and EF, n. 1. Fig. 56. one half less, that the Scheme may take up less room) make BE=x, as the first or chief unknown quantity; then will BF=x+c, and its □xx+2cx +cc: And since as BE to BC=AD so BF to AF a fourth, which will be 〈 math 〉〈 math 〉 and its square = 〈 math 〉〈 math 〉. Where∣fore if this square be substracted from the square of BF, there will remain the square of BA, i. e.

〈 math 〉〈 math 〉; i. e. all being reduced to the same denomination, 〈 math 〉〈 math 〉;

or according to the forms of Cartes and Baker, 〈 math 〉〈 math 〉

Therefore the Central Rule (putting again a for 1 and the L. R.) will be 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH; or by our Reduction, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD; and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉=DH.

Geometrical Construction. Having therefore described a para∣bola (Fig. 56. n. 5.) and drawn the diameter Ay, make A1=a and 1, 2=〈 math 〉〈 math 〉, so you'l have the point D; make moreover D{powerof3} or 2, 3=c, and backwards 3, 4=〈 math 〉〈 math 〉 (we here omit to express the Geometrical determination of these quantities 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 as being very easy) and you'l have the point H, &c. and there will come out the quantity sought NO; which since it is equal to half BE n. 4. the business will be done; which Baker's form will also give, exactly the same, but after a more tedious process.

PROBLEM IX.

IN any Triangle ABC (the scheme whereof see n. 1. Fig. 57.) suppose given the Perpendicular AD, and the differences of the least side from the two others EC and FC to find all the three

sides. i. e. Chiefly the least side AB which being found, the others will be so also.

SOLƲTION.

Make AD=a, EC=b, and FC=c, AB=x; then will BC=x+b and its square be xx+2bx+bb, and AC=x+c and its square be xx+2cx+cc; and BD will be 〈 math 〉〈 math 〉, and DC 〈 math 〉〈 math 〉. But the □ BC may also be obtain'd otherwise, and the Equation also, if □□BD+DC+2▭ of BD by DC be added into one sum according to Prop. 4. Lib. 2. Eucl. viz.

〈 math 〉〈 math 〉, multiplyed by BD √xx−aa, gives the rectangle of the segments 〈 math 〉〈 math 〉 and this doubled i. e. multiplyed by √4, gives the quanti∣ty which is contain'd under the radical sign in the Equa∣tion]

Therefore turning all over on the left hand which are be∣fore the sign √ to the right hand, prefixing to them the contrary signs, you'l have 〈 math 〉〈 math 〉 and taking away the Vinculum on the left hand, and squaring on the right 〈 math 〉〈 math 〉

and adding and substracting on both sides, as much as can be, 〈 math 〉〈 math 〉 and transferring all to the left, 〈 math 〉〈 math 〉 and dividing all by 3, 〈 math 〉〈 math 〉

Note, I sought this Equation also after two other ways; 1. By a comparison of the □ AC with the two squares AB+BC, after 2 □ □ CBD thence substracted, according to Prop. 13. Lib. 2. Eucl which is the 46. Lib. 1. Math. Enucl. and I form'd the same with the present. 2. By putting at the beginning y for x+b and z for x+c, and going on after the former me∣thods, 'till you have this Equation, 〈 math 〉〈 math 〉 in which▪ when afterwards I substituted the values answering the quantities yy and zz, &c. This same last Equation came out a little easier, but (NB) with all the contrary signs.

Now to form the Central Rule, and thence make the Geo∣metrical Construction, we must determine first each of the quantities p, q, r and S, that we may know whether they are negative or positive; and you'l find (n. 2. Fig. 57.) p=G2 positive, q=H{powerof4} negative, and K{powerof4}=S also negative; and that by help of the quantities LM=b or 1, MN and

LO=a, OP=aa, MQ and LR=c, RS=cc, and also LT=cc, TV and LX=c{powerof3}, XY=c{powerof4}. Wherefore the form of the last Equation will be like this, x{powerof4}+px{powerof3}−qxx−rx−S=o, and so the Central Rule (taking here b for 1 and the L. R.) 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH.

Wherefore, having now described a parabola (as may be seen n. 4.) having found the diameter Ay transfer upon it first Ab=½LM (n. 2) and then bc=DP (n. 3.) i. e. 〈 math 〉〈 math 〉; and thirdly cD=〈 math 〉〈 math 〉 i. e. ½H{powerof4} (n. 2.) moreover from D to e put off MB (n. 3.=〈 math 〉〈 math 〉, and from e to f put off DR=〈 math 〉〈 math 〉, and from f to g put off CF=〈 math 〉〈 math 〉; and from g backwards to h put off half the quantity r, or I5 (n. 2.) and having done the rest as usual, you'l have NO, the side required of the Trian∣gle to be described; the description whereof will be now easy (n. 5.) having all the three sides known. This may serve for an Examen, if having described a semi-circle AGB upon AB=NO, you apply the given line AD, and from B thro' D draw indefinitely BDC: Then at the interval AB having described the Arches AE and BF, add the given line EC to BE, for thus having joined the points A and C, FC ought to be equal to FC before given.

SCHOLIƲM.

WE have here omitted our Reduction, because it would be too tedious, and would express the quantities AD and DH (especially the latter) in terms too prolix. For AD would be = 〈 math 〉〈 math 〉 (viz. because 〈 math 〉〈 math 〉 is

found = 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 taken in it self = 〈 math 〉〈 math 〉; but here [where by vertue of the Central Rule it is taken positively, when it is in it self negative] un∣der contrary signs it is = 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 or 〈 math 〉〈 math 〉 yet more contractedly (because b is unity) AD = 〈 math 〉〈 math 〉; which parts may be expressed without any great difficulty on the Diameter Ay, by its portions A1, 1, 2; 2, 3; 3D: But the other quantity DH, or the definition of the Center H, would also have some tedi∣ousness, as because 〈 math 〉〈 math 〉

If you take away out of the quantities 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉 (since this latter is to be substracted, and so left, as it is, under the sign −; but the other, also negative in it self, but here posi∣tively expressed in the Central Rule, must have all the con∣trary signs) I say, if you take out of these quantities those which destroy one another, and add the rest with the two for∣mer quantities, they will be 〈 math 〉〈 math 〉 = = DH. or a little more contracted (because b is 1) 〈 math 〉〈 math 〉 = DH.

But now if any one has a mind to illustrate this by a numeral Example and try the truth, &c. of the quantities found; they may make e. g. a=12, b=1, and c=2; and they

will easily find that in the last Equation the quantity p will be 4, and q−190, r−388, S−195: Secondly in the Central Rule of Baker they'l find 〈 math 〉〈 math 〉=½, 〈 math 〉〈 math 〉=2, and 〈 math 〉〈 math 〉= 95, and so the whole line AD=97 ½; and further 〈 math 〉〈 math 〉=1, 〈 math 〉〈 math 〉=4, 〈 math 〉〈 math 〉=190, and 〈 math 〉〈 math 〉, and so the whole line DH=195−194 i. e.=1. Thirdly, likewise in our Reduction (if we proceed by each part corresponding to Baker's) 〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉. The sum for AD 97 ½; but further 〈 math 〉〈 math 〉; and 〈 math 〉〈 math 〉 to be sub∣stracted; and so the sum for DH=195−194=1. Which same quantities will fourthly come out, if the quanti∣ties AD and DH contracted, as they are expressed in letters a∣bove, be resolved into numbers.

PROBLEM X.

YOU are to build a Fort on the given Polygons EAF (see Fig. 58. n. 1.) whose capital line AB shall equal the ag∣gregate of the gorge and flank, and the squares of these added together shall be equal to the square of the given line GH, and the solid made by the multiplication of the square of the flank by the gorge, shall be equal to the cube of the given line IK.

[illustration]

LVIII

LIX

LX

SOLƲTION.

Make the Gorge AC=z, whose square zz substracted from bb the square of the given line b, will leave the square of the flank DC=bb−zz. Now this square being mul∣tiplyed by the gorge AC or z will give bbz−z{powerof3}=g{powerof3}, the cube of the given line IK; and adding to both sides z{powerof3}, and substracting bbz, o=z{powerof3}*−bbz+g{powerof3}.

Therefore if we take g or IK for 1 and L{powerof3}, g{powerof3} will be the line g, and

The Central Rule: 〈 math 〉〈 math 〉 = AD. and 〈 math 〉〈 math 〉=DH. i. e. according to our Reduction, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 DH.

Geometrical Construction. Having described a Parabola (n. 2. Fig. 58.) make on its ax A1=½IK and 1, 2, s. viz. 1D=½MN (from n. 1.) and DH=½IK. Then having de∣scribed a circle from H, and found the true root NO upon the given angle EAF (n 3.) make AC=NO, and having e∣rected the perpendicular CD divide it by AD=GH (n. 1.) and make AB=AC+CD; and the Fort will be drawn.

PROBLEM XI.

IN a right-angled Triangle ABC (which we denote by n. 1. Fig. 59) having given the greater side AC, and made the less side AB= to the segment CE, which shall cut off from the base BC a perpendicular let fall from the right angle A; to find these lines AB or CE, and consequently the whole trian∣gle.

SOLƲTION.

Make AC=a, CE or AB=x. Therefore, 1. you'l have aa−xx=□AE. And because the ▵ ▵ BEA and CAE are similar, you'l have as AC to CE so BA to AE 〈 math 〉〈 math 〉

And so, 2. □AE=〈 math 〉〈 math 〉. Therefore 〈 math 〉〈 math 〉=aa−xx; and multipl. by aa, x{powerof4}=a{powerof4}−aaxx; or, according to the form of Cartes and Baker, transposing all to the left, x{powerof4}*+aaxx*−a{powerof4}=o i. e. x{powerof4}*+qxx*−S=o.

Therefore (taking a for 1 and L. R.) the Central Rule will be 〈 math 〉〈 math 〉 i. e. o=AD and 〈 math 〉〈 math 〉 i. e. o=DH; that H may fall on the vertex A.

Geometrical Construction. Since a is assumed for unity and L, the quantity S also and Latus Rectum i. e. AI and AK and consequently the mean proportional AL and the radius HL will be = = to the given side AC, and conse∣quently at that interval having described a circle, thro' the Pa∣rabola rightly delineated, you'l have NO the value of the quantity x, i. e. of the lesser side AB. Having drawn there∣fore NA, which is = to AC by Construction, if you draw to it the perpendicular AB cutting NO produc'd to B, you'l have the Triangle sought ABC, and AB (which will be a sign of a true Solution) will be found = NO or CE.

Another Construction. Since in the Equation above found there is neither x{powerof3} nor x, it may be look'd upon as a quadra∣tick, and constructed after the same way, as several other like it among the Examples n. 4. viz. because 〈 math 〉〈 math 〉; according to case 2 of affected quadraticks,

〈 math 〉〈 math 〉, and 〈 math 〉〈 math 〉 i.e. 〈 math 〉〈 math 〉 Wherefore (n. 3. Fig. 59.) if AC be made = a and CD〈 math 〉〈 math 〉a, the mean proportional CG will be = 〈 math 〉〈 math 〉, and taking hence GF=½a, there will remain FC = 〈 math 〉〈 math 〉: And now between this or CH equal to it, and unity AC, having found another mean proportional CE it will be the value of the quantity x sought=NO (n. 2.)

PROBLEM XII.

IN a right-angled Triangle ABC (Fig. 60. n. 1.) there is given the Perpendicular BA, a segment of the Hypothenu∣sa BD, and a segment of the Base EC, from C to the perpendi∣cular DE let fall from the end of the segment BD; to find AE DC, and consequently the Base AC and the Hypothenusa BC, and so the whole Triangle.

SOLƲTION.

Make AB=a, BD=b, and EC=c, and DC=x; which being given, the rest cannot be wanting: Therefore xx−cc=□DE. But the same □ DE may be had, if you infer as BC to BA so DC to DE 〈 math 〉〈 math 〉

And then square the quantity DE, the square will be 〈 math 〉〈 math 〉; and multiplying both sides by 〈 math 〉〈 math 〉 and substracting also 〈 math 〉〈 math 〉 〈1 page missing〉〈1 page missing〉

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Pag. 89.

LXI

so PV= to it, that PX may be 〈 math 〉〈 math 〉; and lastly Py=½q, that PZ may be 〈 math 〉〈 math 〉. These being thus prepar'd, if (n. 4) Ab be made = ½AB (n. 1.) bc=RT, and cd backwards = ½PQ we shall light on the point D; and, if we make De=½BD, ef=PX (n. 5. which interval was too big to be represented in the Paper) and from f you put backwards fg=PZ, and on from g beyond to the right hand gh=2RS; we shall light on the point H, &c. Which of these two methods is the shortest and fittest for practice, any one, never so little experienc'd, may here see; and first Learners may take notice if they would construct by Baker's form, in the Diagrams n. 3. n. 5. and the like, they must take care to make the angles PLN, SPT pretty large; which we have here represented the less to save charges in cutting on Copper.

PROBLEM XIII.

HAving given the Diameter of a Circle CD (n. 1. Fig. 61.) and the line BG, which falls on it perpendicularly, (which we have here only rudely delineated) to find the point A. from which a right line AC being drawn shall so cut the line BG in F, that AF, FG, GD shall be three continual Proportionals.

SOLƲTION.

If CF be found, the point A will be also had, and the se∣ction of the line BG will be made. Make therefore CF=x, and (because the perpendicular BG is given, there will also be given the segments of the diameter CG and GD) make GD=b, and CG=c; then will BG=√bc▪ and CD=b+c. Since therefore the ▵ ▵ CAD and CGF are right-angled, and have the angle at C common, they will be simi∣lar.

Therefore, as CF to CG so CD to CA 〈 math 〉〈 math 〉

Therefore AF = 〈 math 〉〈 math 〉 and FG 〈 math 〉〈 math 〉.

But by the Hypothesis, as AF to FG so FG to GD 〈 math 〉〈 math 〉.

Therefore the rectangle of the extremes will be equal to that of the means, 〈 math 〉〈 math 〉; and multiplying by 〈 math 〉〈 math 〉; and transposing all to the left, 〈 math 〉〈 math 〉; i. e. by the Carte∣sian form,

〈 math 〉〈 math 〉.

Therefore (taking b for 1 and L.) the Central Rule will be, 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 = DH. or according to our Reduction, 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = AD and 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉 = DH.

Geometrical Construction. Having described upon the gi∣ven line CD (n 2.) a semi-circle, and apply'd in it the gi∣ven perpendicular BG, as the figure shews, you'l have the seg∣ments of the Diameter GD=b, and to the quantity p in Ba∣ker's form, and CG=c, which (n. 3. where LM=b, LO and MN=c) will give OP=cc and to the quantity q. Wherefore having describ'd a Parabola (n. 4.) and the line VZ=2 ½b, having cut off the fourth part of XZ, and the eighth of YZ (whereof the one will be = 〈 math 〉〈 math 〉, viz. 〈 math 〉〈 math 〉, and

the other to 〈 math 〉〈 math 〉) if you transfer A1=XZ upon the diame∣ter of the Parabola Ay, and moreover 1, 2 or 1 D= to half OP (n. 3.) and transversly D{powerof3}=YZ and backwards 3, 4=¼ OP, as also 4, 5=½CG (n. 2.) you'l have the center H, and having describ'd a Circle at the interval HA, the root NO must be transferr'd from (n. 2.) C to F, and continued to A the point sought. In Baker's Form (because the quantity p is = b or 1) 〈 math 〉〈 math 〉 is = 〈 math 〉〈 math 〉 and 〈 math 〉〈 math 〉=〈 math 〉〈 math 〉, and the quantity q or cc=OP, (n. 3.) make therefore in the Diameter of the Para∣bola Ab=½GD, and bc=⅛GD, (n. 2.) and lastly cd=½OP (n. 3.) and you'l have the point D the same as before. Make moreover De=¼GD and cf=〈 math 〉〈 math 〉CG, and then back∣wards fg=¼OP, and lastly gh=½GD, and you'l have the same center H, and the coincidence of the parts in both forms will be pleasant to observe; which otherwise seldom happens.

Other Solutions of the same Problem.

Carolus Renaldinus, from whose Treatise de Resol. & Com∣pos. Math. Lib. 2. we have the present Problem, proceeds to solve it in another way, changing it plainly into another Pro∣blem: viz. he observes, 1. That the angles FAD (see n. 1. of our 61. Fig.) and FGD, since both are right ones on the same common base FD, are in circle. Hence he infers, 2. (by vertue of the Coroll. of the 26. Prop. 3. Eucl.) that the □ □ DCG and ACF are equal, and consequently CD, CA, FC and CG are four continued proportionals. Then he observes, 3. That GD is the excess of the first of these proportionals above the fourth CG, and AF is the excess of the second AC above the third CF; and so, since 4. the rectangle of AF and GD is = to the square of the mean proportional FG (for AF, FG, GD, are supposed to be continual proportionals) and this □ FG is the excess, by which the square of the third CF ex∣ceeds the square of the fourth CG; now the present Problem will be 5. reduc'd to this other: Having two right lines (CD and CG) given to find two such mean proportionals (AC and FC) that the ▭ of the excess of the first above the fourth (viz. of FA into GD) shall be equal to the excess,

by which the square of the third (FC) exceeds the square of the fourth (CG) viz. by the square FG.

Wherefore instead of the former he solves this latter Pro∣blem, putting for CG, b, for GD, c, so that the first of the given lines CD shall be = b+c, and the other GD=b; calling the first mean proportional AC, x; and thence deno∣minating the latter 〈 math 〉〈 math 〉 (viz. multiplying the fourth by the first, and dividing the product by the second) and more∣over he finds the excess of the first (b+c) above the fourth (b) to be c, and the excess of the second (x) above the third 〈 math 〉〈 math 〉 to be 〈 math 〉〈 math 〉 i. e. 〈 math 〉〈 math 〉; so that the □ □ of these two excesses is 〈 math 〉〈 math 〉, and because the □ of the third FC is = 〈 math 〉〈 math 〉, having substra∣cted bb=□GC, there is given the □FG = 〈 math 〉〈 math 〉 = ▭ of the excesses we just now found. So that now you'l have the Equation 〈 math 〉〈 math 〉 &c.

We also endeavour'd to find another Solution, by finding an Equation from the line BD (n. 1. Fig. 61.) as which might be twice obtain'd by means of the two right-angled Triangles FAD and FGD, since it is the hypothenusa of both. But here, besides the former denominations of our Solution we must first give a denomination to the line AD, by making as CF to FG so CD to AD 〈 math 〉〈 math 〉, &c.

But whosoever shall prosecute this Solution of ours, or that of Renaldinus to the end, will find much more labour and dif∣ficulty in either, than in the first we have given.