C The constant value of is denoted by 2 R 416. SCHOLIUM. the Greek letter π. It is an irrational number, which can only be found approximately, but with any degree of precision. Ex. 949. Find a circumference equal to the sum of two given circumferences. Ex. 950. Draw three circumferences which are to each other as 4:5:6. PROPOSITION XV. THEOREM 417. The area of a regular polygon is equal to onehalf the product of its apothem and perimeter. D E Hyp. S is the area, P the perimeter, and R the apothem of the polygon ABCDE, having n sides. To prove Proof. S = P× R. Draw radii OA, OB, etc., dividing the polygon into area ABO = 1⁄2 AB × R. S = n × area ABO = 1⁄2n × AB × R. n × AB = P, S=PX R. (Why?) (Why?) Q.E.D. P 418. DEF. A sector of a circle is a figure bounded by two radii and their intercepted arc. 419. DEF. Similar arcs, segments, and sectors are those which correspond to equal central angles. PROPOSITION XVI. THEOREM 420. The area of a circle is equal to one-half the product of its circumference and radius. Hyp. S is the area of a circle of radius R and circumference C. Proof. Circumscribe a regular polygon about the circle, and let P be its perimeter, A its area. A = PX R. Let the number of sides be increased indefinitely, then 421. COR. 1. The area of a circle is equal to times the square of its radius. For SCR and C = 2 TR. .. S=πR2. 422. COR. 2. The areas of circles are to each other as the squares of their radii. 423. COR. 3. The area of a sector is equal to one-half the product of its radius and arc. 424. COR. 4. Similar sectors are to each other as the squares of their radii. 425. REMARK. Since the areas of circles have the same ratio as the squares of their radii, a great many constructions referring to squares may be applied to circles. Ex. 951. Construct a circle equivalent to three-fifths of a given circle. Ex. 952. Construct a circle equivalent to three times a given circle. Ex. 953. Construct a semicircle equivalent to a given circle. Ex. 954. Construct a circle equivalent to the sum of two given circles. Ex. 955. Construct a circle equivalent to the difference of two given circles. Ex. 956. Construct a circle equivalent to the area bounded by two concentric circumferences. * Ex. 957. If upon the three sides of a right triangle semicircles be drawn as indicated in the diagram, the area of the right triangle is equivalent to the sum of the two crescent-shaped areas, bounded by the semicircles. (Hippocrates' Theorem.) * Ex. 958. If C is a point in the straight line AB, the three semicircles, drawn respectively upon AB, AC, and CB as diameters, bound an area equivalent to a circle whose diameter is the perpendicular CD, D being in the largest semicircle. A PROPOSITION XVII. THEOREM 426. Similar segments are to each other as the squares of their radii. S Hyp. S and S' are two similar segments, having the radii R and R' respectively. Proof. Sector AOB: sector A'O'B' = R2: R'2. (Why?) Or, if K and m denote two constant numbers, and sector AOB = KR2, then sector A'O'B' K. R2. Similarly, if ▲ AOB = m · R2, then ▲ A'O'B' = m · R'2. = (268) (372) Subtracting member from member, (K — m) R2 and segment S' = (K— m)R'2. Ex. 959. To construct a segment similar to two given similar segments and equivalent to their sum. Ex. 960. If upon three sides of a right triangle arcs of 90° be drawn, so that those upon the arms lie within, and that upon the hypotenuse lies without the triangle, then the area of the right triangle is equal to the area bounded by the three arcs. PROPOSITION XVIII. PROBLEM 427. The side and the radius of a regular polygon being given, to compute the side of a regular polygon of the same radius and double the number of sides. D B Given. AB equal to a side s of a regular polygon with radius OA = R. Required. To compute AC, the side of a regular polygon of double the number of sides, inscribed in the same circle. or Construction. Draw AO and CO meeting AB in D. AC OA+OC-2 OC x OD, = (Why?) (Why?) |