33. In a given triangle to inscribe a square.—Let A BC Fig. 28. (fig. 28) be the given triangle. Draw CD perpendicular (4) to AB, and draw .CE parallel to A B, making CE equal to CD. Join AE by a right line cutting BC at F. Draw FI parallel to CD, and FG parallel to AB and cutting A C at 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 Fig. 29. angle CAB equal to the given angle (7). Make A B and AC equal respectively to the given sides, draw CD parallel to AB, BD parallel to AC (13); then the figure ACDB is the parallelogram required. By drawing the diagonals A D 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 A BFE is constructed which is equal in area to the first drawn parallelogram, and to double that of either of the triangles ACB or A DB. The perpendicular CH dropt upon A B is equal to A E, and represents the height of the vertex of the triangle ACB 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. 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. Fig. 30. Take A B and AC 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 a given parallelogram. Fig. 31. Let ABCD (fig. 31) be the given parallelogram. Produce one side A B to H, and draw D E and CF at right angles to A H (4). Then the rectangle CDEF is equal in area to the given parallelogram (34). Produce CF to G, making 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 in area to a given rectangle.—Let ACDE (fig. 32) be the Fig. 32. given rectangle. Produce E A 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 FA and produce it to meet DE in G. Draw GKH parallel to E B, meeting FH in E. Produce C A 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 given line. - 38. To construct a rectangle equal in area to a given Fig. 33. 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 Fig. 34. E A B. 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 A B, BC, &c., but also the diagonals EB, EC. First draw one of the sides as A B, then from A as a centre, with AE 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 C D 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 E AD at right angles to A B (3). Bisect A B in C and make A E equal to AC; join E B, and make EAD equal to EB. Cut off AF equal to AD, and the rectangle under A B 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 A B 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 A D. 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) be the given side. On AB Fig. 35. 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 O. Join 0 C, and from O as a centre, with OC for radius, describe a circle, which will pass through each vertex of the triangle A BC. Produce 0 Q and OR to meet the circle in E and D. Join CD, CE, EB, AD; then the figure A BECD is the pentagon required, each side being equal to A B. The ratio of A B to O P is very nearly as 1 : •688, and that of A B to B C 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 A B (fig. 36) be the given side. Construct (25) upon A B an isosceles triangle A O B having each angle 0 A B and O B A double the angle A OB. From O as a centre, with O A as radius, describe |