## Trivia

Subject:
Re: Correct!
Response To:
Correct! ()

Larry Barrett
My way probably more complicated than your's or Henry's solutions.

Label the corners of the square, starting in NW corner and moving CW: A, B, C, D Label each side s and label the midpoint of side AB F.

Draw a diagonal from D to B.
Draw another line from C to F and label the intersection of this line with line DB E.

We want to know the area of triangle DEC with respect to the area of the square, which is s^2.

Area of DEC = s^2 -area of DAB - area of CEB.
Area of DAB = (1/2)s^2.

Area of CFB = (1/2)bh = (1/2)((1/2)s)s = (1/4)s^2.
Area of CFB also = area of EFB + area of CEB.

Let x be the altitude of EFB (from E to line FB). Observe that x is also the altitude of CEB since each altitude is perpendicular to the perpendicular sides and each is drawn from a point (E) on the diagonal to each of the perpendicular sides.
So 1(/4)s^2 = (1/2)((1/2)s)x + (1/2)sx = (3/4)sx
So x=(1/3)s
So area of CEB = (1/2)s(1/3)s = (1/6)s^2.

Therefore area of DEC = s^2 - (1/2)s^2 - (1/6)s^2. 