# Math Help - Rectangle with given sides (in terms of x) and diagonal.

1. ## Rectangle with given sides (in terms of x) and diagonal.

I have a rectangle of sides x and y
x+y=5
The diagonal of the rectangle is 4
Show that 2x^2-10x+9=0

I did $\sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10$

so i tried doing a simultaneous equation by cancelling out the y^2 first
I did the x+y=10 times (x-y)
$10x-10y=x^2-y^2$
$-2x^2+10x+16=10y$

Now I'm stuck. What have I done wrong?

2. Originally Posted by Mukilab
I have a rectangle of sides x and y
x+y=5
The diagonal of the rectangle is 4
Show that 2x^2-10x+9=0

I did $\sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10$

so i tried doing a simultaneous equation by cancelling out the y^2 first
I did the x+y=10 times (x-y)
$10x-10y=x^2-y^2$
$-2x^2+10x+16=10y$

Now I'm stuck. What have I done wrong?
x+y=5 or x+y=10 ?

3. sorry, x+y=5, got confused because it previously asked me to show that x+y=5 (2x+2y=10=perimeter)

4. Originally Posted by Mukilab
I have a rectangle of sides x and y
x+y=5
The diagonal of the rectangle is 4
Show that 2x^2-10x+9=0

I did $\sqrt{x^2+y^2}=4\longrightarrow x^2+y^2=16, x+y=10$

so i tried doing a simultaneous equation by cancelling out the y^2 first
I did the x+y=10 times (x-y)
$10x-10y=x^2-y^2$
$-2x^2+10x+16=10y$