d. bc = 12 and ef = 16
step-by-step explanation:
we are given that δabc : δdef such that
.
also, it is given that ![\frac{ab}{de}=\frac{3}{4}](/tex.php?f=\frac{ab}{de}=\frac{3}{4})
so, from the options, we need that
.
a. ![\frac{bc}{ef}=\frac{6}{9}=\frac{2}{3}](/tex.php?f=\frac{bc}{ef}=\frac{6}{9}=\frac{2}{3})
b. ![\frac{bc}{ef}=\frac{4}{6}=\frac{2}{3}](/tex.php?f=\frac{bc}{ef}=\frac{4}{6}=\frac{2}{3})
c. ![\frac{bc}{ef}=\frac{8}{12}=\frac{2}{3}](/tex.php?f=\frac{bc}{ef}=\frac{8}{12}=\frac{2}{3})
d. ![\frac{bc}{ef}=\frac{12}{16}=\frac{3}{4}](/tex.php?f=\frac{bc}{ef}=\frac{12}{16}=\frac{3}{4})
as, we can see that only option d gives the same ratio as
i.e.
.
hence, the possible lengths are bc = 12 and ef = 16.