we have x2- AZX = - a2; and therefore x = az × x + √2x-4; which, by fub ftituting the value of %, becomes — x + √ & + √ √ 5 − / PROBLEM 2 XXIV. To draw a right-line DF from one angle D of a given rhombus ABCD, fo that the part thereof FG intercepted by one of the fides including the oppofite angle and the other fide produced, may be of a given length. A Let DE be perpendicular to AB; and let AB (AD) a, AE b, FG c, and AF=x: then DF2 (= AF2 +AD2—2ÀE×AF) =xx + aa — 2bx; G F and, by fimilar triangles, xx+ aa2bx(DF3);xx(AF3) :: cc (FG1) : x—a2 (BF2); and confequently xx+aa. 2bx × xx — 2ax+aa = ccxx. Make ma = b, and na= c; fo fhall our equation become xx+aa—2max × xx-2ax+aa = n2a2x2: which, die m and n. From whence the value of x will be alfo known i x known; for + being, we have, by reduc tion, x2 - AZX x aa; and therefore x = -4. 2 PROBLEM XXV. The diagonals AC, BD, and all the angles, DAB, ABC, BCD, and CDA, of a trapezium ABCD, being given; to determine the fides. Let PQRS be another trapezium fimilar to ABCD, whofe fide PQ is unity; and let QP and RS be produced till they meet in T: alfo let PR and QS be drawn, and make Rʊ and Sw perpendicular to TQ Let the (natural) fine of the given angle STP, to the radius 1, be put m; that of TSP, or PSR, = n; that of TRQp; the co-fine of SPQ=r; that of RQv=s; AC = a; BD = b; and PT=x. Then (by plane trigonometry) nm; ; x : PS = and I: ; rmx mx n mx n (PS) ::r:Pw= : whence, by Euc. 13. 2, QS2 (=QP2 η + PS2 2PQ ×Pw) = 1 + nn 2rmx n Again (by trigonometry) p: m : : 1 + × (TQ ) : m+mx m + mx ; and I :s: ms + msx (QR): Qv = And therefore PR2 (= PQ + QR3- because of the fimilar figures ABCD, PQRS, it will be, AC2: BD2 : : PR2 : QS2, that is, a2: b2 : : whence SQ will also be known: and then, again, by reafon of the fimilar figures, it will be as QS: QP (unity) :: BD: AB; which therefore is known, likewife: from whence the reft of the fides BC, CD, and DA will all be found by plane trigonometry. The laft problem is indeterminate in that particular cafe, where the trapezium may be infcribed in a circle, or where the fum of the two oppofite angles is equal to two right ones; for, then, there can but one diagonal be given, in the question, because the value of the other depends intirely upon that. Suppofing BOD to be a quadrant of a given circle; to. find the femi-diameter CE, or CL, of the circle CEGL, infcribed therein; and likewife the femi-diameter of the little circle nFmP, touching both the other circles DLB, LmE, and also the right-line OB, Let BQ, Pn, and CE, be perpendicular to BO'; join C, n and O, n; and draw OC meeting BQ in Q, and nr parallel to BO, meeting CE in r: put OB (= BQ) = 1, OQ(=V2 by Euc. 47. 1.) b, and Pn (nm)=x. Then, by reafon of the fimilar triangles OBQ, OEC, it will be, OQ: BQ:: OC; CE; whence, by compofition, OQ + BQ : BQ :: OL (OC+ CE): CE; that is, b + 1 : 1 :: 1; CE ➡ b- I b - 1 (= b+ 1xb - I b2-I I; which let be denoted by a, then we shall have Cn + Cr2a, and Cn Cr = 2x ; and therefore nr ; -2x, thence will PE + OE there arifes I (= √ Cn+Crx CnCr) = 2Vax. Moreover On + Pn being = 1, and On · Pn= I OP = √ 1 — 2x; which also being (2/ax+a) we therefore have I 2x=2√ax +a; whereof both fides being squared, 2x=4ax + 4ay ax + a2, or I 4ax= 4a√ax; which, because I aa is 2a, will be a →→ I + 2a 1+2a x x = 2ay ax: this, fquared, gives a2 — × 2ax+1 + 2a)2 ×x2 = 4a3x; whence 1 + 2a2 ×x2 -- I + 2a × 2ax — 4a3× — — aa; which, by writing 2x י. 26 - I 76—9±√/863 +296* — 1126+78; which, by writ 26-I2 ing 2 for b, becomes 7 2-9±√136—96 √/2 other being manifeftly too large: but this value will be 49 7 confequently BH = 4BQ; from whence we have the following construction. In the tangent BQ, take BH = & BO; draw HO, cutting the circumference BDL in F, and make the angle OFPOHB, and draw Pn parallel to BQ, meeting OH in n, the center of the leffer circle required. SCHOLI U M. In the preceding solution it was required, not only to extract the fquare root of the radical quantities 136 96 |