Circular paper disc

by **Chikis** » Fri Jan 18, 2013 12:48 pm

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.

by **svetoslav80** » Mon Jan 28, 2013 5:50 pm

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 $\pi R^2 = \pi (\frac{2}{5 }r) ^2= \frac{4}{25 }\pi r^2$
$\frac{S1}{S2 } = \frac{4}{25 }\pi r^2 : \pi r^2 = \frac{4}{25 }$

Circular paper disc

Circular paper disc

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