In figures of this kind, the one with the more sides, of that the Area is the larger: which always being equal to the rectangle described by the semi-perimeter of the polygon, and the Radius of the circle inscribed in the same.
Let the Perimeter of the regular figures be isoperimetric subsequently with 4; they will be:

If from two figures of this kind, the area of the one has been given, the places of the remaining one will be, by these Logarithms, in this way [i.e. by scaling]. Let the Square be given, of which the area shall be 49: I desire to know the Area of the Decagon isoperimetric with the square. The difference of the Logarithms of the Square and the decagon being taken of this situation here, to this being added the Logarithm of the square given, the total will be the Logarithm of the required Decagon. So:

If the area of the given Triangle shall be 49 parts; and the area of the Decagon required; for the difference of the Logarithms, the sum should be taken, ( because the Logarithm of the triangle is deficient [ i.e. a negative log.] ), 0,20390,5843. Which added to the Logarithm of 49 given, gives the Logarithm of the required Decagon, 1,89410,1923.
Or by Chapter 15:

After the area of the polygon has been found or given, if the side or perimeter of the of the same you desire to know, you should consult Ch. 28.
And, if the area is divided by the semi-perimeter, the quotient will be the radius of the inscribed circle