Hello,

You basically have got a simultaneous equation here. Im struggling to see what you have done wrong but i think its in your pythagorous, probaly in the use of brackets. See if this helps-

Firstly the area of the rectangle is length x bredth so

area = (2√x)(√x) = 2x

Then using pythagorous like you said to form a second equation thus:

Hyp^2 = Adj^2 + Opp^2

(√45)^2 = (√x)^2 + (2√x)^2

45 = x + 4x

45 = 5x

x = 9

The area then equals 2 x 9 = 18cm2

Hope that helps! Not sure about part i as there is no mention in the question of a square?