882

THEORY AND PRACTICE OF ENGINEERING.

Book II.

Fig. 1132.

To reduce two Squares into One. — If two squares touchat one angle, and the two sides of one angle are in a linewith the two sides of another angle, as at A B G and E B C,draw a right line G C, on which form the square C G H 1;its area will be equal to that of the two squares; but if thetwo squares do not touch, as KLMN and O P Q, R, drawa right angle S T V, and set off the side VKLon T S, fromT to X, and the side O P on TV, from T to Y; draw theright line XY, on which erect the square XYZ, &c., andit will be equal to K L INI N and O P Q, R.

Euclid, in his 47th prop, of Book I. demonstrates, that inany right-angled triangle the square which is described uponthe side subtending the right angle, is equal to the squaresdescribed upon the sides which contain the right angle.

To reduce a Square into several equal Squares , as that ofA B C D into two equal squares : divide one side into twoequal parts, as A B, in E; from E with a radius A E,describe the circle AGB: elevate a perpendicular EFfrom E, observe where it cuts the demicircle, as at G,in order to draw the lines A G and G B; the squaresconstructed on these lines will contain half A B C D.

In any right-angled triangle the square which is describedupon the side subtending the right angle is equal to thesquares described upon the sides which contain the right angle.

Let GAB be a right-angled triangle, having the rightangle AGB; the square described upon the side A Bis equal to the squares described upon AG, GB. OnA B describe the square A D C B, and on A G, G B,the squares I A, KB, and through G draw GL parallelto A D or B C, and join G D, H B. Then, because eachof the angles AGB, A GI, is a right angle, the twostraight lines G B, G I, upon the opposite side G I,make with it at the point G the adjacent angles, equalto two right angles; therefore, B G is in the samestraight line with GI: for the same reason G A andG K are in the same straight line; and because theangle D A B is equal to the angle HAG, each of thembeing a right angle, add to each the angle GAB, andthe whole angle D A G is equal to the whole HAB:and because the two sides G A, A D are equal to thetwo H A, A B, each to each, and the angle DAG equalto the angle HAB; therefore the base G D is equal to the base II B, and the triangleG A D to the angle H A B. Now the parallelogram A L is double of the triangle GAD,because they are upon the same base A D, and between the same parallels AD, G L; andthe square I A is double of the triangle HAB, because these also are upon the same baseH A, and between the same parallels HA,

IB. But the double of equals are equalto one another, therefore the whole squareA DOB is equal to the two squares I A,

K B, and the square A D C B is describedupon the straight line A B, and the squaresI A, K B upon AG, G B. Wherefore thesquare upon the side A B is equal to thesquares upon the sides AG, G B.

To reduce several Figures to a Rectangleof a given length , as the trapezoid A B C Dand the pentagon EFGHI to a rect-angle of the width K I.: first reduce thefigures to triangles of the same height,and twice the breadth KL, as are thetwo triangles P C Q, and R S O : set offthe base C Q from K to T, and S O fromT to V; at the point T elevate the per-pendicular T X, equal in height to P C Q,and draw the right line K X and X V;reduce the triangle X K V to a rectangle,which will be equal to the two figuresA B C D and E F G H I.

Fig. 1183.

Fig. 1185.

Fig. 1184.