need help about geometry?

Let's call one side of the ABCD square = x

The diagonal AC = sqrt(2) * x

The shape ABED is two triangles (ABE and AED) of equal area.

AB has length x and AE has length x * ( sqrt(2) -1 ). The angle BAE is 45 degrees.

The area for a Side / Angle / Side triangle is .5 * Side1 * Side2 * SIN (Angle)

= .5 * x * x * ( sqrt(2) -1 ) * SIN(45) and we know this = .5 (half of ABED = 1)

This simplifies to X^2 * (sqrt(2)-1) * .707 = 1

This simplifies to X^2 = 3.4142 (and X^2 is the area of ABCD).

You can keep the SQRT and SIN in the equation if you want.

I'll get you started

Draw a sketch of square

D C

A B

Draw the diagonal AC

point E on AC, so that AC = AD = AB

ACE and ABE are isosceles triangles with AC = AE = AB

and

DE = BE

You should know how to get the area of an isosceles triangle with equal legs "a"

If you forgot, draw a line from point A to the middle of CE at point F, and from point A to the middle of BE at point G

Now you have right triangles

1/2 AD = EF

1/2 AD = DF

1/2 AD = EG

1/2 AD = BG Use Pythagoras for AF, then get area of right triangle. ( four times that will be area of ABED = 1)

take it from there!

glty