2. If the side of a rhombus is 67 feet, and one of the an gles 73°, what is the area? Ans. 4292.7 feet.

6. When the dimensions are given in feet and inches, the multiplication may be conveniently performed by the arithmetical rule of Duodecimals; in which each inferior denomination is one twelfth of the next higher. Considering a foot as the measuring unit, a prime is the twelfth part of a foot; a second, the twelfth part of a prime, &c. It is to be observed, that, in measures of length, inches are primes; but in superficial measure they are seconds. In both, a prime is of a foot. But of a square foot is a parallelogram, a foot long and an inch broad. The twelfth part of this is a square inch, which is of a square foot.

144

Ex. 1. What is the surface of a board 9 feet 5 inches, by 2 feet 7 inches,

2. How many feet of glass are there in a window 4 feet 11 inches high, and 3 feet 5 inches broad?

Ans. 16F. 9′7′′, or 16 feet 115 inches.

7. If the area and one side of a parallelogram be given, the other side may be found by dividing the area by the given side. And if the area of a square be given, the side may be found by extracting the square root of the area. This is merely reversing the rule in art. 4. See Alg. 520, 521.

Ex. 1. What is the breadth of a piece of cloth which is 36 yds. long, and which contains 63 square yds. Ans. 13 yds.

2. What is the side of a square piece of land containing 289 square rods?

3. How many yards of carpeting 14 yard wide, will cover a floor 30 feet long and 224 broad?

Ans. 30 X 22 feet=10X71-75 yds. And 75-11=60. 4. What is the side of a square which is equal to a parallelogram 936 feet long and 104 broad?

5. How many panes of 8 by 10 glass are there, in a window 5 feet high, and 2 feet 8 inches broad?

8. RULE I. MULTIPLY ONE SIDE BY HALF THE PERPenDICULAR FROM THE OPPOSITE ANGLE. Or, multiply half the side by the perpendicular. Or, multiply the whole side by the perpendicular, and take half the product.

The area of the triangle ABC (Fig. 5.) is equal to PCX AB, because parallelogram of the same base and height is equal to PCXAB, (Art. 4.) and by Euc. 41, 1, the triangle is half the parallelogram.

Ex. 1. If AB (Fig. 5.) be 65 feet, and PC 31.2, what is the area of the triangle? Ans. 1014 square feet. 2. What is the surface of a triangular board, whose base is 3 feet 2 inches, and perpendicular height 2 feet 9 inches? Ans. 4F. 4' 3", or 4 feet 51 inches.

9. If two sides of a triangle and the included angle, are given, the perpendicular on one of these sides may be easily found by rectangular trigonometry. And the area may be calculated in the same manner as the area of a parallelogram in art. 5. In the triangle ABC (Fig. 2.)

R: BC::sin B: CH

And because the triangle is half the parallelogram of the same base and height,

As radius,

To the sine of any angle of a triangle;

So is the product of the sides including that angle,
To twice the area of the triangle. (Art. 5.)

Ex. If AC (Fig. 5.) be 39 feet, AB 65 feet, and the angle

at A 53° 7' 48", what is the area of the triangle?

Ans. 1014 square feet.

9. b. If one side and the angles are given; then

As the product of radius and the sine of the angle opposite the given side,

To the product of the sines of the two other angles;
So is the square of the given side,

To twice the area of the triangle.

If PC (Fig. 5.) be perpendicular to AB.

R: sin B::BC: CP

sin ACB sin A::AB: BC
Therefore (Alg. 390, 382.)

Rx sin ACB six sin B::ABX BC: CP XBC:: AB: ABXCP=twie area of the triangle.

Ex. If one side of a triangle be 57 feet, and the angles at the ends of this side 50° and 60°, what is the area?

Ans. 1147 sq. feet.

10. If the sides only of a triangle are given, an angle may be found, by oblique trigonometry, Case IV, and then the perpendicular and the area may be calculated But the area may be more directly obtained, by the following method.

RULE II. When the three sides are given, from half their sum subtract each side severally, multiply together the half sum and the three remainders, and extract the square root of the product.

If the sides of the triangle are a, b, and c, and if h=half their sum, then

The area=√h×(h− a)× (h—b)× (h—c)

For the demonstration of this rule, see Trigonometry, Art. 221.

If the calculation be made by logarithms, add the logarithms of the several factors, and half their sum will be the logarithm of the area. (Trig. 39, 47.)

Ex. 1. In the triangle ABC (Fig. 5.) given the sides a 52 feet, b 39, and c 65; to find the side of a square which has the same area as the triangle.

(a+b+c)=h=78
h-a=26

h-b=39

h-c=13

Then the area=V78 x 26 x 39 x 13=1014 square feet.

2. If the sides of a triangle are 134, 108, and 80 rods, what is the area?

Ans. 4319.

3. What is the area of a triangle whose sides are 371, 264, and 225 feet?

11. In an equilateral triangle, one of whose sides is a, the expression for the area becomes

√h×(h− a) × (h—a)× (h—a)

But as h=3a, and h-a=3a-a=a, the area is

{a ×a×a×a=√ √3⁄4a1 =4a2√3 (Alg. 271.)

That is, the area of an equilateral triangle is equal to the square of one of its sides, multiplied into the square root of 3, which is 1.732.

Ex. 1. What is the area of a triangle whose sides are each 34 feet? Ans. 500 feet. 2. If the sides of a triangular field are each 100 rods, how many acres does it contain?

PROBLEM III.

To find the area of a TRAPEZOID.

21. MULTIPLY HALF THE SUM OF THE PARALLEL SIDES

INTO THEIR PERPENDICULAR DISTANCE.

The area of the trapezoid ABCD (Fig. 4.) is equal to half the sum of the sides AB and CD, multiplied into the perpendicular distance PC or AH. For the whole figure is made up of the two triangles ABC and ADC; the area of the first of which is equal to the product of half the base AB into the perpendicular PC, (Art. 8.) and the area of the other is equal to the product of half the base DC into the perpendicular AH or PC.

Ex. If AB (Fig. 4.) be 46 feet, BC 31, DC 38, and the angle B 70°, what is the area of the trapezoid?

R: BC: sin B: PC-29.13. And 42 x29.13=12234.

2. What are the contents of a field which has two parallel sides 65 and 38 rods, distant from each other 27 rods?

PROBLEM IV.

To find the area of a TRAPEZIUM, or of an irregular POLY

GON.

13. DIVIDE THE WHOLE FIGURE INTO TRIANGLES, BY DRAWING DIAGONALS, AND FIND THE SUM OF THE AREAS

OF THESE TRIANGLES. (Alg. 519.)

If the perpendiculars in two triangles fall upon the same diagonal, the area of the trapezium formed of the two triangles, is equal to half the product of the diagonal into the sum of the perpendiculars.

Thus the area of the trapezium ABCH (Fig. 6.) is

BHX AL+BH×CM=1BH×(AL+CM.) Ex. In the irregular polygon ABCDH (Fig. 6.)

BH=36, and the perpendiculars CM 9.3

if the diagonals {CH=32,

DN=7.3

14. If the diagonals of a trapezium are given, the area may be found, nearly in the same manner as the area of a parallelogram in Art. 5, and the area of a triangle in Art. 9.

In the trapezium ABCD (Fig. 8.) the sines of the four angles at N, the point of intersection of the diagonals, are all equal. For the two acute angles are supplements of the other two, and therefore have the same sine. (Trig. 90.) Putting, then, sin N for the sine of each of these angles, the areas of the four triangles of which the trapezium is composed, are given by the following proportions; (Art. 9.)