EXAM. XXIV.-Wanting to know my distance from an inaccessible object O, on the other side of a river; and having no instrument for taking angles, but only a chain or cord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a direct line from the object O 100 yards, viz. AC and BD each equal to 100 yards; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object O from each station A and B?

Ans. {40 536.25

BO 500-09

MENSURATION OF PLANES:

THE area of any plane figure, is the measure of the space contained within its extremes or bounds; without any regard to thickness.

This area, or the content of the plane figure, is estimated by the number of little squares that may be contained in it; the side of those little measuring squares being an inch, a foot, a yard, or any other fixed quantity. And hence, the area or content is said to be so many square inches, or square feet, or square yards, &c.

Thus, if the figure to be measured be the rectangle ABCD, and the little square E, whose side is one inch, be the measuring unit proposed: then, as often as the said little square is contained in the rectangle, so many square inches the rectangle is said to contain, which in the present case is 12.

3

D

4

C

A

B

E

PROBLEM L

To find the area of any parallelogram, whether it be a square, a rectangle, a rhombus, or a rhomboid.

Multiply the length by the perpendicular breadth, or height, and the product will be the area. *

• The truth of this rule is proved in the Geometry, Theor. 81, Cor. 2.

The same is otherwise proved thus: Let the foregoing rectangle be the figure proposed; and let the length and breadth be divided into equal parts, each equal to the lineal measuring unit, being here 4 for the length, and 3 for the breadth; and let the opposite points of division be connected by right lines. Then, it is evident that these lines divide the rectangle into a number of little squares, each equal to the square measuring unit E; and farther, that the number of these little squares, or the area of the figure, is equal to the number of lineal measuring units in the length, repeated as often as there are lineal measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 X 3 or 12.

And it is proved (Geometry, Theor. 25, Cor. 2), that a rectangle is equal to any oblique parallelogram, of equal length and perpendicular breadth. Therefore, the rule is general for all parallelograins whatever.

EXAMPLES.

Ex. 1.-To find the area of a parallelogram, whose length is 12-25, and height 8.5.

12.25 length

8.5 breadth

6125

9800

104 125 area

Ex. 2.—To find the area of a square, whose side is 35.25 chains.

Ans. 124 acres, 1 rood, 1 perch.

Ex. 3.—To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 feet. Ex. 4. To find the content of a piece of land, in form of a rhombus, its length being 6.20 chains, and perpendicular height 5:45.

Ans. 3 acres, 1 rood, 20 perches. Ex. 5.-To find the number of square yards of painting in a rhomboid, whose length is 37 feet, and breadth 5 feet 3 inches.

Ans. 21 square yards.

PROBLEM II.

To find the area of a triangle.

RULE 1.-Multiply the base by the perpendicular height, and half the product will be the area.* Or, multiply the one of these dimensions by half the other.

EXAMPLES.

Ex. 1.—To find the area of a triangle, whose base is 625, and perpendicular height 520 links ?

Here 625 X 260 = 162500 square links,

or equal 1 acre, 2 roods, 20 perches, the answer. Ex. 2.-How many square yards contains the triangle, whose base is 40, and perpendicular 30 feet? Ans. 66% square yards. Ex. 3.—To find the number of square yards in a triangle, whose base is 49 feet, and height 25 feet. Ans. 68, or 68 7361. Ex. 4.—To find the area of a triangle, whose base is 18 feet 4 inches, and height 11 feet 10 inches. Ans. 108 feet, 5 inches.

RULE II.-When two sides and their contained angle are given: Multiply the two given sides together, and take half their product: Then say, as radius is to the sine of the given angle, so is that half product, to the area of the triangle.

Or, multiply that half product by the natural sine of the said angle. †

* The truth of this rule is evident, because any triangle is the half of a parallelogram of equal base and altitude, by Geometry, Theor. 26.

+ For, let AB, AC, be the two given sides, including the given angle A. Now AB X CP is the area, by the first rule, CP being perpendicular. But, by Trigonometry, as sine angle P, or radius, is to sine angle A:: AC: CP sine angle A X AC, taking radius 1. Therefore, the area AB X CP is X sin, angle A, to radius 1; or,

AB X AC

A

Р

as radius sin. angle A :: AB X AC: the area.

Ex. 1.-What is the area of a triangle, whose two sides are 30 and 40, and their contained angle 28° 57′ 18"?

HereX 40 X 30 = 600,

Therefore, 1:4841226 nat. sin. 28° 57' 18"

600

290 47356, the answer.

Ex. 2.-How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 21 feet?

Ans. 20-86947.

RULE III-When the three sides are given: Add all the three sides together, and take half that sum. Next, subtract each side severally from the said half sum, obtaining three remainders. Lastly, multiply the said half sum and those three remainders all together, and extract the square root of the last product, for the area of the triangle. *

Ex. 1.-To find the area of the triangle whose three sides are 20, 30, 40.

The root of which is 290 4737, the area.

Ex. 2.-How many square yards of plastering are in a triangle, whose sides are 30, 40, 50?

Ans. 663. Ex. 3-How many acres, &c. contains the triangle, whose sides are 2569, 4900, 5025 links? Ans. 61 acres, 1 rood, 39 perches.

PROBLEM III.

To find the area of a trapezoid.

Add together the two parallel sides; then multiply their sum by the perpendicular breadth or distance between them; and half the product will be the area; by Geometry, theorem 29.

Ex. 1.—In a trapezoid, the parallel sides are 750 and 1225, and the perpendicular distance between them 1540 links: to find the area.

1225
750

1975 × 770 = 152075 square links = 15 acres, 33 perches,

* For, let a, b, c, denote the sides opposite respectively to A, B, C, the angles of the triangle A B C (see last fig.); then by Theor. 37, Geom. we have BC2=AB2+AC2—2AB.AP, or a2=b2+c2—2c.AP b2+c2—a2 .. AP= ; hence we have

.•. 4c2.CP2={(b+c)2—a2} • {a2—(c—b)2}=(a+b+c)(−a+b+c)(a−b+c)(a+b−c)

2

(a+b+c)=half the sum of the three sides.

Ex 2.-How many square feet are contained in the plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches? Ans. 13 feet.

Ex. 3.—In measuring along one side AB of a quadrangular field, that side and the two perpendiculars let fall on it from the two opposite corners, measured as below: required the content.

AP = 110 links.

AQ = 745

AB = 1110

CP = 352

DQ = 595

Ans. 4 acres, 1 rood, 5·792 perches.

PROBLEM IV.

To find the area of any trapezium.

Divide the trapezium into two triangles by a diagonal: then find the areas of these triangles, and add them together.

Note. If two perpendiculars be let fall on the diagonal, from the other two opposite angles, the sum of these perpendiculars being multiplied by the diagonal, half the product will be the area of the trapezium.

Ex. 1.—To find the area of the trapezium, whose diagonal is 42, and the two perpendiculars on it 16 and 18.

Ex. 2.-How many square yards of paving are in the trapezium, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 334 feet ? Ans. 222 yards.

Ex. 3. In the quadrangular field ABCD, on account of obstructions there could only be taken the following measures, viz. the two sides BC 265, and AD 220 yards, the diagonal AC 378, and the two distances of the perpendiculars from the ends of the diagonal, namely, AE 100, and CF 70 yards. Required the area in acres, when 4840 square yards make an acre?

Ans. 17 acres, 2 roods, 21 perches.

PROBLEM V.

To find the area of an irregular polygon.

Draw diagonals dividing the proposed polygon into trapeziums and triangles. Then find the areas of all these separately, and add them together for the content of the whole polygon.

EXAM. To find the content of the irregular figure ABCDEFGA, in which are given the following diagonals and perpendiculars; namely,

PROBLEM VI.

To find the area of a regular polygon.

RULE 1.-Multiply the perimeter of the polygon, or sum of its sides, by the perpendicular drawn from its centre on one of its sides, and take half the product for the area.*

Ex. 1.—To find the area of the regular pentagon, each side being 25 feet, and the perpendicular from the centre on each side is 17.2047737.

Here 25 X 5 = 125 is the perimeter.

And 17-2047737 X 125 = 2150·5967125.
Its half 1075-298356 is the area sought.

RULE II.-Square the side of the polygon; then multiply that square by the area or multiplier set against its name in the following table, and the product will be the area. †

EXAM. Taking here the same example as before, namely, a pentagon, whose side is 25 feet.

Then, 252 being = 625,

And the tabular area 1.7204774;

Therefore, 1.7204774 × 625 = 1075-298375, as before.

Ex. 2. To find the area of the trigon, or equilateral triangle, whose side is 20, Ans. 173.20508.

*This is only in effect resolving the polygon into as many equal triangles as it has sides, by drawing lines from the centre to all the angles; then finding their areas, and adding them all together.

+ This rule is founded on the property, that like polygons, being similar figures, are to one another as the squares of their like sides; which is proved in the Geometry, Theorem 89. Now, the multipliers in the table, are the areas of the respective polygons to the side 1. Whence the rule is manifest.

Note.-The areas in the table, to each side 1, may be computed in the following manner: From the centre C of the polygon draw lines to every angle, dividing the whole figure into as many equal triangles as the polygon has sides; and let ABC be one of those triangles, the perpendicular of which is CD. Divide 360 degrees by the number of sides in the polygon, the quotient gives the angle at the centre ACB. The half of this gives the angle ACD; and this taken from 90°, leaves the angle CAD. Then, as radius is to AD, so is tangent angle CAD to the perpendicular CD. This multiplied by AD, gives the area of the triangle ABC; which, being multiplied by the number of the triangles, or of the sides of the polygon, gives its whole area, as in the table.

A D B