An elementary treatise on geometry : simplified for beginners not versed in algebra . ently, also, to the squares upon CD, CB, and AB(the square upon AC being equal to the squares upon CB and AB) ;and finally the square upon AE is equal to the sum of the squaresupon ED and AD; or, which is the same, to the sum of the squaresupon DE, CD, CB, and AB Problem XXXII. To describe a square which shallheequal to the difference of two given squares. Solution. Let AB, AC ^ be the sides of fwo squares. /^^^^ 1. Upon the greater side, // >v \ AB, as a diameter, describe /X x. \ a semicircle. ^^ -^^ 2.
An elementary treatise on geometry : simplified for beginners not versed in algebra . ently, also, to the squares upon CD, CB, and AB(the square upon AC being equal to the squares upon CB and AB) ;and finally the square upon AE is equal to the sum of the squaresupon ED and AD; or, which is the same, to the sum of the squaresupon DE, CD, CB, and AB Problem XXXII. To describe a square which shallheequal to the difference of two given squares. Solution. Let AB, AC ^ be the sides of fwo squares. /^^^^ 1. Upon the greater side, // >v \ AB, as a diameter, describe /X x. \ a semicircle. ^^ -^^ 2. From A, within the . A. C semicircle, draw the line AC, equal to the given line AC, and join BC; then CBis the side of the square sought. Demon. The triangle ABC is right-angled in C, and in everyright-angled triangle, the square upon one of the sides, whichinclude the right angle, ie equal to the difference between thesquares upon the hypothenuse and the other no GEOMETRY. Problem XXXIII. To transform a given figurein such a wai/, that it may be similar to another n ti Solution. Let X be the given figure, and ABCDEFthe one to which it is to be similar. 1. Convert the figure ABCDEF into a square (see theremark, page 166), and let its side be mw, so that the areaof the square upon mn is equal to the area of the figureABCDEF ; convert, also, the figure X into a square, andlet its side be i^q, so that the area of the square upon pgrshall be equal to the area of the figure X. 2. Take any side of the figure ABCDEF, say AF;and to the three lines, mn^ pq, AF, find a fourth propor-tional (Problem X), which you cut oif from AF. LetA/be this fourth proportional, so that we have the pro-portion mn : pq = AF : A/. 3. Then draw the diagonals AE, AD, AC, and the linesfe, ed, dc, cb, parallel to the lines FE, ED, DC, CB ;then AbcdefwiW be the required figure, which in area isequal to the figure X, and is similar to the figure ABCDEF. Demon. It is easily proved, that the figur
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Keywords: ., bookauthorgrundfrancisjfrancisjoseph18051863, bookcentury1800
