PROBLEM XXII. 31. To construct a parallelogram, having the sum of its base and altitude given, which shall be equivalent to a given square. On A B, the given sum, as a diame D C ter, describe a semicircumference. At any point, as B, in AB draw the perpendicular BC equal to a side of the given A : B E square; through C draw CD parallel to A B, cutting the circumference in D; draw DE perpendicular to AB. AE, E B are one the base and the other the altitude of the parallelogram required (26). 32. Scholium. If the side of the square is greater than half the sum of the base and altitude, the construction is impossible. PROBLEM XXIII. 33. To construct a parallelogram having the difference between its base and altitude given, which shall be equivalent to a given square. D On A B the given difference, as a diameter, describe a semicircumference. At A draw the perpendicular A D equal to a side of the given square; join D with the centre C, and produce DC to E. DF, DE are one the A base and the other the altitude of the parallelogram required (III. 64). F C B E PROBLEM XXIV. 34. To construct a square equivalent to a given parallelogram. Find a mean proportional between the altitude and base of the given parallelogram (26), and it will be a side of the required square. PROBLEM XXV. 35. To construct a square equivalent to a given triangle. Find a mean proportional between the base and half the altitude (26), and it will be a side of the required square. PROBLEM XXVI. 36. To construct a square equivalent to a given circle. Find a mean proportional between the radius and the semicircumference, and it will be a side of the required square. PROBLEM XXVII. 37. To construct a square equivalent to the sum of two given squares. Construct a right triangle (9) with the sides adjacent to the right angle equal respectively to the sides of the given squares; the hypothenuse will be a side of the required square (II. 27). 38. Scholium. By continuing the same process we can find a square equivalent to the sum of any number of given squares. PROBLEM XXVIII. 39. To construct a square equivalent to the difference of two given squares. Construct a right triangle (11), taking as the hypothenuse a side of the greater square, and for one of the sides adjacent to the right angle a side of the other square; the third side of the triangle will be a side of the required square (II. 28). PROBLEM XXIX. 40. To construct a triangle equivalent to a given polygon. E B C D F Let AD be the polygon. Draw BD cutting off the triangle BCD; through C draw CF parallel A to BD; join BF, and a polygon ABFE is formed with one side less than the given polygon and equivalent to it. For the triangles BCD and BFD, having the same base B D, and the same altitude, are equivalent; adding to each the common part A B DE, we have ABCDE equivalent to ABFE. In like manner a polygon with one side less can be found equivalent to ABFE, and by continuing the process the sides may be reduced to three, and a triangle obtained equivalent to the given polygon. 41. Scholium. Since by (35) a square can be found equivalent to a given triangle, by (40) and (35) a square can be found equivalent to any polygon. PROBLEM XXX. 42. On a given line to construct a polygon similar to a given polygon. Let AD be the given poly- B gon and ML the given line. C H I DG K GLM equal respectively to AFE and AEF, and a triangle GLM will be formed similar to AEF. In like manner on GL construct a triangle similar to A DE; on GK one similar to ACD; on GI one similar to ABC; and the polygons A D, KG, being composed of the same number of similar triangles similarly situated, are similar (II. 75). PROBLEM XXXI. 43. Two similar polygons being given, to construct a similar polygon equivalent to their sum, or to their difference. Find a line whose square shall be equivalent to the sum (37), or to the difference (39), of the squares of any two homologous sides of the given polygons, and this will be the homologous side of the required polygon (II. 31). On this line construct (42) a polygon similar to the given polygons. PROBLEM XXXII. 44. To construct a square which shall be to a given square in a given ratio. E AE B GFC G F On any line AC, as a diameter, describe a semicircumference ABC; divide the line AC at the point D so that AD: DC in the given ratio. Perpendicular to A C draw D B meeting the circumference at B; join BA, BC, and on BC, produced if necessary, take BF = a side of the given square. Through F draw EF parallel to AC, meeting BA in E, and BE is a side of the required square. For as B is a right angle (III. 23), we have (II. 72) But as E F is parallel to A C, we have (II. 47) 4 therefore (Pn. 11) BE2: BF2 = AD : DC PROBLEM XXXIII. 45. To inscribe a square in a given circle. Draw two diameters AC, BD at right angles to each other, and join AB, BC, CD, DA; ABCD is the required square (III. 23; III. 12). 46. Corollary. By bisecting the arcs AB, BC, CD, DA; and drawing the chords of these smaller arcs, a regular A octagon will be inscribed in the circle. By continuing this bisection regular polygons can be inscribed having the number of their sides 16, 32, 64, and so on. PROBLEM XXXIV. 47. To inscribe a regular hexagon in a given circle. Take A B equal to the radius of the given circle, and it will be a side of the hexagon required (III. 33). 48. Corollary. By drawing AC, CD, DA an equilateral triangle will be described in the circle. By bisecting the arcs AB, BC, &c., and continuing this A B D C E besection as in (46), and drawing the chords of these smaller arcs, regular polygons can be inscribed having the number of their sides 12, 24, 48, 96, and so on. PROBLEM XXXV. 49. To inscribe a regular decagon in a given circle. Divide the radius AB in extreme and mean ratio at the point D (28), and take BC = AD, the greater segment, and it will be the side of the required decagon. |