ANSWER
[tex]3\pi sq.\: in.[/tex]
EXPLANATION
Let R be the radius of the bigger circle and r, be the radius of the smaller circle.
Then their ratio is given as,
[tex]R:r=3:1[/tex]
We can rewrite it as fractions to get,
[tex] \frac{R}{r} = \frac{3}{1} [/tex]
We make R the subject to get,
[tex]R = 3r[/tex]
The area of the bigger circle can be found using the formula,
[tex]Area=\pi {r}^{2} [/tex]
This implies that,
[tex]Area=\pi ({3r})^{2} [/tex]
[tex]Area=9\pi {r}^{2} [/tex]
But it was given in the question that, the area of the bigger circle is 27π.
[tex]27\pi=9\pi {r}^{2} [/tex]
We divide through by 9π to get,
[tex]3 = {r}^{2} [/tex]
This means that,
[tex]r = \sqrt{3} [/tex]
The area of the smaller circle is therefore
[tex] = \pi {( \sqrt{3}) }^{2} [/tex]
[tex] = 3\pi[/tex]
