TO GEOMETRICAL PROBLEMS. 
279 
then tiie equation will stand thus, a 4 — 2fx 2 _ g : 
whence x is found rr V f + %/ J' z + 
If, instead of the difference, the sum of the sides had 
been given, ip order to tind the ditference, the method 
of operation would have been the very same, onl}', in 
stead of finding the value of x in terms of«, by means 
of the equation .v\t + — 2s z a~x z -p s z a* — 2p z x z — 2cpff 2 
+ 2p z a z p C 2c/ra\ that of a must have been found*, in 
terms of .r, from the same equation. * 
PROBLEM XXIII 
Having one leg AB of a right-angled triangle ABC ; 
to find the other leg BC, so that the rectangle under their 
difference (BC —AB) and the hypothenusc AC, may he 
equal to the area of the triangle. 
Put ABzra,and BC—t; so shall AC — %/ aa P ; 
and =r x—a. s/aa -f xx, by the conditions of the 
problem. By squaring both sides O 
of this equation we have ga z x* = / 
x- — °ax P a z X a a -p xx : in / 
which the quantities x and« being / 
concerned exactly alike, the solu- / 
tion will therefore be brought out / 
from the general method for ex- / 
tracting the roots^ of these kinds of / 
according to which, having di- A B 
la 1 a 
vided the whole by a z x z , we get — — — 2 P 
X ~ + —; which, by making zz—-P —; will be 
x a ax 
reduced down to ^ — z —: 2 X z, or d — 2z — ^: 
y— . X d 
whence z is given — l + v But since P —- —z t 
I. 
T 4 
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