Triangle Proportionality Theorem Practice Worksheet with eight geometric problems.

To solve the problems using the Triangle Proportionality Theorem, we need to use the fact that if a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally. This theorem can be expressed as:

\[
\frac{a}{b} = \frac{c}{d}
\]

where \(a\) and \(b\) are segments on one side of the triangle, and \(c\) and \(d\) are segments on the other side.

Let's solve each problem step by step.

---

Problem 1:

\[
\begin{array}{c}
12 \\
x \\
6 \quad 8
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{12} = \frac{6}{6+8} = \frac{6}{14} = \frac{3}{7}
\]

Solving for \(x\):

\[
x = 12 \cdot \frac{3}{7} = \frac{36}{7}
\]

So, the solution is:

\[
\boxed{\frac{36}{7}}
\]

---

Problem 2:

\[
\begin{array}{c}
x \\
15 \quad 3 \\
5
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{15} = \frac{5}{3+5} = \frac{5}{8}
\]

Solving for \(x\):

\[
x = 15 \cdot \frac{5}{8} = \frac{75}{8}
\]

So, the solution is:

\[
\boxed{\frac{75}{8}}
\]

---

Problem 3:

\[
\begin{array}{c}
8 \\
x \\
20 \quad 35
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{8} = \frac{20}{20+35} = \frac{20}{55} = \frac{4}{11}
\]

Solving for \(x\):

\[
x = 8 \cdot \frac{4}{11} = \frac{32}{11}
\]

So, the solution is:

\[
\boxed{\frac{32}{11}}
\]

---

Problem 4:

\[
\begin{array}{c}
q \\
3 \quad 5 \\
x
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{q} = \frac{3}{3+5} = \frac{3}{8}
\]

We also know that the other side is divided in the same ratio:

\[
\frac{5}{q} = \frac{5}{8}
\]

From this, we can see that \(q = 8\). Substituting \(q = 8\) into the first proportion:

\[
x = 8 \cdot \frac{3}{8} = 3
\]

So, the solution is:

\[
\boxed{3}
\]

---

Problem 5:

\[
\begin{array}{c}
15 \\
30 \quad 20 \\
x
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{15} = \frac{20}{20+30} = \frac{20}{50} = \frac{2}{5}
\]

Solving for \(x\):

\[
x = 15 \cdot \frac{2}{5} = 6
\]

So, the solution is:

\[
\boxed{6}
\]

---

Problem 6:

\[
\begin{array}{c}
2 \\
6 \quad 8 \\
x
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{2} = \frac{8}{6+8} = \frac{8}{14} = \frac{4}{7}
\]

Solving for \(x\):

\[
x = 2 \cdot \frac{4}{7} = \frac{8}{7}
\]

So, the solution is:

\[
\boxed{\frac{8}{7}}
\]

---

Problem 7:

\[
\begin{array}{c}
4 \\
15 \quad 10 \\
x
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{4} = \frac{10}{10+15} = \frac{10}{25} = \frac{2}{5}
\]

Solving for \(x\):

\[
x = 4 \cdot \frac{2}{5} = \frac{8}{5}
\]

So, the solution is:

\[
\boxed{\frac{8}{5}}
\]

---

Problem 8:

\[
\begin{array}{c}
4 \\
3 \quad 5 \\
x
\end{array}
\]

Here, the line segment \(x\) is parallel to the base of the triangle, dividing the sides proportionally. We can set up the proportion:

\[
\frac{x}{4} = \frac{5}{3+5} = \frac{5}{8}
\]

Solving for \(x\):

\[
x = 4 \cdot \frac{5}{8} = \frac{20}{8} = \frac{5}{2}
\]

So, the solution is:

\[
\boxed{\frac{5}{2}}
\]

---

Final Answers:

\[
\boxed{\frac{36}{7}, \frac{75}{8}, \frac{32}{11}, 3, 6, \frac{8}{7}, \frac{8}{5}, \frac{5}{2}}
\]