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 measures 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 surfaces 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 by 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 a to R, fig. 3, the number of them in the parallelogram o R is the lengths; and, if the length of each were the same, the areas would be as the breadths. That is, And Therefore, A:a:: L: 7, when the breadth is given; A: a:: Bb, when the length is given : A:a:BXL:6 X 7, 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 proved concerning the equality of certain rectangles, may be applied to the product of the lines 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 A BN M, fig. 5, is M N X AD, or A B X E C. For ABX B C is the area of the Fig. 5. Fig. 6. D same as the number of linear inches in the side QR: 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 several 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 OR. 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 BC. It is therefore said concisely, that the area of a parallelogram is 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 for the area is A B X BC; that is, putting a for the area, @ABX BC. It must be remarked, however, that when A B stands for a line, it contains only linear measuring units; but 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 less satisfactory perhaps to some. Let a, fig. 4, represent a square inch, foot, rod, or other measuring unit: and let b and be two of its sides. Also, right-angled parallelogram ABCD; and by Euclid 36, 1, parallelograms upon equal bases, and between the same parallels, are equal; that is, A B CD is equal to A B N 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 (A B)2, that is, a = (A B)2. ABX AB = (AB)a. For the area is equal to A BX BC. But AB BC, therefore, AB X BC The area of a triangle is equal to half the product of the base and height. Thus the area of the triangle A B G, fig. 7, is equal to half ▲ B into G H or its equal в C, that is, a = ABX BC. For the area of the parallelogram A B C D is ABX 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 haif 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. ACX BL, ACE ACX EH, Then transposing x, ECD ECX DG. 1. By Euclid 47, 1, The area of the whole figure is, therefore, equal to (ACX BL) + ( A ©× BH) + ( ECX DG). The expression for the superfices has here been derived from that of a line or lines. It is frequently necessary to reverse this order; to find a side of a figure, from knowing its area. If the number of square inches in the parallelogram A B C D, fig. 3, whose breadth BC is 3 inches, be divided by 3, the quotient will be a parallelogram, A BEF, 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 ▲ B. The number of square inches in one is the same as the number of linear inches in the other. If, therefore, the area of the large parallelogram be represented by a, the side A B = a BC that is, the length of a parallelogram is found by dividing the area by the breadth. If a be put for the area of a square whose side is A B, Then a = (AB)3, And extracting both sides, 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 base of a triangle, and B c its perpendicular height; Then And dividing by BC, a = BCX AB 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 cube; and that the side of a cubic inch is a square inch; the side of a cubic foot, a square foot, etc. Let ABCD fig. 3, represent the base of a parallelopiped, five inches long, three inches broad, and one inch deep. It is 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 parallelopiped, instead of being one inch, is four inches. Its contents must be four times as great. If, then, the length be AB, the breadth в c, and the depth c o, the expression for the solid contents will be, AB X BCX CO. By means of algebraical notation, a geometrical_demonstration 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 parts, 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=a+b; That is, s the square of the whole line, is equal to a and b, the squares the two parts, together with 2ab, 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, fig. 9, to find the perpendicular. Let the base The perpendicular The sum of hyp. and perp. (B C)2+(AB)2= (AC)2 721 2. That is, by the notation, x2+b2 = (a—x)2=a2 —2ax+x2 And, x= a22a =BC, the side required. Hence, Fig. 9. B In a right angled-triangle, the perpendicular is equal to the square of the sum of the hypothemuse and perpendicular, diminished by the square of the base, and divided by twice the sum of the hypothenuse and perpendicular. the letters a and b. Thus if the base is 8 feet, and the sum of It is applied to particular cases by substituting numbers for the hypothenuse and perpendicular 16, the expression 2a 162-82 becomes 2 X 16 =6, the perpendicular; and this subtracted 10, the length of the hypothenuse. from 16, the sum of the hypothenuse and perpendicular, leaves and perpendicular of a right-angled triangle, to find the perProb. 2. Given the base and the difference of the hypothenuse pendicular. B Fig. 10. Let the base, fig. 10, The given difference of A c and 4. Therefore 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 the length of the perpendicular? 50 rods, and the base is to the perpendicular as 4 to 3, what is Prob. 5. Having the perimeter and the diagonal of a parallelogram A B CD, fig. 11, to find the sides. Let the diagonal Half the perimeter being given, and the sides of a parallelogram inscribed in it, to Prob. 7. The three sides of a right-angled triangle ABC, fig. 13, being given, to find the segments made by a perpendicular, drawn from the right angle to the hypothenuse. Prob. 10. If the sum of two of the sides of a triangle be 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 D E F G, on the base A B, fig. 14. Prob. 14. Two sides of a triangle, and a line 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 manner as the parallelogram BEDF, fig. 12, be 12; what are the lengths of the other two sides of the triangle? Ne paraissez jamais ni plus sage ni plus savant que ceux avec qui vous étes. Portre votre savoir comme votre montre, dans une poche particulière. que vous ne tirez point, et que vous ne faites point sonner uniquement pour nous faire voir que vous en avez une.-Lord Chesterfield. coming having come Past Participle: venuto, come INDICATIVE MOOD. Present. venimmo, we came Véngo or végno, I come viéni, thou comest viene, he comes veniamo or vegnámo, we come Imperfect. Veniva or venía, I was coming veniste, you came vénnero or veniro, they came Verrò, I shall or will come Conditional Present. Verrei or verría, I should or would come verrésti, thou wouldst come verrébbe or verria, he would come ver rémmo, we would come terreste, you would come verrebbero, they would come IMPERATIVE MOOD. [No First Person.] Vieni, come (thou) venga, let him come Present. veniamo, let us come venite, come (ye or you) véngano, let them come SUBJUNCTIVE MOOD. Present: pióvere, to rain Past avere piovúto, to have Ci or vi sia, or staci or síavi, | Ci or vi siano, or sianci or sían rained let there be vi, let there be 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 ando or endo; as— Amándo, loving 2. The Past Participle ends as follows in the regular verbs: Amato-a, amati-e, loved Creduto-a, crediti-e, believed 3. The Participle Future is not so often used. It is as Avére ad amare, éssere per amáre, being about to love Avére a credere, éssere per crédere, being about to believe Avére a servíre, éssere per servire, being about to serve The Italians are accustomed to syncopate several past participles of the first conjugation, as may be seen in the following list: INDICATIVE MOOD. Present. Future. Bisogno, it was necessary dried accustomed laden fitted deprived fatigued trónco troncato cut off trováto found vólto voltáto turned vuoto emptied THE ADVERB. disengaged dirtied The Adverb is a word generally joined to a verb, participle, or an adjective, to express some circumstance, quality, degree, or manner of its signification. FORMATION OF ADVERBS. Italian adverbs are formed from adjectives in three ways, Di continuo, continually viz. : 1. By uniting the substantive mente to the feminine of the adjectives ending in a; as— Dotto or dótta, learned saggio or saggia, wise váno or vána, vain Dottamente, learnedly saggiamente, wisely Di di in dì, di giórno in giórno, Di giorno, by day Di quando in quando, di tempo Fin dove, till where Sù in alto, di sópra, above or Fin a quándo, till when Di sótto, abbasso or giù, under D' intorno, all around In giro, álla vólta, by turns after another 2. By adding the substantive mente to the adjectives ending Fin adésso, fin a quest óra, fin A uno a uno, one by one A vicenda, alternately Of Quantity and Number. Quanto, how much Molto, much Troppo, too much Un pochettino, but a little Niente affatto, not at all Non molto, not much Per metà, by half Davvantaggio, di vantaggio, di più, some more Abbastanza, enough Un poco di méno, a little less Un po' troppo, a little too much Túnte vólte, so many times Dappertutto, in ogni párte, every | Due vólte, twice way Molte volte, several times |