1

sin (B+C)

: and this ex

pression by means of the formula for 4 arcs (art 30 ch. iii,) becomes BE. DE. Sin (C+D). Hence, collecting the terms, and arranging them in the order of the sides, they become.

AB. BC. Sin B

+DC.DE sin D.

•

Cor. Taking away from this expression, the 1st, 2d, and

4th terms, which together

make double the trapezium ABCD,

there will remain

AB. BC. sin B
+AB.CD.sin(B+C)
+AB. DE. Sin (B+C+D)
+BCCD. sin c

+BC.DE. sin (c+D)

(+CD.DE. Sin D.

AP.EF. Sin (B+C+D+E)

+DP.EF.Sin (D+E)
+DE.EF. sin E.

AB. EF. Sin (B+C+D+E)

+DC.EF. Sin (D+E)

+DE. EF. sin E

+BP.EF. Sin (B+C+D+E)

AEF

(+CP.EF. Sin (D+E).

Now, writing for BP, CP, their respective values,

sin (B+C) sin (B+C)

the sum of the last two expressions,

in the double areas of AEF, will become

sinc.sin (B+C+D+E)+sin B. Sin (D+E).

BC EF

•

and this, by means of the formula for 5 arcs (art. 30 ch. iii)

•

becomes BC EF Sin (C+D+E). Hence, collecting and properly arranging the several terms as before, we shall obtain

4. In a similar manner may the area of a heptagon be determined, by finding the sum of the areas of the hexagon and the adjacent triangle: and thence the area of the octagon, nonagon, or of any other polygon, may be inferred; the law of continuation being sufficiently obvious from what is done

above, and the number of terms

n-1n-2

when the

12'

number of sides of the polygon isn: for the number of terms is evidently the same as the number of ways in which n-1 quantities can be taken, two and two; that is, (by the nature

n-1 n-2

of Permutations) = "..

1

2

Soholium.

!

Scholium.

This curious theorem was first investigated by Simon Lhuil lier, and published in 1789. Its principal advantage over the common method for finding the areas of irregular polygons is, that in this method there is no occasion to construct the figures, and of course the errors that may arise from such constructions are avoided..

In the application of the theorem to practical purposes, the expressions above become simplified by dividing any proposed polygon into two parts by a diagonal, and computing the surface of each part separately.

Thus, by dividing the trapezium ABED into two triangles, by

the diagonal AC, we shall have

Twice area

AB. BC. sin B

}={+:

trapezium +CD. AD. Sin D.

The pentagon ABCDE may be divided into the trapezium ABCD,

+DE. AF. sin E

Twice area of

Thus again, the hexagon may be divided into two trapeziums, by a diagonal drawn from a to D, which is to be the line

excepted in the theorem; then will

AB. BC. sin B

And lastly, the heptagon may be divid-
ed into a pentagon and a trapezium, the
diagonal, as before, being the excepted B
line: so will the double area be express-
ed by 9 instead of 15 products, thus :

AB. BC. sin в

The same method may obviously be extended to other poly

gons, with great ease and simplicity.

It

It often happens, however, that only one side of a polygon can be measured, and the distant angles be determined by intersection; in this case the area may be found, independent of construction, by the following problem.

PROBLEM I.

Given the Length of One of the Sides of a Polygon, and the
Angles made at its two extremities by that Side and Lines
drawn to all the Other Angles of the Polygon: to find an
Expression for the Surface of that Polygon.

And,

...

sin (b'+"): PQ :: sin b": PB =

But, triangle APB=AP, PB,

Hence, surface A APB=PQ2.

sin a"

sin (a+a) Pc.

sin b'

sin (b'+b'") PQ.
Sin APB=AP.PB. sin (a−b).
sin a. sin b". sin (a - b)
sin (a' + a'). sin sin (b' (b' - b")
sinh". sin c". sin (b' - c'
'sin (b'+6"). sin (c'+c"
sin c". sin d". sin (c' - d')
▲ CPD=PQ sin (c+c"). sin (d+d")

In like manner, A BPC- PQ

DPQ =

2

&c. &c. &c.

QP. PD. sin DPQ = PQ

sin d"

•

sin (d+d"). PQ. sin d

sin (d'+d"). Consequently,

sin a". sin b". sin (a−b')

sin (a + a"). sin (b'+6")

sin b". sin c". sin (b' - c') + sin (b'+b"). sin (c + c2)

+

+

sin d'. sin d"

sin (d'+d")

!

The

The same method manifestly applies to polygons of any number of sides: and all the terms except the last are so perfectly symmetrical, while that last term is of so obvious a form, that there cannot be the least difficulty in extending the for mula to any polygon whatever.

PROBLEM II.

Given, in a Polygon, All the Sides and Angles, except three;
to find the unknown Parts.

This problem may be divided into three general, cases, as shown at the beginning of this chapter: but the analytical solution of all of them depends on the same principles; and these are analogous to those pursued in the analytical investigations of plane trigonometry. In polygonometry, as well as trigonometry, when three unknown quantities are to be found, it must be by means of three independent equations, involving the known and unknown parts. These equations may be deduced from either theorem 1, or 3, as may be most suited to the case in hand; and then the unknown parts may each be found by the usual rules of extermination.

For an example, let it be supposed

D

that in an irregular hexagon ABCDEF,
there are given all the sides except

AB, BC, and all the angles except в ; to

E

determine those three quantities,

1

The angle B is evidently equal to (2n - 4) right angles (A+C+D+E+F); n being the number of sides, and the

angles being here supposed the interior ones.

Let
Let AB = x, BC =y: then by th. 1,
x=y. COS B + DC COS ABACD + DE
+ EF. COS AB EF + AF

y=x.cos B + AF

•

•

COS ABED,
COS ABAAF;

COS BCAFE.

!

COS BC AF + FE
+ DE, COS BCADE + DC. COS BCACD.

In the first of the above equations, let the sum of all the
terms after y. cos B, be denoted by c; and in the second the
sum of all those which fall after x. cos B, by d; both sums
being manifestly constituted of known terms: and let the
known co-efficients of x and y be m and n respectively. Then
will the preceding equations become

x=ny + c....y=mx + d.
Substituting for y, in the first of the two latter equations, its
value in the second, we obtain x =mnx + nd + c.

there will readily be found

Whence

nd + c

mc+d

,

and y=

1-mn

-mn

Thus

Fi