GEOMETRY. 



Then 



AC 2 = AB 2 BC 2 ; 



.-. AC > VAB 2 BC 2 = ^2500 1600 = 30, 



or the side AC is found by taking the square root 

 of the difference of the squares upon the given 

 sides. This proposition en- 

 ables us to find the diagonal 

 of a rectangle when two 

 adjacent sides are given. 

 For, as before, 



AB 2 = AC 2 + BC 2 , 

 or AB = VAC 2 +BC 2 . 



We can also find the diagonal of a square when 

 the side is given ; and conversely, we can find 

 the side when the diagonal is 

 given. Thus A 



AB a = AC a + BC 2 . 



But the figure being a square, 



AC 2 = BC 2 ; 

 .'. AB 2 = 2AC 2 ; 

 .'. AB =AC^/2, 

 or AB : AC : : ^2 : i 



that is, the diagonal is to the side as V 2 is to i. 

 But as V 2 cannot be expressed exactly by a 

 limited number of figures, we say that the diag- 

 onal and side are incommensurable. Then since 

 the diagonal is found by multiplying the side by 

 */2, it follows that the side is found by dividing 

 the diagonal by *Jz. 



Again, in the equilateral tri- 

 angle, if the side is given, we 

 can find the perpendicular on 

 it from the opposite angle. 



Thus AD 2 = AB a BD 2 , 

 or AD 2 = AB 2 JAB 2 = f AB 2 ; 



or AD : AB : ^3 : 2. 



Hence the perpendicular is found by multiplying 

 the half of the side by y^, an< i consequently when 

 the perpendicular is given, we find the side by 

 dividing twice the perpendicular by ^3- 



PROP. II. Having given the three sides of a 

 triangle, to find the length of the perpendicular 

 from any of the angles 

 upon the opposite side. 



In the triangle ABC, 

 let BC = a, AC = 6, 

 AB = c ; find the 

 length AD of the per- 

 pendicular from A or 

 BC. 



If s be taken to denote half the sum of the sides, 

 it can be shewn that 



AD 



- a)(s - t>)(s - 



This result is deduced from some of the foregoing 

 propositions by a somewhat intricate algebraic 

 process, which we here omit. The formula itself 

 is given, as it is essential in the practical rule of 

 finding the area of a triangle when the three sides 

 are given. 



PROP. III. To divide a line into extreme and 

 mean ratio. 



Let AB = a be the 4 9 B 



given line, and let C be 



the point of division. Then AB BC = AC 2 ; it is 



required to find AC. 



Let AC = x = the greater of the two segments ; 

 .'. a(a x) = x 3 , or a' ax = x* ; 

 /. x 3 + ax = a\ 



THE CIRCLE. 



PROP. IV. Having given the height of an arc, 

 and the radius of the 

 circle, to find the length 

 of the chord. 



Let DC, the height of 

 the arc ACB = h, and 

 radius of circle r; to 

 find the length / of the 

 chord AB. In the right- 

 angled triangle AOD, the 

 hypotenuse AO = r, the 

 perpendicular OD = r 



.'. AD' = r= (r h)* 

 .'. AD = 



.'. AB = / = 2 *j2rh h*. 

 We might also have obtained AB thus : 



AD 2 = HD DC (Prop. XXIII., p. 620) 



= (zr K)h = -zrh h*, 

 AD = Jzrh h' ; 



.'. / = 2 *j2rh h*, as before. 

 Interpreting this formula, we have the following 

 rule for determining the chord : Multiply twice 

 the radius by the height ; from the product sub- 

 tract the square of the height, and take twice the 

 square root of the remainder. 



Remark. It may here be noticed that, in the 

 above formula, when 



h = o, the chord is zero, 

 and when h = r, the chord becomes the diameter. 



PROP. V. To find the side of a regular decagon 

 inscribed in a circle. 



Divide the radius AB in extreme and mean 

 ratio (Prop. III., p. 633) at 

 the point C ; make the chord 

 BD equal to AC, and join 

 AD. It has already been 

 shewn that the angle BAD 

 at the centre of the circle 

 is an angle of one-fifth of 

 two right angles, or one- 

 tenth of four ; ten such 

 angles can therefore be 

 made round the centre, con- 

 sequently the line BD is the side of a regular 

 decagon inscribed in a circle. 



633 



