384 



THE POPULAR EDUCATOR 



of the base of the triangle say, for example, the extremity D of 

 the half C D of the base with a radius equal to x, describe an 

 arc cutting p Q in H; join D H, and through C draw C K parallel 

 to D H, and meeting p Q in K, thus completing the parallelogram 

 H D c K ; or, at the point c in the straight line D c we must 



Fig. 42. 



make the angle D c K equal to the given angle Y, and through 

 D draw D H parallel to c K in order to complete the parallelogram 

 as before. 



This process will only be found practicable for the construction 

 of a square equal in area to a triangle when the triangle is a 

 right-angled isosceles triangle ; but for any other description of 

 triangle it will be found necessary first to construct a rectangle 

 equal in superficial area to the given triangle, and then to draw 

 a square equal to the rectangle thus obtained. How to do this 

 will be shown presently in Problem XXXI. 



By Problem XXX. we are enabled to construct a square equal in 

 area to any number of given squares. Thus, suppose wa wsh to 

 construct a square equal in superficial extent to the five squares 

 of which the length of the sides of each is represented by the 

 straight lines A, B, c, D, E respectively (Fig. 44). Draw any 

 straight line, F o, equal to A, and at its extremity, O, draw G H 

 at right angles to it equal to B. Join F H : the square described 



on F H is by Pro- 

 blem XXX. equal to 

 the squares de- 

 scribed on F Q and 

 G H. Next draw 

 H K equal in length 

 to the given line c, 

 at right angles to 

 H F. Join K F. The 

 square described on 



Fig. 43. 



F K is equal to the squares described on K H, H F, or to the 

 squares described on K H, H G, a F, since the square described 

 on H F is equal to the squares described on H G, G F. By con- 

 tinuing this process we at last obtain the straight line M F. The 

 square described on this line is equal to the sum of the squares 

 described on the given straight lines, A, B, c, D, E. Now let us 

 see how far this is of practical value to the artisan. Let us 

 suppose that a cabinet-maker has a number of small squares of 

 veneering of several kinds of choice wood, each square being 

 of a different size, and he wishes to use up this wood in veneer- 

 ing a table or the panel of a cabinet without wasting a single 

 scrap of it. By following the process just described it is manifest 

 that ho possesses the means of readily ascertaining the exact 

 area of the square that these pieces will cover, and after find- 

 ing this, he can, if it be desirable, by Problem XXXI. draw a 

 rectangle equal in area to the square if he prefer this form for 

 using up his squares of veneering, and then arrange his pattern 

 in such a manner that his squares 

 may be worked up without waste. 

 It is also a process that is useful 

 to the maker of floors in parque- 

 try, or to a stonemason who wishes 

 to know how large a square he 

 can pave with a number of smaller 

 squares of stone or slate of dif- 

 ferent sizes. Of course in such 

 cases the operator would work to 

 a given scale, and the process 



might be used as a test of the correctness of the result of the 

 operation by which the whole content of the squares may be 

 found arithmetically, or as one which is far more certain and 

 involves far less trouble than the arithmetical operation, which 

 would be a long and tedious one. 



PROBLEM XXXI. To draw a square that shall be equal in 

 superficial area to a given rectangle. 



Let A B c D (Fig. 45) be the given rectangle ; it is required 

 to draw a square equal in superficial area to the rectangle 

 A B c D. Produce c D indefinitely in the direction of E, 



Fig. 44. 



Fig. 45. 



and on the straight line r> E set off D F equal to the side 

 D A, or B c of the rectangle A B c D. Bisect c F in G, and 

 from G as centre with the radius G c or G F describe the semi- 

 circle c H F. Produce D A until it meets the arc c H F in K. 

 Then along the straight line D c set off D L equal to D K, and 

 through the points K, L, draw the straight lines K M, L M 

 parallel to c F, D K respectively, and meeting in the point M. 

 The figure D L M K is a square, and it is equal in area to 

 the given rectangle ABCD. If F N o L had been the 

 given rectangle, the same process would have been followed, p l> 

 would have been produced in the direction of L, and L c set off 

 on it equal to the side L o of the rectangle L o N F ; c F bisected 

 in G ; the semicircle c H F described as before, and L o produced 

 to meet the circumference c H F in p. The square drawn on L p 

 is equal in area to the rectangle F L o N. 



If it be required to draw 

 a square equal in area to a 

 given parallelogram, we have 

 only to construct a rectangle 

 equal to the given parallelo- 

 gram, and proceed as above. 

 This will be seen from Fig. 

 45, in which the rectangle 

 A B c D is equal to the 

 parallelogram D c Q B. 



PROBLEM XXXII. -To draw a rectangle that shall be equal 

 to a given square, and have one of its sides equal to a given 

 straight line. 



Let ABCD (Fig. 46) bo the given square, and x the given side 

 of the required rectangle, and in this case let x represent the 

 shorter of the two pairs of sides by which the rectangle is enclosed. 

 First produce C D indefinitely both ways towards E and F, and 

 along c E set off C G equal to x, and also along c B set off c H 

 equal to x. Join B G, bisect it in K, and through K draw K L per- 

 pendicular to B G, meeting E F in L. Then from the point L as 

 centre, with the radius L G, describe the semicircle G B M. 

 Through the point M draw M N parallel to A D or C B, and 

 through H draw H N parallel to A B or E F, and let the lines 

 H N, M N meet in N. The rectangle c H N M is equal in area 

 to the square ABCD. 



When the longer of the two pairs of sides that enclose the 

 rectangle is given, as Y in Fig. 46, produce c D indefinitely 

 both ways as before, and set off c M along c F equal to Y. 

 Join B M, bisect B M in O, and through the point O draw O L at 

 right angles to B M, meeting E F in L. Then from L as centre, 

 with the distance L M, describe the semicircle M B G. Set off 

 along c B the straight line c H equal to c G, and complete the 

 rectangle c H N M by drawing H N, M N through the points H and 

 M parallel to c M and c H respectively. 



The learner must remember that tho side of a square is a 

 mean proportional be- 

 tween the sides of any 

 rectangle that is equal to 

 it in superficial area ; and, 

 therefore, that to find the 

 length of the side of a. 

 square equal to a given 

 rectangle, we must set off 

 on the same straight line, 

 but in opposite directions, 

 two lines equal in length 

 to the sides of the given 

 rectangle, bisect the line thus obtained, describe a semicircle on 

 it, and find the mean proportional to the two lines of which it is 

 composed, by drawing a perpendicular from their point of junc- 

 tion to meet the semicircle, as in Problem XIII. (page 192); 

 while, to find the lengths of the sides of a rectangle that shall be 

 equal to a given square, we must draw a straight line at right 

 angles to a line equal in length to the side of the square, and 

 from a point in this line on either side of the line that repre- 

 sents the side of the given square, draw a semicircle with a 

 radius equal to the straight line joining the point that is used 

 as the centre of the semicircle and the more remote extremity 

 of the line that represents the length of the side of the given 

 square. The lines intercepted between the other extremity of 

 this line and the extremities of the arc of the semicircle will 

 be equal in length to the sides of a rectangle, having a super- 

 ficial area equal to that of the given square. 



Fig. 46. 



