(3.1), join FG, make EL FG, EH=C, join HL, take EN= C22 = ) Æ2 + B2 + C2; and because EN=LH, and EM=D, ':' MN)2=(EN2+EM)2=LH2+Do=) A2 + E2 +C2+D2, which was to be shewn; and in the same manner any number of squares may be added together, that is, a square may be found equal to their sum. 158. To find a square equal to the difference of the squares of two given unequal straight lines. F Let A and B be two unequal straight lines, whereof 4 is the greater; it is required to find a square equal to the excess of the square of A above the square of B. In any straight line CH take CD A B D E H (3. 1.) from D as a centre with the K distance DC describe the circle CKF, from E draw EF perpendicular to CH (11.1), and join DF; EF will be the side of the square required. Because FD=(DC=) A, DE=B, and DEF is a right angle, '' (47. 1.) FD2=(DE2+ EF)2=) B+EF2, that is A=B2+ EF; take B from each of these equals, and A-B2— EF}2 ; that is, EF is the side of the square, which is the difference required. 159. Hence, if any two sides of a right angled triangle be given, the third side may be found. (See the preceding figure.) For since DE+EF=DF, DF-DE=EF, DF2-EF2=DE, DE+ER=DF. EXAMPLES.-1. If the base DE of a right angled triangle be 6 inches, and the perpendicular EF 8 inches, required the longest side, or hypothenuse DF‹? DF. Here DE+EF)2= √/C2+82= √√/36+64= √✓/100=10= 2. Given the hypothenuse 20, and the base 11, to find the perpendicular? Thus √20-11)2= √√/400—121= √/279=16.703293=the perpendicular required. 3. Given the hypothenuse 13, and the perpendicular 10, to find the base? Thus √13)2 — 10} ° = √✓/169—100= √/69=8.3066239=the base required. 4. Given the base 7, and the perpendicular 4, to find the hypothenuse? Ans. 8.0622577. 5. Given the hypothenuse 12, and perpendicular 10, to find the base? Ans. 6.6332496. 6. Given the hypothenuse 123, the base 99, to find the perpendicular? ON THE SECOND BOOK OF EUCLID'S ELEMENTS. 160. The second Book of Euclid treats wholly of rectangles and squares, shewing that the squares or rectangles of the parts of a line, divided in a specified manner, are equal to other rectangles or squares of the parts of the same line, differently divided : by what rectangle the square of any side of a triangle exceeds, • In a right angled triangle the longest side, (viz. that opposite the right angle) is called the hypothenuse, the other two sides are called legs, that on which the figure stands called the base, and the remaining leg the perpendicular. or falls short of the sum of the squares of the other two sides, &c. 161. Rectangles and squares may in every case be represented by numbers or letters, as well as by geometrical figures, and frequently with greater convenience; thus, one side of a rectangle may be called a, and its adjacent side b, and then the rectangle itself will be expressed by ab; if the side of a square be represented by a, the square itself will be represented by aa or a2; and since in this book, the magnitudes and comparisons only, of rectilineal figures are considered, its object may be attained by algebraic reasoning with no less certainty and with much greater facility than by the geometrical method employed by Euclid; we will therefore shew, how the propositions may be algebraically demonstrated. 162. Def. 1. Euclid tells us what " every right angled parallelogram is said to be contained by," but he has not informed us either here, or in any other part of the Elements, what we are to understand by the word rectangle, although this seems to be the sole object of the definition; instead then of Euclid's definition, let the following be substituted. "Every right angled parallelogram is called a rectangle; and this rectangle is said to be contained by any two of the straight lines which contain one of its angles." 163. Prop 1. Let the divided line BC=s, its parts_BD=a, DE=b, and EC=c; then will s=a+b+c. Let the undivided line A=x, then if the above equation be multiplied by x, we shall have sx=(a+b+c.x=) ax+bx+cx; "that is, the rectan gle sx contained by the entire lines s and x, is equal to the seve ral rectangles ax, bx, and cx, contained by the undivided line 1 and the several parts a, b, and c, of the divided line s." Q. E. D. Cor. Hence, if two given straight lines be each divided into any number of parts, the rectangle contained by the two straight lines will be equal to the sum of the rectangles contained by each of the parts of the one, and each of the parts of the other. Thus, let s=a+b+c, as before. And xyz. Then sx=(a+b+c.y+z=)ay+by+cy+az+bz+cz. The rectangle contained by two straight lines AB, BC, is frequently called "the rectangle under AB, BC;" or simply "the rectangle AB, BC.” Prop. 2. Let AB=s, AC=a, and CB=b. n a+b=s, multiply these equals by s, and as+bs=ss; he rectangle contained by the whole line s and the part er with that contained by the whole line s and the other e equal to the square of the whole line s. Q. E. D. s proposition is merely a particular case of the former, if the line s be divided into the parts a and b, and the d line r=s, we shall have sx=ax+bx, become ss=as+ this proposition. Prop. 3. Let AB=s, AC=a, and CB=b, then will s≈ d sb=(a+b.b=) ab+bb; in like manner sa=(a+b.a=) that is, in either case the rectangle contained by the and either of the parts a or b, is equal to the rectangle ined by the two parts a and b, together with the square foresaid part a, or b as the case may be. Q. E. D. is proposition is likewise a particular case of the first, in the undivided line is equal to one of the parts of the line. Prop. 4. Let AB=s, AC=a, and BC=b, then will. b; square both sides, and ss=(a+b)2=) aa+2 ab+bb; the square of the whole line s, (viz. ss) is equal to the the squares of the parts a and b, (viz. aa+bb) and twice tangle or product of the said parts, (viz. 2 ab.) Q. E. D. Prop. 5. Let AC=CB=a, CD=x, then will AD=a+ DB=a-x, and their rectangle or product a+x.a—x= ; to each of these equals add xx, and a+x.a−x+xx=aa, , the rectangle contained by the unequal parts, together he square of (x) the line between the points of section is to the square of (a) half the line. Q. E. D. n the corollary, it is evident that CMG=the difference or of CF above LG, that is, of the square of (CB, or) AC the square of CD; but CMG is =AH=(AC+CD× CD=) AD× DB, therefore (CB-CD2, that is) AC=AD × DB, or as we have shewn above aa-xx=a+x. 2 Euclid's demonstration there is no necessity to prove the figure CGKB gular in the manner he has done; it may be shewn thus, "because B is a parallelogram, and the angle CBK (the angle of a square) a right , therefore all the angles of CGKB are right angles by Cor. 46. 1. 292 or falls short sides, &c. 161. Recta by numbers be represe 16 paral form we: to de t 1 2sb+aa=ss+bb, by taking 2 sb from - › — — sb+bb; that is, the square of the differ..B and (b) CB, is less than the sum of the 5) CB, by twice the rectangle (2 sb) v those lines. B=s, AC, CB=b, then s=a+b, or -ss-2s to each of these equals →...=ss+2 sb+bbs; that is, (4 sb, or) e contained by the whole s, and one part ne square of the other part a, is equal to of the straight line made up of the whole D. C=CB=a, CD=r, then will the greater in the less segment DB=a-x. bach=3a+2xx=2.aa+rr —4—2=2.aa+rr, or the sum of the qual parts (a+x and a-x) is equal to double and of the part x between the points |