The squares of the numbers 1, 2, 3, &c., are 1, 4, 9, &c. So likewise, the geometrical square constructed on a double line is evidently four times as great as the square on a single one; on a triple line it is nine times as great,

&c.

PROPOSITION V. THEOREM.

The area of a parallelogram is equal to the product of its base and altitude.

Let ABCD be any parallelogram, and BE its altitude: then will its area be equal to ABXBE. Draw AF, and complete the rectangle ABEF.

F D E C

The parallelogram ABCD is equivalent to the rectangle ABEF (P.1, s.); but this rectangle is measured by ABX BE (P. 4, s. 1); therefore, ABX BE is equal to the area of the parallelogram ABCD.

A

B

Cor. Parallelograms of equal bases are to each other as their altitudes; and parallelograms of equal altitudes are to each other as their bases. For, let C and D denote the altitudes of two parallelograms, and B the base of each: then, BXC BXD :: : C: C D (B. II., P. 7). If A and B are the bases, and C the altitude of each, we shall have,

and parallelograms, generally, are to each other- as the products of their bases and altitudes.

PROPOSITION VI. THEOREM.

The area of a triangle is equal to half the product of its base and altitude.

Let BAC be a triangle, and AD perpendicular to the base: then will its area be equal to one-half of BC ×AD.

A

E

B D

For, draw CE parallel to BA, and AE parallel to BC, completing the parallelogram BE. Then, the triangle ABC is half the parallelogram ABCE, which has the same base BC, and the same altitude AD (P. 2); but the area of the parallelogram is equal to BC×AD (P. 5); hence, that of the triangle must be BCXAD, or BC×AD.

Cor. Two triangles of equal altitudes are to each other as their bases, and two triangles of equal bases are to each other as their altitudes. And triangles generally, are to each other, as the products of their bases and altitudes.

PROPOSITION VII. THEOREM.

The area of a trapezoid is equal to the product of its altitude, by half the sum of its parallel bases.

Let ABCD be a trapezoid, EF its altitude, AB and CD its parallel bases: then will its area be equal to EFX. (AB+CD).

Through I, the middle point of the side BC, draw KL parallel to the opposite side AD; and produce DC till it meets KL.

DE C K

H

A

F

L B

In the triangles IBL, ICK, we have the side IB=IC, by construction; the angle LIB=CIK (B. I., P. 4); and since CK and BL are parallel, the angle IBL=ICK (B. I., P. 20, c. 2); hence, the triangles are equal (B. I., P. 6); therefore, the trapezoid ABCD is equivalent to the parallelogram ALKD, and consequently, is measured by EFXAL (P. 5).

But we have AL-DK; and since the triangles IBL and KCI are equal, the side BL=CK: hence AB+CD= AL+DK=2AL; hence, AL is the half sum of the bases AB, CD; hence, the area of the trapezoid ABCD, is equal to the altitude EF multiplied by the half sum of the bases AB, CD, a result which is expressed thus:

Scholium. If through I, the middle point of BC, the line IH be drawn parallel to the base AB, it will bisect AD at H. For, since the figure ALIH is a parallelogram, as also, HIKD, their opposite sides are parallel, and we have AH-IL, and DH IK; but since the triangles LBI, IKC, are equal, we have IL=IK; therefore, AH=HD. AB+CD

But since the line HI-AL, it is also equal to

2

hence, the area of the trapezoid may also be expressed by EFX HI; consequently, the area of a trapezoid is equal to its altitude multiplied by the line which connects the middle points of its inclined sides.

PROPOSITION VIII. THEOREM.

The square described on the sum of two lines is equivalent to the sum of the squares described on the lines, together with twice the rectangle contained by the lines.

Let AB, BC, be any two lines, and AC their sum; then will

AC2 or (AB+BC)2 — AB2 + BC2 + 2AB× BC.

On AC describe the square ACDE; take AF=AB; draw FG parallel to AC, and BH parallel to AE.

F

E

A

.H

D

I

G

B C

The square ACDE is made up of four parts; the first ABIF is the square described on AB, since we made AF= AB: the second IGDH is the square described on IG, or BC; for, since we have AC AE and AB=AF, the dif ference, AC-AB must be equal to the difference AE-AF, which gives BC = EF; but IG is equal to BC, and DG to EF, because of the parallels; therefore, IGDH is equal to a square described on BC. Now, if these two squares be taken away from the large square, there will remain the remain the two rectangles BCGI, FIHE, each of which is measured by AB×BC: hence, the square on the sum of two lines is equivalent to

the sum of the squares on the lines, together with twice the rectangle contained by the lines.

Cor. If the line AC were divided into two equal parts, the two rectangles FH, BG, would become squares, and the square described on the whole line would be equivalent to four times the square described on half the line.

Scholium, This property is equivalent to the property demonstrated in algebra, in obtaining the square of a binomial; which is expressed thus:

(a+b)2=a2+2ab+b2.

PROPOSITION IX. THEOREM.

The square described on the difference of two lines, is equivalent to the sum of the squares described on the lines, diminished by twice the rectangle contained by the lines.

Let AB, BC, be two lines, and AC their difference; then, AC2, or (AB-BC)

AB+B02-2ABX BC.

A

C B

Since KD=AB, and BC=KL, the two rectangles CI, KG, are each measured by AB×BC: the whole figure ABILKEA, is equivalent to AB+BC2; take from each the two rectangles CI, KG, and there will remain the square ACDE, equivalent to AB2+BC2 dimin ished by twice the rectangle of ABX BC.

Scholium. This proposition is equivalent to the algebraical formula,

(a—b)2=a2—2ab+b2.

PROPOSITION X. THEOREM.

The rectangle contained by the sum and the difference of two lines, is equivalent to the difference of their squares.

Let AB, BC, be two lines; then will

(AB+BC)×(AB—BC)~~AB2—BC2.

E

Upon AB and AC, describe the F squares ABIF, ACDE; prolong AB till BK is equal to BC; and complete the rectangle AKLE, and prolong CD to G.

A

C B

B K

The base AK of the rectangle AL is the sum of the two lines AB BC; and its altitude AE is their difference; therefore, the rectangle AKLE is equivalent to

(AB+BC)×(AB—BC).

Again, DHIG is equal to a square described on CB; and since BH is equal to ED, and BK to EF, the rectangle BL is equal to the rectangle EG: hence, the rectangle AKLE is equivalent to ABHE plus EDGF, which is precisely the difference between the two squares AI and DI described on the lines AB, CB: hence, we have (A.1.), (AB+BC)×(AB−BC)—— AB2-BC2.

Scholium. This proposition is equivalent to the algebraical formula,

(a+b)x(a−b)=a2—b2.

PROPOSITION XI. THEOREM.

The square described on the hypothenuse of a right-angled triangle is equivalent to the sum of the squares described on the two other sides.

Let BCA be a right-angled triangle, right-angled at A: then will the square described on the hypothenuse BC be equivalent to the sum of the squares described on the other two sides, BA, AC.