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Vesnalui [34]
3 years ago
13

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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Opposite sides are corresponding (AB = DC).

Opposite angles are corresponding (D = B).

Consecutive angles are supplementary (A + D = 180°).

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The diagonals of a parallelogram hemisect each other.

Step-by-step explanation:

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