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liq [111]
3 years ago
8

Kayla wants to find the width, AB, of a river. She walks along the edge of the river 65 ft and marks point C. Then she walks 25

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.
(a) Can Kayla conclude that Δ and Δ are similar? Why or why not?

(b) Suppose DE = 15 ft. What can Kayla conclude about the width of the river?

Mathematics
1 answer:
AlladinOne [14]3 years ago
8 0

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 39\ ft

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

\frac{BC}{DC}=\frac{AB}{DE}

substitute the values and solve for AB

\frac{65}{25}=\frac{AB}{15}

AB=15*(\frac{65}{25})=39\ ft

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Step-by-step explanation:

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Required

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f(x + \triangle x) \approx f(x) +\triangle x \cdot f'(x)

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To calculate f'(x);

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Rewrite as:

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Split

f'(x) = \frac{1}{3} \cdot \frac{x^\frac{1}{3}}{x}

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Substitute 216 for x

f'(216) = \frac{216^\frac{1}{3}}{3*216}

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f(216) \approx 6  +1 \cdot \frac{3}{324}

f(216) \approx 6  + \frac{3}{324}

f(216) \approx 6  + 0.0093

f(216) \approx 6.0093

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