Question

(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.

Answer 1

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
`83.3\ ft`

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

`(BC)/(DC)=(AB)/(DE)`

substitute the values and solve for AB

`(62)/(93)=(AB)/(125)`

`AB=125*((62)/(93))=83.3\ ft`