Area of Rectangle

A triangle has side lengths $x,z$, and $z$. The area of the rectangle with side lengths $x$ and $z$ is a two-digit number, with both digits equal to $z$. The sum of the digits in $x$ is one-third the value of $z$. What is the area of the rectangle?
Source: NCTM Mathematics Teacher 2008

SOLUTION 1

$zz$ has three possible values $33,66$, or $99$.
If $zz=33$
$x\cdot 3=33$
$x=11$
$1+1\neq \dfrac{3}{3}$ not a solution
If $zz=66$
$1+1=\dfrac{6}{3}$ possible solution
If $zz=99$
$1+1\neq \dfrac{9}{3}$ not a solution
Area of rectangle
$xz=11\cdot 6=66$

SOLUTION 2
Area of rectangle
$xz=10z+z$
$=11z$
Divide both sides by $z\neq 0$
$x=11$
$\dfrac{z}{3}=1+1$
$z=6$
Area of rectangle
$xz=11\cdot 6=66$

Answer: $66$ square units