ltd MENSURATION. 
regard to any refult deduced from thofe principles; but 
in the new method of confidering the fubjeft, thegreateft 
poflible care was neceffary in order to avoid error, and fre¬ 
quently this was not fufficient to guard againft erroneous 
conclulions. But the facility and generality which it 
polfieffed, when compared with any other method then 
difcovered, led many eminent mathematicians to adopt 
its principles; and of thefe, Huygens, Dr. Wallis, and 
James Gregory, were the mod confpicuous, being all very 
fortunate in their application of the theory of indivifibles. 
Huygens, in particular, mull always be admired for his 
folid, accurate, and mafterly, performances in this branch 
of geometry. The theory of indivifibles was however 
disapproved of by many mathematicians, and particularly 
by Newton, who, amongft his numerous and brilliant dif- 
coveries, has given us that of the method of fluxions, the 
excellence and generality of which immediately fuper- 
feded that of indivifibles, and revived fome hopes of 
fquaring the circle, and accordingly its quadrature was 
again attempted with the greatelt eagernefs. The qua¬ 
drature of a fpace, and the rectification of a curve, was 
now reduced to that of finding the fluent of a given 
fluxion ; but Hill the problem was found to be incapable 
of a general folution in finite terms. The fluxion of 
every fluent was found to be always aflignable, but the 
converfe propofition, viz. of finding the fluent of a given 
fluxion, could only be effe&ed in particular cafes, and 
amongft thefe exceptions, to the great difappointment and 
regret of geometricians, was included the cafe of the cir¬ 
cle, with regard to all the forms of fluxions under which 
It could be obtained. At length, all hopes of accurately 
fquaring the circle and fome other curves being aban¬ 
doned, mathematicians began to apply themfelves to find¬ 
ing the moll convenient feries for approximating towards 
Hheir true lengths and quadratures; and the theory of 
menfuration then began to make rapid progrefs towards 
.perfedion. 
As the menfuration of furfaces and folids is generally 
•cffeded by the fluxional calculus, the reader will find, 
•ainder the article Fluxions, vol. vii. the principal bufi- 
enefs already accomplifhed ; and all that feems requifite in 
this place is to make a feledion of fuch problems as may 
be more particularly ufeful to the pradical mechanic and 
•artificer. In refped to the meafurement of altitudes and 
diftances, thefe will be treated of under the article Tri¬ 
gonometry. For fome other particulars which may 
feem to belong to the fubjed of Menfuration, fee the 
articles Geometry and Conic Sections. 
Of Right-lined Plane Figures. 
The meafure of the fpace or furface contained within 
the boundaries of any plane figure is called its area, or 
fuperficial content. This is eftimated in acres, fquare 
yards, fquare feet, or fome other fixed or determinate 
meafure. Thus, if we fuppofe A B C D, fig. i, to reprelent 
the top of a rectangular table whofe length D C is 5 feet, 
and breadth DAr:;; then the upper furface will contain 
5X3 or 15 fquare feet; a fquare foot being the unit or 
integer by which the area is eftimated. But, if the di- 
meniions are taken in yards, its length will be if, and 
breadth 1 yard; and the fuperficial content =i§X i = i§ 
fquare yards;, for AO PD is the fquare yard, and the 
redangle O C is f of a fquare yard: in this cafe a fquare 
yard is the meafuring unit. And, when the length and 
breadth are denoted in inches, a fquare inch becomes the 
meafuring unit or integer; and the area will be 60X 36= 
2160 fquare inches. 
To find the Area of a Parallelogram of any kind. —Mul¬ 
tiply the length by the perpendicular breadth, or the bafe 
by the height, and the produd will be the area. 
Example. 'What is the content of the parallelogram 
DGLC, fig. 2. whofe length D C is 5 feet, and breadth 
C B is 3 ? — Let D A be perpendicular to D C ; then the 
parallelogram GLCD is equal to fke redangle ABCD ; 
and the area of the latter is DC XCB, or 5X3=15 fquare 
leet, the anfwcr. 
To find the Area of the Triangle. —Multiply the bale by 
the perpendicular height, and half the produd will be the 
area. Or multiply the bafe by half the height, or the 
height by half the bafe. For a triangle is equal to half a 
parallelogram of the fame bafe and altitude. 
Ex. How many acres are contained in a triangular 
field, one fide being 470 yards, and the perpendicular on 
that fide =396 yards ?—Anfwer, 19^. 
When two fides of a triangle and their included angle 
are given, multiply the produd of the given fides by the 
fine of the included angle, and half this lalt produd will 
be the area. 
Let B AC, fig. 3, be the given angle, and AC, AB, the 
including fides ; alfo fuppofe B D is perpendicular to A C. 
Then, as the radius or fine of the angle ADB 93 0 : A B :: 
fine D AB : DB the perpendicular. Therefore, when the 
firft term of the proportion or the radius is 1, DB the 
4th term will be = fine DABx AB; and fine DABxAB 
X AC = twice the area of the triangle; confequently the 
continued produd of one fide, £ the other, and the fine 
of the included angle, will be the area. 
Ex. Let ABC be a triangular field, and fuppofe tire 
angle B AC taken with a theodolite is 40 0 5'; required 
the content in acres when AC=224, and AB=i88, 
yards ?—When the radius is i, the natural fine of 40 0 f 
will be '6439. Therefore 224X9+X'6439=13557-9584, 
the area in fquare yards, equal to 2-80123 acres, nearly. 
But the operation is ftiorter by logarithms: 
224 = log. 2-350248 
94 = log. 1-973128 
45 0 5' log. fine = 9-808819 
log. of 13558 yards* 
the fame as before nearly. 10 is rejeded in the fum of 
the indices of the three logarithms. 
But the area may be found without letting fall a per¬ 
pendicular, by the following rule: Subtrad each fide from 
half the perimeter; then the area of the triangle will be 
equal to the fquare-root of the continued produd of th# 
faid half-perimeter and the three remainders. 
Ex. Let A B D, fig. 4. be a triangle, C the centre of 
the inferibed circle; and let the radii or perpendiculars 
CP, CO, CR, be drawn to the fides. Then^ADxCP 
is the area of the triangle ACD; £ABxCO that of the 
triangle ACB; and £BDxCR the area of BCD. But 
CP, CO, CR, are equal; therefore the three halves to¬ 
gether, or half the perimeter of the triangle multiplied, 
by the radius of the inlcribed circle is the area of the 
triangle. About G in B G, the line bifeding the angle 
ABD, fuppofe an arc of a circle is deferibed to touch 
BS, BT, and AD; and draw the perpendiculars GS, 
GT. Then BS and BT will each be equal to half the 
perimeter of the triangle A B D. ForAS=AI, and DT 
=DI; therefore B S and BT together are equal to the 
fum of three fides. And confequently DT or D I is the 
difference of half the perimeter (BT) and the fide BD; 
and AI or A S the difference of half the perimeter (B S) 
and AB. But PD —RD, RB=OB, and OA=PA; there¬ 
fore 2PD + 2OB-P2OA is the perimeter, or PD-J-OB-j-OA 
equal to half the perimeter. But OB-jpOA=AB; there¬ 
fore PD=AS=AI; and therefore AP(AO)=DI. And 
becaufe GS=AD, therefore OB is the difference of half 
the perimeter and the fide AD. Confequently AS, AO, 
O B, are the three differences between half the perimeter 
and the three fides of the triangle. 
In the quadrilaterals ASGI, OAPC, the angles at S 
and I, and O and P, are right ones; therefore OCP-f- 
O AP are equal to two right angles; and fince OAP+SAI 
make two right angles, SAI=OCP, therefore SGI =0 AP, 
and the quadrilaterals are equiangular; and becaufe GI= 
GS, and CO=CP, they are alfo limilar. And fince the 
triangles BOC, BSG, are funilar, we have OA:OC:: 
SG : SA, and BO : BS:: OC : SG. Therefore, OAxBO : 
OCxBS :: SGXOC : SAxSG :: OC : SA :: OCxBS : SA 
XBS; or, AOX-BO : OCxBS:: OCxBS: SAxBS. Con- 
iequently, OCX-bS, the area of the triangle, is a,mean 
proportional 
