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h₁ = ?
a₂ = 16 cm
h₂ = ?
h₁ + h₂ = 21 cm
(1/2)*h₁*12 = (1/2)*h₂*16 (z pola trójkąta)
Rozwiązujemy układ równań
h₁ + h₂ = 21
(1/2)*h₁*12 = (1/2)*h₂*16 /*2
h₁ + h₂ = 21
12h₁ = 16h₂ /:4
h₁ + h₂ = 21
3h₁ = 4h₂ /:3
h₁ + h₂ = 21
h₁ = (4/3)h₂
Wstawiam h₁ = (4/3)h₂ do pierwszego i mam
(4/3)h₂ + h₂ = 21
(4/3)h₂ + (3/3)h₂ = 21
(7/3)h₂ = 21 /:(7/3)
h₂ = 21 * (3/7)
h₂ = 9 cm
zatem h₁ = (4/3) * 9 = (4/3) * (9/1) = 36/3 = 12 cm
zatem pole jest równe
P = (1/2) * a₁ * h₁ = (1/2) * 12 * 12 = 6 * 12 = 72 cm²
z drugiej strony
P = (1/2) * a₂ * h₂ = (1/2) * 16 * 9 = 8 * 9 = 72 cm²