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

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.

[illustration]

XXXVIII

XXXIX

XL

XL

XLI

XLI

XLII

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.