APPLICATION OF ALGEBRA. TO THE SOLUTION OF GEOMETRICAL PROBLEMS. A problem is a question which requires a solution. A geometrical problem is one, in which certain parts of a geometrical figure are given or known, from which it is required to de termine certain other parts. When it is proposed to solve a geometrical problem by means of Algebra, the given parts are represented by the first .etters of the alphabet, and the required parts by the final letters, and the relations which subsist between the known and unknown parts furnish the equations of the problem. The solution of these equations, when so formed, gives the solution of the problem. No general rule can be given for forming the equations. The equations must be independent of each other, and their number equal to that of the unknown quantities introduced (Alg. Art. 103.). Experience, and a careful examination of all the conditions, whether explicit or implicit (Alg. Art. 94,) will serve as guides in stating the questions; to which may be added the following particular directions. 1st. Draw a figure which shall represent all the given parts, and all the required parts. Then draw such other lines as will establish the most simple relations between them. If an angle is given, it is generally best to let fall a perpendicular that shall lie opposite to it; and this perpendicular, if possible, should be drawn from the extremity of a given side. 2d. When two lines or quantities are connected in the same way with other parts of the figure or problem, it is in general, not best to use either of them separately; but to use their sum, their difference, their product, their quotient, or perhaps another line of the figure with which they are alike connected. 3d. When the area, or perimeter of a figure, is given, it is sometimes best to assume another figure similar to the proposed, having one of its sides equal to unity, or some other known quantity. A comparison of the two figures will often give a required part. We will add the following problems.* *The following problems are selected from Hutton's Application of Algebra to Geometry, and the examples in Mensuration from his treatise on that subject PROBLEM I. In a right angled triangle BAC, having given the base BA, and the sum of the hypothenuse and perpendicular, it is required to find the hypothenuse and perpendicular. Put BA=c=3, BC=x, AC=y and the sum of the hypothenuse and perpendicular equal to S=9 and x2=y2+c2 (Bk. IV. Prop. XI.) From 1st equ: x=s—y and x2=s2—Qsy+y2 By subtracting, 0=s2-2sy-c2 2sy=s2—c2 B or hence, Therefore x+4=9 or x=5=BC. PROBLEM II. In a right angled triangle, having given the hypothenuse, and the sum of the base and perpendicular, to find these two sides. Put BC=a=5, BA=x, AC=y of the base and perpendicular=s=7 and the sum By completing the square y2—sy+1s2=1a2—¡s2 or Hence y=‡s±√+a2—+s2=4 or 3 x={s=√ {a2=+s2=3 or 4 PROBLEM III. In a rectangle, having given the diagonal and perimeter, to find the sides. Ꭰ or Let ABCD be the proposed rectangle. From which equations we obtain, y={a±√1⁄2d2—a2=8 or 6, and x={a= √√ď2—↓a2=6 or 8. PROBLEM IV. Having given the base and perpendicular of a triangle, to find the side of an inscribed square. Let ABC be the triangle and HEFG the inscribed square. Put AB=b, CD=a, and HE or GH-x: then CI=a-x. We have by similar triangles or Hence, or AB: CD:: GF: CI b: a:: x: a- X ab-bx-ax F x=- the side of the inscribed square; ab a+b = which, therefore, depends only on the base and altitude of the triangle. PROBLEM V. In an equilateral triangle, having given the lengths of the three perpendiculars drawn from a point within, on the three sides to determine the sides of the triangle. 2x; hence AH=x, and CH=√AC2—AH2= √/4x2—x2 Now since the area of a triangle is equal to half its base into the altitude, (Bk. IV. Prop. VI.) AB × CH=≈ × ≈ √3=x2 √3=triangle ACB But the three last triangles make up, and are consequently equal to, the first; hence, x2 √3=ax+bx+cx=x(a+b+c); REMARK. Since the perpendicular CH is equal to x√3, it is consequently equal to a+b+c: that is, the perpendicular let fall from either angle of an equilateral triangle on the opposite side, is equal to the sum of the three perpendiculars let fall from any point within the triangle on the sides respectively. PROBLEM VI. In a right angled triangle, having given the base and the difference between the hypothenuse and perpendicular, to find the sides. PROBLEM VII. In a right angled triangle, having given the hypothenuse and the difference between the base and perpendicular, to determine the triangle. PROBLEM VIII. Having given the area of a rectangle inscribed in a given triangle; to determine the sides of the rectangle. PROBLEM IX. In a triangle, having given the ratio of the two sides, together with both the segments of the base made by a perpendic ular from the vertical angle; to determine the triangle. PROBLEM X. In a triangle, having given the base, the sum of the other two sides, and the length of a line drawn from the vertical angle to the middle of the base; to find the sides of the triangle. PROBLEM XI. In a triangle, having given the two sides about the vertical angle, together with the line bisecting that angle and terminating in the base; to find the base. PROBLEM XII. To determine a right angled triangle, having given the lengths of two lines drawn from the acute angles to the middle of the opposite sides. PROBLEM XIII. To determine a right-angled triangle, having given the perimeter and the radius of the inscribed circle. PROBLEM XIV. To determine a triangle, having given the base, the perpendicular and the ratio of the two sides. PROBLEM XV. To determine a right angled triangle, having given the hypothenuse, and the side of the inscribed square. PROBLEM XVI. To determine the radii of three equal circles, described within and tangent to, a given circle, and also tangent to each other. |