Deltoid Geometry: Calculate Area Ratio with 30cm Main Diagonal and 4:2 Division

To find the ratio of the areas of the two isosceles triangles $\triangle ABD$ and $\triangle BCD$, we need to calculate their areas using the segments of the main diagonal that acts as heights, and the secondary diagonal that acts as the base.

The main diagonal $AC = 30$ cm is divided into $AD = 20$ cm and $DC = 10$ cm due to the given ratio of 4:2.

Both triangles share the same base $BD = 11$ cm (the secondary diagonal).

Let's calculate each area:

• The area of $\triangle ABD$ using base $BD$ and height $AD$:
  $\text{Area}_{ABD} = \frac{1}{2} \times BD \times AD = \frac{1}{2} \times 11 \times 20 = 110 \, \text{cm}^2$.
• The area of $\triangle BCD$ using base $BD$ and height $DC$:
  $\text{Area}_{BCD} = \frac{1}{2} \times BD \times DC = \frac{1}{2} \times 11 \times 10 = 55 \, \text{cm}^2$.

Therefore, the ratio of the areas is $\frac{110}{55} = 2:1$.

The solution to the problem is $2:1$.