Kayla wants to find the width, AB, of a river. She walks along the edge of the river 90 ft and marks point C. Then she walks 72 ft further and marks point D. She turns 90° and walks until her location, point A, and point C are collinear. She marks point E at this location, as shown.

Can Kayla conclude that ∆ABC and ∆EDC are similar? Why or why not? Suppose DE = 54 ft. What is the width of the river? Round to the nearest foot. Provide all work.

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vijayalalitha vijayalalitha · Answer 1 · 2020-03-15T14:31:27+01:00

Part a: The triangles ∆ABC and ∆EDC are similar by AA similarity rule.

Part b: The width of AB is 67.5 feet

Explanation:

Part a: We need to prove that the two triangles ABC and EDC are similar.

To prove the triangles are similar, then their angles must be similar.

Thus, we have,

∠DCE and ∠BCA are similar (vertical angles)

∠CDE and ∠CBA are similar (right angles)

∠B and ∠A are similar

Hence, the triangles ∆ABC and ∆EDC are similar by AA similarity rule.

Part b: We need to determine the width of AB

Since, the triangles are similar, then their corresponding lengths are proportional.

Thus, we have,

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

where [tex]DE=54 ft[/tex], [tex]DC=72 ft[/tex] and [tex]CB=90ft[/tex]

Substituting these values, we get,

[tex]\frac{54}{AB} =\frac{72}{90}[/tex]

Multiplying both sides by 90, we get,

[tex]\frac{4860}{AB}=72[/tex]

[tex]\frac{4860}{72}=AB[/tex]

Dividing, we have,

[tex]67.5=AB[/tex]

Thus, the width of the river AB = 67.5 feet