Deltoid Diagonal Problem: Finding Secondary Diagonal Length with 6:1 Ratio

To find the length of the secondary diagonal, let's break down the problem:

• Step 1: Understand the Segmentation of the Main Diagonal
  The main diagonal $AC = 42$ cm is divided by the secondary diagonal $BD$ in a 6:1 ratio. Let's denote the segments of the main diagonal as $AE$ and $EC$ where $E$ is the intersection point. Therefore, if $AE : EC = 6:1$, this means that $AE = 6x$ and $EC = x$. The sum is $6x + x = 42$, so $7x = 42$. Therefore, $x = 6$.
• Step 2: Calculate the Segments $AE$ and $EC$
  Substituting back, $AE = 6x = 6 \times 6 = 36$ cm and $EC = x = 6$ cm.
• Step 3: Use the Triangle Area to Find the Height
  The small triangle's area with base $EC = 6$ cm is $18$ cm². Using the area formula $\frac{1}{2} \times \text{base} \times \text{height}$, we have$\frac{1}{2} \times 6 \times \text{height} = 18$.
  Solving for the height, we have $3 \times \text{height} = 18$, so the height is $\text{height} = 6$ cm.
• Step 4: Conclude the Length of the Secondary Diagonal
  The total length of the secondary diagonal $BD$ is the sum of the heights from triangles on each side of $BD$, as both equilateral triangles will have the height equal to the $6$ cm calculated since they are symmetrical and divide diagonals equally in a geometric deltoid. Hence, $BD = 6$ cm.

Therefore, the length of the secondary diagonal is $6$ cm.