# Errata

**Author(s):** Anonymous  
**Year:** 1699  
**Journal:** Philosophical Transactions (1683-1775)  
**Volume:** 21  
**Pages:** 2 pages  
**Identifier:** jstor-102652  
**JSTOR URL:** <https://www.jstor.org/stable/102652>  

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Lines $ax$, $xb$, whose Squares together shall be equal to the Square given $gg$.

Let $axb$ whose height is $xy$ be the Triangle required. Bisect $ab$ in $m$ and draw $mx$.

**Analysis.**

Let therefore $axa + xbx = gg$

But by the 13th. of the Intro. $axa + xbx = 2ama + 2mxm$

Therefore $gg = 2ama + 2mxm$

or $gg - 2ama = 2mxm$

Therefore the Problem is solv'd, but the Length of $mx$ being given and not its Position, it is evident that it may be the Semidiameter of a Circle whose Circumference shall be the Locus of the point $x$.

**Construction and Demonstration.**

From the Square given $gg$ Subtract the double Square of $am$, the Square root of half the remainder shall be the line $mx$, with the Center $m$ and distance $mx$, describe the Circle $pxd$. I say that any point $x$ taken in its Circumference resolves the Problem.

For since the double of the Squares of $am$ and $xm$ is equal to the Square $gg$, by the Construction, and by the 13th. Proposition of the Introduction to the Squares $ax$ and $xb$: The two Squares $ax$ and $xb$ together will be equal to the Square $gg$. Which was to be done.

**FINIS.**

**ERRATA.**

Page 355. l. r. for IV. r. III. p. 356. l. 26. for III. r. IV. and for subtraction, subtraction, &c. r. subtraction, &c. p. 357. l. 33. r. Soligenes.