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+EFDF,

2

DF-DE2=EF2,

DE+EF=DF.

√DF-DE2=EF.

DF2-EFDE, ··· DF2 - EF2DE.

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= √/6° +82= √/36+64= √✓/100=10=

=

2. Given the hypothenuse 20, and the base =11, to find the perpendicular?

Thus √202-11)2= √/400—121=/279=16.703293=the perpendicular required.

3. Given the hypothenuse 13, and the perpendicular 10, to find the base?

Thus 132-10)2= √/169—100= √√/69=8.3066239= the base required.

4. Given the base 7, and the perpendicular 4, to find the hypothenuse? Ans. S.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 is 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 f."

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 sa contained by the entire lines s and x, is equal to the several rectangles ax, bx, and cx, contained by the undivided line ¤, 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 salled "the rectangle under AB, BC;" or simply " the rectangle AB, BC."

164. Prop. 2. Let AB=s, AC=a, and CB=b.

Then a+b=s, multiply these equals by s, and as+bsss; that is, the rectangle contained by the whole line s and the part a, together with that contained by the whole line s and the other part b, are equal to the square of the whole line s. Q. E. D.

This proposition is merely a particular case of the former, in which if the line s be divided into the parts a and b, and the undivided line x=s, we shall have sx=ax + bx, become ss=as+ bs, as in this proposition.

165. Prop. 3. Let AB=s, AC=a, and CB=b, then will s= a+b,and sb=(a+b.b=) ab+bb; in like manner sa=(a+b.a=) aa+ab; that is, in either case the rectangle contained by the whole s, and either of the parts a or b, is equal to the rectangle ab contained by the two parts a and b, together with the square of the aforesaid part a, or b as the case may be. Q. E. D.

This proposition is likewise a particular case of the first, in which the undivided line is equal to one of the parts of the divided line.

166. Prop. 4. Let AB=s, AC=a, and BC=b, then will s=a+b; square both sides, and ss=(a+b) aa+2ab+bb; that is, the square of the whole line s, (viz. ss) is equal to the sum of the squares of the parts a and b, (viz. aa+bb) and twice the rectangle or product of the said parts, (viz. 2 ab.) Q. E. D.

167. Prop. 5. Let AC=CB=a, CD=x, then will AD=a+ x, and _DB=a—x, and their rectangle or product a+x.a-x= aa-xx ; to each of these equals add xx, and a+x.a−x+xx=aa, that is, the rectangle contained by the unequal parts, together with the square of (x) the line between the points of section is equal to the square of (a) half the line. Q. E. D.

In the corollary, it is evident that CMG=the difference or excess of CF above LG, that is, of the square of (CB, or) AC above the square of CD; but CMG is = AH=(4C+CD× AC-CD=) ADX DB, therefore (CB-CD, that is) ACCD2=AD × DB, or as we have shewn above au-xx=a+x.

a-x.

In Euclid's demonstration there is no necessity to prove the figure CGKB rectangular in the manner he has done; it may be shewn thus, " because CGKB is a parallelogram, and the angle CBK (the angle of a square) a right angle, therefore all the angles of CGKB are right angles by Cor. 46. 1.

169. Prop. 6. Let AC=CB=a, BD=x, then will AB=2 a, and AD=2a+x; then the rectangle contained by AD and DB will be 2a+x.x=2 ax+xx, to these equals let aa (the square of half AB) be added, and 2 a+x.x+aa == (aa+2 ax+xx =) a+x)2 ; that is, the rectangle contained by the line produced and part produced, together with the square of half the line bisected, is equal to the square of the line made up of the half, and part produced. Q. E. D.

Cor. Hence, if three lines a, a+x, and 2a+x be arithmetically proportional, the rectangle contained by the extremes (x.2a+x) together with the square of the common difference a, (or aa) is equal to (a+x2) the square of the middle term.

169. Prop. 7. Let AB=s, AC=a, CB=b, then s=a+b, and ss=(a+b)2=aa+2ab+bb=) 2 ab+bb+aa, to these equals add bb, and ss+bb= (2 ab+2 bb+aa=2.a+b.b+aa=) 2sb+aa; that is, the square of the whole line, (or ss) and the square of one part b (or bb,) is equal to twice the rectangle contained by the whole s, and that part b, (or 2 sb,) together with (aa) the square of the other part. Q. E. D.

Cor. Hence, because 2 sb+aa=ss+bb, by taking 2 sb from both, we have aa=ss—2 sb+bb; that is, the square of the difference of two lines (s) AB and (b) CB, is less than the sum of the squares of (s) AB and (b) CB, by twice the rectangle (2 sb) 2.AB.CB contained by those lines.

170. Prop. 8. Let AB=s, AC=a, CB=b, then s=a+b, or a=s—b, aa—(s—b)2==) ss — 2 sb+bb, to each of these equals add 4 sb, and 4 sb+aa=ss+2 sb+bb=s+b2; that is, (4 sb, or) four times the rectangle contained by the whole s, and one part b, together with (aa) the square of the other part a, is equal to (s+or) the square of the straight line made up of the whole s, and the part b. Q. E. D.

171. Prop. 9. Let AC=CB=a, CD=x, then will the greater segment AD=a+x, and the less segment DB=a-x.

2

2

That is, a+ x)2 + a−x' =2.aa+xx, or the sum of the squares of the unequal parts (a+x and a-x) is equal to double the square of the half a, and of the part x between the points

of section; or, which is the same thing, "the aggregate of the squares of the sum and difference of two straight lines a and x is equal to double the squares of those lines." Q. E. D.

172. Prop. 10. Let AC=CB=a, BD=x, then will AD= 2 a+x, and CD=a+x.

Now 2a+x=4 aa+4 ax + xx

Add xx to this, and the sum is 4 aa+4 ax + 2 xx

Also a +x=aa+2 ax+xx, add aa to this, and it becomes 2 aa+2 ax + xx; now the former of these sums is double of the latter, that is 4 aa+4 ax+2xx=2.2 aa+2 ax+xx; or, the square of the produced line 2 a+x, together with the square of the part produced r, is double the square of a half the line, and the square of a+d the line made up of the half and the part produced. Q. E. D.

173. Prop. 11. This proposition is impossible by numbers, for there is no number that can be so divided, that the product of the whole into one part, shall equal the square of the other part; the solution may however be approximated to as follows:

Let AB=2 a, AH=x, HB=y, then by the problem x+y= 2a, and 2 ay=xx; from the first equation y=2 a−x, this value being substituted for y in the latter equation, we shall have 4aa-2ax=xx, or xx+2ax=4 aa, this solved (by Art. 97. part. 3.) gives x=±√5 aa—a, and y=(2 a—x=2 a−√√5 aa—a=) 3a— √5 aa, or which is the same x=1.236068, &c. xa, and y= 763931, &c. X a.

174. Prop. 12. Let AB=a, BC=b, CD=x. and AD=z; Then (47. 1.) AB=BD +DA)2=b+x)2+zz=

2

2

bb+2 bx+xx+zz

bb * +xx+zz

2 bx * *

That is, the square of AB, 'the side subtending the obtuse angle, is greater than the sum of the squares of CB and AC, the sides containing the obtuse angle, by (2 bx) twice the rectangle BC, CD. Q. E. D.

175. Prop. 13. Let AB=a, CB=b, AC=c, AD=d, BD=m, DC=n; then the first case of this proposition is proved as follows:

First, bb+mm=2bm+nn (7.2.) To each of these equals add