Bonjour,
Problème 1 :
On pose DM = MC = x
Puisque ABCD est un carré, on a alors AD = DC = 2x.
On a :
(AD × DM)/2 = Aire du triangle rectangle AMD
(2x × x)/2 = 4
2x²/2 = 4
2x² = 8
x² = 4
x = √4
x = 2 cm
On a donc DM = MC = 2 cm
Or AD = DC = 2x
Donc AD = DC = 2×2 = 4cm
Aire ABCD = AD × DC = 4 × 4
Aire ABCD = 16 cm²
Problème 2
CD = 2 × CN = 2 × 4 = 8 cm
CB = 2 × CM = 2 × 5 = 10 cm
Aire ABCD = CD × CB = 8 × 10 = 80 cm²
Aire CMN = (CM × CN)/2 = (5×4)/2 = 10 cm²
Aire coloré = Aire ABCD – Aire CMN = 80 – 10 = 70 cm²
