Answer
Floor: 5 meters x 2 meters
Pool: 2 meters x 2 meters
Work Step by Step
Let's note:
$f$=the side of the floor
$p$=the side of the pool
We can write the system:
$\begin{cases}
f^2-p^2=21\\
4f-p+3p=24
\end{cases}$
$\begin{cases}
f^2-p^2=21\\
4f+2p=24
\end{cases}$
We will use the substitution method. Solve Equation 2 for $p$ and substitute the expression of $p$ in Equation 1 to eliminate $p$ and determine $f$:
$\begin{cases}
f^2-p^2=21\\
p=\dfrac{24-4f}{2}
\end{cases}$
$\begin{cases}
f^2-p^2=21\\
p=12-2f
\end{cases}$
$f^2-(12-2f)^2=21$
$f^2-144+48f-4f^2=21$
$-3f^2+48f-144-21=0$
$-3f^2+48f-165=0$
$-3(f^2-16f+55)=0$
$f^2-16f+55=0$
$f^2-5f-11f+55=0$
$f(f-5)-11(f-5)=0$
$(f-5)(f-11)=0$
$f-5=0\Rightarrow f_1=5$
$f-11=0\Rightarrow f_2=11$
Substitute each of the values of $f$ in Equation 2 to determine $p$:
$p=12-2f$
$f_1=5\Rightarrow p=12-2(5)\Rightarrow p_1=2$
$f_2=11\Rightarrow p=12-2(11)\Rightarrow p_2=-10$
The system's solutions are:
$(5,2),(11,-10)$
As $p, f$ positive, the solution is:
$f=5$ meters
$p=2$ meters
