Part II. viz. the greater and leffer Roots, the Sum of both being always to the Coefficient b. For DC (ba) . . D B (p) :: D B (p) . . DA=4, per 13. 6. Eucl. El. Ergo, haaa = PP, as above. Let AB S Within a given Square A B CD, to infcribe 2 long Square EFGH, whofe Sides E F, of GH, fhall be in Proportion to EH, or F G as R to S, and parallel to the Diameters A C BD of the Square. Solution. AD 12 4 = √2aa (=EB9: +BFq) = EF a√2 b. = = b — a xv 2 (=FC q: +CG 9|2) FG. By 47. 1. Encl. El. 3 5 By the Prob. confeq. Equat. 8 6R 7 8+ Ra Rb Ra+ Sa. 9 R.b 9÷R+S =a=4 R+S The PROBLEM II. The Perimeter, (viz. the Sum of the 3 Sides, A B, AC, BC) of any Ld A A B C, and its Area being given; thence to find each Side. being drawn, fhall be in a given Proportion of m (the greater) to n (the leffer.) Inquifition. Suppose it done, and C the Center of the Circle: Then, because the Points A, B are given, the Line A B is given; and because of m to n, its Point E, this being one of the Points D; therefore A E and B E are given. Put A Eb, then m. n:: b n 2bBE. Supposing then from D on AB produc'd, if Need be, a 1 DP, and taking (on oppofite Sides of Á B) AD Ad, and BD A =Bd; the As on the oppofite Sides of A B will be fimilar and equal; and therefore DPd one ftreight Line, bifected in P, and at Ls to AB; In which therefore is the Center C, and the Diameter ECF, which muft lie towards the Side B, not A, otherwife BF would be AF, that is here BDA D. Now fuppofe EC=r, and Ep =x; then Ap = b + x, and Bp="box; then (by the Property of a Circle) n m 2rxxx = PDq: And (by the 47. 1. Eucl. El.) 27X- ·xx+bb +2 2b x + xx = ADq: = 2rx + 2bx+ bb: Now (by Hyp.) m..n :: AD.. BD, confequently m2 .. n2:: ADq.. BDq: that is :: 2rx + 2bx + bb : n2 b2 = 2n2rx + 2n bx+n' b', and m'rnmbnr+ n2b. and by Tranfpofition m'rn'rmnb + nb, and by dividing each Part of the laft Step by m+n you'll have − n x r = nb. And lastly, by dividing each Part by m-n, you'll have nb The As A B D and DB C, are fimilar; Wherefore a ..DB:: DB..c 2. Therefore acDBq: 3. By 47. 1. Eucl. El.. bb 4. By Tranfp. bb — aa + ac aaac 5. By Comp.: bb + & cc = aa + ac + & cc 6. By Evol. bb + 3/2 cc 2/2 = a + 1/2 c 7. By Tranfp. bb + 1⁄2 cc 1⁄2 - 1⁄2 c = a Canon. LEMMA to PROBLEM V. The Area of any Trapezium POQR, whofe opposite Sides O P and Q R are parallel, is had by multiplying the Sum of the Sides OP and Q R by half the common Height S P or TO. Suppofe P SOT=b, RQ= 1, POST=m, and let R S be=n; then T Q will be = l m n. R I fay, 1 + m × 4b Area of Trapezium O P R Q And by adding the three foregoing Areas into one Sum you'll have n+1-m―n + 2 m x 1⁄2 b = l + m × 1⁄2 b Area of the Trapezium P O QR. QED. The Sides of the) BC c given Trapezium,DA=f The Area of a Trapezium and its Sides, feverally being given, to make a Trapezium within the former, in fuch Manner that the Sides of the one may be every where feparated from the Sides of the other, by an equal parallei Diftance; and that the Space lying between both Trapezia may be equal to The Area of the Trapezium A B C D = s =t Trapezia, Then the Area of the infcrib'd Tra-' pezium E F G H, -} = The equal parallel Distance a = ? Data. Preparation. |
