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September 1, 2014

September 1, 2014

Posted by **Jujube** on Monday, January 21, 2008 at 12:05pm.

- math PLEASE HELP -
**Reiny**, Monday, January 21, 2008 at 2:19pmthe diagonals of a rhomus bisect each other at right angles, forming 4 equal right-angled triangles

the hypotenuse of each is 96/4 or 24 and one side is 32/2 or 16.

so by Pythagoras, the other side is sqrt(320)

The centre of the incsribed circle is at the intersection of the diagonals, so I drew a height inside one of the triangles.

by some basic geometry I found the r^2 = 1280/9

so the area of the circle is pi*1280/9 but that is given to be k*pi/w

so k*pi/w = pi*1280/9

since 1280/9 is relatively prime, k+w = 1280+9 = 1289

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