Questions: Trouve l'aire de la figure ABCD.

Solution Steps

The equation of line $d$ is $x - 2y + 2 = 0$. Line $d_1$ is vertical and passes through point A. The x-coordinate of A is the x-intercept of $d_1$. Since $d$ and $d_1$ intersect at A, the x-coordinate of A can be found by setting $y=0$ in the equation of $d$. Thus $x + 2 = 0$, so $x = -2$. The coordinates of A are $(-2, 0)$. Line $d_2$ is vertical and passes through point B. Since B is the intersection of $d$ and $d_2$, and E(5,6) is on $d_2$, the x-coordinate of B is also 5. Substituting $x=5$ into the equation of $d$ gives $5 - 2y + 2 = 0$, which simplifies to $2y = 7$, so $y = 3.5$. The coordinates of B are (5, 3.5).

Point D has coordinates (4.8, 0) and line segment CD is perpendicular to the x-axis. Since B lies on the same vertical line as C, the x-coordinate of C is also 5. Since D and C have the same y-coordinate, the y-coordinate of C is 0. The coordinates of C are (5, 0).

The figure ABCD is a trapezoid. The area of a trapezoid is given by the formula: $Area = \frac{1}{2} (base_1 + base_2) \times height$. In this case, $base_1 = AD = 4.8 - (-2) = 6.8$, $base_2 = BC = 3.5 - 0 = 3.5$, and $height = CD = |5-4.8| = 0.2$. Therefore, the area of trapezoid ABCD is $\frac{1}{2}(6.8 + 3.5)(0.2) = \frac{1}{2}(10.3)(0.2) = 1.03$.

Final Answer: