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(50 points to best answer) (asap please and thank you)
Kayla wants to find the distance, AB, across a creek. She starts at point B and walks along the edge of the river 62 ft and marks point C. Then she walks 93 ft further and marks point D. She turns 90° and walks until her final location and marks point E. Point E, point A, and point C are collinear.

Can Kayla conclude that ∆ABC and ∆EDC are similar? Why or why not?

Suppose (DE) ̅=125 ft. Calculate the distance of (AB) ̅ to the nearest tenth of a foot. Show your work. Don’t forget to label your answer.

50 points to best answer asap please and thank you Kayla wants to find the distance AB across a creek She starts at point B and walks along the edge of the rive class=

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calculista

Answer:

Part A) The triangles ABC and EDC are similar by AAA, because the three internal angles are equal in both triangles  

Part B) The width of the river is about [tex]83.3\ ft[/tex]

Step-by-step explanation:

we know that

If two triangles are similar, then the ratio of its corresponding sides is equal and its corresponding angles are congruent

Part A) we know that

In this problem , triangles ABC and CDE are similar by AAA, because its corresponding angles are congruent

so

m<DCE=m<ACB -----> by vertical angles  

m<EDC=m<ABC -----> is a right angle

m<DEC=m<CAB -----> the sum of the internal angles must be equal to 180 degrees

Part B) we know that

The triangles ABC and EDC are similar -------> see Part A

therefore

[tex]\frac{BC}{DC}=\frac{AB}{DE}[/tex]

substitute the values and solve for AB

[tex]\frac{62}{93}=\frac{AB}{125}[/tex]

[tex]AB=125*(\frac{62}{93})=83.3\ ft[/tex]

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