Find the area of the trapezoid whose vertices have coordinates (5; 2) (5; 11) (15; 9) (15; 13).

To find the area of ​​a given trapezoid, it is necessary to find the length of the bases AB and CD using the coordinates of these points according to the formula for finding the distance between two points:

AB = √ ((5 – 5) ^ 2 + (11 – 2) ^ 2) = √ (0 ^ 2 + 9 ^ 2) = √81 = 9;

CD = √ ((15 – 15) ^ 2 + (9 – 13) ^ 2) = √ (0 ^ 2 + 4 ^ 2) = √16 = 4;

We hold the height of the ВK. K is the middle of the CD base.

We find the coordinates of the point K by the formulas for finding the midpoint of the segment.

X = (15 + 15) / 2 = 30/2 = 15;

Y = (9 + 13) / 2 = 11;

K (15; 11);

We find the length of the ВК height by the formula for finding the distance between two points:

ВK = √ ((5 – 15) ^ 2 + (11 – 11) ^ 2) = √ ((-10) ^ 2 + 0 ^ 2) = √100 = 10;

Hence, the area of ​​this trapezoid is:

S = ((AB + CD) / 2) * BK = ((9 + 4) / 2) * 10 = 6.5 * 10 = 65.

Answer: 65.