feet, we must multiply the number of linear feet in one side by the number of feet in the contiguous side. This shows us what is meant by the product of two lines; and we therefore say-The area of a rectangle is expressed by the product of its base multiplied by its height.

160. As any parallelogram is equivalent to a rectangle of the same base and height-The area of any parallelogram is the product of the base multiplied by the height. And therefore (153)—The area of a triangle is half the produci of the base multiplied by the height. Consequently any two triangles are to cach other as the products of their bases by their heights.

161. If we designate the height of a rectangle by a, and the base by b, the area will be expressed by a × b; or a.b; and the rectangle is said to be contained by the two lines ɑ and b. If the base of the rectangle is equal to its height, the figure is a square (62), and its area is expressed by a.a, or a2 (73). Hence the second power of a quantity is frequently called its square.* A triangle, whose height is a, and whose base is b, will be repre

sented by a.b, or

a.b

2

162. Problem. To construct upon a given line c (fig. 88) a rectangle equivalent to the given rectangle Fig. 88. a.b. Find a fourth proportional (97) to the three given lines c, a,b; let d be this fourth proportional; we have C b

X

Multiply both sides by a and also by d, we have

c × d = a × b; the rectangle contained by the two lines c and d is the rectangle required.

163. Problem. To find a square equivalent to a given rectangle. Let a and b be the two sides of the given rectangle; find a mean proportional between these two lines (98); let c be this mean proportional; we have

C ; multiply both by c and b, we have a xbc2;

b

c is the side of the required square.

* The expression "a square," or "the square of a, may be used when, as in the present case, we wish to designate the area of a rectangle, each of whose sides is expressed by a; but when the product a X a, or a2, is not considered as representing such an area, it should never be read " a square."

Fig. 89.

164. Problem. To find a square equivalent to a given triangle. Let a be the height of the triangle and the Find ca mean proportional between a and 1⁄2 b.

base.

both the denominators we obtain a × b = c2; that is, c is the side of the square sought.

165. Any polygon may be changed into an equivalent polygon of a less number of sides, by using the principle that triangles of the same base and height are equivalent. Suppose we have the pentagon ABCDE. Draw the diagonal AD (fig. 89); through the vertex E draw EF parallel to this diagonal and meeting CD produced in F; and draw AF; the pentagon ABCDE will be transformed into the equivalent quadrilateral ABCF. In the same manner all polygons may be reduced to equivalent triangles.

166. If we would ascertain the measure of a trapezoid, Fig. 90. as ABCD (fig. 90), AB and CD being the parallel sides; draw the diagonal AD; this will divide the trapezoid into two triangles whose bases are respectively the two parallel sides of the figure, and whose common height is the perpendicular distance AF of these two sides. Each triangle will have for its measure the height multiplied by half the base (160). Therefore the sum of the two triangles will have for its measure the product of the common height into half the sum of their bases. We therefore say-The measure of a trapezoid is the product of half the sum of the parallel sides, multiplied by their perpendicular distance.

If we take a for the height of a trapezoid, and b and b' for the two parallel sides, the area will be expressed by a. (b + b') a. (b+b'), or 2

167. Let us now compare two similar triangles, as Fig. 91. M and M', (fig. 91). Let their heights be a and a',

their bases b and b'; this will give

leaving out the common factor, we

;

and

2

M 1 (a.b)

or

M

1 (a' . b')

M

a.b

have

M a'.b' we have (91)

b

b'

for its equal

M b2

We see, there

or which is the same thing,
fore, that the two triangles are to each other in the ratio
of the second powers of their bases. Any other homolo-
gous sides being taken for bases, would give a similar
result. And as all homologous lines are proportional in
similar triangles, we have this general rule-Similar tri-
angles are to each other as the second powers of their ho-
mologous dimensions; or, as the squares described upon
their homologous lines.

168. As two similar polygons of any number of sides (93) are composed of the same number of similar triangles of proportional dimensions, this result may be readily applied to them.

The two similar pentagons (fig. 92) are divided by the Fig. 92. diagonals, each into three triangles similar to the corresponding triangles in the other (93). The homologous diag

onals a, b, and a', b', give the proportion

a

a'

b

(92).

b'

The similar triangles, in the two figures, give the propor

L a2 M

tions

L' a'2' M'

N b2

N

a2

and

L+M+ N

(167); therefore (84) L+M+N

L' M' N' a/2

-The areas of similar polygons are to each other as the second powers of their homologous dimensions ;

squares described upon their corresponding lines.

or as the

169. Suppose we have the two triangles ABC, AEF, (fig. 93) having the common angle A; draw the perpen- Fig. 93. diculars BD and EG to the side AC; the similar triangles

we multiply these two proportions in order, and suppress

the common factor

BD
EG'

ABC АВХАС

we shall have

AEF AEXAF

That is-If two triangles have an angle of the one equal

Fig. 94.

Fig. 95.

Fig. 96.

to an angle of the other, their areas will be as the products of the sides containing the equal angles.

170. If we take CE (fig. 94) equal to the sum of the two lines a and b, and construct upon it the square CEFG; and upon the part CD, equal to a, construct the square CDIH, and produce the sides HI and DI to L and K; we shall have the square of CE (the sum of the two lines a and b) divided into four parts, viz. CDIH, IKGL, DELI, and HIFKI. Because CEGF is a square, CF is equal to CE; and because CDIH is a square, CH is equal to CD; therefore FH is equal to DE, or b; and HI being one side of the square CD or a, the rectangle HFKI is a rectangle contained by two lines, one equal to ɑ, and the other equal to b; DELI is is a rectangle of the same dimensions; and ILGK is a square whose side is equal to DE, or b; so that the whole figure, which is the square of CE (the sum of the two lines, a and b), contains two squares, which are the squares of the parts ɑ and b; and two equal rectangles each of which is contained by these two parts. This gives the equation (a + b) 2 a2 + b2 + 2 (a.b), which may be enunciated as follows-If a line is equal to the sum of two other lines, the square described upon this line is equal to the sum of the squares of the two lines, plus twice the rectangle contained by the two lines.

171. Take (fig. 95) CD the difference between two lines a and b; construct upon BD, equal to ɑ, the square ABDI; and upon CD, equal to the difference of the two lines, another square CDEF; produce EF to G, making FG equal to b; and complete the square FGHI, which will be the square of b; so that the whole figure is equal to the square of a plus the square of b; but ABCK is a rectangle contained by the two lines a and b; EGIIK is an equal rectangle; if we take away these two rectangles, we shall have left the square CDFE, (the square of a-b). This gives the equation (ab)2=a2 + b2 -2 (a.b), which gives this general rule-The square of the difference of two lines, is equal to the square of the greater, plus the square of the less, minus twice the rectangle contained by the two.

2

172. Let a represent the hypothenuse of a right-angled triangle whose two sides are b and c. From the vertex of the right-angle draw AD (fig. 96) perpendicular to the hypothenuse. We have (101) the two equations, CD=

2

equations, we have CD+DB = CB = a =

and multiplying both sides by a, we have a2 = b2 + c2,
that is-The square described upon the hypothenuse of a
right-angled triangle, is equivalent to the sum
squares described upon the other two sides.

173. We may demonstrate this truth from the areas immediately, without referring the lines to numbers, as in the above proof. Upon the hypothenuse CB (fig. 97) Fig. 97. construct the square CF; and upon the inside of the other sides CA, AB, construct the two squares CII and AL; from the vertex A of the right-angle draw AD perpendicular to the hypothenuse and produce it to G; and produce ML and IIN till they meet the periphery of the square CEFB: the points of meeting will be the vertices F and E; for the triangle LBF has its sides respectively perpendicular to those of the original triangle ABC, its angles are therefore equal; and having LB equal to AB, being a side of the same square, the two triangles are therefore equal by the third case, their hypothenuses are consequently equal, and ML produced must meet BF in the point F; in the same manner we may show that IIN produced will fall upon the point E.

The square AL and the parallelogram AKFB, are upon the same base AB, and have the same height BL, they are therefore equivalent (151); BK and BG are parallelograms upon the same base BF, and have the same height BD, they are therefore equivalent; that is, the square upon the side AB is equivalent to the rectangle BG in the same manner we may show that the square upon CA is equivalent to the rectangle CG therefore the sum of the squares b2 + c2 is equivalent to the sum of the rectangles CG and BG; but these two rectangles taken together make the square of the hypothenuse a2. This gives us the equation a2 = b2 + c2, the same result as that obtained by the other process.

:

2

174. Remark. This is one of the most important propositions in elementary geometry. A method of proving it geometrically was long a desideratum with the ancient geometers. It was at length discovered by Pythagoras. It is known by the name of the Pythagorean proposition; and is the 47th of the 1st book of Euclid.