QUERY II.

D

F

If a parallelogram ABEF stands on the same basis, AB, as a rectangle, ABCD, and has its height equal to the height of that rec- A tangle, what relation do the areas of these two figures bear to each other?

B

H

A. The area of the parallelogram ABEF is equal to the area of the rectangle ABCD; therefore I can say that the parallelogram ABEF is equal to the rectangle ABCD, (see remark 3d. Introd. to Sect. III.)

Q. How do you prove it?

A. The right-angular triangle ACF, has the hypothenuse AF and the side AC equal to the hypothenuse BE and the side BD in the rightangular triangle BDE, each to each, (AF and BE, AC and DB, being opposite sides of the parallelogram ABEF, and the rectangle ABCD); therefore these two triangles are equal (page 55); and by taking from each of the two equal triangles ACF, BDE, the part DGF common to both, the remainders AGDC, BGFE, are also equal (truth IV); and then by adding again to each of the equal remainders the same triangle ABG, the sums, that is, the rectangle ABCD

and the parallelogram ABEF are also equal to one another, (truth III.)

૨. What important truths can you infer from the one you have just demonstrated?

A. 1st. All parallelograms, which have equal bases and heights, are equal to one another; for each of them is equal to a rectangle upon the same basis and of the same height. (Truth 1.)

2dly. Parallelograms upon equal bases and between the same parallels are equal to one another; for if they are between the same parallels their heights must be equal. (Query 12. Sect. I.)

3dly. The area of a parallelogram is found by multiplying the basis, given in rods, feet, inches, &c., by the height, expressed in units of the same kind. Because the area of the rectangle upon the same basis and of the same height to which it is equal, is found in the same manner.*

4thly. The areas of parallelograms are to each other, as the products obtained by multiplying the length of the bases of the parallelograms by their heights, because these products are the areas of the parallelograms.

* The area of a rhombus or lozenge is found like that of a parallelogram; a lozenge being only a peculiar kind of parallelogram.

The parallelogram ABCD, for instance, is to the parallelogram GHEF, as the answer obtained by multiplying the length of the basis AB, by the height MN, is to the answer obtained by multiplying the length of the basis GH, by the height OP; because AB multiplied by MN is the area of the parallelogram ABCD, and GH multiplied by OP is the area of the parallelogram GHEF. This proportion may be expressed thus:

Parallelogram ABCD: parallelogram GHEF = AB × MN : GH × OP.

5thly. Rectangles or par

allelograms, which have

CM D

E 0 F

equal bases, are to each

other as their heights. A NB G PH For if in the above proportion the basis AB is equal to the basis GH, I can write AB instead. of GH, and thereby change it into

Parallelograms ABCD: parallelograms GHEF AB X MN: AB X OP,

=

that is, the parallelogram ABCD is to the parallelogram GHEF, as AB times the height MN is to AB times the height OP; or, which

is the same, as the height MN alone is to the height OP alone; which is written thus: Parallelogram ABCD: parallelogram GHEF =MN: OP.

6thly. In precisely the same manner it may be proved, that if the heights MN and OP are equal to each other, the parallelograms, ABCD, GHEF, are to each other as their bases AB and GH; which may be expressed thus: Parallelogram ABCD: parallelogram GHEF =AB: GH.

QUERY III.

If two triangles ABC, ABE, stand on the same basis AB, and have equal heights CK, EG, what relation do the areas of these triangles bear to each other?

A. The areas of these triangles are equal to one another.

* For let the length of the basis AB be expressed by any number you please, 10 for instance; then it is evident that ten times the height MN is in the same ratio to ten times the height OP, as the height MN alone is to the height OP alone. The ratio of 3 to 6, for instance, is as 1 to 2; because 3 is twice contained in 6; and the ratio of 10 times 3 to 10 times 6 (30 to 60) is also as 1 to 2; and so is the ratio of 20 times 3 to 20 times 6 (60 to 120) as 1 to 2; the ratio of 50 times 3 to 50 times 6 (150 to 300) as 1 to 2, and so on.

Q. How can you prove it?

A. Draw the line AD parallel to CB; BF parallel to AE; and through the two vertices C and E, the line CF parallel to AG (which is possible since the heights CK and EG are equal) and the area of the parallelogram ABCD will be equal to the area of the parallelogram ABEF (query 2, Sect. IV.); and as the triangle ABC is half of the parallelogram ABCD-for the diagonal AC divides the parallelogram ABCD into two equal parts (query 12, Sect. II.)—and the triangle ABE is half of the parallelogram ABEF; therefore the areas of these two triangles must also be equal to one another; for if the wholes are equal, the halves are also equal; and the same can be proved of triangles, which have equal bases and heights.

Q. What consequences follow from the principle just advanced?

A. 1st. Every triangle is half of a parallelogram upon equal basis and of the same height. (This is evident from looking at the figure, and from Query 12, Sect. II.)

2d. The area of a triangle is half of the area of a parallelogram upon the same basis and of the same height. Thus the area of a triangle is found by multiplying the length of the basis by the height, and dividing the product