Question

(60 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

Remember, similar triangles can have different side lengths.

In order for two triangles to be similar, the angles must be the same.

Since ∠DCE and ∠BCA are vertical angles, they have the same measure.

Since ∠CDE and ∠CBA are right angles, they have the same measure.

And since two of the angles are the same, the third angle must be the same,

so △ABC is similar to △EDC by the Angle-Angle similarity rule.

(b)

To solve this, we have to know that △ABC is similar to △EDC .

Because △ABC is similar to △EDC , we can say that

22/100 = 32/x , where x is the length of AB in feet.

x = 3200/22 ≈ 145.45

So the width of the river is about 145.45 feet, just as you found! smiley

Explanation:

'cause it is

Answer 2

Answer:
`83.\overline{3}\text{ feet}`

Explanation:

Since, In the given diagram,

The sides of the triangles ABC and EDC are,

BC = 62 feet, DC = 93 feet and DE = 125 feet(given)

And, angles B and D are right angles,

⇒
`\angle ABC\cong \angle EDC`

`\text{Also, } \angle ACB\cong \angle ECD` ( Vertically opposite angles )

By AA similarity postulate,

`\triangle ABC\sim \triangle EDC`

The sides of similar triangles are in same proportion ,

⇒
`(AB)/(ED)=(BC)/(DC)`

⇒
`AB * DC = BC * ED`

⇒
`AB = (BC.ED)/(DC)`

By substituting the values,

We get,

⇒
`AB = (62* 125)/(93)`

⇒
`AB = (7750)/(93)=83.\overline{3}\text{ feet}`