# Circular paper disc

### Circular paper disc

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.

Chikis

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Joined: Tue Jan 31, 2012 9:39 am
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### Re: Circular paper disc

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

svetoslav80

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Joined: Tue Jan 22, 2013 8:46 pm
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### Re: Circular paper disc

Let initial radius = $r^{1}$
Final radius = $r^{2}$

Then
$\frac{2\pi r^{1}}{2\pi r^{2}} = \frac{2}{5}$

$\frac{r^{1}}{r^{2}} = \frac{2}{5}$

Area = 4/5

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