Suppose M is the midpoint of  FG⎯⎯⎯⎯⎯.
  Find the missing measure.

 

FM=8a+1, FG=42

Given that \( M \) is the midpoint of segment \( FG \), it divides the segment into two equal parts. Therefore, the length of \( FM \) is equal to the length of \( MG \).

### Problem Information:
- \( FM = 8a + 1 \)
- \( FG = 42 \)

Since \( M \) is the midpoint:
\[
FM = MG = \frac{FG}{2}
\]

So, we have:
\[
8a + 1 = \frac{42}{2}
\]
\[
8a + 1 = 21
\]

### Solve for \( a \):
\[
8a = 21 - 1
\]
\[
8a = 20
\]
\[
a = \frac{20}{8} = 2.5
\]

### Final Answer:
The value of \( a \) is \( 2.5 \).

Would you like more details or have any questions?

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Here are five related questions you might consider:
1. How would you solve for \( a \) if \( FM = 10a + 5 \) and \( FG = 50 \)?
2. What is the length of \( MG \) if \( FM = 8a + 1 \) and \( a = 3 \)?
3. How would the equation change if \( M \) is not the midpoint?
4. How can the midpoint formula be used to find the coordinates of \( M \)?
5. What is the geometric significance of a midpoint in a triangle?

**Tip:** When dealing with midpoint problems, always remember that the midpoint divides the segment into two equal parts. This concept can simplify many geometry problems!

Suppose M is the midpoint of FG. Find the missing measure. FM=8a+1, FG=42

Learn how to solve for the variable 'a' in a midpoint problem where FM = 8a + 1 and FG = 42. This problem explores key algebraic and geometric concepts suitable for students in Grades 6-8.

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