EXAMPLES.

Ex. 1. To find the area of a parallelogram, the length being 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. 9 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 I. MULTIPLY the base by the perpendicular height, and take half the product for the area.* Or, multiply the one of these dimensions by half the other.

in the length, repeated as often as there are linear measuring units in the breadth, or height; that is, equal to the length drawn into the height; which here is 4 X3 or 12.

And it is proved. (Geom. theor. 25, cor. 2), that any oblique parallelogram is equal to a rectangle, of equal length and perpendicular breadth. Therefore the rule is general for all parallelograms whatever.

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

EXAMPLES.

EXAMPLES.

Ex. 1. To find the area of a triangle, whose base is 625, and perpendicnlar height 520 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. 663 square yards. Ex. 3. To find the number of square yards in a triangle, whose base is 49 feet, and height 254 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, for the area.*

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

By Natural Numbers.

First X 40 X 30= 600,

then, 1 600 :: 484046 sin. 28° 57'

Answer

600

By Logarithms.

log. 9-684887

2-778151

290-4276 the area answering 2-463038

Ex. 2. How many square yards contains the triangle, of which one angle is 45°, and its containing sides 25 and 211 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. Then 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.*

ed in K ; which will pass through G, because & is a right angle, and through 14 because, by means of the parallels, AIFB DF, therefore HD Ha, and HF

HI=AB.

half their =AC+ CB

Hence HA or HD is half the difference of the sides AC, CB, and HC = sum or = AC+ CB; also HK = нI = IF or AB; conseq. CK=

AB half the sum of all the three sides of the triangle ABC, or CKs, calling s the sum of those three sides. Again HK HI = theref. CL CK - KL = s— AB, and AK = CK — CA =38 —

CX CDs S- CB.

AB

AB Or KLAB;

AC,

and ALDK

Now, by the first rule, AG. CG the ▲ ACE, and AG. FG the ▲ ABE, theref. A6. CFA ACB. Also by the parallels, AG: CG:: DF OF IA: CF, theref. AG. CG. DF, conseq. AG. CF. CG, DF =2 ACB.

CF =(A ACB =) CG. IA

But CG. CF CK. CLs. 8 AB, and AG. DF ak .'al =

BC; theref. AG. CF, CG. DF =▲2 ACBS. S —AB. 18—AC. 8-BC

is the square of the area of the triangle ABC. 2. E. D.

Otherwise.

Because the rectangle AG. Cr the ▲ ABC, and since CG: AG :: CF : DF, drawing the first and second terms into CF, and the third and fourth into AG, the prepor. becomes cG. CF: AG. CF :: AG. CF: AG. DF, or CG. CF: A ABC:: A ABC: AG. DF, that is, the ▲ ABC is a mean proportional between co. CF and ag . df, or betweens.

— AB and 18 — AC. S→ BC, Q. E. Á.

Ex. 1..

Ex. 1. To find the area of the triangle whose three sides

Then 45 X 25 X 15 X 5 = 84375,

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, feet?

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 the distance between them; and take half the product for the area. By Geom. theor. 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 X 770

152075 square links 15 acr. 33 perc.

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

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 toge ther.

Or thus, let fall two perpendiculars on the diagonal from the other two opposite angles; then add these two perpendi culars together, and multiply that sum by the diagonal, taking half the product for 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, Here 16+18= 34, its half is 17. Then 42 X 17 714 the area.

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 33 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 cr 70 yards. Required the construction of the figure, and 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 trape. ziums and triangles. Then find the areas of all these separately, and add them together for the content of the whole polygon

EXAMPLE.