Encyclopaedia Britannica, or, a Dictionary of arts, sciences, and miscellaneous literature : enlarged and improved. Illustrated with nearly six hundred engravings > Volume 13, MAT-MIC

MENSURATION.
Of Plar.e
Figures.
Rule III.
Fig. ip-
When the three fides are given, add together the
three fides, and take half the fum. Next, fubtraft
each fide feverally from the laid half fum, thus obtain¬
ing three remainders. Lailly, multiply the laid half
fum, and thofe three remainders all together, and ex-
traft the fquare root of the lalt product for the area of
the triangle.
This pra&ical rule is deduced from the following
geometrical theorem. The area of a triangle is a
mean proportional between two rellang/cs, one cf which
is contained by half tM perinLter of the triangle, and
the exefs of half the perimeter above any one of its
fdes ; and the other is contained by the exceffes of half
the perimeter above each of the other two fides. As this
theorem is not only remarkable, but alfo of great utili¬
ty in menfuration, we tliall here give its demonftra-
tion.
Let ARC then be any triangle ; produce AB, any
one of its fides, and take BD, and B d, each equal to
BC ; join CD and C d, and through A draw a line pa¬
rallel to BC, meeting CD and C d produced in E and
e ; thus the angle AED will be equal to the angle
BCD, (Geometry, Sedl. I. Theor. 21.), that is, to
the angle BDC or ADC, (Seft. I. Theor. 11.) 5 and
hence AErrAD (Se£L I. Theor. 12.) •, and in like
manner, becaufe the angle A e r/ is equal to the angle
BC d, that is, to the angle B r/ C, or A r/ e, therefore
A e~A d.
On A as a centre, at the diftance AD or AE, de-
feribe a circle meeting AC in F and G ; and on the
fame centre, with the diftance A r/ or A e, deferibe an¬
other circle meeting AC in f and £•, and draw BH
and B h perpendicular to CD and C d. Then, becaufe
BD, BC, B d are equal, the point C is in the circum¬
ference of a circle, of which D r/ is the diameter, there¬
fore CD and C d are bifebted at FI and h (Sedft. II.
Theor. 6.) and the angle DC d is a right angle, (Se£E
II. Theor. 17.), and hence the figure CHB h is a reft-
angle, fo that B/z :r: CH CD, and BFI—C/zz=
^Cd.
Join BE, and B e, then the triangle BAG is equal
to each of the triangles EEC, B e C (Seft. IV. Theor.
2. Cor. 2.) •, but the triangle BEC is equal to -§• EC X
BH (Se6t. IV. Theor. 2.), that is to ^ EC d ; and
in like manner the triangle B e C is equal to ^ C x
B //, that is to ^ e C X CD, therefore the triangle ABC
is equal to EC X C z/, and alfo to ^ e C X CD.
Now fince CD : C</:: CE X CD : CEX Cz/y Se<ft. IV.
and alfo CD : Cz/:: C e X CD : C e X Cz/ £ Theor. 3.
ThereforeCE X CD : CE X C z/:: C X CD : Cz=X Cd;
that is, becaufe CE x CD—EC x CG, and C e x C d—
fC xC g (Seft. IV. Corollaries to Theor. 28. and 29.).
FCxCG : CExCz/: CexCD ^CxC^;
which laft proportion (by taking one-fourth of each of its
terms, and fubftituting the triangle ABC for its equi¬
valent values ^ CE X C z/ and C e X CD) gives us
4 FC x^CG : trian. ABC :: trian. ABC : |/C x4Cg.
Now, if it be confidered that the radius of the circle
DGE is AB-|-BC, and that the radius of the circle
g d e is AB—BC, it will readily appear that, putting
2 r for the perimeter of the triangle ABC, we have
FC(=AB+ BC+AC)=2r
CG(=AB-f BC — AC ) — is—2 AC,
/C (=AC-f-^ AB—BC j- j —2x—2 BC,
gC{x=.AC—AB—BC^ ) — 2s—2AB.
Put now a, b, c for the fides AC, BC, and ABrcfpec-
tively, then FCzrr, 4 GC—j—a,\fC—s—b, ^ Cg
—s—c • thus the laft proportion becomes
r X (•i—-«) : trian. ABC :: ZrA/?. ABC: (j—X (r—c),
which conclufion, when exprefled in words at length, is.
evidently the proportion to be demonttrat d.
And as a mean proportional between two quantities
is found by taking the fquare root of the product, it
follows that the area of the triangle ABC, which is a
mean between s x (•!—a) and (r—£) X (r—c), is
equal to
*/^sX 0—a) x (r—b) x (s—c) "|-
which formula, when expreffed in words at length, gives
the preceding rule.
Example. Required the area of a triangle whofe
three fides are 24, 36, and 48 chains refpedlively.
Here 24-{-36-1-48—108 — the fum of the three fides.
And — — rqzr half that fum.
2
Alfo 54—24 = 30, the firft remainder 5 54—36=
18, the fecond remainder j and 54—48=6, the third
remainder.
The produdft of the half fum and remainders is
54 X 3° X 18 X 6= 174960.
And the fquare root of this product is
\/(I7496o):=4i8.28 fiq. ch. the area required.
Problem IV.
To find the area of a trapezoid.
Rule.
Add together the two parallel fides, then multiply
their fum by the perpendicular breadth, or diftance be¬
tween them, and half the product will be the area.
This rule is demonltrated in Geometry, Sect. IV.
Theor. 7.
Example. Required the area of the trapezoid AB
CD, whofe parallel fides AB and DC are 7.5. and
I 2.25 chains, and perpendicular breadth DE is 1 c.4
chains.
The fum of the parallel fides is 7.5-1-12.23 = 19.755
which multiplied by the breadth is
• 9-7 i X 15.4=304.15 5
and half this produdt is
304.1c
—:— = i52-°75/?* c}l'— *Sac- 33-2T°*
the area required.
3 T 2
Problem