Find the area of a trapezoid whose vertices have (1; 3), (10; 3), (6: 8), (3; 8).

Let’s consider the problem. The formula for the area of ​​a trapezoid: S = 1 / 2h (a + b), where h is the height, a, b are the base of the trapezoid.
Let’s roughly imagine how the trapezoid looks on the coordinate line, where it is located. The larger base is located like this: (1; 3), (10; 3). Its length can be calculated when viewed relative to the Ox axis. Let us mentally draw perpendiculars from these points to the Ox axis, and now we calculate the length of the base: a = 10-1 = 9. (the first point does not go out of the coordinate (0; 0) or (0; y), so we subtract less). We will do the same with a smaller base having coordinates (6: 8), (3; 8): b = 6-3 = 3. Now we will find the height, for this we need to lower the perpendiculars from points to the Ok axis and calculate the difference: h = 8-3 = 5. We can find the area of ​​the trapezoid: S = 1 / 2h (a + b) = 1/2 * 5 * (3 + 9) = 30