Answered

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In the diagram of △△ADC below, EB∥∥DC, AE=2, AB=10, and BC=45. What is the length of AD?

In The Diagram Of ADC Below EBDC AE2 AB10 And BC45 What Is The Length Of AD class=

Sagot :

Answer:

11 units

Explanation:

Given that lines EB and DC are parallel, we use the proportional division theorem:

[tex]\frac{AE}{ED}=\frac{AB}{BC}[/tex]

Substitute the given values:

[tex]\begin{gathered} \frac{2}{ED}=\frac{10}{45} \\ \text{Cross multiply} \\ ED\times10=2\times45 \\ \text{Divide both sides by 10} \\ \frac{ED\times10}{10}=\frac{2\times45}{10} \\ ED=9 \end{gathered}[/tex]

Next, find the length of AD:

[tex]\begin{gathered} AD=AE+ED \\ =2+9 \\ =11\text{ units} \end{gathered}[/tex]

The length of AD is 11 units.

Alternate Method

[tex]ED=AD-2[/tex]

So, we have that:

[tex]\frac{AE}{ED}=\frac{AB}{BC}\implies\frac{AE}{AD-2}=\frac{AB}{BC}[/tex]

Substitute the given values:

[tex]\frac{2}{AD-2}=\frac{10}{45}[/tex]

Cross multiply:

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