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GEOMETRY, CONTINUED.

Having now explained so much of the principles of Algebra and Proportion as may be requisite to elucidate the subsequent propositions, we again proceed with the ELEMENTS OF GEOMETRY. The geometric definitions, &c. have been given, generally, in pages 10 to 14; but to those already explained are to be added several which follow, as it now becomes necessary that they, also, should be known.

132. EQUIVALENT FIGURES are such as have equal surfaces, without regard to their form.

133. IDENTICAL FIGURES are such as would entirely coincide, if the one be applied to the other.

134. In EQUIANGULAR FIGURES, the sides which contain the equal angles, and which adjoin equal angles, are homologous.

135. Two figures are similar, when the angles of the one are equal to the angles of the other, each to each, and the homologous sides are proportionals.

136. In two CIRCLES, similar sectors, similar arcs, or similar segments, are those which have equal angles at the centre.

Thus, if the sector ABC be similar to the sector DEF, then the angle ABC will be equal to the angle DEF; or, if the arc AC be similar to the arc DF, then the angle at B will be equal to the angle at E. Also, if the segment GMH be similar to the segment KNL, the angle I will be equal to the angle R.

137. The AREA of a figure is the quantity of surface, containing a certain

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THEOREM 48.

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138. Parallelograms which have equal bases and equal altitudes are equal. Take the two parallelograms, ABCD,

Fig. 1.

Fig. 2. and ABEF, upon the same base, AB, and P between the same parallels, AB and DE; these parallelograms are equal.

For, in the parallelogram ABCD (fig. 1), the opposite sides CD and AB are equal; and in the parallelogram ABEF, the opposite sides EF and AB are equal; therefore EF is equal to CD (84)*. Again, in the parallelogram ABCD (fig. 2), the side CD is equal to AB; and, in the parallelogram ABEF, the side FE is equal to AB ; therefore EF is equal to CD. Now, in fig. 1, since CD is equal to EF, add CF to both; then will DF be equal to CE. In fig. 2, take away the common part CF, and there will remain DF equal to CE. Therefore, in each of these figures, the three straight lines AD, DF, FA, are respectively equal to the three straight lines BC, CE, EB; and, consequently, the triangle ADF is equal to the triangle BCE: therefore, from the quadrilateral ABED take away the triangle BCE, there will remain the parallelogram ABCD; and, from the same quadrilateral, take away the equal triangle ADF, and there will remain the parallelogram ABEF: therefore the parallelogram ABCD is equal to the parallelogram ABEF.

139. COROLLARY.—Every parallelogram is equal to a rectangle, of the same base and altitude.

THEOREM 49.

140. Any triangle is equal to half a parallelogram of the same base and altitude.

For the triangle ABC is equal to the triangle ACD, and the parallelogram ABCD is equal to the sum of both triangles; and, consequently, double to one of them.

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• The figures thus inserted in a parenthesis refer to a preceding or a following paragraph ; as, in this instance, to 84, on page 29.

141. COROLLARY 1.-Hence every triangle is half a rectangle, having the same base and altitude.

142. COROLLARY 2.—Triangles which have equal bases and equal altitudes are equal.

THEOREM 50.

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143. Rectangles, of the same altitude, are to one another as their bases.

Let ABCD, AEFD, be two rectangles, which have a common altitude AD; they are to one another as their bases AB, AE.

For, suppose that the base AB contains seven equal parts, and that the base AE contains four similar parts; then, if a AB be divided into seven equal parts, AE will contain four of them. At each point of division draw a perpendicular to the base, and these will divide the figure ABCD into seven equal rectangles (138); and, as AB contains seven such parts as AE contains four, the rectangle ABCD will also contain seven such parts as the rectangle AEFD contains four; therefore the bases AB, AE, have the same ratio that the rectangles ABCD, AEFG, have.

THEOREM 51.

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144. Rectangles are to one another as the products of the numbers which express their bases and altitudes.

Let ABCD, AEGF, be two rectangles, and let some u line taken, as a unit, be contained m times in AB, the base of the one, and n times in AD, its altitude; also

P times in AE, the base of the other, and q times in AF, its altitude; the rectangle ABCD shall be to the rectangle AEGF, as the product mn is to the product pq.

Let the rectangles be so placed that their bases AB, AE, may be in a straight line; then their altitudes AD, AF, shall also form a straight line (48). Complete the rectangle EADH; and, because this rectangle has the same alti

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tude as the rectangle ABCD, when EA and AB are taken as their bases; and the same altitude as the rectangle AEGF, when AD, AF, are taken as their bases ; we have the rectangle ABCD : ADHE :: AB: AE :: m :po... (143) But....

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:: mn: pn therefore, ABCD : ADHE:: mn: pn. In like manner, ADHE : AEGF :: pn: pq But, placing the terms of these two sets of proportionals alternately, we have,

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:: ABCD: mn and...

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:: AEGF : pa therefore, by equality, ABCD : :: AEGF : pa therefore, alternately, ABCD: AEGF ::

: pq 145. OBSERVATION.-If ABCD, one of the rectangles, be a square, having the measuring unit for its side ; this square may be taken as the measuring unit of its surfaces; because the linear unit, AB, is contained p times in EF, and 9 times in EH, by the proposition.

1xl: pq :: ABCD : EFGH. Hence the rectangle EFGH will contain the superficial unit ABCD, as often as the numeral product pq contains unity.

Consequently, the product pq will express the area of the rectangle, or will indicate how often it contains the unit of its surfaces.

Thus, if EF contains the linear unit AB four times, and EH contains it three times, the area EFGH will be 3 x 4=12: that is, equal to twelve times a square whose side AB is =1.

In consequence of the surface of the rectangle EFGH being expressed by the product of its sides, the rectangle, or its area, may be denoted by the symbol EF x FG, in conformity to the manner of expressing a product in arithmetic.

However, instead of expressing the area of a square, made on a line AB, thus, AB x AB; it is thus expressed AB?.

146. NOTE.-A rectangle is said to be contained by two of its sides, about any one of its angles.

THEOREM 52.

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

For the parallelogram ABCD is equal to the rectangler ABEF, which has the same base AB, and the same altitude (138), and this last is measured by AB x BE, or by AB AF; that is the product of the base of the parallelogram and its altitude.

148. COROLLARY.–Parallelograms of the same base are to one another as their altitudes ; and parallelograms of the same altitudes are to one another as their bases.

For, in the former case, put B for their common base, and A, a, for their altitudes; then we have B x A:B xa :: A:a. And, in the latter case, put A for their common altitude, and B, b, for their bases; then B A: A :: B : b.

THEOREM 53.

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

For the triangle ABC is half the parallelogram ABCE, which has the same base, BC, and the same altitude AD (140); but the area of the parallelogram is BC ~ AD (147), therefore, the area of the triangle is BC ~ AD, or B BC RAD.

150. COROLLARY.—Two triangles of the same base are to one another as their altitudes ; and two triangles, of the same altitude, are to one another as their bases.

THEOREM 54. 151. The area of every trapezoid, ABCD, is equal to the product of half

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