Find the area of the triangle ABC if: A (3; 0), B (0; 0), C (0; 4)

First, we determine the size of the sides by the coordinates of the points:

A (3; 0), B (0; 0), C (0; 4). When calculating, we write down only the values of the coordinates of the points.

AB = √ [(3 – 0) ² + (0 – 0) ²] = √ (3 – 0) ² = √ (3²) = √9 = 3,

BC = √ [(0 – 0) ² + (4 – 0) ²] = √ (4 – 0) ² = √ (4) ² = √16 = 4.

AC = √ [(3 – 0) ² + (4 – 0) ²] = √ [(3) ² + (4) ²] = √ (9 + 16) = √25 = 5.

Perimeter P = 3 + 4 + 5 = 12. Half of the perimeter n = P / 2 = 12/2 = 6.

Area C = √p * (p – AB) * (p – BC) * (p – AC) = √ [6 * (6 – 3) * (6 – 4) * (6 – 5)] = √ [6 * 3 * 2 * 1] = √36 = 6.

Answer: the area is 6.