Find the area of the trapezoid whose vertices have coordinates: (2, -2) (8, -4) (8,8) (2,10).

1. Let’s denote the vertices of the trapezoid:

A (2; -2);
B (8; -4);
C (8; 8);
D (2; 10).
2. As you can easily see, the parallel sides, that is, the bases, of the trapezoids are AD and BC, since the abscissas of their vertices are equal, respectively:

x (A) = x (D) = 2;
x (B) = x (C) = 8.
3. Lengths of bases:

AD = y (D) – y (A) = 10 + 2 = 12;
BC = y (C) – y (B) = 8 + 4 = 12.
The bases of the trapezoid are equal, which means it is a parallelogram.

4. Height ABCD:

H = x (B) – x (A) = 8 – 2 = 6.

5. Area ABCD:

S (ABCD) = AD * H = 12 * 6 = 72.

Answer: 72.