F K D C G H B A Having placed the two rectangles so that the angles at A may be vertical and opposite, produce the sides CD, GF until they meet in K; the two rectangles ABCD, AFKD, having the same altitude AD, are to each other as their bases AB, AF. In like manner, the two rectangles AFKD, AFGH, having the same altitude AF, are to each other as their bases, AD, AF. Hence we have these two proportions: ABCD AFKD::AB: AF. AFKD: AFGH:: AD: AH. Multiplying the corresponding terms of these proportions together, and observing to omit AFKD, since it will occur in an antecedent and consequent, we shall have ABCD: AFGH:: AB-AD: AH-AF. Scholium. Hence we are at liberty to take for the measure of a rectangle, the product of its base by its altitude, provided we understand by this product the same as the product of two numbers, which numbers denote the linear units in the base and altitude respectively. This measure, however, is not absolute, but only relative. It supposes that the area of any other rectangle is estimated in a similar manner, by measuring its sides by the same linear unit; we shall thus obtain a second product, and the ratio of these two products is the same as that of the two rectangles, in accordance with this proposition. For example, if the base of the rectangle A is ten units, and its height three, the rectangle will be represented by the number 10× 3=30, a number which, thus isolated, has no signification; but if we have a second rectangle B, whose base is twelve units and height seven, this second rectangle will be represented by the number 12 × 7=84. From which we conclude that the two rectangles A and B are to each other as 30 to 84. If we take the rectangle A as the unit of measure of surfaces, the rectangle B will have for its measure ; that is, it will be of our superficial units. It is more common and more simple to take a square for the unit of surface, and we choose a square whose side is a unit of length. In this case, the measure which we have regarded as relative becomes absolute; for example, the number 30, which measured the rectangle A, represents 30 superficial units, or 30 squares, each side of which is a unit long. In geometry, we frequently confound the product of two lines with that of their rectangle, and this expression is even employed in arithmetic to denote the product of two unequal numbers; but we use the term square to denote the product of a number multiplied by itself. The squares of the numbers 1, 2, 3, etc., are 1, 4, 9, etc. From which we see that the square formed on a line of double length is quadruple; on a line of triple length it is nine times as great, and so of other squares. THEOREM XXII. The area of any parallelogram is equal the product of its base by its height. G D F с For the parallelogram ABCD is equivalent to the rectangle ABFG, which has the same base AB, and same altitude BF (T. XVIII., C.). But the rectangle ABFG is measured by AB × BF, consequently the parallelogram is measured by AB × BF. A B Cor. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; and in general, parallelograms are to each other as the products of their bases multiplied by their altitudes. THEOREM XXIII. The area of a triangle is equal to the product of its base by half its altitude. For, the triangle ABC is half the parallelogram ABCF (T. XIX.). The parallelogram is measured by the base BC multiplied into the altitude AD; therefore, B D C the triangle is measured by the base BC into half the altitude AD. Cor. I. Triangles of the same altitude are to each other as their bases; and triangles of the same or equal bases are to each other as their altitudes. Cor. II. The area of a trapezoid ABDC is equal to the product of half the sum of the bases, AB, CD, into the altitude IK; or, it is equal to the product IK into the straight line EF drawn at equal distance between the bases. In effect this trapezoid is composed of two triangles CAB, BCD; and by the Theorem, we have And as we have proved (B. I., T. XXXIII.) that EF = The area of a triangle is equal to half the product of its perimeter by the radius of its inscribed circle. Let K be the centre of the inscribed circle. Draw KD, KE, KF respectively perpendicu lar to the sides AB, BC, AC. Also draw KA, KB, KC. We obviously have ABC= AKB+BKC+AKC. But (T. XXIII.) we have C F E K A D B (AB+BC+AC) × KD THEOREM XXV. с The area of a triangle is equal to the product of the three sides divided by twice the diameter of its circumscribed circle. Let K be the centre of the circumscribed circle. Draw the diameter CKD, and the chord AD, and the perpendicular CH, which will be the altitude of the triangle. D K B H The two triangles CAD, CHB are rightangled, the one at A (B. II., T. X., Cor. I.), and the other at H; moreover, the angles at D and B are equal, being inscribed in the same segment ADBC. Hence these triangles are similar, and their homologous sides being proportional, we have in which, substituting the value of CH just found, we have ABC= ABX CBX AC 2CD RATIOS BETWEEN THE AREAS OF SIMILAR FIGURES THEOREM XXVI. The areas of two triangles which have an equal angle, are proportional to the rectangles of the sides containing the equal angle. A ABC: ADE:: AB× AC: AD × AE. For, joining D and C, we have, since triangles of the same altitude are to each other as their bases, ABC: ADC::AB: AD, ADC: ADE:: AC: AE. Multiplying together the corresponding terms of these proportions, and omitting the common term ADC which enters into the antecedent and consequent of the first couplet, we have ABC: ADE:: AB x AC: AD x AE. Cor. Whenever the side DE of the second triangle is parallel to BC of the first, the triangle ADC is a mean proportional between the triangles ABC and ADE. In effect we have, in this case (T. III.), AB: AD:: AC: AE; so that the two first proportions of the preceding Theorem have equal ratios, and give ABC: ADC::ADC: ADE. THEOREM XXVII. The areas of similar triangles are proportional to the squares of their homologous sides. These triangles being similar, are equiangular (T. V.). Hence, by the preceding Theorem, we have B ABC: A'B'C':: AB x AC: A'B' x A'C'; but since the triangles are similar, we also have AC: AC'::AB: A'B'. Multiplying the corresponding terms of these two proportions, omitting the common factor AC which enters into the antecedents, and the common factor A'C' which enters into the consequents, we have ABC: A'B'C' :: AB2 : A'B'. |