Whence, multiplying the means and extremes, and squaring the result, there will arise bx2 {b(a2+c2)+2acx}=a2 c2 (b2 —x2). Or, by collecting the terms together, and dividing by the coefficient of the highest power of x, ab ac bc 2c 26 2a' abc From which last equation x may be determined, and thence the sides of the triangle (d). PROBLEM XX.. Given the three sides AB, BC, CD, of a trapezium abed, inscribed in a semicircle, to find the diameter, or remaining side AD. Let AB α, BC=b, CD=c, and AD=x; then, by Euc. VI, D, AC XBD=AD XBC+AB XCD=bx+ac. But ABD, ACD, being right angles, (Euc. III, 31,) we shall ve AC=√(AD2 —DC2), or √(x2 Whence, by substituting these two values in the former expression, there will arise (d) This, and the following problem, cannot be constructed geometrically, or by means only of right lines and a circle, being what the ancients usually denominated solid problems, from the circumstance of their involving an equation of more than two dimensions; in which cases they generally employed the conic sec tions, or some of the higher orders of curves. √(x2 — c2)× √(x2 —a2)=bx+ac. Or, by squaring each side, and reducing the result, x3-(a3+b2+c3)x=2abc. From which last equation the value of x may be found, as in the last problem (e). MISCELLANEOUS PROBLEMS. PROBLEM I. To find the side of a square, inscribed in a given semicircle, whose diameter is d. Ans. d✓5 PROBLEM II. Having given the hypothenuse (13) of a right angled triangle, and the difference between the other two sides (7), to find these sides (f). Ans. 5 and 12 PROBLEM III. To find the side of an equilateral triangle, inscribed in a circle, whose diameter is d; and that of another circumscribed about the same circle. PROBLEM IV. Ans. d/3, and d/3 To find the side of a regular pentagon, inscribin a circle, whose diameter is d. Ans. 1d/(10-2/5) (e) Newton, in his Universal Arithmetic, English edition, 1728, has resolved this problem in a variety of different ways, in order to show, that some methods of proceeding, in cases of this kind, frequently lead to more elegant solutions than others; and that a ready knowledge of these can only be obtained by practice. (f) Such of these questions as are proposed in numbers, should first be resolved generally, by means of the usual symbols, and then reduced to the answers above given, by substituting the numeral values of the letters in the results thus obtained. PROBLEM V To find the sides of a rectangle, the perimeter of which shall be equal to that of a square, whose side is a, and its area half that of a square. Ans. a+a2 and a-1/2 PROBLEM VI. Having given the side (10) of an equilateral triangle, to find the radii of its inscribed and circumscribing circles. Ans. 2.8868 and 5.7736 PROBLEM VII. Having given the perimeter (12) of a rhombus, and the sum (8) of its two diagonals, to find the diagonals. Ans. 4+2 and 4-2 PROBLEM VIII. Required the area of a right angled triangle, whose hypothenuse is 3, and the base and perpendicular x2* and x*. PROBLEM IX. Ans. 1.029085 Having given the two contiguous sides (a, b) of a parallelogram, and one of its diagonals (d), to find the other diag Ans. (2262 — d2) PROBLEM X. Having given the perpendicular (300) of a plane triangle, the sum of the two sides (1150), and the difference of the segments of the base (495), to find the base and the sides. Ans. 945, 375, and 780 PROBLEM XI. The lengths of three lines drawn from the three angles of a plane triangle to the middle of the opposite sides, being 18, 24, and 30, respectively; it is required to find the sides. Ans. 20, 28.844, and 34.176 2. PROBLEM XII. In a plane triangle, there is given the base (50), the area (796), and the difference of the sides (10), to find the sides and the perpendicular. Ans. 36, 46, and 33.261 . PROBLEM XIII. Given the base (194) of a plane triangle, the line that bisects the vertical angle (66), and the diameter (200) of the circumscribing circle, to find the other two sides. Ans. 81.36587 and 157.43865 PROBLEM XIV. The lengths of two lines that bisect the acute angles of a right angled plane triangle, being 40 and 50 respectively, it is required to determine the three sides of the triangle. Ans. 35.80737, 47.40728, and 59.41143 PROBLEM XV. Given the altitude (4), the base (8), and the sum of the sides (12), of a plane triangle, to find the sides. Having given the base of a plane triangle (15), its (45), and the ratio of its other two sides as 2 to 3, it is required to determine the lengths of these sides. Ans. 7.7915 and 11.6872. PROBLEM XVII. Given the perpendicular (24), the line bisecting the base (40), and the line bisecting the vertical angle (25) to determine the triangle. From which the other two sides may be readily found. |