33. In a given triangle to inscribe a square.-Let ABC

A Cat G; draw G H parallel to CD. Then FGHI is a square inscribed in the given triangle.

34. To construct a parallelogram of given sides and given angle.-Draw a right line A B (fig. 29) and construct the

By drawing the diagonals AD and BC, the figure is divided into four triangles CA B, ABD, BDC, DCA, each of which is equal in area to one-half that of the parallelogram itself.

If the lines A E, BF are drawn perpendicular to AB and meeting CD in E and F, a rectangle ABFE is constructed which is equal in area to the first drawn parallelogram, and to double that of either of the triangles ACB or ADB.

The perpendicular CH dropt upon A B is equal

to A E, and represents the height of the vertex of the triangle A CB above the base A B; hence, when a triangle and a parallelogram are upon the same base and have the same vertical height, the area of the parallelogram is double that of the triangle.

Fig. 30.

35. To construct a square equal in area to the sum of the areas of two given squares.-Let A B and AC (fig. 30) be two lines drawn at right angles. Take A B and A C equal in length to the sides of the given squares respectively; join CB; then BC is the side of the square which equals in area the sum of the squares upon A B and A C.

36. To construct a square which shall be equal in area to

FG equal to FE; bisect CG in I. From I as a centre and with IG as radius, describe a circle cutting A H in the point H. Then the square constructed on the line FH will equal in area the rectangle CDEF, and therefore will be equal to the given parallelogram.

37. Upon a given right line to construct a rectangle equal

rectangle.-Let ACDE (fig. 32) be the given rectangle. Produce EA to B, making A B equal to the given line.. Through B draw FBH parallel to CA (13), and produce DC to meet FH in F. Draw the diagonal F A and produce it to meet DE in G. Draw G KH parallel to EB, meeting FH in H. Produce CA to meet G H in K. Then the figure ABHK is the required rectangle equal in area to the given one and drawn upon the

- 38. To construct a rectangle equal in area to a given

Fig. 33.

Fig. 34.

irregular rectilinear figure. - Let ABCDE (fig. 33), be the given figure. Draw the lines BE, EC so as to divide the figure into triangles. Draw AF perpendicular to BE (4), BG and DH to EC. Construct (34) the rectangle P (fig. 34) having the side IK equal to AF, and the side KL equal to half the

line EB; then the area of P is equal to that of the triangle EAB. In the same manner construct a rectangle Q equal to the triangle EBC, and draw upon KL (37) a rectangle KLNM equal in area to Q. Draw a rectangle R equal to

the triangle EDC and construct upon MN a rectangle

equal to R. Then the whole rectangle IT is equal in area to the given figure.

39. To construct a rectilinear figure of a given irregular form.—Let ABCDE (fig. 33) be the given figure, as for example, the plan of a field. In order to lay this down on paper we must not only measure the sides AB, BC, &c., but also the diagonals EB, EC. First draw one of the sides as A B, then from A as a centre, with A E as radius, describe a circle; and from B as a centre, with BE as radius, describe an arc cutting the circle in E; then the positions of two sides A B and AE are fixed. Next, from E as a centre, with EC as radius, describe a circle; and from B as a centre, with B C as radius, draw an arc cutting the last circle in C: join BC. From C as a centre, with CD as radius, describe a circle; and from E as a centre, with ED as radius, draw an arc cutting the last circle in D; join CD, ED, and the figure is completed.

In this manner the outline of any irregular figure can be laid down by dividing it into as many triangles, less two, as the figure has sides.

40. To divide a given finite right line into two parts, so that the rectangle under the whole line and one of the parts shall be equal in area to the square upon the other part.Let A B (fig. 17) be the given line. Draw EAD at right angles to A B (3). Bisect AB in C and make AE equal

to AC; join EB, and make E A D equal to EB. Cut off AF equal to AD, and the rectangle under AB and BF is equal to the square on A F.

41. To produce a given right line so that the rectangle under the whole produced line and the part produced shall

be equal in area to the square upon the given line.-Let A B (fig. 18.) be the given line. Bisect AB in C and draw CD at right angles to A B (3), making CD equal to A B. Join A D, and produce A B to E, making CE equal to AD. Then the rectangle under A E and EB is equal in area to the square upon A B.

42. To construct a regular pentagon, or figure of five sides, the length of a side being given.-Let A B (fig. 35)

Fig. 35.

be the given side. On A B construct (25) an isosceles triangle ACB in which each angle CAB and CBA is double the angle ACB. Bisect the sides of the triangle in the points P, Q, R; and draw the lines PO, QO, RO, at right angles to the sides respectively. These perpendiculars will all meet in one point 0.

Join OC, and from O as a centre, with OC for radius, describe a circle, which will pass through each vertex of the triangle ABC. Produce OQ and OR to meet the circle in E and D. Join CD, CE, EB, AD; then the figure ABECD is the pentagon required, each side being equal to A B.

The ratio of AB to OP is very nearly as 1: 688, and that of A B to BC is very nearly as 1:1-618.

43. To construct a regular decagon, or figure of ten sides, the length of a side being given.-Let AB (fig. 36) be the given side. Construct (25) upon AB an isosceles triangle AOB having each angle OA B and OBA double the angle AOB. From O as a centre, with OA as radius, describe