ZADANIE Z TEMATU POLA FIGOR PODOBNYCH

z.4 a) L1 - obwód F1; L2 - obwód F2 L1 = (3/4) L --> L1/L = 3/4 = k zatem P1/P = k² = (3/4)² = 9/16--> P1 = (9/16) P P - P1 = P - (9/16) P = (7/16) P [(7/16)P] : P = 7/16 ≈ 0,4375 = 43,75 % -------------------------------------------------- b) L1 = 0,8 L zatem L1 / L = 0,8  = k P1 / P = k² = 0,8² = 0,64 --> P1 = 0,64 P P - P1 = P - 0,64 P = 0,36 P 0,36 P : P = 0,36 = 36 % ------------------------------------- c) L = (5/3) L1  ----> L1 = (3/5) L L1 ? l = 3/5 = k P1/P = k² = (3/5)² = 9/25  --> P1 = (9/25) P P - P1 = P - (9/25) P = (16/25) P (16/25) P : P = 16/25 = 64/100 = 64 % -------------------------------------------------------- d) L = 1,4 L1 = (14/10) L1 = (7/5) L1  ----> L1 = (5/7) L L1/L = 5/7  = k P1/P = k² = (5/7)² = 25/49 --> P1 = (25/49) P P - P1 = P - (25/49) P = (24/49) P (24/49) P  : P = 24/49 ≈ 0,4897 = 48,97 % ---------------------------------------------------========================= z.5 Mamy d1 = 4 cm  oraz L2 = 4 dm = 40 cm --> a2 = 10 cm a1*√2 =  4 cm --> a1 = 4/√2  cm a) a1 : a2 = [4/√2  cm] : 10 cm = √2/5 ======================================== b) P1 = (a1)² = [4/√2 cm]² = 16/2  cm² = 8 cm² P2 = (a2)² = (10 cm)² = 100 cm² P1 : P2 = 8 cm² : 100 cm² = 2/25 =========================================== z.6 a) L1 - obwód  I kola L2 - obwód II koła L1 = 12 π  oraz L2 = 84 L1 = 2 π r1 = 12 π ---> r1 = 6 zatem  d1 = 2 r1 = 2*6 = 12 L2 = 2 π r2 = 84 --> r2 = 42/π zatem d2 = 2 r2 = 84/π d1 : d2 =  12 : [84/π] = 12 *[π/84] = π/7 ====================================== b) P1 = π (r1)² = π *6² = 36 π P2 = π (r2)² =  π * [42/π]² = π *[1764/π²] = 1764/π P1 : P2 = 36 π : [1764/π] = 36 π *[π/1764] = π²/49 ==================================================