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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.
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.