455. THEOREM. Of all regular polygons having a given area, that one which has the greatest number of sides has the least perimeter. P R Ꭱ Given regular polygons P and Q such that P = Q while Q has the greater number of sides. To prove that perimeter of Q< perimeter of P. : Proof Construct a regular polygon R having the same number of sides as P and the same perimeter as Q. (§ 323) Therefore, perimeter of R< perimeter of P. But R was constructed with a perimeter equal to that of Q. Hence perimeter of Q< perimeter of P. 1. Of all inscribed polygons of a given number of sides in a given circle, that one which is regular has the maximum area and the maximum perimeter. 2. Of all circumscribed polygons of a given number of sides about a given circle, that one which is regular has the least perimeter and the least area. 3. Of all triangles having two given sides, that in which these sides include a right angle has the maximum area. SUGGESTION. Show that the altitude is greatest when the angle is a right angle. C C1 C2 A B Major and minor arcs Maxima and minima 149, 158, 175, 179, 226, 228, 233, etc. | Preliminary theorems, Mean proportional Means Measurement, approximate, 235, 306 of angles of line-segments. of the circle exact Median of a triangle. Methods of attack 232-237 Proportional division 349 Proposition, geometric 232, 301, 304 Protractor. 18, 33 140, 162, 167 293 Proof 242 by exclusion 17, 202 direct, indirect 59, 391, 392 91, 175, 176, 394 242 242 |