High School

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3. Cho

Given that triangles ABC and DEF are similar, with AB = 18cm, DE = 12cm, DF = 10cm, and EF = 16cm, calculate the lengths of the segments AC and BC.

Use the ratio:
\( \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \)

Answer :

To solve the problem of finding the lengths of segments [tex]AC[/tex] and [tex]BC[/tex] in triangles [tex]\triangle ABC[/tex] and [tex]\triangle DEF[/tex] which are given to be similar, we use the concept of similarity ratios.

Similarity Ratios

Given that the triangles are similar, the corresponding sides are proportional. The given ratio of similarity is:

[tex]\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}[/tex]

Step-by-Step Solution


  1. Given Measurements:


    • [tex]AB = 18 \text{ cm}[/tex]

    • [tex]DE = 12 \text{ cm}[/tex]

    • [tex]DF = 10 \text{ cm}[/tex]

    • [tex]EF = 16 \text{ cm}[/tex]



  2. Calculate the Ratio:

    Using [tex]\frac{AB}{DE}[/tex] which is [tex]\frac{18}{12} = \frac{3}{2}[/tex].


  3. Find [tex]BC[/tex]:

    Using [tex]\frac{BC}{EF} = \frac{3}{2}[/tex], we have:

    [tex]\frac{BC}{16} = \frac{3}{2}[/tex]

    Cross-multiplying gives:

    [tex]2 \times BC = 3 \times 16[/tex]

    [tex]2 \times BC = 48[/tex]

    Solving for [tex]BC[/tex]:

    [tex]BC = \frac{48}{2} = 24 \text{ cm}[/tex]


  4. Find [tex]AC[/tex]:

    Using [tex]\frac{AC}{DF} = \frac{3}{2}[/tex], we have:

    [tex]\frac{AC}{10} = \frac{3}{2}[/tex]

    Cross-multiplying gives:

    [tex]2 \times AC = 3 \times 10[/tex]

    [tex]2 \times AC = 30[/tex]

    Solving for [tex]AC[/tex]:

    [tex]AC = \frac{30}{2} = 15 \text{ cm}[/tex]



Conclusion

The lengths of [tex]BC[/tex] and [tex]AC[/tex] are 24 cm and 15 cm, respectively. This solution uses the properties of similar triangles to find the unknown side lengths by leveraging the given side ratios.

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