Answer
$AC=49$
$CB = 168$
$AB=175$
Work Step by Step
We have a right angled triangle. Where $CB = x$; $AB=x+7$; $Perimeter = 392$
We can write an expression:
$AC+CB+AB=392$
$AC+x+(x+7)=392$
To find each side of the lot we have to find $x$ value. For that, at first we have to express $AC$ with $x$, then we can calculate the expression above.
We will express AC with $x$ using the Pythagoras' theorem:
$AC = \sqrt{(x+7)^2-x^2}=\sqrt{x^2+14x+49-x^2}=\sqrt{14x+49}$
Now put this value into the sum above and we will have an equation with one integer:
$\sqrt{14x+49}+x+(x+7)=392$
$\sqrt{14x+49}=385-2x$ //To simplify we have to disappear the root, so we will move root to one side and everything else to another and square both sides.
$14x+49=148225-1540x+4x^2$
$4x^2-1554+148176=0$ //To simplify, we will divide by $2$
$2x^2-777x+74088=0$
$D=603729-592704=11025$
$x_{1}=\frac{777-105}{4}=168$
$x_{2}=\frac{777+105}{4}=220.5$
Now we face a little problem. We got two values for $x$ and since we have restrictions (about what perimeter and sides can be) it is not possible to have different values for these sides. So we have to test both of these numbers by inputing them and see which one fits in all those restrictions:
If $x=168$ [Correct]
$AC=\sqrt{14*168+49} = \sqrt{2401}=49$
$CB = 168$
$AB=168+7=175$
$P=49+168+175=392$
If $x=220.5$ [Incorrect]
$AC=\sqrt{14*220.5+49} = \sqrt{3136}=56$
$CB = 220.5$
$AB=220.5+7=227.5$
$P=56+220.5+227.5=504$
