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

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.