We will determine if the sets are proportional, if they are, then those will be the sets of lengths that would belong to triangle z, that is:
a.
[tex]\frac{4}{8}=\frac{7}{11}=\frac{14}{10}\Rightarrow\frac{1}{2}\ne\frac{7}{11}\ne\frac{7}{5}[/tex]
So, set a is not one.
b.
[tex]\frac{4}{10}=\frac{7}{17.5}=\frac{10}{25}\Rightarrow\frac{2}{5}=\frac{2}{5}=\frac{2}{5}[/tex]
So, set b could be a set of lengths that belong to triangle z.
c.
[tex]\frac{4}{6}=\frac{7}{9}=\frac{10}{11}\Rightarrow\frac{2}{3}\ne\frac{7}{9}\ne\frac{10}{11}[/tex]
So, set c is not one.
d.
[tex]\frac{4}{6}=\frac{7}{10.5}=\frac{10}{15}\Rightarrow\frac{2}{3}=\frac{2}{3}=\frac{2}{3}[/tex]
So, set d could be a set of lengths that belong to triangle z.
e.
[tex]\frac{4}{8}=\frac{7}{14}=\frac{10}{20}\Rightarrow\frac{1}{2}=\frac{1}{2}=\frac{1}{2}[/tex]
So, set e could be a set of lengths that beling to triangle z.
So, the options that could be candidate are: b, d & e.
