439 
GEOMETRY. 
Thtrefore tlie inacccHible dirtance RQ, is found by mul¬ 
ti plying tlie fquareof the fide of the rhombus by OP, and 
dividing that produtl by the product of OD and PD. 
APPLICATION OF ALGEBRA to GEOMETRY. 
If it be propoi'ed to refolve a geometrical problem 
algebraically, or by algebra, it is proper, in the fiifi; 
place, to draw a figure that Iliall reprelent the feveral 
parts or conditions of the problem, an'd to fuppofe that 
figure to be the true one. Then, having confidered at¬ 
tentively the nature of the problem, the figure is next 
to be prepared for a folution, if necefiary, by producing 
or drawing fuch lines in it, as appicar mod conducive to 
that end. This done, the ulual fymbols or letters for 
known and unknown quantities, are employed to denote 
the given parts of the figure, or as many of them as 
necelfary, as alfo fuch unknown line or lines as may be 
ealieft found, whether required or not. Then proceed 
to the operation, by obferving the relations that the fe¬ 
veral parts of the figure have to each other; from 
which, and the proper theorems in the foregoing ele¬ 
ments of geometry, make out as many equations, inde¬ 
pendent of oach other, as there are unknown quantities 
employed in them ; the refolution of which equations, 
in the fame manner as in arithmetical problems, will de¬ 
termine the unknown quantities, and refolve the problem 
propofed. 
As no general rule can be given for drawing the lines, 
and felefting the fittefl: quantities to fubfiitute for, fo as 
always to bring out the mod fimple conclufipns, becaufe 
different probletns require diderent methods of folution; 
the bed way to gain experience in this matter is, to try 
the folution of the fame problem in difi'erent ways, and 
then apply that which fucceeds bed to other cafes of 
the fame kind, when they afterwards occur. The fol¬ 
lowing indruftions, however, may be of fome ufe to the 
learner, viz. 
In preparing the figure, by drawing lines, let them 
be either parallel or perpendicular to other lines in the 
figure, or fo as to form fimilar triangles. And if an 
angle be given, it will be proper to let the perpendi¬ 
cular be oppofite to that angle, and to fall from one end 
of a given line, if poflible. 
In feledting the quantities proper to fubditute for, 
thofe are to be chofen, whether required or not, which 
lie neared the known or given parts of tlie figure, and 
by means of which, the next adjacent parts may be ex- 
preffed by addition and fubtradtion only, without tlie 
intervention of furds. 
When two lines or quantities are alike related to other 
parts of the figure or problem, the bed way is, not to 
make life of either of them feparately, but to fubditute 
for their fiim, or difference, or redtangle, or the lum of 
their alternate quotients, or for fome line or lines, in 
the figure, to which they have both the fame relation. 
Wlien the area; or the perimeter, of a figure, is given, 
or fuch parts of it as have only a remote relation to the 
parts required; it is foraetimes of ufe to aOume anotlier 
figure fimilar to the propofed one, liaving one fide 
equal to unity, or fome other known quantity. For, 
from hence the other parts of the figure may be found, 
by the known proportibns of the like Tides, or parts, 
and fo an equation be obtained.—The following pro¬ 
blems may ferve to illudrate wliat has been dated above. 
zi 6 . In a right-angled triangle, having given the bafe, and 
C the fum of the hypothenuje and perpendicular ; 
to find both thefe two fides .—Let ABC repre- 
fent the propofed triangle, right-angled at 
B. Put the bafe AB=:3ii;^, and the fum 
AC-j-BC of the hypothenufe and perpen- 
. dicular ; alfo, let x denote the per- 
B pendicular BC; hence it follows that the 
hypothenul'e AC will be expreffed by s — x. 
But AC2~AB®-)-BC« (by theor. 34, Geom.); that 
is, (j —, or j*—or 
=2sx} hence _ L.~4—BC, and s—. =:s 
2S 2S 
— AC, tile perpendicular and hypothemifs as required. 
217. In a right- angled triangle, having given the hypothe- 
nuje, and the fum of the bafe and perpendicular ; to find both 
thefe two fides. —Let ABC reprefent the propofed trian¬ 
gle, riglu-angled at B. Put the given liypotlienufe 
AC=5=:a, and the fum AB-pBC of the bafe and per¬ 
pendicular alfo, let X denote the bafe AB ; 
hence it follows that the perpendicular BC will be ex- 
preded by s— x. 
But AC 2 =e:AB 2 -J-BC 2 by tlie nature of right-angled 
triangles; that is, a^—x^-\-{s — x)^—x^-\-s- — 2sx-{-x'^, 
or 2x^ — 2Sx~a^ — s~, or — sx=ila'^ — ^s-•, hence, by 
completing the fquare, &c. is found a—J- v— 
—3=AB; and s—BC ; the bafe 
and perpendicular required. 
Or, let z denote the diderence of the bafe and per¬ 
pendicular, tlie fum being denoted by s, as above. 
Then, becaufe fliat in any two quantities half their 
diderence added to half the fum gives the greater quan¬ 
tity, and the fame fubiraffed gives the Icfs. quantity, it 
follows, that gr-j-gz and -^s —will denote the two 
legs, or the perpendicular and bafe. Hence then a^ — 
+ : or 2-=z2a^—s^, and 
z=:^2a'^ — s^. Then the half of this being added and 
luhtradled with the half fum, gives 
20'^ — S-, and 
^s—lz=\s—r\f2a^s^, the fame as before. 
2i'8. In a reBangk, having given the diagonal, and the 
perimeter, or fum ofi all the jour fides \ to -jy q 
find each of the fides fieverally. — Let 
ABCD be tlie propofed rettangle; and 
put the diagonal ACrxio—i, and lialt 
the perimeter AB-|-BC or AD-{-DC= 
142=5 ; alfo put one fide AB222X ; then .2^ B 
the other fide BC will be denoted by a—x. Hence, by 
right-angled triangles, AC^ — AB^-^-BC^, tliat is, 
2=2x2—2ax-^-a2 ; confeqiiently x^—ax 
— id^ —ia2, and by completing the fquare, &c. x— \cr 
-j-iv' 25 ; 2 _a 2 '—g—AB, and a—xz=.\a —\V 2d'^ — 
=BC, the two fides; and the fame for the other two, 
AD, DC. 
Or, by fubftltuting for the fum and difference of the 
two lides, AB, EC, this problem may be relolved, like 
the latter method of the loiegoing one, by a liiiiple qua¬ 
dratic, and confeqiiently without completing the fquare. 
219. Having given the baje and perpendicular ef any tri¬ 
angle-, to fiind the fide oj a /quare in- 
ficribed in the fiame. —Let ABC reprefent 
tlie given triangle, and EFGH its in- 
Icribed fquare. Put the bafe AB=^, 
the perpendicular CD2225, and the fide 
of the fquare GF or GH=DI=x; 
then.will ClnrCD— DI=a —x. 
Then, becaule the like linos in the 
fimilar triangles ABC, GPC, are pro¬ 
portional, A"B : CD :: GF : Cl, that ri, b a x : a —.v. 
Hence ab — bx—ax, or abzzax-\-bx, and confeqiiently 
Y— ■ —GF or GH, the fide of the inlcribed fquare ;. 
a-{-b 
which therefore is of the fame m.agnitude, whatever 
the fpecies or the angles of the triangle may be. 
220. In an equilateral triangle, having given the lengths of 
the three perpendiculars, drawn from a cer¬ 
tain point within, on the three fides ; to de¬ 
termine the Let ABC reprefent 
the equilateral triangle, and DE,,DF, 
DG, the given perpendiculars from 
the point D. Draw the lines DA, 
DB, DC, to the three angular points; 
and let fall , the perpendicular CH on 
