The image shows a triangle $\triangle BCD$ with a midsegment $QR$. A midsegment of a triangle connects the midpoints of two sides, and according to the midsegment theorem, the length of the midsegment is half the length of the third side.

From the image, we are given that:

• $QR = x - 2$ (the length of the midsegment),
• $BD = x + 7$ (the third side of the triangle).

Since $QR$ is a midsegment, we know from the midsegment theorem that:

$QR = \frac{1}{2} BD$

Substituting the expressions for $QR$ and $BD$:

$x - 2 = \frac{1}{2}(x + 7)$

Now, let's solve for $x$.

1. Multiply both sides by 2 to eliminate the fraction:

$2(x - 2) = x + 7$

2. Distribute the 2 on the left side:

$2x - 4 = x + 7$

3. Subtract $x$ from both sides to get:

$x - 4 = 7$

4. Add 4 to both sides:

$x = 11$

Thus, the value of $x$ is $\boxed{11}$.

Would you like any further explanation or have any questions about this solution? Here are 5 related questions for you to explore:

1. What is the length of $QR$ when $x = 11$?
2. How does the midsegment theorem help in solving triangle problems?
3. What other properties do midsegments of triangles have?
4. How would the problem change if QR were not a midsegment?
5. Can the midsegment theorem be applied to other polygons, and if so, how?

Tip: When working with geometry problems involving midsegments, always remember the key properties such as the midsegment theorem and the relationships between the sides of the triangle.