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HELP 40 POINTS!!!

In the image, segments AB and DE are parallel. If AC = 9.2, BC = 3.2, and DC = 8.4, what is the length of CE? Round your answer to the nearest hundredth of necessary.

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HELP 40 POINTSIn the image segments AB and DE are parallel If AC 92 BC 32 and DC 84 what is the length of CE Round your answer to the nearest hundredth of neces class=

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alicelovestea

Answer: 24.15

I'm going to assume this topic is on similarity so,

We know there are 2 triangles made by the lines

Triangle ABC is given AC = 9.2 , BC = 3.2

Triangle CDE is given CD = 8.4

Identify which side corresponds to which

AC corresponds to CE , and BC corresponds to CD

So we can write

CD/BC = CE/AC

which is

8.4/3.2 = CE/9.2

To find CE, bring 9.2 to the other side of the equation

(8.4/3.2)(9.2) = CE

CE = 24.15

msm555

Answer:

CE = 24.15 units

Step-by-step explanation:

Here:

[tex] \triangle ABC \sim \triangle EDC [/tex] by AAA axiom.

Since the corresponding sides of similar triangles are proportional, so we can set up the proportions based on corresponding sides.

From the given information:

[tex]\sf \dfrac{AC}{CE} = \dfrac{BC}{DC} = \dfrac{AB }{DE} [/tex]

Taking two of them.

[tex]\sf \dfrac{AC}{CE} = \dfrac{BC}{DC} [/tex]

Substituting the known values:

[tex]\sf \dfrac{9.2}{CE} = \dfrac{3.2}{8.4} [/tex]

We can solve for [tex]\sf CE [/tex] by cross-multiplication:

[tex]\sf 3.2 \times CE = 9.2 \times 8.4 [/tex]

[tex]\sf 3.2CE = 77.28 [/tex]

Divide both sides by [tex]\sf 3.2 [/tex]:

[tex]\sf CE = \dfrac{77.28}{3.2} [/tex]

[tex]\sf CE \approx 24.15 \textsf{(in nearest hundredth)}[/tex]

So, the length of [tex]\sf CE [/tex] is approximately [tex]\sf 24.15 [/tex] units.

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