We assume point C is the unlabeled point at lower right. By the Pythagorean theorem, (lengths in meters)
CD² = 1.5² - BD² = 2² - (2.5 - BD)²
2.25 - BD² = 4 - (6.25 -5·BD +BD²) . . . . expand the square
2.25 - BD² = -2.25 + 5·BD - BD² . . . . . . eliminate parentheses
4.50 = 5·BD . . . . . . . . . . . . . . . . . . . . . . . add 2.25 + BD²
0.90 = BD
We now recognize ΔBDC is a 3-4-5 right triangle with a scale factor of 0.3. Hence CD = 0.3·4 = 1.2
Area ABC = (1/2)·(2.5 m)·(1.2 m) = 1.50 m²
Area BDC = (1/2)·(0.9 m)·(1.2 m) = 0.54 m²
Area DCA = (1/2)·(1.6 m)·(1.2 m) = 0.96 m²
