Measurement of Segments (Level 4) | Examples III

Measurement of Segments (Level 4) | Examples III
In this last post on measurement of segments we will go over 2 challenging examples involving congruent segments and plenty of algebra to solve. Let’s go ahead and try the first example.

In this problem we are provided with a figure that has 4 line segments. We are asked to determine the length of segment RS.

Notice that the figure contains two line segments with matching single tick marks and two lines segments with matching double tick marks. These marks tell us that line segment TS is congruent to line segment PR and that line segment PT is congruent to line segment RS.

We are also told that the perimeter of the figure is 10 more than 5 times the length of segment RS. We can translate this fact into an equation as the perimeter equals 10 plus 5 times the length of segment RS.

We are also told that the length of segment PR is 26. This means that the length of segment TS is also 26 since both line segments are congruent. Next we need to determine what our unknowns are in this case we are trying to find the length of line segment RS so we will represent this length with the variable x since line segment PT is congruent to this segment we can also represent the length of segment PT with the variable x.

Now it is just a matter of finding a variable expression representing the perimeter of the figure in this case the perimeter will be equal to the sum of the lengths of all 4 line segments. Now that we have this relation let’s substitute the numerical and variable expression for each line segment, then we simplify the expression by collecting like terms. Doing that we obtain the following variable expression for the perimeter.

The last step is to set the variable expression representing the perimeter of the figure equal to 10 plus 5 times the length of segment RS, recall that the length of segment RS can be denoted with the variable x. From here it is just a matter of solving for x doing that we obtain x equals 14 and this is the length of segment RS.

It is important to use geometric relations and algebra together when solving many of the problems that you will encounter in this course. This is why it is important to understand both the geometric concepts and that you are well versed with algebra.

Alright let’s end the post with the final example.

In this problem we are provided with a figure that contains various line segments. We are also given various geometric and algebraic relations. We need to use these relations to solve for the variable r and m.

Let’s first take a look at the geometric relations; we know that line segment AD is congruent to line segment CB so let’s draw single tick marks on these segments on the figure shown. We also know that line segment AC is congruent to line segment DB so let’s draw double tick marks on these segments. Now that we can visually see which segments are congruent, let’s figure out a way to solve for the variables r and m.

We know from the given information that segment AD is congruent to segment CB this means that there lengths are equal in other words the measurement of segment AD is equal to the measurement of segment CB, since we have an algebraic expression for the length of each segment let’s go ahead and set these algebraic expressions equal to one another. Notice that we now have an equation with a single variable m. So we go ahead and solve for m, solving for m we obtain five thirds.

We now do that same for segment AC and segment DB, equating the algebraic expressions and solving for r we obtain 2. So r equals 2 and m equals five thirds.

It is important to set up the geometric relations by using the information given in the problem once this is done it is just a matter of using those relations to solve for the unknown variables.

Alright and this ends the measurement of segments series. In our next series of posts we will focus on measurement of angles.