1. geometry problem

how to solve part (c)?

2. Originally Posted by helloying
how to solve part (c)?
Use proportions: (I assume that x = 10 is correct)

$\dfrac{z+k}{k} = \dfrac{10}6~\implies~z=\dfrac{10}7 k - k~\implies~z=\dfrac23k$

3. Originally Posted by earboth
Use proportions: (I assume that x = 10 is correct)

$\dfrac{z+k}{k} = \dfrac{10}6~\implies~z=\dfrac{10}7 k - k~\implies~z=\dfrac23k$

Thank you. Your ans is correct but i don understand why.Can u explain to me please?

4. Originally Posted by helloying
Thank you. Your ans is correct but i don understand why.Can u explain to me please?
You have 2 similar triangles: DEA and BCA.

You know that $DE \parallel BC$

Then the segments of sides must correspond to the lengths of the parallels:

The side AB correspond to BC
The side AD correspond to DE

$\dfrac{AB}{BC} = \dfrac{AD}{DE}$

Since AB = z + k and the measures of the parallels are known to you this proportion becomes:

$\dfrac{z+k}{10} = \dfrac{k}{6}$

Multiply both sides by 60 to get rid of the denominators:

$6(z+k)=10k~\implies~6z = 4k~\implies~z=\dfrac23k$