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PROBLEM XV. In any given Triangle, as ABD, to describe a Circle that fall

touch all its Sides. (4. 6. 4.) Bifect any two Angles of the Tri

Р angle, as A and B, and where the bifecting Lines meet (as at C) will be the Center of the Circle required ; -and its Radius will be the nearest Distance to the Sides of the

SD Triangle.

PROBLEM XVI.
To describe a Circle about any given Triangle. (5. 1. 4.)
This Problem is perform’d in all respects like the Ninth, viz.
·by bisecting any Two Sides of the given Triangle ; the Point,
where those bisecting Lines meet, will be the Center of the Circle
required!

PRO B L E M XVII.
To describe a Square about any given Circle. (7.6.4.)
Draw two Diameters in the given

FI E
Circle (as D A and EB) crossing at
Right Angles in the Center C; and,
with the Circle's Radius C A, describe
from the extream Points of those Dia-

A meters, viz. A, B, D, E, cross Arches, as at F, G, H, K; then join those Points where the Arches cross with Right Lines, and they will form the H

B
Square required.

PROBLEM XVIII.
In any given Circle, to describe the largesi Square it can

contain(6. 6. 4.)
Having drawn the Diameters, as D A and E B, bisecting each
other at Right-angles in the Center C, (as in the last Scheme) ;
then join the Points A, B, D, and E, with Right lines, viz.
A B, B D, DE, E A, and they will be sides of the Square re-
quired,

PRO

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PROBLEM XIX.

с

Upon any given Right-line, as AB, to describe a regular Pentagon,

or Five-sided Polygon.
Make the given Line Radius, and upon each End of it de-
scribe a Circle ; and through those

K
Points where the Circles cross each
other (as at G x) draw the Right-
line Gex: Upon the Point G with
the same Radius describe the Arch
HAB D, and laying a Ruler up-
on the Points D, e, mark where
it crosses the other Circle, as at F.
Again, lay the Ruler upon the Points
H, e, and mark where it crosses the

H н

D other Circle, as at C: Then from the Points F and C (with the same Radius as before) describe cross Arches, as at K : Join the Points AF, FK, KC, and CB, with Right-lines, and they will form the Pentagon required, viz. AF-FK - KC=CBAB; and the Angles at A, B, C, K, F will be equal.

B

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PROBLEM XX.

In any given Circle, to describe a regular Pentagon.

(1.1.4. & 1o. e. 3.) Or, in general Terms, to describe any regular Polygon in a

Circle,

Draw the Circle's Diameter DA, and divide it into as many equal Parts as the proposed Polygon bath

C..
Sides ; then make the whole Diameter
a Radius, and describe the two Arches
CA and CD. If a Right-line be drawn
from the Point C, through the Second
of those equal Parts in the Diameter, as
at 2, it will assign a Point in the oppo-

D

A fite Semicircle's Periphery, as at B. Join

3 4
DB with a Right-line, and it will be
the side of the Pentagon required.

B
Q4 2

These

These Twenty Problems are sufficient to exercise the young Practitioner, and bring his Hand to the right Management of a Ruler and Compasses, wherein I would advise him to be very ready and exact.

As to the Reason why such Lines must be so drawn as directed at each Problem, that, I presume, will fully and clearly appear from the following Theorems ; and therefore I have (for Brevity's Sake) omitted giving any Demonstrations of them in this Chapter, defiring the Learner to be satisfied with the bare Knowledge of doing them only, until he hath fully considered the Contents of the next Chapter; and then I doubt not but all will appear very plain and easy.

CHA P. III.

A Collection of most - useful Theorems in plain Geometry

Demontrated. Note, In order to shorten several of the following Demonstrations,

it will be necesary to premise, that 1. THE Periphery (or Circumference) of every Circle (whe

ther great or small) is suppos’d to be divided into 360 equal Parts, called Degrees; and every one of those Degrees are divided into 60 equal Parts, call’d Minutes, &c.

2. All Angles are measured by the Arch of a Circle describ'd upon the Angular Point (See Defin. 9. Page 287.) and are esteěm'd greater or less, according to the Number of Degrees contain's in chat Arch.

3. A Quadrant, or Quarter part of any Circle, is always 90 Degrees, being the Measure of a Right-angle (Defin. 6. P. 287.) and a Semicircle is 180 Degrees, being the Measure of iwa Right-angles.

4. Equal Arches of a Circle, or of equal Circles, measure equal Angles.

To thofe five general Axioms already laid down in Page 146, (which I here suppose the Reader to be very well acquainted with) it will be convenient to understand these following, which begin their Number where the other ended.

Arioms.

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6. Every whole Thing is Greater than its Part.
That is, the whole Line AB is
greater than its Part A c, &c.

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A-

B
The same is to be understood of Superficies's and Solids.

7 Every Whole is Equal to all its Parts taken together. That is, the whole Line A B is equal

d to its Parts AC +cdt de te B. SA

-B The fame is also true in Superficies's and Solids.

8. Those Things which, being laid one upon another, do agree or meet in all their Parts, are equal one to the other,

But the Converse of this Axiom, to wit, that equal Things being laid one upon the other will meet, is only true in Lines and Angles, but not in Superficies's, unless they be alike, viz. of the same Figure or Form: As for Instance, a Circle may be equal in Area to a Square; but if they are laid one upon the other, 'tis plain they cannot meet in all their Parts, because they are unlike Figures. Allo, a Parallelogram and a Triangle may be equal in their Area's one to another, and both of them may be equal to a Square; but if they are laid one upon the other, they will not meet in all their Parts, &c. Note, Besides the Characters already explain'd in Part I, and in

other Places of this Tract, these following are added. Viz. 5 denotes an Angle in general, and 55 signifies Angles; A fignifies a Triangle; Ö fignifies a Square, and denotes a Parallelogram. And when an Angle is denoted by any three Letters (as, A B C) the middle Letter (as B) always denotes the Angular Point; and the other two Letters (as A B, and B G) denote the Lines or Sides of the Triangle which includes that Angle.

These Things being premised, the young Geometer may proceed to the Demonstrations of the following Theorems; wherein he may perceive an absolute Necessity of being well versed in several Things that have been already deliver'd: And also it will be very advantageous to store up several useful Corollaries and Lemma's, as they become discover'd Truths : For it often happens, that a Prom position cannot be clearly demonstrated a priori, or of itself, without a great Deal of Trouble; therefore it will be useful to have Recourse to those Truths that may be afafting in the Demonstration then in Hand.

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THEOREM I.
If a Right line stand upon (or meet with) another Right-line, and make

Angles with it, they will either be two Right-angles, or two Angles
equal to two Right-angles. (13. e. 1.)

Demonftration. Suppose the Lines to be A B and DC, meeting in the Poiré at C: Upon C describe any Circle at pleasure ; then will the Arch A D be the Measure of the sb, and the Arch D B the Measure of se; but the Arches AD+DB=180°,

ble

С viz. they compleat the Semicircle. Consequently the sb to se=180°. Which was to be prov'd.

Corollaries. 1. Hence it follows, that if the <b= 90° then se= 90°; but if < b be obtufe, then the se will be acute, &c.

From hence it will be easy to conceive, that if several Rightlines stand upon, or meet with any Right-line at one and the fame Point, and on the fame Side, then all the Angles taken together will be = 180°, viz. Two Right-angles.

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THEOREM II.

if two Angles intersect (i. e. cut or cross) each other, the two ofa

pofite Angles will be equal. (15. 6. 1.)

Demontration.
Let the two Lines be A B and
DE, intersecting each other in the
Center C.
Then sbt ge=180° 2
And Fót =180} per laft.

A b B
Consequently sbt respot

ac sa, per Axiom 5.

Subtract bon both sides of the. Æquation, and it will leave Se=ca.

Again, <bt se=180°, as before ; and set C= 180°, consequently seta=sot se Subtract 5e, and then <C=Sb. Q. E. D.

Corel.

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