ABCDE is a regular hexagon. Area = 60 cm squared. Find the area of triangles ABC and ACD.

Since the hexagon is regular, its large diagonals divide it into six equilateral, equal-sized triangles.

Then Sod = S6 / 6 = 60/6 = 10 cm2.

The area of ​​the triangle СОD is equal to: Sсod = СD^2 * Sin60 / 2 = 10.

СD ^ 2 = 10 * 2 / (√3 / 2) = 40 / √3 cm2.

In triangle ABC we define its area by the cosine theorem.

Savs = AB ^ 2 + BC ^ 2 – 2 * AB * BC * Cos120 = 40 / √3 + 40 * √3 + 40 * √3 = 120 / √3 = 40 * √3 cm2.

The area of ​​a quadrilateral ABCD is equal to half the area of ​​a hexagon. Savsd = 60/2 = 30 cm.

Then Ssd = Savsd – Ssdo = 30 – 10 = 20 cm2.

Answer: The area of ​​the ABC triangle is 40 * √3 cm2, the area of ​​the ASD triangle is 20 cm2.