Bonjour
1) a) P carré = 4x
b) AF = x + 2
P triangle = 3(x + 2) = 3x + 6
c) On cherche x tel que P carré = P triangle
4x = 3x + 6
4x - 3x = 6
x = 6
2) a) P = AE + EF + FB + BC + CD + DA
= 2(x + 2) + 2 + 3x = 4x + 4 + 2 + 3x = 7x + 6
b) A = A carré + A triangle
A carré = x²
A triangle = b × h / 2 = (x + 2) × h / 2
Théorème de Pythagore pour trouver h :
(AF/2)² + h² = AE²
h² = AE² - AF²/4 = (x + 2)² - (x + 2)²/4 = (3/4) × (x + 2)²
h = √ ( (3/4) × (x + 2)² ) = (x + 2) × (√3) / 2
A triangle = (x + 2)(x + 2)(√3)/(2) × (1/2) = (x + 2)² (√3)/4
A =
[tex]x {}^{2} + \frac{(x + 2) {}^{2} \sqrt{3} }{4} [/tex]
