Find the area of a quadrangle ABCD whose vertices are given by their coordinates A (2: 2), B (3; 5) C (6; 6) D (5; 3).

The area of ​​an arbitrary quadrilateral can be found as the half-product of the lengths of its diagonals, multiplied by the sine of the acute angle α between them. From the condition of the problem it is known that the quadrangle ABCD has vertices, which are given by the coordinates A (2; 2), B (3; 5), C (6; 6) and D (5; 3), we obtain:

(6 – 2; 6 – 2) = (4; 4) – coordinates of the AC vector;

√ (4² + 4²) = √32 – the length of the AC vector;

(5 – 3; 3 – 5) = (2; – 2) – coordinates of the vector BD;

√ (2² + 2²) = √8 – the length of the vector BD;

4 ∙ 2 + 2 ∙ (- 2) = 0 scalar product of vector AB by vector BD, hence vector AB значит BD and sin α = 1.

Then the area of ​​this quadrangle S (ABCD) = (AC ∙ BD) / 2 = 0.5 ∙ √ (32 ∙ 8) = 8.

Answer: the area of ​​this quadrangle S (ABCD) is 8 square units.