(8) If the measure of m yards be c, what is the measure of n feet? 8. Now suppose the measures of the sides of a right-angled triangle to be p, q, r respectively, the right angle being subtended by that side whose measure is r. Then since the geometrical property of such a triangle, established by Euclid 1. 47, may be extended to the case in which the sides are represented by numbers or symbols standing for numbers, p2 + q2 = r2. If any two of the numerical quantities involved in this equation are given, we can determine the third. For example, if r = 5 and q = 3, p2 + 9 = 25, .. p2 = 16, .. p=4. EXAMPLES.-II. (1) The hypothenuse being 51 yards, and one of the sides containing the right angle 24 yards, find the other side. (2) The sides containing the right angle being 8 feet and 6 feet, find the hypothenuse. (3) A rectangular field measures 225 yards in length, and 120 yards in breadth; what will be the length of a diagonal path across it? (4) A rectangular field is 300 yards long and 200 yards broad; find the distance from corner to corner. (5) A rectangular plantation, whose width is 88 yards, contains 2 acres; find the distance from corner to corner across the plantation. (6) The sides of a right-angled triangle are in Arithmetical Progression and the hypothenuse is 20 feet; find the other sides. (7) The sides of a right-angled triangle are in Arithmetical Progression; shew that they are proportional to 3, 4, 5. (8) A ladder, whose foot rests in a given position, just reaches a window on one side of a street, and when turned about its foot, just reaches a window on the other side. If the two positions of the ladder be at right angles to each other, and the heights of the windows be 36 and 27 feet respectively, find the width of the street and the length of the ladder. (9) In a right-angled isosceles triangle the hypothenuse is 12 feet, find the length of each of the other sides. (10) What is the length of the diagonal of a square, whose side is 5 inches? (11) The area of a square is 390625 square feet; what is the diagonal? (12) Each side of an equilateral triangle is 13; find the length of the perpendicular dropped from one of the angles on the opposite side. (13) If ABC be an equilateral triangle and the length of AD, a perpendicular on BC, be 15; find the length of AB. (14) The radius of a circle is 37 inches; a chord is drawn in the circle: if the length of this chord be 70 inches, find its distance from the centre. (15) The distance of a chord in a circle from the centre is 180 inches; the diameter of the circle is 362 inches: find the length of the chord. (16) The length of a chord in a circle is 150 feet, and its distance from the centre is 308 feet; find the diameter of the circle. 9. We now proceed to treat of the ratio of the circumference of a circle to the diameter of that circle. 10. It is evident that a straight line can be compared as to its length with a circular arc, and that consequently the ratio between such lines can be represented in the form of a fraction. 11. We must assume as an axiom that an arc is greater than the chord subtending it: that is, if ABD be part of the circumference of a circle cut off by the straight line AD, the length of ABD is greater than the length of AD. B D 12. A figure enclosed by any number of straight lines is called a polygon. 13. A regular polygon is one in which all the sides and angles are equal. 14. The perimeter of a polygon is the sum of the sides. Hence if AB be one of the sides of a regular polygon of n sides the perimeter of the polygon will be n. AB. 15. The circumference of a circle is greater than the perimeter of any polygon which can be inscribed in the circle, but as the number of sides of such a polygon is increased the perimeter of the polygon approaches nearer to the circumference of the circle, as will appear from the following illustration. Let AB be the side of a regular hexagon ABDEFG inscribed in a circle. Then AB is equal to the radius of the circle. EUCL. IV. 15. Now the arc ACB is greater than AB, and the circumference of the circle is therefore larger than the perimeter of the hexagon. Hence the circumference is greater than six times the radius, Now and greater than three times the diameter. suppose C to be the middle point of the arc AB. Join AC, CB. These will be sides of a duodecagon or regular figure of 12 sides inscribed in the circle. Now AC, CB are together greater than AB: but AC, CB are together less than the arc ACB. Hence the perimeter of the duodecagon will be less than the circumference of the circle, but will approximate more nearly to the circumference of the circle than the perimeter of the hexagon. So the larger the number of sides of a polygon inscribed in a circle, the more nearly does the perimeter of the polygon approach to the circumference of the circle; and when the number of sides is infinitely large, the perimeter of the polygon will become ultimately equal to the circumference of the circle. 16. To shew that the ratio of the circumference of a circle to the radius is the same for all circles. Let O and o be the centres of two circles. Let AB, ab be sides of regular polygons of n sides inscribed in the circles, P, p the perimeters of the polygons, and C, c the circumferences of the circles. Then OAB, oab are similar triangles. .. OA: oa :: AB : ab :: n.AB: n. ab :: P: p. Now when n is very large the perimeters of the polygons may be regarded as equal to the circumferences of the circles; .. OA oa :: C: c. Hence it follows that the circumference of a circle varies as the radius of the circle. |