PROPOSITION VIII. THEOREM. 378. If the number of sides of a regular inscribed polygon be increased indefinitely, the apothegm will be an increasing variable whose limit is the radius of the circle. B In the right triangle OC A, let O A be denoted by R, OC by r, and AC by b. (a is the shortest distance from a point to a straight line). (one side of a ▲ is greater than the difference of the other two sides). § 52 $97 By increasing the number of sides of the polygon indefinitely, A B, that is, 2 b, can be made less than any assigned quantity. .. b, the half of 2 b, can be made less than any assigned quantity. .. R—r, which is less than b, can be made less than any assigned quantity. .. lim. (R − r) = 0. .. R— lim. (r) = 0. .. lim. (r) = R. 199 Q. E. D. PROPOSITION IX. THEOREM. 379. The area of a regular polygon is equal to one-half the product of its apothegm by its perimeter. B C E Let P represent the perimeter and R the apothegm of the regular polygon ABC, etc. We are to prove = the area of ABC, etc., RX P. Draw O A, OB, O C, etc. The polygon is divided into as many A as it has sides. The apothegm is the common altitude of these ▲, and the area of each is equal to R multiplied by the base. § 324 .. the area of all the A is equal to R multiplied by the sum of all the bases. But the sum of the areas of all the A is equal to the area of the polygon, and the sum of all the bases of the A is equal to the perimeter of the polygon. .. the area of the polygon = R × P. Q. E. D. PROPOSITION X. THEOREM. 380. The area of a circle is equal to one-half the product of its radius by its circumference. Let R represent the radius, and C the circumference of a circle. We are to prove the area of the circle = RX C. Inscribe any regular polygon, and denote its perimeter by P, and its apothegm by r. Then the area of this polygon=rXP, $379 (the area of a regular polygon is equal to one-half the product of its apothegm by the perimeter). Conceive the number of sides of this polygon to be indefinitely increased, the polygon still continuing to be regular and inscribed. Then the perimeter of the polygon approaches the circumference of the circle as its limit, the apothegm, the radius as its limit, § 378 and the area of the polygon approaches the O as its limit. But the area of the polygon continues to be equal to onehalf the product of the apothegm by the perimeter, however great the number of sides of the polygon. = .. the area of the RX C. § 199 Q. E. D. In the equality, the area of the O=RX C, That is, the area of a ○ = π times the square on its radius. 382. COR. 2. The area of a sector equals the product of its radius by its arc; for the sector is such part of the circle as its arc is of the circumference. 383. DEF. In different circles similar arcs, similar sectors, and similar segments, are such as correspond to equal angles at the centre. PROPOSITION XI. THEOREM. 384. Two circles are to each other as the squares on their radii. Let R and R' be the radii of the two circles Q and Q'. R2 and Then (the area of a = π times the square on its radius), Q' = π R2, = π R2. R2 § 381 381 Q. E. D. 385. COROLLARY. Similar arcs, being like parts of their respective circumferences, are to each other as their radii; similar sectors, being like parts of their respective circles, are to each other as the squares on their radii. PROPOSITION XII. THEOREM. 386. Similar segments are to each other as the squares on their radii. C C Let A C and A'C' be the radii of the two similar segments A BP and A' B' P'. The sectors A C B and A' C' B' are similar, § 383 (having an of the one equal to an of the other, and the including sides § 284 Now proportional). (similar sectors are to each other as the squares on their radii); $ 385 $ 342 Hence (similar are to each other as the squares on their homologous sides). or, $271 (if two quantities be increased or diminished by like parts of each, the results will be in the same ratio as the quantities themselves). Q. E. D. |