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Conftru&tion.

On the longest Leg AB of the given Triangle defcribe a Semicira cle, and in it apply AC the other E Leg, drawing alfo CB; fo will

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CBg: beddcc: Then on CB defcribe another Semicircle, and therein apply CB the Bafe of the Triangle given, drawing liked Wife The Line B6: Then will & Bq be — dd — cc - bb And confequently & B = √ dd - cc - 66: And lince the firft part of the eighth Step is divided by 2b, make as BE = 2 b .. BB = d2 = c2 = b2 | 1⁄2 : : B F = BB. ¿D=DC the Segment fought

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CHA P. II.

A Problem producing a Simple Quadratick Equation.

Having given the Square B

AD, and a Right Line N

BN, you are to produce the
Side AC to E, fo that EF A
drawn from E towards B, fhall
be equal to B N.

It will be evident, if you imagine a Semicircle to paß thro the Points B and E, that the most commodious way will be to find

B

C

N

H

the Line DG, that you may have the Diameter BG, upon which having afterwards defcrib'd a Semicircle, there will be need of no other Operation to fatisfy the Problem, than to produce the Side AC till it occur the preferib'& Periphery.

Solu

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the ABFD,

GEH and 7b (BD) y (B F) : : b (EH) •• y (E G.)

BEG are

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8b+x(BG)..y (EG): :j+e(BE) ..b (EH)

Equation 9 bb + bx=yy+yc.

47. 1. Eucl. El. 10+2 6 x + x x (BG q) = yy (E Gq :) {bb 2b xx +y+246 + cc (BE q :) 11bx+xx=yy + yo+cc.

10 9

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- cc 12 bx + x x

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12=9 13 bx+xxcc (= y y + y c) = b b + b si 13-bx + c2 14 x x = b b + co.

14 w

12/15│x=√bb+cc=5.

=

Conftruction.

Having produc'd the Side of the Square BA to N, so that BN fhall be the given Right Line BN; then fince BD is b, and B Nc, the Hypothenufe DN will be = bb + c c√ z = x : Having therefore made DG DN, and defcrib'd a Semicircle upon the whole Line B G, if AC be prolong'd until it occur the Periphery in E, you'll have done that which was requir'd.

CHA P. III.

CHA P. III.

Of PROBLEMS producing Adfected Quadratick Equations.

PROBLEM I.

N the LA ABC, the Per

Ipendicular, BC, and the alter

nate Segment of the Hypothenuse, (made by a 1 let fall from the L) viz. A D being given, to find the other Segment D C, &c.

Solution.

Let IBC=p the I= 45
22bAD the alternate Segment = 48

Andfuppofe 3 a DC

By 4. 6.
Eucl. El.

=

46. BD BD ....a

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Aq.

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By 47. I.

Eucl. El.

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aa

S — =

5= 6 7 ba

7 + aa 8 a2 + ba = pp. Cafe 1. Of Adfected Qua

dratick Equations. Comp. 9a+ba + 2 bb = pp + 4 bb

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90210 a + b = √ pp + 4 bb

10 — 1611a: = √ pp + & bb - 1 b = 27.

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Points of the; then compleat the Diameter, by produBbb

cing

Part II. cing A D to O and C; then will DC, or O A, be = a for

OD (a+b).. DL (p) :: DL (p). DC=a per 13. 6 Eucl. El. confequently aa + ba = pp; as in the 8th Step.

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PROBLEM II.

The Difference A E between the Bafe A B and Perpendicular BC of the LABA C, and the Perpendicular BD let fall from the right Angle ABC upon the Hypothenufe A C being given, thence to B find the Hypothenufe AC.

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fuppofe 3a=

and 4 e

4. 6: Eucl.

Sd+c (AB). p (BD) a (AC)

..

El. 352 BC= e
Ag. 6 de + ee = pa

47.1.Encl.7 5 dd + 2de + ee

El.

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7

dd + 2de +

6 x 28 2de + 2ee = 2pa

9

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ee (ABq:+BCq:) 2ee = aa

ོ༡ -- 8 dd
Cafe 2.
гра.
=aa
9 + pp íc dd + pp = aa = 2pa + pp

10211 V dd + pp

=a-p

10 + pl2p + v dd + BP = 9 = 75

N

M

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With L A 2p, and A E d, makethed LAEN, and upon M, the middle Point of L A, defcribe fuch a Semicircle as will pafs thro' N and E, the emoteft Points of the then compleat the Diame

ter, by producing LA to O and C; and CA LO will bea for CL (a-2p). LN (d): LN (d)

a, per 13. 6. Eucl. El. Whence aa —

~.LO
as in the 9th Step.

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PROBLEM III.

The Hypothenufe A C of any d AA BC, and the Perpendicular BD, let fall from the LA BC, upon the Hypothenufe AC being gi ven, to find A D, the greater Seg ment of the Hypothenufe.

Solution.

2pa=dd;

D

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1/2 1/2

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4

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IOW211 a

11 + 1/2 b 12 a =

bb

PP

=

Cafe 3:

4 bb

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b 1

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± √ 4 bb — pp = 48 or 27

The Geometrical Conftruction of Cafe 3. viz. ha

a app, may be thus perform'd.

Draw a right Line of

any convenient Length, as
AZ, and near its Mid-
dle erect a Perpendicular
DBp: from the Top,
or upper End B of that A

M

Perpendicular fet off BMb; and upon the Point M, where B M touches A Z with the Radius M B, defcribe a Semicircle A BCM; then will its Diameter A C =b be cut by the Perpendicular B D into two Segments, AD and DC, which are the two Values of the Root a, B b b 2

viz.

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