the length taken for the unit, the product is a like part of the multiplicand. Thus, if one of the factors is 6 inches, and the other half an inch, the product is 3 inches. Instead of referring to the meusures in common use, as inches, feet, etc., it is often convenient to fix upon one of the lines in a figure, as the unit with which to compare all the others. When there are a number of lines drawn within and about a circle, the radius is commonly taken for the unit. This is particularly the case in trigonometrical calculations. The observations which have been made concerning lines, may be applied to turfaces and solids. There may be occasion to multiply the area of a figure by the number of inches in some given line. But here another difficulty presents itself. The product of two lines is often spoken of as being equal to a surface; and the product of a line and a surface, as equal to a solid. But if a line has no breadth, how can the multiplication, that is, the repetition, of a line, produce a surface? And if a surface has no thickness, how can a repetition of it produce a solid? In answering these inquiries it must be admitted that measures of length do not belong to the same class of magnitudes with superficial or solid measures; and that none of the steps of a calculation can, properly speaking, transform the one into the other. But, though a line cannot become a surface or a solid, yet the several measuring units in common use are so adapted to each other, that squares, cubes, etc., are bounded hy lines of the same name. Thus the side of a square inch is a linear inch; that of a square rod, a linear rod, etc. The length of a linear inch is, therefore, the same as the length or breadth of a square inch. If then several square inches are placed together, as from Q to a, fig. 3, the number of them in the parallelogram oK is the Fig. 3. D _ C same as the number of linear inches in the side Q B: and if we know the length of this, we have of course the area of the parallelogram, which is here supposed to be one inch wide. But if the breadth is severml inches, the larger parallelogram contains as many smaller ones, each an inch wide, as there are inches in the whole breadth. Thus, if the parallelogram A c, fig. 3, is 5 inches long and 3 inches broad, it may be divided into three such parallelograms as on. To obtain, then, the number of squares in the large parallelogram, we have only to multiply the number of squares in one of the small parallelograms, into the number of such parallelograms contained in the whole figure. But the number of square inches in one of the small parallelograms is equal to the number of linear inches in the length A B. And the number of small parallelograms is equal to the number of linear inches in the breadth sc. It is therefore said concisely, that the area of a parallelogram it equal to its length multiplied into its breadth. We hence obtain a convenient algebraical expression for the area of a right-angled parallelogram. If two of the sides perpendicular to each other are A B and B C, the expression fur the area is A B X Bc; that is, putting a for the area, a ~ A B X BC. It must be remarked, however, that when A B stands for a line, it contains only linear measuring units; bat when it enters into the expression for the area, it is supposed to contain superficial units of the same name. The expression for the area may also be derived by another method more simple, but lesB satisfactory perhaps to some. Let a, fig. 4, represent a square inch, foot, rod, or other unit: and let b and I be two of its sides. Also, lengths; and, if the length of each were the same, the would be as the breadths. That is, a : a : ; L : Z, when the breadth is given; And A : a : : s : b, when the length is given: Therefore, A:o::bxl:*x'( when both vary. That is, the area is as the product of the length and breadth. Hence, in solving problems in geometry, the term product is frequently substituted for rectangle. And whatever is there prosed concerning the equality of certain rectangles, may be applied to the product of the hues which contain the rectangles. The area of an oblique parallelogram is also obtained by multiplying the base into the perpendicular height. Thus the expression for the area of the parallelogram it.i>, fig. o, is n N x A s, or A B x B c. For A B X B C is the area of lae Fig. 5. Kg. 6 D C M D N C right-angled parallelogram A Bod; and by Euclid 36, 1, parallelograms upon equal bases, and between the same parallels, are equal; that is, A B c D is equal to A B K M. The area of a square is obtained by multiplying one of the sides into itself. Thus the expression for the area of the square A c, fig. 6, is (aB)3, that is, a ■— (a B)'. For the area is equal to A B X Bc. But Ab = Bc, therefore, Abxbc = Abxab = (ab)3. The area of a triangle is equal to half the product of the 1 and height. Thus the area of the triangle A Bo, fig. 7> is equal to half A B into o H or its equal B C, that is, a = Jil X »c. For the area of the parallelogram Abcdisab X Bc. And by Euclid 41,1, if a parallelogram and a triangle are upon the same base, and between the same parallels, the triangle is half the parallelogram. Hence, an algebraical expression may be obtained for the area of any figure whatever which is bounded by right lines. For every such figure may be divided into triangles. Kg. 7. Kg. 8. D G C D Thus the right-lined figure, Abcbb, fig. 8, is compeeed of the triangles A B C, A C B ar.d BCD. The area of the triangle That of the triangle That of the triangle Jci=1ao)( Eh, The area of the whole figure is, therefore, equal to {i AC X Bl) + (i ACX ") + (jec X T)0). The expreaaion for the superfices has here been derived from that of a line or lines. It is frequently necessary to rtvtrse this order; to find a side of a figure, from knowing its area. If the number of square inches in the parallelogram Aicd, fig. 3, whose breadth B c is 3 inches, be divided by 3, the quotient will be a parallelogram, Ibii, one inch wide, and of the same length with the larger one. But the length of the small parallelogram is the length of its side A B. The number of square inches irfone ia the same as the number of linear inches in the other. If, therefore, the area of the large a parallelogram be represented by a, the side Ab=-p that is, the length of a parallelogram is found by dividing tfte area by the breadth. If a be put for the area of a square whose side is A B, That is, the side of the square is found, by extracting the square root of the number of measuring units in its area. If A B be the bate of a triangle, and B C its perpendicular height; Then «=licxu That is, the base of a triangle is found, by dividing the area by half the height. As a surface is expressed by the product of its length and breadth, the contents of a solid may be expressed by the product of its length, breadth and depth. It is necessary to bear in mind, that the measuring unit of solids is a cubs; and that the side of a cubic inch is a square inch; the aide of a cubic foot, a square foot, etc. Let Aicd fig. 3, represent the base of a parallelepiped, five inches long, three inches broad, and one inch deep. It evident there must be as many cubic inches in the solid, as there are square inches in its base. And as the product of the lines A B and B c gives the area of this base, it gives, of course, the contents of the solid. But suppose that the depth of the paTallelopiped, instead of being one inch, is four inches. Its contents must be four times as great. If, then, the length be A B, the breadth B c, and the depth c o, the expression lor the solid contents will be, A B X B c X o o. By means of algebraical notation, a geometrical demon stration may often be rendered much more simple and concise than in ordinary language. The proposition, (Euclid 4, 2,) that when a straight line is divided into two parte, the square of the whole line is equal to the squares of the two parts, together with twice the product of the parts, is demonstrated, by squaring a binomial. Let the side of a square be represented by s; And squaring both sides, s> = a2 -{- lab -J- b-. That is, s1 the square of the whole line, is equal to a' and »*, the Bquares the two parts, together with lab, twice the product of the parts. Algebraical notation may also be applied, with great advantage, to the solution of geometrical problems. In doing this, it will be necessary, in the first place, to form an algebraical equation from the geometrical relations of the quantities given and required; and then by the usual reductions, to find the value of the unknown quantity in this equation. Prob. 1. Given the base, and the sum of the hypothenuse and perpendicular, of the right-angled triangle A B c, tig. 9, to find the perpendicular. Let the base A B = 4 The perpendicular ic = i 2 X 16 from 16, the sum of the hypothenuse and perpendicular, leave* 10, the length of the hypothenuse. Prob. 2. Given the base and the difference of the hypothenuse and perpendicular of a right-angled triangle, to find the perpendicular. Let the base, fig. 10, A B = 4 = 20 The perpendicular ic = t Then will the hypothenuse A C S= * + d. 1. Then by Euclid 47, 1, (a C)2 = (a B)2 + (b O)» 2. That is, by the notation, (x + d)2 = 42 -f- a2 3. Expanding (* + d)2, x- + 2dx + <P = t* + a* i. Therefore % — V~? = 15. 2a Prob. 3. If the hypothenuse of a right-angled triangle is 3 feet, and the difference of the other two sides 6 feet, what is the length of the base? Prob. 4. If the hypothenuse of a right-angled triangle is 50 rods, and the base is to the perpendicular as 4 to 3, what is the length of the perpendicular? Prob. 5. Having the perimeter and the diagonal of a paraU D, fig. 11, to find the sides. lelogram A B c D. Let the diagonal Half the perimeter Prob. G. The area of a right-angled triangle 11c, fig. 12, being given, and the sides of a parallelogram inscribed in it, to find the side B c. Prob. 7. The three sides of a right-angled triangle A B 0, fig. 13, being given, to find the segments made by a'peTpendicular, drawn from the right angle to the hypothenuse. Prob. 10. If the sum of two of the sides of a triangle he 1155, the length of a perpendicular drawn from the angle included between these to the third side be 300. and the difference of the segments made by the perpendicular be 495; what are the lengths of the three sides ■ Prob. 11. If the perimeter of a right-angled triangle be 720, and the perpendicular falling from the right angle on the hypothenuse be 144; what are the lengths of the sides? Prob. 12. The difference between the diagonal of a square and one of its sides being given, to find the length of the sides. Prob. 13. The base and perpendicular height of any plane triangle being given, to find the side of a square inscribed in the triangle, and standing on the base, in the same manner as the parallelogram j, K F a, on the base A B, fig. 14. Prob. 14. Two sides of a triangle, and aline bisecting the included angle being given, to find the length of the base or | third side, upon which the bisecting line falls. \ Prob. 15. If the hypothenuse of a right-angled triangle be | 35, and the side of a square inscribed in it, in the same man-' ner as the parallelogram Bbdf, fig. 12, be 12; what are the lengths of the other two sides of the triangle r Ne i>srai>foz jamais ni plus sage ni plus savant que ceux avec qui vous etes. Portre votre eavnir comme votre montre. dans une j»oche particullere. que vous lie tirez point, et que vou» ne failea point sonner uniqutment pour nous faire voir que vous en avez une.—Lord Chester/kid. Pretenire, to anticipate senses IMPERSONAL VERBS. Impersonal verbs are those which have only the third person singular, and whose subject is unknown and cannot be supplied by a noun. The following are impersonal:— THE PARTICIPLE. The Participle is a word which possesses the qualities both, of the verb and the adjective. Participles are of three kinds :—Present, Past and Future. 1. The Present Participle terminates in dndo or indo; as— Amdndo, loving 2. The Past Participle ends as follows in the regular verbs: Anuito—a, amiiti—e, loved S. The Participle Future is not so often used. It is as Avere ad amdre, issere per amdre, being about to love The Italians are accustomed to syncopate several past participles of the first conjugation, as may be seen in the following list:— Except— Male, bad, which makes malamente, badly. Adverbs are of Of Time Present. Ora, adesso, ai presènte, at present Oggi, to-day In quésto instante, in questo pùnto, in tempo, this moment Or óra, directly Quésta mane, questa mattina, sta mane, sta mattina, this morning Quésta séra, sta séra, this even- Quésta nòtte, sta nòtte, to-night Of Time Past. Anticamente, prima, anciently ler l'àltro or avantieri, the |