...equals altitude into |(AB + DC), = altitude into ab. QED FIG. 228. PROPOSITION raí. 326. Theorem. — The area of a regular polygon is equal to onehalf the product of its apothem into its perimeter. DEM. — Let ABCDEFC be a regular polygon whose apothem is Oa; then is its area...

...altitude into i (AB + DC), — altitude into ab. q. ED Fra. PROPOSITION VIII. 326. Ttieorem. — T/ie area of a regular polygon is equal to onehalf the product of its apothem into its perimeter, DEM. — Let ABCDEFC be a regular polygon whose apolhem is Oa; then is its area...

...altitude into i(AB + DC), = altitude into ab. QED n- B FIG. 228. PROPOSITION VIII. 326. Theorem.—The area of a regular polygon is equal to onehalf the product of its apothem into its perimeter. DEM.—Let ABCDEFC be a regular polygon whose apothem is Oa; then is its area equal...

...polygons of 12, 24, etc.; from a pentedecagon, figures of 30, 60, etc. Proposition 9. Theorem.—The area of a regular polygon is equal to onehalf the product of its perimeter and apothem. Let ABC be a polygon and DE its apothem; then the area of AB C is equal to $(AB+BF...

...polygon, and denote its perimeter by P, and its apothem by r. Then the area of this polygon = lr XP, §379 (the area of a regular polygon is equal to one-half the product of its apothem by the perimeter). Conceive the number of sides of this polygon to be indefinitely increased, the polygon...

...assigned quantity. .-. Um. (R — r) = 0. .-.R — lim.(r) = Q. §199 .-.lim. (r) = R. a. ED REGULAR POLYGONS AND CIRCLES. PROPOSITION IX. THEOREM. 379....regular polygon ABC, etc. We are to prove the area of ABC, etc., = \ RX P. Draw 0 A, OB, 0С, etc. The polygon is divided into as many A as it has sides....

...than any assigned quantity. ."". Urn. (R — r) = Q. .-.R—lim.(r) = 0. §199 .-. lim. (r) = R. QED PROPOSITION IX. THEOREM. 379. The area of a regular polygon is equal to one-half the product of Us apothegm by its perimeter. B ED Let P represent the perimeter and R the apothegm of the regular...

...DC) ; and area ABCD. which equals £ (AB + DC) x IK, = ab x IK. QED PROPOSITION IX. 351. Theorem. — The area of a regular polygon is equal to one-half the product of its apothem into its perimeter. DEMONSTRATION. Let ABCDEFG be a regular polygon, whose perimeter is AB + BC + CD...

...opposite. Whence Km = oD. In like manner, we may show that Cp = nB. PROPOSITION IX. 351. Theorem.—The area, of a regular polygon is equal to one-half the product of its apothem into its perimeter. DEMONSTRATION. Let ABCDEFG be a regular polygon, whose perimeter is AB + BC + CD...

...trapezoids thus formed, and their sum will be the area required of the polygon. FIG. 47. 425. THEOREM. The area of a regular polygon is equal to one-half the product of its perimeter by its apothem. Thus, in the regular octagon ABC, etc., if we draw from the centre O lines...