Let initial radius = [tex]r^{1}[/tex]
Final radius = [tex]r^{2}[/tex]

Then
[tex]\frac{2\pi r^{1}}{2\pi r^{2}} = \frac{2}{5}[/tex]

[tex]\frac{r^{1}}{r^{2}} = \frac{2}{5}[/tex]

Area = 4/5

Let the radius of the first circle is r, and the radius of the trimmed circle is R. The circumference of the first circle was [tex]2\pi r[/tex], and the second circle had circumference [tex]2\pi R[/tex]. After the trimming the area of the new circle was [tex]\frac{2}{5 } 2\pi r = \frac{4}{5 } \pi r => 2\pi R = \frac{4}{5 } \pi r => R = \frac{2}{5 }r[/tex] The area of the first circle was [tex]\pi r^2[/tex], and the area of the trimmed circle is [tex]\pi R^2 = \pi (\frac{2}{5 }r) ^2= \frac{4}{25 }\pi r^2[/tex]
[tex]\frac{S1}{S2 } = \frac{4}{25 }\pi r^2 : \pi r^2 = \frac{4}{25 }[/tex]

If a circular paper disc is trimmed in such a way that it circumference is reduce in the ratio 2:5, in what ratio is the surface area reduced?

Please I need help in tackling this question.