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2. For a pentagon, as ABCDE. Its area is obviously equal to the sum of the areas of the trapezium ABCD, and of the triangle ADE. Let the sides AB, DC, as before, meet when produced at P. Then, from the above, we have

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+AB. DC. sin (B+c)
+ BC DC. sin c.

And, by the preceding corollary,

Twice triangle


That is, twice

triangle DAE



BC. sin c

Now, BP =


sin (B+c)'

two terms become


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AP.DE.Sin (P+ D) or sin (B+C+D)

+ DP.DE. sin D.

AB. DE. sin (B+C+D)

+ DC. DE. sin D

+BP. DE. sin (B+C+D)
+CP. DE. sin D.

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sin (B+c) therefore the last

BC. DE. sin c. sin (B + C + D) BC.DE.sin B. sind
sin (B+c)

sin B. sin D + sin c. sin (B+C+D)

sin (B+ c)

sin (B+C)

: and this expression,

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

Twice the area'

of the penta



AB. BC . sin B

+ AB. DC. sin (B+c) ·

AB. DE sin (B+C+D)
+ BC. DC. sin c'

+ BC. DE sin (C+D)
+ 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

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3. For a hexagon, as ABCDef. double area will be found, by supposing it divided into the pentagon ABCDE, and the triangle AEF. For, by the last rule, and its corollary, we have,


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13.3. s B


34 sin B+C+D' -43-27-sin.s+c+p+s)

-3C.. sin c

-3C.DE.sin (C+D)
-C.EF.sin C+D+B)

CD DE. sin D

→CD . EF . sin (D+8)

+DE. B. sin È.

4:r mer may the area of a heptagon be á Lining he sum of the areas of the hex ** de arrange: and thence the area of the v MOLT I ET her polygon, may be interes of figu

are the municer of terms =

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rine fhe polygon is n: for the muc”, te time as the number of ways ***** Ken, two and two; that ***

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This curious theorem was first investigated by Simon Lhuillier, 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 ABCD into two triangles, by the diagonal AC, we shall have

Twice area?


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AB. BC. sin B +CD. AD. sin D.

The pentagon ABCDE may be divided into the trapezium ABCD, and the triangle ADE, whence

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AB. BC . sin B
+AB. DC. sin (B+C)

+ BC DC. sin c

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+DE. AE. sin E.

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

Twice area of
hexagon S

AB. BC. sin B
+AB. DC. sin (B+c)
+BC. DC. sin c

+DE. EF . sin E

+DE. AF. Sin (E+F)
+EF. AF sin F.


And lastly, the heptagon may be divided into a pentagon and a trapezium, the diagonal, as before, being the excepted line: so will the double area be expressed by 9 instead of 15 products, thus : AB . BC . sin B +AB. CD. sin (B+C)

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+EF. GA. Sin (F+G)

L+FG. GA. Sin G.

The same method may obviously be extended to other

polygons, with great ease and simplicity.

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.


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.

Here we suppose known PQ; also
AQPa", BQP = b", cap=c", DQP=d".
Now, sin PAQ = sin (a+a"); sin PBQ =
sin (b'+b").

Therefore, sin (a' + a"): PQ :: sin a": PA =

And,. . . sin (b' + b′′) : PQ :: sin b′′ : P3 =


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sin a"


sin (a'+a")
sin "


sin (+b)

But, triangle APB=AP. PB. 1⁄2 sin APB=÷AP. PB.sin (a ́—b'). .

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▲ DPQ = QP. PD. sin DPQ = PQ.

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sin d'. sin d"
sin (d'+d') *


Surface PABCDQ = PQ2.

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PQ. sind


sin d"
sin (d+d")

sin a". sin " . sin (a'-b')
sin (a+a"). sin (l' + b′′)
sin b". sin c", sin (b'-c')
sin (b' + b). sin (c' + c')
sin c". sin d'. sin (c'-d')

+ sin (c′ + c′′). sin (d'+d′′)

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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 formula to any polygon whatever.



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 that

in an irregular hexagon ABCDEF, there
are given all the sides except AB, BC, and F
all the angles except B; to determine
those three quantities.




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 AB = x, BC=y: then, by th. 1,

y = x

x = y. cos B + DC. COS AB CD + DE. COS ABAED + EF. COS ABAEF+ AF. COS ABAF; COS BAF. COS BCAAF + FE. COS BCAFE 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 r. cos B, by d; both sums being manifestly constituted of known terms: and let the known coefficients 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. Whence there will readily be found

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-- mn

Thus AB and BC are determined. Like expressions will serve for the determination of any other two sides, whether contiguous or not: the coefficients of r and y being designated by different letters for that express purpose; which would have been otherwise unnecessary in the solution of the individual case proposed.


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