2.7372294

The diff. is the log. of 546.0462 the area

2. Lay down the following trapezium, and find its area; AE measuring 1125, AB 3243, DE 1168, DF 1216, DB 2418, and CF 610 links.

Note. This figure is divided into two triangles, because two perpendiculars cannot be taken upon either of the diagonals.

Here 3243 x 1168237878242 = 1893912

14740802737490,

the area of the triangle ABD. Also, 2418 x 610 2 the area of the triangle BCD. Then 18939127374902681402 square links= 26a. 1r. 10p. the area of the trapezium ABCD.

3. How many square yards of paving are there in a trapezium whose diagonal is found to measure 126 feet 3 inches, and perpendiculars 58 feet 6 inches, and 65 feet 9 inches? Ans. 871.47569 yards. 4. In taking the dimensions of a trapezium, I found the first perpendicular to rise at 568, and to measure

835 links; the second at 1865, and to measure 915 links; the whole diagonal measured 2543 links; what is the area of the trapezium? Ans. 22a. 1r. Op.

5. Lay down a trapezium, and find its area from the following dimensions; namely, the side AB measures 345, BC 156, CD 323, DA 192, and the diagonal AC 438, feet. Ans. 52330.33406 feet.

6. The sides of a trapezium, two of whose opposite angles are together equal to 180 degrees, measure 30, 321, 35, and 37 feet; what is its area?

Ans. 1131 ft. 2 in. 9 pa.

PROBLEM VIII.

To find the area of a trapezoid.

RULE.

Multiply the sum of the parallel sides by the perpendicular distance between them, and half the product will be the area.

Or, half the sum of the sides multiplied by their distance will give the area.

EXAMPLE.

1. What is the area of the trapezoid ABCD, the parallel sides AD and BC of which are 25 and 18; and AB, the perpendicular distance between them, 38 feet?

Here 25+ 18 x 38 43 x 38 = 1634; and

1634

2

=817 feet, the area required.

2. Required the area of the trapezoid ABCD, whose parallel sides AB and DC measure 46 feet 10 inches, and 28 feet 4 inches; DE, the perpendicular distance between them, 26 feet 9 inches; and AE 12 feet 61 inches.

=

Here 46 feet 10 in. 46.83333, and 28 feet 4 inches = 28.33333; then 46.83333 + 28.33333 × 26.75 75.16666 × 26.75 2010.708155, half of which is = 1005.3540775 feet, the area required.

3. The parallel sides of a piece of ground measure 856 and 684 links, and their perpendicular distance 985 links; what is its area? Ans. 7a. 2r. 13 p.

4. If the parallel sides of a garden be 65 feet 6 inches, and 49 feet 3 inches, and their perpendicular distance 56 feet 9 inches; what did it cost, at £325 10s. per acre? Ans. £24 6s. Tid.

5. The hipped roof of a square building is flat at the top; the length of the eaves, from hip to hip, is 54 feet 6 inches; the side of the square, at the top, is 30 feet 9 inches; and the nearest distance from the top to the eaves, is 18 feet 3 inches; how many square yards of slating are contained in the four sides of the roof? Ans. 345.73611 yards.

PROBLEM IX.

To find the area of an irregular polygon of any number of sides,

F

RULE.

Divide the figure into triangles and trapeziums in the most convenient manner; and find the area of each separately; then the sum of these areas will be the area of the polygon.

Note. In calculating the content of an irregular polygon, it is sometimes more eligible to find the double area of each figure into which it is divided; and half the sum of these double areas, will be the area of the whole polygon.

EXAMPLES.

1. It is required to lay down the irregular figure ABCDEFGA, and to find its area from the following dimensions.

CONSTRUCTION. Draw the line AB, which make 35.5 and lay off 25 from A to m, at which point

erect the perpendicular Cm = 8; join AC and BC, and you will have triangle ABC.

With C as a centre, and the radius Cn describe an arc; and with A as a centre, and the radius An, describe another arc cutting the former in n. Through n draw the diagonal AD = 36, upon which lay off Ao 12. At o erect the perpendicular Go 7.5; join CD, DG, and GA, and the trapezium ACDG will be completed.

The trapezium DEFG may be constructed in a similar manner.

CALCULATION.

Here 35.5 x 8 284, double the

area of the triangle ABC.

Again, 7.58.6 x 36 16.1 × 36 = 579.6, double the area of the trapezium ACDG.

Also, 10.49.2 × 34 = 19.6 × 34 = 666.4 double the area of the trapezium DEFG.

Then,

284579.6

2

666.4 1530
2

the area of the irregular polygon required.

= 765 feet,

2. It is required to lay down a pentangular field, and find its annual value at £2. 5s. per acre; the first side measuring 926, the second 536, the third 835, the fourth 628, and the fifth 587 links; and the diagonal from the first angle to the third 1194, and that from the third to the fifth 1223 links. Ans. £18. 10s. 74d. Note. This field is divided into three triangles, the arcas of which may be found by Problem V.

PROBLEM X.

Given the side of a regular polygon, to find the radius of its inscribed or circumscribing circle.

RULE.

Multiply the given side by the number standing opposite to the name of the polygon, in the third, or fourth column of the following table, as the case requires; and the product will be the radius of the inscribed or circumscribing circle.