You mean "diagonal of the rectangle".

?? Ah! a= "adjacent side" and "0= Opposite side". A very good reason NOT to use "O" as a variable is that it is easy to mistake it for zero: "0". (And, in fact, you used "0" rather than "O" or "o"!). Now, what do you mean by "i did" that? It would be far better to say that you recognized that the diagonal of the rectangle forms the hypotenuse of a right triangle with the two sides of the rectangle as legs.i) find the area of the rectangle

ii) the area of the square is twice the area of the rectangle

find the length of a side of the square.

for part i) i did the a^2 + 0^2 = h^2

I assume you mean that you took one leg to be but how do you know that x= 5?then in the calc. i did 2√x got 2√5 then sq it and got 10 ,...then i did √x got √5

[/quote] then sq it and got 5 , added them and got 15 so then i did the hypotenuse thing and √45 and got 20, yeah i know you are prob wondering how i got that and to be honest so am i lol because i used a new calc. for this and when i repeated it on an older calc. of mine i got a different answer[/quote]

45= 9(5). [tex]]\sqrt{45}= 3\sqrt{5}= 6.708...[tex]. Even without a calculator you should have been able to see that = (20)(20)= (2)(2)(10)(10)= 400, nowhere near 45!

The correct answer to what problem? You appear to be saying that x= 9 cm^2 which is impossible because "x" is a length, not an area. The first question asked for the area of the rectangle not "2x" nor "x".so the CORRECT answer was :

2x=18cm^2

part 2 was= 6cm, but i got the answer wrong for i) so i was confused[/QUOTE]

It is close to impossible to figure out what you are saying here but I presume that you used the Pythagorean theorem to say that or [tex] 4x+ x= 5x= 45[tex] and so x= 45/5= 9, NOT 5. Then so the two sides have length and [tex]sqrt{x}= 3[tex]. Now, what is the area of a rectangle with side lengths 6 cm and 3 cm?