:

XVI. Right Lined Figures, are fuch Figures as are contained under Right Lines; of which none can consist of less than Three, and those 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 such 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 Isofceles Triangle, as the the Figure O. Or, (3.) All the Three Sides may be unequal, and such a Triangle is called a Schalenum Triangle, as the Figure P. And these are the Distinctions in relation to their Sides.

Now for the Distinction of Triangles in relation to their Angles, they are Three also: 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 Care Acute, and fuch a Triangle is called Orthogonium. Or, (2.) If the Triangle have all the Angles Acute, as the Figure R; all whose Angles at D E 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, whose Angle at I is Obtufe, and the other Two at G and H are Acute; fuch a Triangle is called Ambligonium. And these are the Denominations in relation to their Angles.

XVII. Of Four-fided Figures (or Quadrilaterals) a Quadrat, or
Square, is that whose Sides are Equal, and Angles Right.

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 Correspondent (or Oppofite) and the Angles all Right Angles, as the Figure P; such a Figure is called a Parallelogram, or Long Square, (and sometimes) a Rectangle. But, (3.) If all the Sides be Equal, but the Angles Unequal, that is, Two Acute, and Two Obtufe, each to his Correfpondent (asthe Figure R) such Figure is called a Rhombus or Diamond Form. And farther, (4.) If such a Figure have Two Sides Longer, and Two Sides Shorter; and also Two Angles Acute, and Two Obtufe, each

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each to it opposite corresponding, as the Figure Q; such 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 Sand T; such Figures are called Trapezia, or Table Forms.

PRAPRACTICAL

PROBLEMS,

GEOMETRICAL.

7

PROBLEM I.

To divide a Right Line A B, into Two Equal Parts A E and
BE, and at Right Angles.

Pratice. 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 describe the obscure Arch bc, -and (with the same Distance) One Foot set in B, with the other describe the obscure 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 CEB on the other Side.

PROB. I.

Upon any Point (as O) taken in the Right Line QR,
to eret a Perpendicular O S.

Practice.

First, Open the Compaffes to any small Distance

and setting 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 Compasses to any Distance, greater than the for- Fig. VIH..

mer; and setting one Foot in T, with the other describe the Arch bh. -Also, with the fame Distance, set one Foot in V, and with the other describe the Arch gg, croffing the former Arch bh in the Point S.

Fig. X.

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 eret a Perpendicular Z A.

Practice.

First, The Compafles being opened to any small Di

stance, set one Foot in Z, and with the other describe the Arch BCD, and upon it set the same Distance from B to C, and from C to D.

2. The Compasses still continuing at the fame Distance, set one Foot in C, and with the other describe the Arch FD: Also fet one Foot in D, and with the other describe 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.

one Foot in H, and with the other describe the Arch

I K, and set that fame Distance from I to K.

2. Upon K (with the same Distance) describe the Arch I L

MN.

3. Upon this Arch set the fame Distance from I to L, from L to M, and from M to N.

4. A Line drawn from H, through N, shall be Perpendicular to the Line G H.

Fig. XII. 1.

A Third Way.

THE Compaffles opened to any fmall Distance, fet one

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 PO in S, and also describe 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, shall be a Perpendicular to PO.

PROB. PROB. V.

From a given Point above, as P, to let a Perpendicular fall upon a given Right Line under it, NO.

N the Performance there are Two Varieties: For, (1.) The

I given Point may be scituate over (or about) the Middle; or

over (or near) the End of the given Line

Practice.

L

In the First Cafe.

ET NO be the Right Line given; and let P be
the Point above, from whence the Perpendicular

is to be let fall.

1. Open the Compasses to a Distance greater than is the nearest Distance between the Point above P, and the Line upon which the Perpendicular is to fall; and with that Distance, setting one Foot in P, with the other describe 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 first 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 N O.

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 Compasses in S, and opening the other to any convenient Distance, describe an Arch yy; and removing the Compaffes to R, describe the Arch z z, croffing the former in A; and fo, a Line drawn from A, through the given Point, shall be Perpendicular to the given Line N O.

In the Second Cafe.

Let V be the Point given,

Praffice. First, From any Part of

the given Line N O, as from T, a ight Line to the given Point V, which divide into Two equal Parts in X.

2. Set one Foot of the Compasses in X; and with the Distance X T, describe the Semicircle V OT, cutting the given Line in O, so shall O be the Point in the Line NO; from which, if you draw a Line to V, it will be Perpendicular to N O.

C

PROBL.

Fig.XIII.

Fig. XIV.