428. COR. If R = 1, and the side of a polygon of n sides equals s, then Ex. 961. The side of an inscribed square is equal to R√2. Compute the side of the regular octagon. PROPOSITION XIX. PROBLEM 429. To compute the ratio of the circumference of a circle to the diameter. 824 = √2 − √4 −(0.51764)2 = 0.26105 6.26526 $48 = 2-√4(0.26105)2 = 0.13081 6.27870 86=√2-√4-(0.13081)2 : √2 − √4 — (0.13081)2 = 0.06534 6.28206 $192 S384 $768 = :√2 −√4 − (0.06534)2 = 0.03272 6.28291 Taking the last perimeter as the approximate value of the circumference, 430. REMARK. Most computations referring to the circle involve the use of the two formulæ C = 2πR and S = TR2. If R is not given, compute R first. (421.) Ex. 962. Find the circumference of a circle whose radius is 5. Find the circumference of a circle whose area equals S. Ex. 969. The circumference of a circle equals 10. Find the circumference of a circle having twice the area of the given circle. Ex. 970. Two concentric circles have their circumferences equal to 30 and 40 respectively. Find the area bounded by the two circumferences. Ex. 971. Find a semicircle equivalent to an equilateral triangle whose side equals 5. Ex. 972. Find the area of a sector whose radius equals 5, and whose central angle equals 40°. Ex. 973. A square is inscribed in a circle of radius 10. of a segment bounded by a side of the square. Find the area Ex. 974. From a point without a circle, two tangents are drawn to a circle whose radius equals 10. Find the area bounded by the two tangents and the circumference, if the tangents include an angle of 120°. Ex. 975. Find the area bounded by three arcs of 60° and radius 5, if the concave sides of the arcs are turned toward the area. Ex. 976. Find the area bounded by three arcs of 60° and radius 5, if the convex sides of the arcs are turned toward the area. Ex. 977. A circle has an area of 60 sq. in. of 40°. Find the length of an arc Ex. 978. Find the central angle of a sector whose perimeter is equal to the circumference. Ex. 979. Find the central angle of a sector whose area is equal to the square of the radius. MISCELLANEOUS EXERCISES Ex. 980. If in a circle two chords intersect at right angles, and circles are constructed on the segments of the chords as diameters, the area of the given circle is equivalent to the sum of the areas of the four circles. Ex. 981. The side of the regular circumscribed triangle is equal to twice the side of the regular inscribed triangle. Ex. 982. The square of a side of an inscribed equilateral triangle is equal to three times the square of the inscribed regular hexagon. Ex. 983. If the radius of a regular octagon be equal to r, prove that its side equals rV2 √2. Ex. 984. If the radius of a regular decagon be equal to r, prove that Ex. 985. If the radius of a regular pentagon be equal to r, prove that its side equals V10 – 2√5. Ex. 986. If the radius of a regular dodecagon be equal to r, prove that its side equals rV2 v3. Ex. 987. If in a circle a regular deeagon and a regular pentagon be inscribed, the side of the decagon increased by the radius is equal to twice the apothem of the pentagon. *Ex. 988. The square of the side of a regular decagon increased by the square of the radius is equal to the square of the side of a regular pentagon, having the same radius. APPENDIX TO PLANE GEOMETRY ALGEBRAIC SOLUTIONS OF GEOMETRICAL PROPOSITIONS. MAXIMA AND MINIMA CONSTRUCTION OF ALGEBRAIC EXPRESSIONS 431. NOTE. In the following propositions, a, b, c, d, etc., denote given lines, while x, y, z, etc., denote required lines. HINT.x is the fourth proportional to c, a, and b. 433. Complex algebraical expressions are constructed means of the eleven constructions of (432). It is quite oft necessary to transform the algebraical expressions, in order make them special cases of (432). Different algebraical tra formations lead to different solutions. i.e. x is the mean proportional between 3 a and b, i.e. find √ab by means of (432, 7), and √ab √3 by means of (Ex. 991) i.e. find the fourth proportional to 4 c, a + b, and a — b. Ex. 994. Ex. 995. x = √a2 - ab. x = √a(a - b). x = √ a2 – bc = √ a2 – (√bc)2, i.e. find the mean proportional between b and c, and construct a right triangle, having one arm equal to the mean proportional, and the hypote |