by 2;* for the area of a parallelogram is equal to the whole product of the length of the basis multiplied by the height.†

3d. The areas of triangles upon the same basis and between the same parallels are equal; because if they are between the same parallels their heights must be equal; and we have the same case as in the last query; namely triangles upon the same basis and of equal heights.

4th. The areas of triangles are to each other as the products obtained by multiplying the length of their bases by their heights; for these products are the areas of the triangles. Thus

the area of the triangle ABC is to the area of the triangle EGP, as the length of the basis AB multiplied by the height CN, is to the length

* Instead of multiplying the basis by the whole height and dividing the product by 2; you may simply multiply the basis by half the height, or the height by half the basis.

If the basis of a triangle is 8 feet and the height 4 feet, the area of the triangle is equal to 32 (4 times 8) divided by 2; that is, 16 square feet; whereas the rectangle upon 8 feet basis and 4 feet high measures 32 (4 times 8) square feet, which is exactly double of the area of the triangles.

of the basis EG, multiplied by the height PM; which may be expressed thus:

Triangle ABC: triangle EGP = AB × CN : EG X PM.

5thly. The areas of triangles upon equal bases are to each other as the heights of the triangles; because the areas of parallelograms upon the same bases and of the same heights are to each other in the ratio of the heights; and their halves (the areas of the triangles) must be in the same ratio.* Thus if the two triangles

ABG, ECF, have their bases AB, EC equal to each other, we have the proportion :

Triangle ABG triangle ECF = CM: FN. 6th. The areas of triangles, which have equal heights, are to each other as the bases of the triangles. This truth follows like the preceding one from the same principle established with regard to parallelograms, of which the triangles are the halves. (Page 108, 6thly.)

* This principle and the following one might have been established immediately from the proportion :

Triangle ABC: triangle EGPABX CN: EG × PM, in precisely the same manner as it has been proved for parallelograms. (Page 107, 5thly.)

the two parallel sides by their distance, and dividing the product by 2.

Quest. How can you prove this?

Ans. By drawing the diagonal AD, the trapezoid ABCD, will be divided into the two triangles ACD and ABD. The area of the triangle ACD is found by multiplying the length of the basis, CD, by the height, AF, and dividing the product by 2; (page 109, 2d.) In the same manner we find the area of the triangle ABD, by multiplying the length of the basis, AB, by the height DE, and dividing the product by 2; and as the height, DE, of the triangle ABD, is equal to the height AF, of the triangle ACD, (because DE and AF are perpendiculars between the same parallels) we can find the area of the two triangles, or of the whole trapezoid, ABCD, at once, by multiplying the sum of the two parallel lines AB, CD, by their distance AF, (which is the common height of the triangles ACD, ABD,) and dividing the product by 2.*

* If you multiply two numbers successively by the same number, and then add the products together, the answer will be the same

QUERY V.

A

How do you find the area B of a polygon ABCDEF, or, in general, of any other rectilinear figure?

Ans. By dividing it by

means of diagonals, (as in

F

E

the figure before us) or by any other means into triangles. The area of each of these triangles is then easily found by the rule given; (page 109, 2d.) and the sum of the areas of all the triangles, into which the figure is divided, is the area of it.

as the sum of the two numbers at once multiplied by that number. If you multiply each of the numbers, 6 and 5, for instance, by 4, and then add the products 24 and 20 together, you will have 44; and adding, in the first place, 6 to 5, and then multiplying the sum, 11, by 4, you will again have 44.

Instead of multiplying the sum of the two parallel sides by their distance, and then dividing the product by 2, you may multiply, at once, half the sum of the two parallel sides by the distance; or the sum of the two parallel sides by half their distance.

REMARK.

In calculating the area of geometrical figures, it frequently occurs that the bases and heights of triangles, parallelograms, &c. are given in feet, inches, seconds, thirds, &c. Most treatises on Arithmetic, teach how to perform such multiplications, but not all give a sufficiently clear explanation of the principle on which this operation is founded. It is simply this:

1. If you multiply feet by feet, the product will evidently be square feet; for you may consider the product as the area of a rectangle, whose basis and height are given in feet. (See the figures to Query 1, Sect. III.) For the same reason, if you multiply inches by inches, the product will be square inches; if seconds by seconds, square seconds, and so on.

2. If you multiply feet by inches, each foot multiplied by an inch will give the area of a rectangle a b c d which is one foot (twelve inches) long, and one inch high, which will therefore measure 12 square inches; and it will take twelve of these rectangles to complete a square foot; (see the figure). Thus, if you multiply 3 feet by 6 inches, your product will be 18 such rectangles, or one square foot and a half.

If you multiply feet by seconds, each foot multiplied by one second, will give you the area of a rectangle, which is one foot long, and one second high, and it will therefore take twelve of these rectangles to complete a rectangle a foot long and one inch high (because there are twelve seconds in an inch). In the same manner, if you multiply feet by thirds, each foot multiplied by one third, will give you the area of a rectangle which is one foot long and one third high; and it will take twelve of these rectangles to complete a rectangle a foot long and one second high; (because 12 thirds make a second) and so on.

3. If you multiply inches by seconds, each inch multiplied by one second will give you the area of a rectangle one inch (12 seconds) long and one second high, which will therefore measure twelve square seconds, and it will take twelve of these rectangles