a , b - długości boków prostokąta
Obw = 2a + 2b ∧ Obw = 18 ⇒ 2a + 2b = 18 /÷2 ⇒ a + b = 9
P = a × b ∧ P = 18 ⇒ a × b = 18
a + b = 9 ⇒ a = 9 - b
a × b = 18 ∧ a = 9 - b ⇒ (9 - b) × b = 18
(9 - b) × b = 18
9b - b² = 18
-b² + 9b - 18 = 0 /×(-1)
b² - 9b + 18 = 0
Δ = (-9)² - 4×1×18 = 81 -72 = 9
√Δ = 3
[tex]b_{1} =\dfrac{9-3}{2} =\dfrac{6}{2} =3\\\\\lor\\\\b_{1} =\dfrac{9+3}{2} =\dfrac{12}{2} =6[/tex]
a = 9 - b ∧ b = 6 ⇒ a = 3 ∨ a = 9 - b ∧ b = 3 ⇒ a = 6
Odp: Długości boków prostokąta są liczbami naturalnymi i wynoszą 6 [j] oraz 3[j].
sprawdzam:
a = 6 ∧ b = 3
2a + 2b = 18 ∧ a = 6 ∧ b = 3
2×6 + 2×3 = 18
12 + 6 = 18
18 = 18
L = P
a ×b = 18 ∧ a = 6 ∧ b = 3
6×3 = 18
18 = 18
L = P
