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

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,

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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.