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3 years ago

Similar Triangles Question w/ Picture

Mathematics

1 answer:

melomori [17]3 years ago

7 0

Answer:

$EF=\dfrac{120}{7}\ ft$

Step-by-step explanation:

Triangles CEF and CAD are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so

$\dfrac{EF}{AD}=\dfrac{CF}{CD}\\ \\\dfrac{EF}{40}=\dfrac{CF}{CD}$

Triangles DEF and DBC are similar right triangles (they have the common angle, so by AA postulate they are similar). Similar triangles have proportional corresponding sides, so

$\dfrac{EF}{BC}=\dfrac{DF}{DC}\\ \\\dfrac{EF}{30}=\dfrac{CD-CF}{CD}\\ \\\dfrac{EF}{30}=1-\dfrac{CF}{CD}$

Substitute the fraction $\frac{CF}{CD}$ from the first equality into the second equality:

$\dfrac{EF}{30}=1-\dfrac{EF}{40}\ [\text{Muliply by 120}]\\ \\4EF=120-3EF\\ \\4EF+3EF=120\\ \\7EF=120\\ \\EF=\dfrac{120}{7}=17\dfrac{1}{7}\ ft$

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zimovet [89]

Hello.

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