Fig. 10.

B.

E

D

Find D the middle of the base, B C, and from P take a course of P D. Again set your instrument at A, and take the same course A E; cause a pole to be set at E, a line or fence from E to P will bisect the farm, which is easily demonstrated from the figure. See Bland.

14th. Again, suppose the well P, to be situated within the farm, and it be required to divide it equally between three occupants, so that each may have the use of the well.

In fig. 11 divide the base B C, into three equal parts in D and E. Set your instrument

at P, and take B

D

Fig. 11.

the courses PD and P E. Remove your instrument to A, and take A F the same course as P D, and A G the same as PE. Cause stakes to be driven at F and E in a straight line between B and C. Fences from F, G, and A, to P, trisect the farm, which is plain from the figure.

15th. To find the area of a Trapezoid Rule, multiply half the sum of the parallel sides by the perpendicular distance between them, and the product is the area.

It is evident the triangles, B F G, and FC H, are similar and equal. (26th Euclid, 1st.) .. E F, half the sum of the sides, multiplied by the perpendicular distance between them, A D, gives the area.

Being surveying on the side of a bog, and wanting four acres to make up a division, and seeing A

Fig. 13.

Pond.

50

B would pass through a pond, I found A C fifty chains, and L C 56°; how far must I measure from C towards B, so that the triangle A B C, may contain four acres.

Since A CX CBX the natl. sine of 56°=4 acres, it follows that 4 acres divided by the product of one half the natl. sine of 56° into A C, gives B C the required side. Thus :

C

50X4=200 perches X, 4145188-82.9; and 640 perches in 4 acres, divided by 82.9=7.72 per the length of B C, and in like manner any other similar case can be done.

17th. Sometimes it is found necessary to obtain the area of a trapezium from having the diagonals and the angle of intersection given:

Rule-Half the product of the diagonals multiplied by the natural sine of the angle of intersection, will be the area.

EXAMPLE.

If the two diagonals of a trapezium be 40.15, and 60.13 chains the of intersection 75° 45', what is the area. of 40.15X60.13-1207.1097= half the product of the diagonals, and 1207.1097× 96923=(natural sine of 75°45')=1169.966934531=: the area, in square four pole chains, or 116.3. 39. 47. Answer.

A. R. P.

18th. To find the area of a trapezium, when each side and the angle of intersection of the diagonals are given.

the trapezium; add

Rule-Square each side of together the squares of each pair of opposite sides; subtract the less from the greater; multiply the difference by the tangents of the angle of intersection. One fourth of the product will be the area.

EXAMPLE.

What is the area of a trapezium, the sides of which are 10, 13, 7.16, 8.32, and 10.05 chains respectively, and the V of intersection of the -diagonals 52° 15'.

2

171.8393-Sum of sqs. of opposite sides.

(15.05)=226.5025

2

( 7.16)= 51.2656

277.7681-Sum of sqs. of other sides.

105.9288 Difference,

Multiplied by .32288= the natural tangent,

34.20290944 or

A 3. 1 .27,23 perches.

For a demonstration of the foregoing, see Gibson's Surveying, by Trotter.

19th. To find the area of a trapezium, when the four sides are severally given, and also the sum of any two opposite angles. Rule-From half the sum of the four given sides, subtract each severally; multiply the four remainders continually together; from the result subtract one half the continual product of the four sides, multiplied by unity, increased by the natural cosine of the sum of the given angles. The square root of the result will be the area.