BOOK XIV. APPLICATIONS OF ALGEBRA TO GEOMETRY. 692. WHEN it is proposed to solve a geometrical problem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required. Represent the known parts by the first letters of the alphabet, and the required parts by the last letters. Then, observing the geometrical relations that the parts of the figure have to each other, make as many independent equations as there are unknown quantities introduced, and the solution of these equations will determine the unknown quantities or required parts. To form these equations, however, no definite rules can be given; but the best aids may be derived from experience, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem. PROBLEM I. 693. In a right-angled triangle, having given the hypothenuse, and the sum of the other two sides, to determine these sides. Let ABC be the triangle, right-angled at B. Put A C =a, the sum A B BCs, AB = x, and BC= y. + 95. What will the diameter of a sphere be, when its solidity and the area of its surface are expressed by the same numbers ? Ans. 6. 96. There is a circular fortification, which occupies a quarter of an acre of ground, surrounded by a ditch coinciding with the circumference, 24 feet wide at bottom, 26 at top, and 12 deep; how much water will fill the ditch, if it slope equally on both sides? Ans. 135483.25 cu. ft. 97. A father, dying, left a square field containing 30 acres to be divided among his five sons, in such a manner that the oldest son may have 8 acres, the second 7, the third 6, the fourth 5, and the fifth 4 acres. Now, the division fences are to be so made that the oldest son's share shall be a narrow piece of equal breadth all around the field, leaving the remaining four shares in the form of a square; and in like manner for each of the other shares, leaving always the remainders in form of squares, one within another, till the share of the youngest be the innermost square of all, equal to 4 acres. Required a side of each of the enclosures. Ans. 17.3205, 14.8324, 12.2474, 9.4868, and 6.3246 chains. 98. Required the dimensions of a cone, its solidity being 282 inches, and its slant height being to its base diameter as 5 to 4. Ans. 9.796 in. the base diameter; 12.246 in. the slant height; and 11.223 in. the altitude. 99. A gentleman has a piece of ground in form of a square, the difference between whose side and diagonal is 10 rods. He would convert two thirds of the area into a garden of an octagonal form, but would have a fish-pond at the centre of the garden, in the form of an equilateral triangle, whose area must equal five square rods. Required the length of each side of the garden, and of each side of the pond. Ans. 8.9707 rods, each side of the garden, and 3.398 rods, each side of the pond. BOOK XIV. APPLICATIONS OF ALGEBRA TO GEOMETRY. 692. WHEN it is proposed to solve a geometrical problem by aid of Algebra, draw a figure which shall represent the several parts or conditions of the problem, both known and required. Represent the known parts by the first letters of the alphabet, and the required parts by the last letters. Then, observing the geometrical relations that the parts of the figure have to each other, make as many independent equations as there are unknown quantities introduced, and the solution of these equations will determine the unknown quantities or required parts. To form these equations, however, no definite rules can be given; but the best aids may be derived from experience, and a thorough knowledge of geometrical principles. It should be the aim of the learner to effect the simplest solution possible of each problem. PROBLEM I. 693. In a right-angled triangle, having given the hypothenuse, and the sum of the other two sides, to determine these sides. Substitute in second equation this value of x, s22sy + 2 y2 = a2. If A C 5, and the sum A B+ BC=7, y = 4 or 3, 694. Having given the base and perpendicular of a triangle, to find the side of an inscribed square. Let ABC be the triangle, C that is, the side of the inscribed square is equal to the product of the base by the altitude, divided by their sum. PROBLEM III. 695. Having given the lengths of two straight lines drawn from the acute angles of a right-angled triangle to the middle of the opposite sides, to determine those sides. Let ABC be the given triangle, CE2+ CB2 BE2, = = A E we have and = x2+4y2 = a2, y2+4x2= b2. B D C by subtracting the first equation from four times the second, which values of x and y are half the base and perpendiculars of the triangle. PROBLEM IV. 696. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within to the three sides, to determine these sides. |