Answer:
[tex](\frac{4}{16})^2[/tex]
[tex]\frac{12}{192}[/tex]
[tex](\frac{3}{12})^2[/tex]
Step-by-step explanation:
we know that
If two figures are similar, the the ratio of its areas is equal to the scale factor squared
In this problem
The scale factor is 1/4
Let
z ---> the scale factor
x ---> the area of the smaller rectangle
y ---> the area of the large rectangle
so
[tex]z^2=\frac{x}{y}[/tex]
we have
[tex]z=\frac{1}{4}[/tex]
substitute
[tex]z^2=(\frac{1}{4})^2 =\frac{1}{16}[/tex]
Verify each option
a) we have
[tex]\frac{4}{16}[/tex]
Compare with  [tex]\frac{1}{16}[/tex]
so
[tex]\frac{4}{16} \neq \frac{1}{16}[/tex]
This option no show the ratio of the area of the smaller rectangle to the area of the larger rectangle
b) we have
[tex](\frac{4}{16})^2=\frac{16}{256}=\frac{1}{16}[/tex]
Compare with  [tex]\frac{1}{16}[/tex]
so
[tex]\frac{1}{16} = \frac{1}{16}[/tex]
This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle
c) we have
[tex]\frac{12}{192}=\frac{1}{16}[/tex]
Compare with  [tex]\frac{1}{16}[/tex]
so
[tex]\frac{1}{16} = \frac{1}{16}[/tex]
This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle
d) we have
[tex](\frac{4}{12})^2=\frac{16}{144}=\frac{1}{9}[/tex]
Compare with  [tex]\frac{1}{16}[/tex]
so
[tex]\frac{1}{9} \neq \frac{1}{16}[/tex]
This option no show the ratio of the area of the smaller rectangle to the area of the larger rectangle
e) we have
[tex](\frac{3}{12})^2=\frac{9}{144}=\frac{1}{16}[/tex]
Compare with  [tex]\frac{1}{16}[/tex]
so
[tex]\frac{1}{16} = \frac{1}{16}[/tex]
This option show the ratio of the area of the smaller rectangle to the area of the larger rectangle
