The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$ My friends recently took a Maths GCSE. In the paper, they came across a very difficult question which we spent a full half-hour train journey trying to figure out. We didn't manage it, so I've come here hoping you can help us out.
Here is the question:

The perimeter of the rectangle is 20.
Show that $x^2-10x+18=0$
Sorry for my appalling rectangle drawing skills. Drawing with Paint is like trying to staple an apple to your face. Painful and difficult.
 A: Consider one triangle formed when diagonal is drawn. Let $x$ be one of the sides, other side will be $10-x$
$(10-x)^2+x^2=k^2$ 
$10^2+2x^2-20x=k^2$ , Looking at the expression $k$ looks like $8$. So, $36+2x^2-10x=0$
A: Use the Pythagorean theorem. The width of the rectangle is $w=x$, and the length is $2l=20-2x$, therefore $l=10-x$. Therefore the diagonal $8^2=w^2+l^2=x^2+(10-x)^2$. Now expand this and show that it is equal to the quadratic you stated.
A: Hint: Let y be the other side. Then $y^2 + x^2 = 64$ and $2(y+x) = 20$. 
A: Let the side not given be $a$. Since the perimeter is 20, we know that $2x+2a=20$ which we can rearrange to obtain $a=10-x$. By the Pythagorean theorem, $x^2+a^2=8^2=64$. Substituting for $a$, we have $x^2+(10-x)^2=64$ which simplifies to $x^2-10x+18=0$; the desired result. 
A: Use the Pythagorean Theorem and simultaneous equations to come up with $x^2 + y^2 = 8^2$
and $2x + 2y = 20$.
Therefore, $x + y = 10$.
x (squared) + (10-x) (squared) = 64
2x(squared) - 20x + 100 = 64
(2x (squared) - 20x + 36)/2 = 0
x - 10x + 18 = 0
A: Let the height = y.
Therefore we can now create 2 simultaneous equations:-
1) X+y=10 (since perimeter of whole rectangle equals 20, X and y together must equal half ie 10)
2) (8^2)-(x^2)=(y^2) (using pythagrus)
So if we change around the first equation, we can get:- y=10-X, now plugging this into the second equation we get;- (8^2)-(x^2)=(10-X)^2, which is the same as - 64-(x^2)=100-20x+(x^2), which equals (ie, bringing everything on the left hand side over to the right) - 0=36-20x+2(x^2), now divide this whole equation by 2, you get:-
0=18-10x+(x^2),
Which of course when rearranged, is the same as:-
(x^2)-10x+18=0
A: Name one side $y$. 
$2x+2y=20$ therefore $x+y=10$ and $y=10-x$
By the Pythagorean Theorem,
$(10-x)²+x²=8²$ and therefore 
$2x²-20x-100=64$, 
$2x²-20x+36=0$,
$x²-10x+18=0$
