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

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.