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 ?

Let AB = x
So the area of square ABCD = x²
Also, AB = EF so EF = x

The angle between BD and DC is 45^o. So the angle between EF and DC is also 45^o since BD and EF are parallel. Therefore, triangle EFC is an isosceles right triangle. Let’s call the legs of triangle EFC y.

So we have y² + y² = x²
y² = x²/2

The area of triangle EFC is simply y*y/2 = y²/2 = (x²/2)/2 = x²/4

The area of region BDFE is the area of triangle BDC minus the area of triangle EFC. So that’s:
x²/2 – x²/4 = x²/4

So the percentage of area that is colored is simply the area of BDFE divided by the area of ABCD.
(x²/4)/(x²)*100% = 25%

Posted by nikki on 2005-02-01 14:53:23