Imagine a square ABCD with a diagonal BD. Now draw a line EF parallel to BD, such that E lies on BC and F lies on CD. Also length of EF = length of AB. Now Colour the space enclosed by BDFE.
Of the square ABCD, what percentage of area is coloured ?
Call the side length of the square s. Then length of DB is s√2 .
Triangles FEC and DBC are similar, so |DB|/|BC| = |EF|/|EC| and we are
told |EF| = |AB| = s. Solve this to find |EC| = s/√2 which by symmetry
is also |FC|.
Thus A(ΔFEC) = sē/4, A(ΔDBC) = sē/2 so A(BDFE) = sē/4 which means 25% of the square is colored.
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Posted by Sean
on 2005-02-01 14:44:27 |