may serve however, as the ground work of, or for correcting Military sketches."

For temporary use, it will sometimes be necessary to measure a distance by pacing, in order to adapt a scale to the plan or sketch (167).

N

M

D

254. The Compass will also be found useful in reconnoitring a country with a map or plan when the direction of the meridian is laid down, and we know the magnetical variation. Let SN be the direction of the true meridian on a map; and suppose the wind-mill O bears NE 68°, and the spire R, NW 36° by the compass; also let the variation be 23° W.

R

S

B

Make the angle MCN 23°, then CM will represent the magnetical meridian: let the angle MCP 68°, and MCD

36°; then if OB, RB are drawn parallel to PC, DC, respectively, the intersection B will be the place of observation on the map or plan. If however, the intersection (B) is very acute or obtuse, the position thus determined may be considerably wide of the truth.

MENSURATION.

Of Right-lined Plane Figures.

of a

255. THE measure of the space or surface contained within the boundaries of any plane figure is called its Area or Superficial Content. This is estimated in acres, square yards, square feet, or some other fixed or determinate measure. Thus, the top if we suppose ABCD to represent rectangular table whose length DC is 5 feet, and breadth DA = 3; then the upper surface will contain 5 × 3 or 15 square feet 89, corol. 2): a square foot being the unit or integer by which the area is estimated.

A. 0

B

Р C

But if the dimensions are taken in yards, its length will be 13, and breadth 1 yard; and the superficial content = 1 × 1 = 1 square yards; for AOPD is the square yard, and the rectangle OC is of a square yard in this case a square yard is the measuring unit. And when the length and breadth are denoted in inches, a square inch becomes the measuring unit or integer, and the area will be 60 × 36 = 2160 square

inches.

To find the area of a Parallelogram of any kind.

256. MULTIPLY the length by the perpendicular breadth, or the base by the height, and the product will be the area.

1. What is the content of the parallelogram DGLC whose length DC is 5 feet, and breadth CB is 3?

D

Ans. 5 x 3 15 square feet.

VOL. I.

XX

For let DA be perpendicular to DC; then the parallelogram GLCD is equal to the rectangle ABCD (82); and the area of the latter is DC x CB.

2. What is the superficial content of a rectangular board, the length being 13f. 5in. and breadth 10 inches?

Ans. 11f. 106 in.

3. How many acres are contained in a square field, the side being 11 chains, 56 links?

Ans. 13ac. 58 1376 poles.

4. How many yards (in length) of matting that is of a yard wide will cover a floor 42 feet long, and 26 broad: And what will be the expence at 1s. 5d. per square yard?

Ans. 1663 yards in length.

Expence 9l. 2s. 572d.

5. What length must be cut off a board which is 16 inches broad, and 4 feet long, so that the part remaining shall be equal to 5 square feet?

Ans. 10 inches.

To find the area of a Triangle.

257. MULTIPLY the base by the perpendicular height, and half the product will be the area. Or multiply the base by half the height, or the height by half the base.

For a triangle is equal to half a parallelogram of the same base and altitude. (82", corol. 1).

EXAMPLES.

1. How many acres are contained in a triangular field, one side being 470 yards, and the perpendicular on that side 396 yards?

Ans. 19

OF PLANES.

2. Required the number of square yards in a triangle whose base is 29, and perpendicular 224 feet?

3.

Ans. 35.

What is the area of a triangle whose base is 15f. 5in.. and perpendicular 9f. 71⁄2in?

Ans. 74f. 274in.

258. When two sides of a triangle and their included angle are given.

Multiply the product of the given sides by the sine of the included angle, and half this last product will be the area.

Demonstration. Let BAC be the given

angle, and AC, AB the including sides; also suppose BD is perpendicular to AC.

A

B

Then (221) as the radius or sine of the angle ADB 90° : AB :: sin. DAB DB the perpendicular; therefore when the first term of the proportion or the radius is 1, DB the 4th. term will be sin. DAB × AB; and sin. DAB × AB x AC twice the area of the triangle; consequently the continued product of one side, the other, and the sine of the

included angle, will be the area.

EXAMPLES.

1. Let ABC be a triangular field, and suppose the angle BAC taken with a Theodolite is 40° 5'; required the content in acres when AC = 224, and AB = 188 yards?

When the radius is 1, the natural sine of 40° 5' will be 6439 (218); therefore 224 × 94 × 6439 = 13557.9584 the area in square yards, equal to 2.80123 acres, nearly.

N. B. 10 is rejected in the sum of the indices of the three logarithms, because when the log. sine is used, the log. of the radius becomes a divisor.

2. If an angle of a triangle is 104° 27' 26", and the including sides 29.46, and 36.9; what will be the area?

259. When the three sides are given, find the segments of the base (228); then the perpendicular may be determined by Case II. Trigonometry, or the square root (83, corol.)

Examp. Suppose CA 42, AB 30, and CB= 22; what is the area of the triangle?

As 42: 30+ 22: 30-22: 9.905 the difference of the segments PA, PC.

B

16018 PC; whence BP = 15.049; and the

area of the triangle=316·029.

260. But the area may be found without letting fall a perpendicular by the following rule:

Subtract each side from half the perimeter; then the area of the triangle will be equal to the square root of the continued product of the said half perimeter and the three remainders.

Investigation. Let ABD be a triangle, C the centre of the inscribed circle; and let the radii or perpendiculars CP, CO, CR be drawn to the sides.