1. **Problem 1: Find $x$ and $y$ in trapezoids QUIK and SLOW** Given trapezoids QUIK and SLOW with sides and angles: - QUIK: $QU=15m$, $QK=20m$, angles $\angle K=84^\circ$, $\angle I=79^\circ$ - SLOW: $SW=24m$, $SL=x$, $WO=y$ Since the trapezoids are similar (implied by the problem), corresponding sides are proportional. 2. **Set up proportions for trapezoids:** $$\frac{QU}{SW} = \frac{QK}{WO} = \frac{UI}{LO}$$ Using known sides: $$\frac{15}{24} = \frac{20}{y}$$ Cross multiply: $$15y = 24 \times 20$$ $$15y = 480$$ Divide both sides by 15: $$y = \frac{480}{15} = 32$$ Similarly, for $x$: $$\frac{QU}{SW} = \frac{UI}{SL}$$ Assuming $UI$ corresponds to $SL=x$, and $UI$ length is unknown, but since $x$ is unknown and $SL$ corresponds to $UI$, we use the ratio: $$\frac{15}{24} = \frac{x}{20}$$ Cross multiply: $$15 \times 20 = 24x$$ $$300 = 24x$$ Divide both sides by 24: $$x = \frac{300}{24} = 12.5$$ --- 3. **Problem 2: Find $x$, $y$, and $z$ in triangles with sides and angles:** Given: - Sides: 6, 9, 12 in one triangle - Sides: 4, $x$, $y$ in the other - Angles: $72^\circ$ at one vertex, $z$ and $y$ unknown angles Triangles are similar, so corresponding sides are proportional. 4. **Set up proportions:** $$\frac{6}{4} = \frac{9}{x} = \frac{12}{y}$$ Calculate $x$: $$\frac{6}{4} = \frac{9}{x} \Rightarrow 6x = 36 \Rightarrow x = 6$$ Calculate $y$: $$\frac{6}{4} = \frac{12}{y} \Rightarrow 6y = 48 \Rightarrow y = 8$$ 5. **Find angle $z$:** Sum of angles in triangle is $180^\circ$. Given one angle $72^\circ$ and another angle $y$ (which is an angle, not side), but from the problem, $z$ is the unknown angle corresponding to $72^\circ$ in the other triangle. Since triangles are similar, corresponding angles are equal: $$z = 72^\circ$$ --- 6. **Problem 3: Similarity correspondence and congruent angles** a. Similarity correspondence: $$ABC \sim MNO$$ b. Pairs of congruent corresponding angles: $$\angle A = \angle M = 85^\circ$$ $$\angle B = \angle N = 72^\circ$$ c. Proportions using pairs of corresponding sides: $$\frac{AB}{MN} = \frac{BC}{NO} = \frac{AC}{MO}$$ Given sides: $$\frac{6}{3} = \frac{9}{?} = \frac{?}{4}$$ --- 7. **Problem 4: Shadow problem** Given: - Man height = 6 ft, shadow length = 8 ft - Tree shadow length = 20 ft, tree height = ? Triangles formed by man and tree are similar. Set proportion: $$\frac{\text{Man height}}{\text{Man shadow}} = \frac{\text{Tree height}}{\text{Tree shadow}}$$ $$\frac{6}{8} = \frac{h}{20}$$ Cross multiply: $$6 \times 20 = 8h$$ $$120 = 8h$$ Divide both sides by 8: $$h = \frac{120}{8} = 15$$ --- **Final answers:** - Problem 1: $x=12.5$, $y=32$ - Problem 2: $a.\ x=6$, $b.\ y=8$, $c.\ z=72^\circ$ - Problem 4: Tree height $=15$ ft