Or, py X= q (2) Now, from equations (1) and (2) we can determine x andy, the sides of the A; and thus the determination has been attained, carefully and easily, step by step. PROBLEM V. Three equal circles touch each other externally, and thus inclose one acre of ground; what is the diameter in rods of each of these circles? Draw three equal circles to touch each other externally, and join the three centers, thus forming a triangle. The lines joining the centers will pass through the points of contact, (Th. 7, B. III). Let R represent the radius of these equal circles; then it is obvious that each side of this A is equal to 2R. The triangle is therefore equilateral, and it incloses the given area, and three equal sectors. As the angle of each sector is one third of two right angles, the three sectors are, together, equal to a semicircle; but the area of a semi-circle, whose radius is R, is R2 expressed by ; and the area of the whole triangle 2 2 must be +160; but the area of the ▲ is also equal to R multiplied by the perpendicular altitude, which is R√3. Hence, R 31.48 + rods, for the required result. PROBLEM VI. In a right-angled triangle, having given the base and the sum of the perpendicular and hypotenuse, to find these two sides. PROB. VII.-Given, the base and altitude of a triangle, to divide it into three equal parts, by lines parallel to the base. PROB. VIII. In any equilateral A, given the length of the three perpendiculars drawn from any point within, to the three sides, to determine the sides. PROB. IX.—In a right-angled triangle, having given the base, (3), and the difference between the hypotenuse and perpendicular, (1), to find both these two sides. PROB. X.-In a right-angled triangle, having given the hypotenuse, (5), and the difference between the base and perpendicular, (1), to determine both these two sides. PROB. XI.-Having given the area of a rectangle inscribed in a given triangle, to determine the sides of the rectangle. PROB. XII.-In a triangle, having given the ratio of the two sides, together with both the segments of the base, made by a perpendicular from the vertical angle, to determine the sides of the triangle. PROB. XIII.-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. -- PROB. XIV. 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. PROB. XV. To determine a right-angled triangle, having given the perimeter, and the radius of the inscribed circle. PROB. XVI.—To determine a triangle, having given the base, the perpendicular, and the ratio of the two sides. PROB. XVII.-To determine a right-angled triangle, having given the hypotenuse, and the side of the inscribed square. PROB. XVIII. To determine the radii of three equal circles inscribed in a given circle, and tangent to each other, and also to the circumference of the given circle. PROB. XIX.-In a right-angled triangle, having given the perimeter, or sum of all the sides, and the perpendicular let fall from the right angle on the hypotenuse, to determine the triangle; that is, its sides. PROB. XX.-To determine a right-angled triangle, having given the hypotenuse, and the difference of two lines drawn from the two acute angles to the center of the inscribed circle. PROB. XXI. To determine a triangle, having given the base, the perpendicular, and the difference of the two other sides. - PROB. XXII. To determine a triangle, having given the base, the perpendicular, and the rectangle, or product of the two sides. PROB. XXIII.—To determine a triangle, having given the lengths of three lines drawn from the three angles to the middle of the opposite sides. PROB. XXIV. In a triangle, having given all the three sides, to find the radius of the inscribed circle. PROB. XXV.-To determine a right-angled triangle, having given the side of the inscribed square, and the radius of the inscribed circle. PROB. XXVI. - To determine a triangle, and the radius of the inscribed circle, having given the lengths of three lines drawn from the three angles to the center of that circle. PROB. XXVII. To determine a right-angled triangle, having given the hypotenuse, and the radius of the inscribed circle. PROB. XXVIII.-The lengths of two parallel chords on the same side of the center being given, and their distance apart, to determine the radius of the circle. PROB. XXIX. The lengths of two chords in the same circle being given, and also the difference of their distances from the center, to find the radius of the circle. PROB. XXX.-The radius of a circle being given, and also the rectangle of the segments of a chord, to determine the distance of the point at which the chord is divided, from the center. PROB. XXXI.-If each of the two equal sides of an isosceles triangle be represented by a, and the base by 2b, what will be the value of the radius of the inscribed circle? PROB. XXXII. —From a point without a circle whose diameter is d, a line equal to d is drawn, terminating in the concave arc, and this line is bisected at the first point in which it meets the circumference. What is the distance of the point without from the center of the circle? It is not deemed necessary to multiply problems in the application of algebra to geometry. The preceding will be a sufficient exercise to give the student a clear conception of the nature of such problems, and will serve as a guide for the solution of others that may be proposed to him, or that may be invented by his own ingenuity. MISCELLANEOUS PROPOSITIONS. We shall conclude this book, and the subject of Geometry, by offering the following propositions, some theorems, others problems, and some a combination of both, -not only for the purpose of impressing, by application, the geometrical principles which have now been established, but for the not less important purpose of cultivating the power of independent investigation. After one or two propositions in which the beginner will be assisted in the analysis and construction, we shall leave him to his own resources, with the caution that a patient consideration of all the conditions in each case, and not mere trial operation, is the only process by which he can hope to reach the desired result. 1. From two given points, to draw two equal straight lines, which shall meet in the same point in a given straight line. Let A and B be the given points, and CD the given straight line. Produce the perpendicular to the straight line AB at its middle point, until it meets CD in G. It is then easily proved that G is the point in CD in which the equal lines from A and B must meet. That is, that AG = BG. If the points A and B were on opposite sides of CD, the directions C for the construction would be the same, and we should have this fig ure; but the reasoning by which we prove AG = BG would be unchanged. A E A D E 2. From two given points on the same side of a given straight line, to draw two straight lines which shall meet in the given line, and make equal angles with it. Let CD be the given line, and A and B the given points. From B draw BE perpendicular to CD, and produce the perpendicular to F, making EF equal to BE; then draw AF, and from the point G, in which it intersects CD, draw GB. Now, BGE= LEGF=\AGC. Hence, the angles BGD and AGC are equal, and the lines AG and BG meet C B ED in a common point in the line CD, and made equal angles with that line. |