Jaka jest skala podobieństwa koła o obwodzie 18 cm do koła o średnicy 6 cm ?
L1 = 18 p(2) pi cm
d2 = 6 p(12) cm = 6 * 2 p(3) cm = 12 p(3) cm
zatem
L2 = d2 *pi = 12 p(3) pi cm
k = L2 : L 1 = [ 12 p(3) pi cm ] : [ 18 p(2) pi cm ] = ( 2/3) * p( 3/2) = (2/3) *[ p(6)/2 ] =
= p(6)/ 3
=========
Odp. k = p(6) / 3
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L1 = 18 p(2) pi cm
d2 = 6 p(12) cm = 6 * 2 p(3) cm = 12 p(3) cm
zatem
L2 = d2 *pi = 12 p(3) pi cm
k = L2 : L 1 = [ 12 p(3) pi cm ] : [ 18 p(2) pi cm ] = ( 2/3) * p( 3/2) = (2/3) *[ p(6)/2 ] =
= p(6)/ 3
=========
Odp. k = p(6) / 3
---------------------------