a) We have been given two similar figures and we have to find ratio of perimeters of these figures.
Let a be given side of larger figure and b be the given side of smaller figure. Let [tex]p_{1} \text{ and } p_{2}[/tex] be the perimeters of two figures and [tex]A_{1} \text{ and } A_{2}[/tex] be the areas of the two figures.
We know that perimeters of two similar figures are in the ratio of corresponding sides.
[tex]\frac{p_{1}}{p_{2}}=\frac{a}{b}[/tex]
We have been given, [tex]a=26\text{ and }b=6[/tex]
Therefore, upon substituting these values in the above equation, we get
[tex]\frac{p_{1}}{p_{2}}=\frac{26}{6}[/tex]
[tex]\frac{p_{1}}{p_{2}}=\frac{13}{3}[/tex]
(b)
Further we know that areas of two similar figures are in the ratio of squares of corresponding sides.
[tex]\frac{A_{1}}{A_{2}}=\frac{a^{2}}{b^{2}}\\ \frac{A_{1}}{A_{2}}=\frac{26^{2}}{6^{2}}\\ \frac{A_{1}}{A_{2}}=\frac{676}{36}\\ \frac{A_{1}}{A_{2}}=\frac{169}{9}\\[/tex]
Therefore, first option is the correct answer.
