1. **Problem 1:** Given a triangle with vertices D, C, B and parallel lines inside, find $x$ where $AB=11$, $BC=x+2$, $AD=7$, and $DC=5$. 2. **Problem 2:** Given a right triangle with vertices A, C, B and segments creating points D and C on the base, find $x$ where $AB=11$, $AC=8$, and $DB=x$. 3. **Problem 3:** In triangle $\triangle PQR$, with $ST \parallel RQ$, given $PT=7.5$, $TQ=3$, and $SR=2.5$, find $PS$. --- ### Step 1: Find $x$ in Problem 1 - Since $AD$ is parallel to $BC$, triangles $ADB$ and $DCB$ are similar by AA similarity. - The ratio of corresponding sides is equal: $$\frac{AB}{BC} = \frac{AD}{DC}$$ - Substitute known values: $$\frac{11}{x+2} = \frac{7}{5}$$ - Cross multiply: $$11 \times 5 = 7 \times (x+2)$$ $$55 = 7x + 14$$ - Isolate $x$: $$7x = 55 - 14$$ $$7x = 41$$ - Divide both sides by 7: $$x = \frac{41}{7}$$ - Show cancellation: $$x = \cancel{\frac{41}{\cancel{7}}}$$ - Final answer: $$x = 5.857$$ --- ### Step 2: Find $x$ in Problem 2 - In right triangle $ABC$, with $D$ on $AB$ and $C$ on $AC$, and $DB = x$. - Since $D$ and $C$ are points on the legs, and segments are perpendicular, use Pythagoras or segment relations. - Given $AB=11$, $AC=8$, and $DB=x$. - Since $D$ lies on $AB$, and $DB=x$, then $AD = AB - DB = 11 - x$. - Using similarity or Pythagoras, but no further data is given, so assume $DB$ is the length to find. - Without more info, cannot solve for $x$ exactly here. --- ### Step 3: Find $PS$ in Problem 3 - Given $ST \parallel RQ$ in $\triangle PQR$, by the Basic Proportionality Theorem (Thales theorem): $$\frac{PS}{SR} = \frac{PT}{TQ}$$ - Substitute known values: $$\frac{PS}{2.5} = \frac{7.5}{3}$$ - Simplify right side: $$\frac{PS}{2.5} = 2.5$$ - Multiply both sides by 2.5: $$PS = 2.5 \times 2.5$$ $$PS = 6.25$$ --- ### Final answers: - Problem 1: $x = \frac{41}{7} \approx 5.857$ - Problem 3: $PS = 6.25$ - Problem 2: Insufficient data to solve for $x$.