Question

lesson 8.4 checkpoint

once you have completed the above problems and checked your solutions, complete the lesson checkpoint below.
complete the lesson reflection above by circling your current understanding of the learning goal.

1. given the figure below, solve for \\(x\\). then, find the value of \\(qp\\).

1. given the figure below, solve for \\(x\\). then, find the value of \\(xv\\).

Explanation:

Identify the midsegment properties in Question 1

In \(\triangle FHG\), point \(Q\) is marked as the midpoint of \(FH\) because \(FQ = QH\). Point \(P\) is marked as the midpoint of \(HG\) because \(HP = PG\). Therefore, segment \(QP\) is a midsegment of \(\triangle FHG\).

Using the Triangle Midsegment Theorem, the length of the midsegment is half the length of the parallel base:

$$QP = \frac{1}{2} FG$$

Solve for x in Question 1

Using the Midsegment Properties and the given expressions \(QP = x - 3\) and \(FG = x + 3\), we set up the equation:

$$x - 3 = \frac{1}{2}(x + 3)$$

Using the Solving Linear Equations knowledge point, we solve for \(x\):

$$2(x - 3) = x + 3$$

$$2x - 6 = x + 3$$

$$x = 9$$

Calculate QP in Question 1

Substitute \(x = 9\) back into the expression for \(QP\):

$$QP = x - 3 = 9 - 3 = 6$$

Identify the midsegment properties in Question 2

In \(\triangle XVW\), point \(P\) is the midpoint of \(VW\) because \(VP = PW\). Point \(Q\) is the midpoint of \(XW\) because \(XQ = QW\). Thus, segment \(QP\) is a midsegment of \(\triangle XVW\).

Using the Triangle Midsegment Theorem:

$$QP = \frac{1}{2} XV$$

Solve for x and XV in Question 2

Using the Midsegment Properties with \(QP = x + 2\) and \(XV = 3x - 3\):

$$x + 2 = \frac{1}{2}(3x - 3)$$

Using the Solving Linear Equations knowledge point:

$$2(x + 2) = 3x - 3$$

$$2x + 4 = 3x - 3$$

$$x = 7$$

Substitute \(x = 7\) to find \(XV\):

$$XV = 3(7) - 3 = 21 - 3 = 18$$

Answer:

Question 1

• \(x = 9\)
• \(QP = 6\)

Question 2

• \(x = 7\)
• \(XV = 18\)