The perimeter of the rectangle is $20$, diagonal is $8$ and side is $x$. Show that $x^2-10x+18=0$

Question

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.

Answer 1

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, $36+2x^2-10x=0$

Answer 2

Hint: Let y be the other side. Then $y^2 + x^2 = 64$ and $2(y+x) = 20$.

Answer 3

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.

Answer 4

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.