Question

proportional parts in triangles and parallel lines date: period:find the missing length indicated.1)2)3)4)solve for x.5)6)find the missing length indicated.7)8)

Explanation:

Step1: Set up proportion for #1

By the Triangle Proportionality Theorem, $\frac{\text{Missing length}}{15} = \frac{4}{14}$

Step2: Solve for missing length #1

$\text{Missing length} = 15 \times \frac{4}{14} = \frac{60}{14} = \frac{30}{7} \approx 4.29$

Step3: Set up proportion for #2

$\frac{\text{Missing length}}{25} = \frac{24}{15}$

Step4: Solve for missing length #2

$\text{Missing length} = 25 \times \frac{24}{15} = 40$

Step5: Set up proportion for #3

$\frac{\text{Missing length}}{18} = \frac{20}{8}$

Step6: Solve for missing length #3

$\text{Missing length} = 18 \times \frac{20}{8} = 45$

Step7: Set up proportion for #4

$\frac{\text{Missing length}}{15} = \frac{2}{12}$

Step8: Solve for missing length #4

$\text{Missing length} = 15 \times \frac{2}{12} = \frac{30}{12} = 2.5$

Step9: Set up proportion for #5

$\frac{5x}{45} = \frac{20}{36}$

Step10: Solve for x in #5

$5x = 45 \times \frac{20}{36} = 25 \implies x = \frac{25}{5} = 5$

Step11: Set up proportion for #6

$\frac{3x-5}{10} = \frac{28}{8}$

Step12: Solve for x in #6

$3x-5 = 10 \times \frac{28}{8} = 35 \implies 3x = 40 \implies x = \frac{40}{3} \approx 13.33$

Step13: Set up proportion for #7

$\frac{\text{Missing length}}{25} = \frac{6}{15}$

Step14: Solve for missing length #7

$\text{Missing length} = 25 \times \frac{6}{15} = 10$

Step15: Set up proportion for #8

$\frac{\text{Missing length}}{77} = \frac{25}{30}$

Step16: Solve for missing length #8

$\text{Missing length} = 77 \times \frac{25}{30} = \frac{1925}{30} = \frac{385}{6} \approx 64.17$

Answer:

1. $\frac{30}{7}$ or $\approx 4.29$
2. $40$
3. $45$
4. $2.5$
5. $x=5$
6. $x=\frac{40}{3}$ or $\approx 13.33$
7. $10$
8. $\frac{385}{6}$ or $\approx 64.17$