Respuesta :

SOLUTION

Step 1 :

In this question, we are given that :

Triangle ABC is similar to Triangle DEF.

We are meant to find the necessary of side FD.

Step 2 :

Using the principles of Similar Triangles, we have that:

[tex]\frac{FD}{CA}\text{ = }\frac{EF}{BC}[/tex]

where:

[tex]\begin{gathered} FD\text{ = ?} \\ CA\text{ = 49} \\ EF\text{ = 5} \\ BC\text{ = 23} \end{gathered}[/tex][tex]\frac{FD}{49}\text{ = }\frac{5}{23}[/tex]

Cross- multiply, we have that:

[tex]\begin{gathered} FD\text{ = }\frac{49\text{ X 5}}{23} \\ FD\text{ = }\frac{245}{23} \\ FD\text{ = 10. 65 } \\ FD\text{ = 10. 7 ( to the nearest tenth)} \end{gathered}[/tex]

CONCLUSION:

The length of FD = 10. 7 ( to the nearest tenth).

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