IMAGE EVALUATION TEST TARGET (MT.3) V. /^/ "'^ f/j V] V^ % l; c*y ^ > A 1/ ^^" "«^^ /; %■ % 1.0 1^128 125 150 "^ 2.2 I.I " 1^ IIIIIM 1.8 • !-25 II 1.4. Ill '•* < 6" — ► Photographic Sdences Corporation 33 WEST MAIN STRUT WEBSTER, N.Y. M580 (716) 872-4503 f/. CIHM/ICMH Microfiche Series. CIHM/ICMH Collection de microfiches. Canadian Institute for Historical Microreproductions / Institut Canadian de microreproductions historiques Technical and Bibliographic Notat/Notas tachniquaa at bibliographiquas Tha Instituta has attamptad to obtain tha bast original copy avaiiabia for filming. Faaturas of this copy which may ba bibliographically unique, which may alter any of the images in the reproduction, or which may significantly change the usual method of filming, are checked below. Coloured covers/ Couvarture de couleur I I Covers damaged/ Couverture endommagia Covers restored a.id/or laminated/ Couverture restauria et/ou pellicul^e Cover title missing/ Le titre de couverture manque Coloured maps/ Cartes gAographiquas en couleur Coloured ink (i.e. other than blue or black)/ Encra da couleur (i.e. autre que bleue ou noire) Coloured plates and/or illustrations/ Planches et/ou illustrations 9n couleur Bound with other material/ Raiii avac d'autres documents D D D Tight binding may causa shadows or distortion along interior margin/ La re liure serr^e peut causer de I'ombre ou de la distorsion He long de la marge int^rieure Blank leaves added during restoration may appear within the text. Whenever possible, these have been omitted from filming/ II se peut que certaines pages blanches ajoutias lors d'une restauration apparaissent dans le texte, mais, lorsqua cela «tait possible, ces pages n'ont pas iti filmAas. Additional comments:/ Commentaires supplAmantairas: L'Institut a microfilm^ la meiileur exemplaire qu'il lui a iti possible de se procurer. Les details de cet exemplaire qui sont peut-dtre uniques du point de vue bibliographique, qui peuvent modifier une image reproduite. ou qui peuvent exiger una modification dans la m^thoda normale de filmaga sont indiquis ci-dessous. n n n n D D D Coloured pages/ Pages de couleur Pages damaged/ Pages endommagies Pages restored and/or laminated/ Pages restaurias et/ou pellicuiies Pages discoloured, stained or foxed/ Pages ddcolories. tachatdes ou piquies Pages detached/ Pages ditach^es Showthrough/ Transparence Quality of print varies/ Quality in«gale de I'impression Includes supplementary material/ Comprend du material supplimentaire Only edition available/ Seuie Mition disponible Pages wholly or partially obscured by errata slips, tissues, etc.. have been retilmed to ensure the best possible image/ Les pages totalement ou partiellement obscurcies par un feuillet d'errata, une pelure, etc., ont M filmAes A nouveau da fa^on i obtenir la meilleure image possible. This item is filmed at tha reduction ratio checked below/ Ce document est film* au taux de reduction indiqu« ci-dessous. 10X_ 1*X 18X 22X y 12X 16X 20X 26X 30X UX 28X 32X The copy filmed here has been reproduced thanks to the generosity of: BIbliothdque natlonale du Quebec L'exemplaire film6 f ut reprodult grAce A ia g*n«rositA de: Bibliothdque nationale du Quebec The images appearing here are the best quslity possible considering the condition and legibility of the original copy and In keeping with the filming contract specifications. Original copies in printed paper covers are filmed beginning with the front cover and ending on the last page with a printed or illustrated impres- sion, or the back cover when appropriate. All other original copies are filmed beginning on the first psge wKh a printed or illustrate ' Impres- sion, and ending on the last page with a printed or Illustrated Impression. The last recorded frame on each microfiche shall contain the symbol —►(meaning "CON- TINUED"), or the symbol y (meaning "END"), whichever applies. Maps, plates, charts, etc., may be filmed at different reduction ratios. Those too large to be entirely included In one exposure are filmed beginning In the upper left hand corner, left to right and top to bottom, as many frames as required. The following diagrams illustrate the method: Les Images sulvantes ont 6t€ reproduites avec le plus grand soin, compte tenu de la condition et de le nettet« de l'exemplaire film*, et en conformity avec les conditions du contrat de fllmage. Les exemplaires origlneux dont le couverture en papier est ImprimAe sent filmfo en commen^ant par le premier plat et en termlnant solt par la derniire page qui comporte une empreinte d'impression ou d'illustration, solt par le second plat, salon le cas. Tous les autres exemplaires origlneux sont filmte en commen9ent par la premiere pege qui comporte une empreinte d'impression ou d'illustration et en termlnant par la dernlAre page qui comporte une telle empreinte. Un des symboles suivsnts apparaftra sur la derniire Image de cheque microfiche, selon le cas: le symbols — ► signifie "A SUIVRE", le symbols V signifie "FIN". Les cartes, plenches, tableeux, etc., peuvent Atre filmAs A des taux de reduction dIffArents. Lorsque le document est trop grand pour Atre reprodult en un seul cllchA, 11 est filmA A partir de i'sngle supArleur gauche, de gauche A drolte, et de haut en bes, en prenent le nombre d'images nAcessalre. Les diagrammes suivants iliustrent la mAthode. 1 2 3 1 2 3 4 5 6 ^^f %'> l-bfj) ALGEBRA APPLIED TO ^91^^^^=^ l^tMt |oIution8 of |JMm«l»[ical jjroMems, FOB BEGINNERS. BY REV. L. P. PAQUIn7 O.MI.I~, Ql^pRO«SSOn OF piYIL ^KOINtERiNO IN THK PoLLEOE OF PtTAWA. s^ "i'V">"<"r;"'v • • « • » « * « • ' ♦ » * • k. i ' ,* » I , • « • • • » » « « • %•• * m ■ • -• i « »■ >■ OTTAWA, CANADA: PRINTED BY JOSEPH LOVEDAY. 1S7A • * « ' * • 4 '^••*t •*••• • • * • • .• » • I > • t> « • • • ' • • • •• ^ im OOPREFACE.CO I <»» ■ The present little work is intended to have a twofold object — to furnish the young Student with useful exercises in applying Algebra to Geometry, and to put within his reach an easy and ready means of obtaining the Solution of the most Prac- j tical Problems of Elementary Geometry, especially of those respecting the construction, [measurement and division of plane figures. A good many of the Formulas presented in the following pages, may be used advantageously in ordinary cases of Land Surveying, as the laying out, the dividing of lands, and the calculating of their content. The Author having in mind not to present an extensive and expensive book on the subject, has omitted to insert the pro- cess of operations which brought him to the General Formulas contained therein. His doing so affords moreover the Student the advantage of verifying by himself the correctness of these Formulas, and therefore of exercising himself in Algebraic Investigations. No Geometrical Figures are drawn ; but the Pi*oblemo and their Solutions are given with sufficient clearness to be plainly understood by the Student, who may very easily construct the figure corresponding to the given Problems, according to the special cases. On account of the practical nature of this little book on the one hand, and its cheapness on the other, the Author is in hopes that it will be received and found convenient in Institu- tions where Mathemathical Sciences are taught. L.P.P. 60358 * £ ||lfl?Jn'» |p))Ii itil racfeal iptomfllri). INTEODUCTION. ft T. When the relative positions of points connected by straight lines are to be indicated, then the connecting lines are, for the sake of convenience, represented by two letters, AD, BO . . . ; bttt when the relations of the magnitude of lines are only under consideration, then these lines are more con- veniently represented by a single letter, or an algebraic sym- bol, a,h,c... Let a and b represent two given lines, then a -f b expresses their sum, a — b expresses their difference, a^ expresses a square constructed on the line a, (a -\-by expresses a square constructed on the sum of these two lines, expresses a rectangle whose a is the base, and b the height, or vice versa, expresses a triangle, whose a is the base, and b tlie height. ^-^ ^ expresses a trapezoid, whose p is the height; m and n the two bases. n. Any linear problem may be solved either numerically or gra- phically; and in both cases an algebraic expression of the ab ab ~2~ / —6— quantity songht furnishes the student with an easy and ready means of obtaining the solution, inasmuch as it indicates either the arithmetical operations, or the graphical construction to bo performed. When a problem is to be solved numerically, any expres- dcs, in vfhich the quantity sought is put in terms of known quantities, will suit; should it be solved graphically, then the expression must be reduced to one of the five following formulas : — (1) X = a -\- b — c Sum and difference of lines. he (2) X = —^ Fourth proportional. (3) ip = V a6 Mean proportional. (4) a? = a/ a^ -f. 6* Hypotenuse. ^) ic = V a^ — 6* Side of a right angle. III. The first formula is evident by itself. he When X = , then it is a fourth proportional to a, 6, c, a because if both members of the equation are multiplied by a^ we obtain ax — h c Hence, a: b : : c : x Hence, in general, any fraction containing two factors of the first degree in the numerator, and one in the denominator^ will expreoa & fourth proportional. "When X = V~a b, it is a mean proportional between a and h, because, if both members of the equation are raised to the second pow&r, -we obtain OS* = ab Hence, a: x : : x : b Hence, in general, any expression, being a radical of the second degree includinff two factors of the first deerree. will represent a mean proportional. K \ — 7- Whcn X = Vd' -j- b', an h3'potenuse is to be drawn on a and b put ut right angle, because, if both members of the equation are raised to the second power, wc obtain a.^ = (0- -f 6« The fifth formuhi is proved in the same wa}'. N.B. — The student is supposed to know how to draw a fourth proportional to three given lines, a mean proportional between two given lines, etc. lY. Let it be illustrated by a few examples, that any expression may easily be reduced to one of the five preceding formulas. ah c Ex. 1. Let X = r s then X = — ^~ X — by decomposing^ s ah Let iha fourth proportional expressed by be drawn and re- presented by ?/, then X = y X — = -^~(2ud formula). S S Ex.2. Let X = a}^ -f- h- c -{- dhm if -j- q^ Let an hypotenuse bo drawn on p find q, and represented by y, then X = ,-! = r ~y' 'V Y' ~ ly ^ U "^ "yV Let each of those last three terms bo reduced as in the Ey. 1, and the three lines found bo represented respectively b}- u, V, z, then X = u -\- V -\- z (1st formula). Ex. o. Lot X =: V 3 Or then X = ^3 a x a (3rd formula). -8— ^x. 4. Let X z= J~ then X = Va xi_« (3rd formula;. p is evident that in this last case, the line a 8hould1)e divided into five equal parts, and a mean proportional drawn betweer. the whole hae and three of the equal parts. Ex. 5. Let X = / a b c ^ Ex. 6. then X _ ^J~- X c by decomposing. making ~- = y (by formula 2nd), then X = Vy^ (3rd formula). Let X = a/ a^* -f 6=* -f- c^-\- d^ Let an hypotenuse y be drawn on a and &, and substituted in the expression, then X ~ \/y -f- c2 -f <^^ making f -\- c^ = z', in the same way, then X = v'I-"'-fir"(4th formula). 1 ■a 'S. — y- ALGEBRAIC SOLUTIONS OF PROBLEMS. m PROBLEM I. To find the side a; of a square x^ whose area shall be equiva- ient to twice tha^ of a given square a^. = 'v/'2rt X a G. S. PROBLEM IL To find the side x of a square x^, whose area shall be jquiva. lent to five times that of a given square a^ = '^Sa X a G. S. PROBLEM III. To find the side a; of a square x^ whose area shall be equiva- lent to or -fifth of that of a given square aK I a? a N. S. G. S. i PROBLEM IV. To find the side x of a square x\ whose area exceeds that of ^ given square a^ by the three-eighths of it. T =7- N. s. 11 a 8 X a G. S. —10— PROBLEAt V. To find the side x of a square x^^ whose area shall be equiva- lent to that of a given square a-' less by the two-sevenths of it. _ / 5 d^ ^ -■\I~T~ :n". s. — /^^ ^ ^ ~ -yj-rf- X a G. S. PROBLEM VI. To find the side x of a square a.^, whose area shall be to that of 'a given square if, in the ratio of m : n. rr V n I m a = ^7«~ ^' S. (See Intro. IT., Ex. 5.) PROBLEM VIL To find the side x of a square x^, equivalent to a given rect- or = Virr' N. and G. S. angle a b. PROBLEM VIJL To find the side x of a square x^, equivalent to a given tri- 1 a h X I a h X - N. S. ^Z G. S. PROBLEM IX. To find the the side x of a squiiro x\ equivalent to a given trapezoid ^J!!l.±Jl} -11— X T / m -^ n make -^ ^ ^ = y by drawing a line parallel to the bases, at eq^ual distance from each j then ^ PROBLEM X. To find tlie side ./■ of a .sqyaro j:% c(iiiivalont to a given regular polygon. Let c = the side of the polygon ; a = the number of sides ; and / = the apothem ; then Va c "IT = 7F K s. KB. — a being a given number, then il c is onlj- one literal factor, and we have the 3rd formula. PROBLEM XL To find the side .r of a square .r-^, equivalent to a given triangle -, reetant»-lc bJ. ii-nnezoid JL}-^ ~, rectangle bd, trapezoid -^^^"i-!^, and regular pentngon 2 5 c J ~2~ ^=J~-h.od~{-p^}-^ilL N. S. Transform the four expressions whicli are under the radical, into equivalent squares, /, f^^ y"i^ y""i^ ^^^the rroblems \U., Vlir., IX., X., then make y' + y"^ = --2 (gee Intro. IV., Ex. 6) then .1; = a/ F+ /'^ If "7^ij making ~-' + y"^ = -'- , tlien ;r = a/7^ + p'^ G. S. '■—12— PROBLEM XII. To find tho side x of a square i^, equivalent to any irregular polygon. Divide the polygon into triangles by drawing diafronalw ; let these triangular parts be calculated separately, and repre- sented respectively hy P, Q, R, S ; then = ^y^+F^+r'+y^' (See Prob. XT.) = A/^q^T^ (. s. PROBLEM XIII. To find the side x of a square i^, equivalent to a given circle. X = WW~ K s. = V'slfx n G. s. N.B.— If a very exact graphical construction is required, the- line R should be taken 3. 14 times, instead of 3 times. PROBLEM XIV. To find the side x of a square x^, which is a mean proportion- al between any tw j given polygons P and R, 4 x= ^ P X R transforming the two polygons P and R into equivalent squares, if, z^ then 4 X = Vl'^ y% X z^ = \/ y N. and G. S. ^4 PROBLEM XV. To construct a triangle -^ on a given base c, equivalent to a given square a^. Altitude X = ^JL __ 2 a X rt N. S. G. S. —13— I ^< PROBLEM XVI. To construct a rectangle c ji- on a given base c, equivalent to a given square a'^. til Altitude J- = -' c __a X a c G. S. PROBLEM XYIL To construct a trapezoid ^^!^Li^^ equivalent to a giv square a^, having given the two bases, m and w. en Altitude p =Y a' 2- (m + ^) Y a X a KS. y (See Problem IX.) Gr. S. PROBLEM XVIIL To construct a trapezoid P (^^^4- rC) equivalent to a given square a^, having given one of the bases m, and the altitude p. the other base n = h P KS. make ^ p m = / by Problem VIII., then a' make a^ n = f f \V by the 5th formula, then n =■ z X z PROBLEM XIX. G. S. To construct a regular polygon ;^, having gi^ Apothom r her ^ , equivalent to a given square a^, having given the side c, and the number of sides h. _2a2 :n-. s. —14^ _ 2 a xa h c G. S. r* ?T^^® factor 6 beinir a number, then h c is onlv ono literal factor, viz., the line c taken b timey. The radius of tlu^ cireumscribed circle is evidently B = ^^' + f :N". and G. S. PROBLEM XX. !Io construct a circle equivalent to a given square a-. Eadius ==: -^ ^ ]\j- g ^ ** ( 1 rob. Alii. > PROBLEM XXI. To transform a triangle ~ into a rectangle 6 .>• having givo the base h. ..... // 7j. N. and CI. 8. Altitude X = — '* 9 26 PROBLEM XXn. To transform an irregular polygon into a triangle ^ I'- havin<." given the base a Let P, q, B, S be the triangular parts of the polvgon. and j;2 a square equivalent to their sum; find .r by the IVob- lem XII., and then Altitude h = --' a 2 .r X X a G. S. By a similar proces.s, any polygon or any number of ditfei-onl polygons may bo easily and readily transformed into an equi- valent one of any kind. PROBLEM xxirr. To construct a square .>'^ having given the exee^.s a. of the diagonal over tlie side .r. X = rt -f V'2 d^ N.H. —15- = ^1+ V2a X a making v"7^~T= 1/ hy the 3j-d formula, then rilOBLEM xxiy. To construct a rectangle .• y, having given the perimeter l> = - -'' + ^ >/, and the surface JS =-. x g. Let a = the side of an equivalent square, then the base a- = ^ + ^J]f^a' ls\ S. maki ng ^^'- «. = ,: i,y ^|^, 5^1^ ^^^.^^^^j^^^ ^^^^^ & , h It is evident that the height y = PnOBLExM XXV. b~ 2x 2 To construct a rectangle, having given the surface S = a: u iind the difference of the adjacent sides x — y = d Let a = the side of an equivalent square; then * the base .r =^^J,^_^^ N.S. d , ' — -. - I r* 9 1^ ~ f (See Problem I xxiy.) the height y=_^+^,,.+ |^ G.S. PROBLEM XXVI. To construct a parallelogram, having given the adjacent sides '^ b, and the difference d of the two diagonals. Let the greater diagonal = .v ^ y 4 .r d = 2 + ^/ y' d 4 N.S. (bj formula 4) a . = 2 + - C^y formula 5) G. S. —IB- PROBLEM XXVJI. To determine a Eight-Angled Triangle, having given the hy^ potenuse a, and the sum of the two sides b -^ c. b = — tits 4- / "^ (6+0^ 2 ^rv 2 ""-4- PROBLEM XXVIir. To determine a Eight-Angled Triangle, having given the base A and the sum of the hypotenuse, and the other side a -{- c. side c = ~~- — j — -— PROBLEM XXLX. T.^f^r'lZ ^ ^'Sht-Angled Triangle, having given the base b, and the difference d between the hypotenuse a and the other side c \ so that a — c = d. b^ — d"" 2d make b^ — d^ = y^ N. S. (by formula 5), Gr. S. then c = -l!- = lULl 2d 2d and then a := d -^ c. PROBLEM XXX. To determine a Eight- Angled Triangle, having given the hy- potenuse a, and the diiference d between the two other sides h ^nd c; so that 6 — c = cf . ^ = — 2 + J a/2"«^^=^ :n-. s. make I V 2 d' — d' = .^. (by formula 5), d G. S. then c = X — z and then b = c -\- d PROBLEM XXXL To determine a triangle, having given the base h, tbe perpen- dicular h to the base from vertical angle, and the' difference d between the two other sides a and c; so that a -~ c = d. 1 a —17— Let X = one of the segments determined on the base bv th. perpendicular; then ^^ ^''® make i^ -^^f ^^y formula 5), then .r = 6 + JiiL^l+ c/^ = i ^JiFEr7^~d — make -ii><_i = ^ . and '^ X ^ ... y - ^ and -_ = 5 (^b^ formula 2), then j; = & -f. VTTip'^ = 6 _|_ VT^qr^F Q^^^ Problem YH ) NT? r. fK- ~.*~^'' (by formula 4). G. s. ±>i.i5.—-ln this case, because the CTranhiVnl Q^i.,*- ™a„^y^tra„sf„™ati„„, the ^^^^^.t^l^^t PROBLEM XXXfl. Having given the three sides a, b, c, of a triin^I^ u. fi , ., radius r of the inscribed circle. ' ^' ^ ^^^^'^"ff'^^ ^^ ^nd the .2\3 C2)- - :Kr. s. PROBLEM XXXIIL To determine a Right-Angled Triangle, having given the hv potenuse ., and the radius r of the inscribed drdc ^" Let b and c be the two other sides a 2 a 2 ^ V ^^ — 4 a;- — 4 r^ N. S. PROBLEM XXXrv. Having given the two equal „ide., . of a„ Weles trianWo and the base *, t„ fl„d the^clius r of the inscribed circle. ^ / 2 b^ —18— PROBLEM XXXV. Let the radius of the circle circumscribed to a poly£:on to be represented hy R ; the radius of the inscribed circle to be represented by ;• ; and the surfiico of the poly^ron to be ex- pressed by 6'.— Then, havin- given the three sides a, h. c of a triangle, and S, to find R and ;•. ah c B = r 4^' 2S KS. PRODLEM XXX Vr. Having given the side r/, of an equilateral triangle, to find U, r and S (See Problem XXXY.). B- /"' ^•^ - V 3 r _ f'ci^ a^ S='! a" V'S r PROBLEM XXXVIL Having given the side a of a regular hexagon, find 7?, y and >Sr (See Problem XXXY.). B = a 2 PROBLEM XXXVIIL Having given the side a ofa^square, to find! 7? and y(Sce XXX Y ) =v =V '' X - 2 a a V '^ X i KS. G.'S. it KS. G. S. —IS— PROBLEM XXXfX. Having given the side a of a Regular pentagon, to find ff r «nd S (See Problem XXXV.\ ' JR = /zzz '^ 1.381G = / ^ V 1.3g _a2 3816 4 2 V i.3yj6 ~4 PROBLEM XL. Having given the side a of a regular octagon, to find B, r and 8 (See Problem XXXY.). J? .5858" .6858" = / a2 r =7— ^ .68; V .5858" 4 PROBLEM XLL Having given the side a of a regular decagon, to findi?, r and S (See Problem XXXV.). R 2 + V^ + a3 >S = 5 « J R'^-f N.B.--In the last [two formulas the value of R, when found 18 to be substituted for R. ' io«na, PROBLEM XLIL Having given the side « of a regular dodecagon, to find R r and 8 (See Problem XXXY.). ~ V 1.072 4 PROBLEM XLIII. To express the side of the following regular 'polygons in terms of the radius R of the circumscribed circle. side of the triangle = R x» 1.732 side of the square ...... = R x 1.414 side of the pentagon ... = R x 1.176 side of the hexagon = R side of the octagon = R x .7653 side of the decagon = R X .618 side of the dodecagon... = R x .5176 PROBLEM XLIV. Having given the hypotenuse a and one side b, of a i-i-ht- angled triangle, to find the surface S. PROBLEM XLV. Having given the base h of an isosceles triangle, and the ob- lique side a, to find the surface. 6 . >S^ = I V 4 a^ — 62 PROBLEM XLVL Having given the three sides a, 6, c, of any triangle, to find t!ie surface. make, for the sake of abbreviation <^-\-^-\~c _, ' 2 " then 8 = Vp~(P — ^U^'^^r^yj^ZZTay —21— • PROBLEM XLVJI. Having given the surface S of a regular polygon inscribed in a circle, to find the surface s of a similar circumscribed poly- gon (See Problem XXXV.). s in vs htr Ob- fitid PROBLEM XLVIIL To express the surface of a circle in terms of the radius B. PROBLEM XLIX. Having given the difference d between the circumferences of two circles, and the ratio of the radii, so that r: Ji:itn:n,lo find the circumference and the radii. md Circ. of the greater = Circ. of the lesser = Bad. of the gi'cater = m — n mn d m^ — mn m d Had, of the lesser = - 2n(m — n) mn d 2 n m^ — ffin PROBLEM L. Let the base of a triangle be ^'(7 = h, the vertical angle be A and then the sides ^ ^ and ^ C be respectively « and .. lo divide this triangle into two equivalent parts by a straight line passing through a given point D on the side A C, Let the distance ADh. = d. The Problem is solved as soon as another point ^on the side A B, is found, through which the dividing line is to pass. Let the distance ^ ^be denoted by x; then '' X = c a Td d N. S. as. PDrvnTTTi^r rT To divide the same triangle into three equivalent parts by two lines passing through the same point D. It is evident that —22 two points jffand H' are to be found on the eide A B. Let the distance A M = x, and the distance A H' = x'. X = X = c a c X 3 « d 2ca ~ d G. S. N. S. G". S. PROBLEM LII. To divide the same triangle (see Pioblem L.) into two equi- valent parts by a line parallel to the base. It is e/idont that the problem is solved as soon as one point D is found on one «ide A C, through which the dividing line is to pass. Let the distance AJ) = x; then X =V 2" =a/c X N. S, G. S. PROBLEM LIII. To divide the same tviangle (see Problem L.) into five equiva- lent parts by lines parallel to the base. Then 4 points B, D', D", -D'" are to be found en the same side A C. Let AD = x; A B' = x'; A D" = x"; A D'" = x'". N. S. G. S. X =V "5 X =V "6- = v^ ^a a 2 a X -5- / 3 a yja X -^- .X-'": V'4 a2 «.! / X -, 4 a 5 —23- PROBLEM LIV. To divide the same triangle (see Problem L.) into two parts which are in the ratio m : n. Vm a ^ i — N. S. y = \ a X m a: m a m -\- n make — r-^ = y (by formula 2). m-f- n ^ ^ '' ^ then X = V ay G. S. PROBLEM LV. To divide a trapezoid ^ (^ + ^) ^^gee Intro. I.), into two A equivalent parts by a line x parallel to the bases m, n. / mi^ -{- r? = V 2 X N.S. make vp? '\-Tf^ — y^ (by formula 4). then X G. S. _ / y xy N.B. — The length of the dividing line being known, its po- sition in the figure will be easily found. ■ • Having giTcm a. cjitQlp; .•who30. yadius; «;?; ^,t<> .find j^the radius r of anothtarjftll'iey^'wltose 'a?^4.iih*|l* fcie^.to:'^that of the first circle in the ratio of m\n. mR n N.S. G. S. PROBLEM LVII. Having a given circle n- -B^ to find the radius r of a concentric _24-- circle, so that the circular ring may be equal to the threo-fifth» of the smaller circle. r = Vn 7 K^ = JJix 5 R ~8' G. S. PROBLEM LVIII. To make the same construction (see Problem LYII.) so that the circular ring may be a mean proportional between the two circles. r =-f±7 E'-{- if2 PROBLEM LIX. To make the same construction so that the interior circle may be a mean proportional between the exterior circle and the circular ring. = ±7-4^+y¥Tf I « f ■* «» t I • • . ' J » ".-■>- t t -J f. i(^^^n7^vy Joupb Loveday, Printer, Ottawa. Canada.