I was solving using similar triangles but couldn’t understand the way forward.
I found out two sides to be root 2 for the bigger traingle and one as 2
Well can you show your labeled diagram?
AC and CE= Root2 2, AE=2
Okay that’s correct. Anyway, the way I had in mind doesn’t require you to explicitly solve for anything.
As a start, if the side length of the square is x, express BD and AE in terms of x.
that’s what I have not been able to solve
Can you send a drawing with all angles and sides labeled?
AG and FE each =x. Therefore 3x=2 and x=2/3
Then we see that BC^2+CD^2=x^2, we get BC=CD= Root(2/3)
Can you tell where am I going wrong?
Looks good to me. Now find the area of \triangle BCD with the side lengths you obtained to actually solve the problem at hand.
Btw, what I was going for was something similar in vein to this:
We are given that \triangle BCD \sim \triangle ACD. The ratio of the areas of two similar triangles is the square of the ratio of their corresponding sides . Since the ratio of corresponding sides is 3, the ratio of their areas must be 3^2 = 9.
Not getting the correct answer through my method and I am unable to understand the method you mentioned. Could you pls elaborate on both?
Area of triangle BCD = \frac 12 (BC) (CD) = \frac 19.
Oh, I just saw this and it should be \frac{\sqrt{2}}{3}. I assumed that was what you meant with some typos, but I guess not
got it, thanks!!