Here is an example of a problem that involves the segment addition postulate, midpoint, and knowing the total length:
Problem: In a line segment AB, point M is the midpoint. The length of segment AB is 12 units. Point N is located on segment AB such that AN is 3 units longer than MN. Find the lengths of AN and NB.
Solution: Let’s assume that the length of MN is x units. Since M is the midpoint, the length of AM is also x units.
According to the segment addition postulate, the total length of AB is equal to the sum of the lengths of AN and NB. Therefore, we have:
AN + NB = AB
Since AB is given as 12 units, we can substitute it into the equation:
AN + NB = 12
We also know that AN is 3 units longer than MN. Since MN is x units, AN can be expressed as (x + 3) units.
Now, we can rewrite the equation using the given information:
(x + 3) + NB = 12
To find the value of NB, we need to isolate it on one side of the equation. Subtracting (x + 3) from both sides, we get:
NB = 12 - (x + 3) NB = 12 - x - 3 NB = 9 - x
So, the length of NB is 9 - x units.
Now, we can substitute the expression for NB into the equation AN + NB = 12:
AN + (9 - x) = 12
To find the value of AN, we need to isolate it on one side of the equation. Subtracting (9 - x) from both sides, we get:
AN = 12 - (9 - x) AN = 12 - 9 + x AN = 3 + x
So, the length of AN is 3 + x units.
Therefore, the lengths of AN and NB are 3 + x units and 9 - x units, respectively.