AAH Math Team
Saturday, June 4, 2011
Summer Problem Solving Marathon Solution #5
We have d =
4
⁄
c
. c = πd, so by substitution d =
4
⁄
πd
. Cross-multiplying, πd
2
=4, and d
2
=
4
⁄
π
.
Since r =
d
⁄
2
, r
2
=
d
2
⁄
4
. So the area of the circle is πr
2
= π(
d
2
⁄
4
) = π(
1
⁄
π
) = 1.
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