XVI. Right Lined Figures, are fuch Figures as are contained under Right Lines; of which none can confift of less than Three; and thofe are called Trilaterals, or Triangles. And Triangles are denominated partly from the Differences of their Sides, and partly from the Differences of their Angles: As for the Differences of their Sides, they may be, (1.) All Equal, and fuch a Triangle is called an Equilateral Triangle, as the Fig. VI. Figure N. Or, (2.) Two Sides may be Equal, and the third unequal, and fuch a Triangle is called an Ijofceles Triangle, as the the Figure O. Or, (3.) All the Three Sides may be unequal, and fuch a Triangle is called a Schalenum Triangle, as the Figure P. And these are the Diftinations in relation to their Sides. Now for the Distinction of Triangles in relation to their Angles, they are Three alfo: For, (1.) If a Triangle have One Right Angle, it will have Two Acute ones, as the Figure Q, where the Angle at A is a Right Angle, and the Angles at B and C are Acute, and fuch a Triangle is called Orthogonium. Or, (2.) If the Triangle have all the Angles Acute, as the Figure R; all whofe Angles at DE and F are Acute; fuch a Triangle is called an Oxogonium Triangle. Or, (3.) If the Triangle have One Obtufe Angle, as the Figure S, whofe Angle at I is Obtufe, and the other Two at G and Hare Acute; fuch a Triangle is called Ambligonium. And thefe are the Denominations in relation to their Angles. XVII. Of Four-fided Figures (or Quadrilaterals) a Quadrat, or And as Triangles have their various Denominations from the Species of their Sides and Angles, fo have Quadrilateral Figures alfo. For, (1.) If all the Sides be Equal, and all the Angles Fig. VII. Right Angles, as the Figure O, fuch a Figure is called a Quadrat or Square. But, (2.) If of the Four Sides, Two be longer than the other, each to its Correfpondent (or Oppofite) and the Agles all Right Angles, as the Figure P; fuch a Figure is called a Parallelogram, or Long Square, (and fometimes) a Rectangle. But, (3.) If all the Sides be Equal, but the Angles Unequal, that Two Acute, and Two Obtufe, each to his Correfpondent (as the Figure R) fuch Figure is called a Rhombus or Diamond Form. And farther, (4.) If fuch a Figure have Two Sides Longer, and Two Sides Shorter, and alfo Two Angles Acute, and Two Obtufe, each each to it oppofite correfponding, as the Figure Q; fuch a Figure is called a Rhomboyades, or Diamond-like Figure. But, (5.) If Quadrilateral Figures have all their Sides, and all their Angles, Unequal, as the Figures S and T, fuch Figures are called Trapezia, or Table Forms. PRA PRACTICAL PROBLEMS, GEOMETRICAL PROBLEM I. To divide a Right Line A B, into Two Equal Parts AE and Practice. First, Open your Compaffes to any Distance greater than half the Length of the given Line A B. 2. With that Distance fet one Foot in A, and with the other defcribe the obfcure Arch bc,-and (with the fame Distance) One Foot fet in B, with the other defcribe the obfcure Arch de, croffing the former Arch in the Points C and D. 3. Through the Points C and D, draw a Right Line C D, which will divide the given Line AB into Two Equal Parts in the Point E, and at Right Angles. The Angle AEC on one Side thereof, being equal to the Angle CE B on the other Side. PROB. I. Upon any Point (as O) taken in the Right Line Q, R, Practice. Flrft, Open the Compaffes to any small Diftance and fetting one Foot in the given Point O, with the other Foot make Marks on either Side of O, as at T and V. 2. Open the Compaffes to any Diftance, greater than the for- Fig. VIII. mer; and fetting one Foot in T, with the other defcribe the Arch bb. -Alfo, with the fame Diftance, fet one Foot in V, and with the other defcribe the Arch gg, croffing the former Arch bb in the Point S. 3. Draw 3. Draw the Line O S, and it will be Perpendicular to the given Line QR. PROB. III. From the End of a Line XZ, to ere& a Perpendicular Z A. Fig. X. Practice. Frift, The Compaffes being opened to any small Di ftance, fet one Foot in Z, and with the other defcribe the Arch B CD, and upon it fet the fame Distance from B to C, and from C to D. 2. The Compaffes ftill continuing at the fame Distance, fet one Foot in C, and with the other defcribe the Arch FD: Alfo fet one Foot in D, and with the other defcribe the Arch CE, cutting the Arch DF in A. 3. From Z draw the Line Z A, which will be Perpendicular to the Line X Z. PROB. IV. Another Way to erect a Perpendicular upon the End of a Line. Fig. XI. Practice. Fl I K, and fet that fame Distance from I to K. 2. Upon K (with the fame Diftance) defcribe the Arch IL M N. 3. Upon this Arch fet the fame Distance from I to L, from L to M, and from M to N. 4. A Line drawn from H, through N, fhall be Perpendicular to the Line G H. A Third Way. Fig. XII. 1. THE Compaffes opened to any small Distance, fet one Foot in O, and pitch the other Foot down at Pleasure, as at Q; and one Foot refting upon Q, turn the other Foot about, till it cross the Line P O in S, and alfo defcribe the small Arch ba. 2. Lay a Ruler from S to Q, and it will cut the Arch ba in R. 3. A Line drawn from O, through R, fhall be a Perpendicular to P O. PROB. PROB. V. From a given Point above, as P, to let a Perpendicular fall upon a given Right Line under it, N 0. N the Performance there are Two Varieties: For, (1.) The given Point may be fcituate over (or about) the Middle; or over (or near) the End of the given Line In the First Cafe. Practice. Lthe Point above, from whence the Perpendicular is to be let fall. 1. Open the Compaffes to a Distance greater than is the neareft Distance between the Point above P, and the Line upon which the Perpendicular is to fall; and with that Diftance, fetting one Foot in P, with the other defcribe the Arch of a Circle, which will cut the given Line in Two Points, R and S. 2. Divide the Space between R and S (by the firft Probl.) into Two equal Parts in the Point Q, then will a Right Line, drawn from P to Q, be Perpendicular to the given Line NO. Note, To avoid the dividing of the Space between R and S into Two equal Parts, to find the Point Q, (if you have room either above or below the given Line) you may fet one Foot of the Compaffes in S, and opening the other to any convenient Distance, defcribe an Arch yy; and removing the Compaffes to R, defcribe the Arch zz, croffing the former in A; and fo, a Line drawn from A, through the given Point, thall be Perpendicular to the given Line NO. In the Second Cafe. Let V be the Point given, Practice. F, draw a Right Line to the given Point V, which Irft, From any Part of the given Line NO, as from divide into Two equal Parts in X. 2. Set one Foot of the Compaffes in X; and with the Distance X T, defcribe the Semicircle V O T, cutting the given Line in O, fo fhall O be the Point in the Line NO; from which, if you draw a Line to V, it will be Perpendicular to Ñ O. C PROBL Fig.XIII. Fig. XIV. |