1. **Problem statement:** Identify the similar triangles and find the unknown side lengths $x$. 2. **Step 1: Identify similar triangles by comparing corresponding sides and angles.** 3. **Problem 16: Triangles CDE and FEG** - Given sides: $CE=15$, $DE=10$, $EF=6$, $FG=x$. - Check ratios: $\frac{CE}{EF} = \frac{15}{6} = 2.5$, $\frac{DE}{FG} = \frac{10}{x}$. - Since triangles are similar, $\frac{DE}{FG} = \frac{CE}{EF}$, so: $$\frac{10}{x} = 2.5$$ - Solve for $x$: $$10 = 2.5x \Rightarrow x = \frac{10}{2.5} = 4$$ 4. **Problem 17: Triangles STQ and RPS** - Given sides: $QR=12$, $RS=20$, $PT=16$, $QS=x$. - Similar triangles imply ratios: $$\frac{ST}{RS} = \frac{SQ}{PR} = \frac{TQ}{PS}$$ - Using given sides, set up ratio: $$\frac{x}{20} = \frac{16}{12} = \frac{4}{3}$$ - Solve for $x$: $$x = 20 \times \frac{4}{3} = \frac{80}{3} \approx 26.67$$ 5. **Problem 18: Triangles WZU and WUY** - Given sides: $WZ=3x-6$, $ZU=x+6$, $WY=40$, $UY=32$. - Similar triangles imply: $$\frac{WZ}{WY} = \frac{ZU}{UY}$$ - Substitute values: $$\frac{3x-6}{40} = \frac{x+6}{32}$$ - Cross multiply: $$(3x-6) \times 32 = (x+6) \times 40$$ $$96x - 192 = 40x + 240$$ - Simplify: $$96x - 40x = 240 + 192$$ $$56x = 432$$ $$x = \frac{432}{56} = \frac{108}{14} = \frac{54}{7} \approx 7.71$$ 6. **Problem 19: Triangles HJK and PNQ** - Given sides: $HJ=4x+7$, $HK=6x-2$, $JK=25$, $PN=8$, $NQ=12$, $PQ=20$. - Check ratios: $$\frac{HJ}{PN} = \frac{HK}{NQ} = \frac{JK}{PQ}$$ - Substitute: $$\frac{4x+7}{8} = \frac{6x-2}{12} = \frac{25}{20} = \frac{5}{4}$$ - Solve first ratio: $$\frac{4x+7}{8} = \frac{5}{4}$$ - Cross multiply: $$4(4x+7) = 8 \times 5$$ $$16x + 28 = 40$$ $$16x = 12$$ $$x = \frac{12}{16} = \frac{3}{4} = 0.75$$ 7. **Problem 20: Triangles ABE and EDC** - Given sides: $AB=25$, $AC=15$, $BE=10$, $ED=x$, angles $B=36^\circ$, $E=54^\circ$. - Since angles sum to $90^\circ$, triangles are similar. - Use ratio: $$\frac{AB}{ED} = \frac{BE}{AC}$$ - Substitute: $$\frac{25}{x} = \frac{10}{15} = \frac{2}{3}$$ - Solve for $x$: $$25 \times 3 = 2x$$ $$75 = 2x$$ $$x = \frac{75}{2} = 37.5$$ 8. **Problem 21: Triangles GDH and GDJ** - Given sides: $GH=7$, $HJ=10$, $GD=2x-2$, $DH=2x+4$, $GJ$ unknown. - Use ratio: $$\frac{GH}{HJ} = \frac{GD}{DH}$$ - Substitute: $$\frac{7}{10} = \frac{2x-2}{2x+4}$$ - Cross multiply: $$7(2x+4) = 10(2x-2)$$ $$14x + 28 = 20x - 20$$ $$28 + 20 = 20x - 14x$$ $$48 = 6x$$ $$x = 8$$ **Final answers:** - Problem 16: $x=4$ - Problem 17: $x=\frac{80}{3} \approx 26.67$ - Problem 18: $x=\frac{54}{7} \approx 7.71$ - Problem 19: $x=0.75$ - Problem 20: $x=37.5$ - Problem 21: $x=8$