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THE POPULAR EDUOATOfL 



Tig. 46. 



common piece B c D K from the trapezium A B c D and the 

 triangle B F c, we have the triangle A K B, the remainder of the 

 trapezium A B c D, arid the triangle K F D, the remainder of the 

 triangle B F c. 



But these triangles are also parts of the triangles A D B, 

 B F D, which are equal in area, since they are on the same base, 

 B D, and between the same parallels A r, B D, and as the triangle 

 K D B is common to both, the triangle A K B is equal to the 

 triangle K p D. Iii the same iminncr, by drawing the diagonal 



A c of the tra- 

 B pezium A B c D, 



producing D c in 

 the direction of 

 G ; drawing B H 

 parallel to A c, 

 and meeting D G 

 in H ; and lastly, 

 joining A H, it 

 may be shown 

 that the triangle 

 A D H is also equal 

 in superficial area to the irregular quadrilateral figure A B c D. 



It will be useful for the student to repeat this construction as 

 an exercise, taking the sides c B, B A, and A D in succession as 

 the base of the trapezium A B c D, or the side on which it 

 stands. 



PROBLEM XXXIV. To draw a triangle tlmt shall be equal in 

 supeificial area to any given multilateral figure or polygon. 



First let us take a five-sided figure, as being next in order to 

 a four-sided figure, as far as the number of its sides are con- 

 cerned, and lot A B c D E (Fig. 47) represent the five-sided figure 

 or pentagon, to which it is required co draw a triangle equal in 

 superficial area. From C, the apex of the pentagon, draw the 

 straight lines c A, C E, to the points A, E, the extremities of the 

 base on which it stands. By doing this we divide the pentagon 

 ABODE into three triangles A B C, C A E, and c E D. Produce the 

 base A E indefinitely both ways in the direction of F and G, and 

 through B and D draw the straight linos B H, D K, parallel to 

 c A, c E respectively, and meeting the base A E produced, in the 

 points H and K. Join C H, C K ; the triangle c H K is equal 

 in superficial area to the pentagon ABODE. That this is 

 ti-uo may be seen as follows : Of the three triangles ABC, 

 c A E, and C E D, into which the pentagon was divided, the 

 triangle c A E is common to both the pentagon and the triangle 

 c H K. Of the remaining portions of the pentagon and triangle, 



the triangle A B c of the 

 former is equal to the tri- 

 angle C H A of the latter, 

 because they are on the same 

 base, A C, and between the 

 same parallels ; and for the 

 same reason the triangle 

 c E D of the pentagon is 

 ~o equal to the triangle c E K 

 of the triangle. 



The learner will find it 



Fig. 47. 



useful to repeat this construction as an exercise, taking the sides 

 A B, B c, c D and D E in succession, as the base on which the 

 pentagon is supposed to stand. 



That the learner may thoroughly understand the process of 

 drawing a triangle equal in superficial area to a polygon having 

 a great number of sides, and see that it is as easy as it is to 

 draw a triangle equal in area to a pentagon, which has only five 

 iides, vre will take the irregular seven-sided figure, or heptagon 



A B c D E F G (Fig. 48), and proceed to construct a triangle equal 

 to it in area. As the figure is complicated, the lines which con- 

 tain the heptagon and the triangle equivalent to it in area have 

 been drawn thicker than the lines which are necessary in working 

 out the process (as in Fig. 47), that the reader may the more 

 readily distinguish the relative areas of the figures in question. 

 The first step is to draw straight lines from A, the apex of 

 the polygon, taking D E to represent its base, to the points 

 c, c, E, F, or to each salient point of the polygon except the two 

 immediately on the right and left of the apex. The straight 

 lines A c, A D, A E, A F divide the polygon A B c D E F G into 

 five unequal triangles, ABC, Ac D, A D E, A E F, and A F o. 

 The reader will note that however many may be the sides of the 

 polygon, it is divided by this process into a number of triangles 

 always less by two than the number of its sides. Thus in the 

 figure below the number of triangles into which it is divided by 

 drawing straight lines from its apex to its salient points is five, 

 the number of its sides being seven ; a dodecagon, or twelve- 

 sided figure, would be divided into ten triangles, and so on. 

 Now beginning with the triangle ABC, the highest triangle 

 on the left side of the apex by producing D C in the direction 

 of p, indefinitely ; drawing B H parallel to A C to meet c D pro- 

 duced in H; and joining A H ; we get a triangle, A H c, equal to 

 the triangle ABC, and by adding the polygon A c D E F G to 

 each of these triangles, we find that we have a hexagon or six- 

 sided figure, A H D E F G, equal in area to the original seven- 

 sided polygon A B c D E F G. By making the triangle A K D 

 equal to the triangle 



A H D by the same 

 construction, which we 

 need not repeat, we 

 get a pentagon, or five- 

 sided figure, A K E F G, 

 equal in area to the 

 hexagon A H D E F G, 

 and consequently to _ i 

 the original heptagon Q K 



A B C D E F G. Con- 



tinuing the process with making the triangle A F L equal to the tri- 

 angle A F G, the highest triangle on the right side of the apex, we 

 get an irregular quadrilateral figure, A K E L, equal to the pen- 

 tagon A K E F G, the hexagon A H D E F G, and the heptagon 

 A B c D E F G. Once more, by making by a similar construction 

 the triangle A E M equal to the triangle A E L, we get at last a 

 triangle, A K M, equal in area to the quadrilateral figure A K E L, 

 and the above-named pentagon and hexagon and the original 

 heptagon A B c D E F G. 



The learner will find the value of this geometrical process in 

 determining the areas of irregular polygons in mensuration. To 

 calculate the area of the heptagon A B c D E F G, it would be 

 divided as in the above figure into five triangles, and by an 

 arithmetical process to be explained hereafter the superficial 

 content of each triangle would be found, and the five results 

 added together to obtain the area of the polygon. By reducing 

 the area of the polygon to a triangle, its area can be found by 

 one calculation instead of Jive, and a sum in compound addition ; 

 or, to ensure accuracy, both processes may bo gone through, each 

 proving a test whereby the correctness of the other may be 

 ascertained. 



As in the preceding propositions, let the learner repeat the 

 above construction as an exercise, taking the sides E F, F G, G A, 

 A B, B C, and c D in succession, as the base on which the polygon 

 is supposed to stand, and the salient point which happens to ba 

 immediately opposite the base in each case as the apex. 



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