B E To make a Parallelogram, under any given Angle, equal to a Rectangle, and having the fame Base. ABCD is the given Rectangle. F Make the Angle DAE equal to the given An A gle, cuting BC in E, by Produce BC; and draw DF parallel to AE.-5 The Parallelogram AEFD is equal to the given Rectangle ABCD. Or, it may be thus demonftrated. P. 4. I 8. I. DEM. AB is equal DC (15. 1.) and BE is equal CF. Ax. 7. COR. 1. Hence it is evident, that two Spaces may have the fame Area, yet differ greatly in compass or circuit. For, the Sides AE and DF, of the Parallelogram AEFD, are greater than A B and DC of the Rectangle ABCD. 12. 1. But, EF is equal BC (Ax. 3.) for each is equal AD.—15. 1. Therefore, the Circuit of the Par. AEFD is greater than that of ABCD; and they contain equal Areas. 18. I. COR. 2. From hence is deduced the general Rule for meafuring all Parallelograms; which is, to multiply the Bafe; i. e. any Side (as AD) by its perpendicular height, (CD) For it gives the Area of the Rectangle ABCD, and confequently of the Par. A EFD, which is equal to the Rectangle. PRO To make a Rectangle equal to a given Trapezium; ABC D. Draw either Diagonal, as AC; to which, I draw the Perpendiculars BE and DF, from the oppofite Angles. G B K Pr. 7. A Bifect the Perpendiculars, BE and FD, in The Rectangle IKLM is equal to the Trapezium ABCD. DEM. For, the Rect. AIKC is equal to the Triangle ABC, Otherwife, Having drawn a Diagonal and Perpendicu in the Points H and L, I and K. I i Then, the Rect. HIKL is equal to the Trap. ABCD. DEM. For, the Rect. GHIC is equal to the Triangle ABC. and the Rect, GLKC is equal to the Triangle ACD. 20. N B. by H N. B. By this Problem and the laft may be performed the 45th of Euclid, without the affiftance of the next, which he makes ufe of, and much readier; making the Parallelogram under a given Angle, as hikl, which is equal to HIKL by the laft. COR. Hence we learn, to find the Area of any Quadrilate ral whatever; the Rule for which is, to multiply the Diagonal (AC) by half the Sum of the two Perpendiculars (Fig. 1.) or, the Sum of the two Perpendiculars (BE and FD) by half the Diagonal (GC, Fig. 2.) or, if we multiply the whole Diagonal by the Sum of the two Perpendiculars, they will give the Area of the Rectangle MNIK; half the Sum, of which, is the Area of the Tiapezium ABCD. It may appear ftrange, to thofe who have not confidered it, that the Area of any Figure fhould be obtained without measuring its Sides; which are of no ufe in this operation. But, having well digefted what has been advanced, they will find, that the whole business of Menfuration is to find a Rectangle equal to any Figure; for (as I fhall make appear hereafter) the multiplication of any two Numbers, applied to measure, denotes a Rectangle of fuch Dimenfions. Now, fince it feldom happens, that a Trapezium has either Right Angles or parallel Sides, it is plain, that they cannot be of any ufe towards obtaining its Area; whereas, the Diagonals and Perpendiculars are at Right Angles with each other. A Diagonal divides any Quadrilateral into two Triangles; and every Triangle is equal to half a Parallelogram having the fame or an equal Bafe, and the fame Altitude. (Prop. 17. 1.) Hence, it is eafy to account for the Rules given for measuring a Trapezium, as two Triangles having a common Bate; which is a Diagonal of the Trapezium. For, every Trapezium is equal to half a Parallelogram which, circumfcribes it, having two Sides parallel to either Diagonal; and all Parallelograms having the fame Bafe and Altitude are equal (18. 1.) confequently, they are equal to a Rectangle of thole Dimenfions. PRO PROBLEM XXIII. 44. I. Euclid. To make a Parallelogram, equal to a Triangle, having an Angle equal to a given one, and a Side equal to a given Right Line. F, of the Parallelogram DEFC, draw HI, cuting DE, produced, in I. EI, is the other Side of the Parallelogram, fought; which may be compleated by Prob. 19. Draw IL parallel to AH, and HL parallel to DI meeting at L; produce EF to G, and CF to K. DEM. The Par. KLGF (having its Angles, at K and G, equal to the given Angle X, and a Side, FG, equal to a given Line, Z) is equal to the Par. DEFC.-P. 19. 1. which, is equal to the Triangle ABC. SCHOL. This is, properly, geometrical Divifion. H 2 I. Pr. 10. G D Let the Rectangle ABCD be any Quantity given, to be di vided; whose Area is 24; the Side AB being 6, and A D 4 equal Parts; it is required to be divided by 3. 24 6 H Produce any Side, as AB; and make BE equal 3 (half AB) alfo produce AD indefinite. Through E and the Angle C, draw EC, and produce it, cuting AD produced, in F; DF will be equal 8 of the divifions, or twice AD, which is the Quotient fought. Compleat the Rect. AEHF; produce BC to I, and DC to G; the Rect, CHAC... For, CI is equal DF, i. e. equal 8; and, CG is equal BE equal 3; 8 multiplied by 3, is equal 24, the Area of CH; which is equal to the Dividend, or the Rectangle ABCD. If the Divifor had been 8, equal D F, the Quotient would have been 3, equal B E. If a fractional Number was given for the Divifor, the Quotient would, moft probably, be fractional likewife; for the Rectangle under the Divifor and the Quotient will always be equal to the given Rectangle, which is the Dividend. As in Divifion, the Quotient multiplied by the Divifor is (when there is no remainder) equal to the Dividend; which proves the work to be true. N. B. It is equal, which way the given Rectangle is fituated, in this operation; or which Side is produced for the Divifor; the Quotient, will always be the fame. To make a Triangle equal to any given Right-lined Draw the Diagonals AC and CE; and produce EA, indefinite. Poft. 3. D. From the Angle B, and parallel to AC (the adjacent Diagonal) draw BF; cuting EA, |