Deltoid Diagonal Problem: Finding Triangle Ratios with 25cm and 9cm Diagonals

To solve this problem, we'll begin by noting that the diagonals in the deltoid (kite) are perpendicular. This allows us to treat one diagonal as the base and the perpendicular segment of the other diagonal as the height in the calculation of triangles' area.

The secondary diagonal, which is 9 cm long, serves as the common base for the two isosceles triangles.

The main diagonal of length 25 cm is divided into two segments by the secondary diagonal, in a ratio of 3:2. Let's determine the lengths of these segments:

• The total segment lengths sum is 25 cm.
• If we let the longer segment be $\frac{3}{5} \times 25$ and the shorter segment be $\frac{2}{5} \times 25$, we get:
• Length of the first segment: $25 \times \frac{3}{5} = 15$ cm
• Length of the second segment: $25 \times \frac{2}{5} = 10$ cm

Now, let's calculate the area of each triangle:

• The area of Triangle 1 (base = 9 cm, height = 15 cm) is: $\frac{1}{2} \times 9 \times 15 = 67.5$ square cm
• The area of Triangle 2 (base = 9 cm, height = 10 cm) is: $\frac{1}{2} \times 9 \times 10 = 45$ square cm

Therefore, the ratio of the areas of these two triangles is:

$\frac{67.5}{45} = \frac{3}{2}$

This implies the ratio of triangle areas is $3:2$.

The question is asking for the ratio of the two isosceles triangles, assuming area calculations align with given dimensions directly, and since we are computing respective height proportions.

Therefore, the ratio of the two isosceles triangles whose common base is the secondary diagonal is $2:3$.