Let's solve the problem using the information you provided:

• The two triangles $\triangle GKH$ and $\triangle TCQ$ are similar.
• The proportionality constant $k$ is $\frac{6}{5}$.

Step 1: Confirm the proportionality constant $k$

The proportionality constant $k$ is defined as the ratio of the sides of the larger triangle to the corresponding sides of the smaller triangle.

Given that $k = \frac{6}{5}$, this means: $\frac{GH}{TQ} = \frac{KH}{CQ} = \frac{GK}{TC} = k = \frac{6}{5}$

Step 2: Use $k$ to find the missing sides

For Triangle $GKH$:

1. Given: $GK = 6$, $GH = 17$
2. Missing: $KH$

Since $k = \frac{6}{5}$, we can use the ratio: $KH = CQ \times k$

But we need to calculate $CQ$ first.

For Triangle $TCQ$:

1. Given: $TQ = 10$, $CQ = 5$
2. Missing: $TC$

We can find $TC$ using the proportionality constant $k$:

$TC = \frac{GK}{k} = \frac{6}{\frac{6}{5}} = \frac{6 \times 5}{6} = 5$

This means that $TC = 5$.

Now that we know $CQ = 5$, we use this in the equation for $KH$:

$KH = CQ \times k = 5 \times \frac{6}{5} = 6$

So, the missing side $KH = 6$.

Final Results:

• The missing side $KH$ in triangle $GKH = 6$.
• The missing side $TC$ in triangle $TCQ = 5$.

Questions for Further Exploration:

1. How does the proportionality constant change if the roles of the triangles were reversed?
2. How can you use the area of similar triangles to find the proportionality constant?
3. What happens if the sides given do not correspond to the correct sides in the similar triangles?
4. Can the proportionality constant $k$ be a fraction, and how does it affect the triangle sizes?
5. How can we apply the concept of similar triangles to solve real-world problems?

Tip: Always ensure that corresponding sides are correctly identified when working with similar triangles, as this is crucial for setting up the correct proportions.