I'm going to do this based on the way I think your drawing should've looked (as it is now, ACEF does not make a square...). It seems like it should look like this, based on your tag about congruent triangles.

F

A B E

D C

If you draw this on paper, you'll see it breaks up into 5 congruent triangles. The **area of square ABCD is twice the area of triangle ABC,** and the area of triangle ACEF is four times the area of triangle ABC. Area of a triangle is half the base times the height.

The area of triangle ABC: 0.5*AB*BC

Area of square ABCD: 2*0.5*AB*BC but we know AB=BC as it's a square. So **Area of square ABCD = 2*0.5*AB*AB = AB^2 **

AB = 1/2*AE, and the **area of ACEF is twice the area of triangle ACE**.

Area of triangle ACE = 0.5*AE*BC, but BC=AB and AE = 2*AB, so area of ACE = 0.5*2*AB*AB = AB^2

**Area of square ACEF = 2*AB^2** which is twice the area of ABCD, so the area of ABCD is half the area of ACEF.

Daniel O.

You could also prove it using pythagoras theorem and the fact that the area of a square is equal to the side length squared.

Area of ABCD = AB^2You could also write the area of ABCD as (using AB = 0.5*AE thus AE = 2*AB): (0.5*AE)^2 = (0.5*2*AB)^2 = AB^2

Area of ACEF = AC^2Use pythag to find length of AC: AC^2 = AB^2+BC^2 = AB^2+AB^2 = 2*AB^2

Note that the

area of ACEF = AC^2 = 2*AB^2 = 2*(area of ABCD)so the area of ABCD is half that of ACEF.11/02/12