Answer
The length and width are 8 inches by 6 inches
Work Step by Step
Let us assume $ L $ be the length and $ W $ be the width of the television viewing area such that:
$\begin{align}
& {{L}^{2}}+{{W}^{2}}=100\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& LW=48\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
The substitution method is used to solve the system. Now, the value of $ L $ from the second equation is:
$ L=\frac{48}{W}$ (III)
Substitute the value of L in equation (I):
$\begin{align}
& \frac{{{48}^{2}}}{{{W}^{2}}}+{{W}^{2}}=100 \\
& 2804+{{W}^{4}}=100{{W}^{2}} \\
& {{W}^{4}}-100{{W}^{2}}+2304=0 \\
& \left( {{W}^{2}}-36 \right)\left( {{W}^{2}}-64 \right)=0
\end{align}$
Now,
${{W}^{2}}-36=0$
Or,
${{W}^{2}}-64=0$ $$
If ${{W}^{2}}=36$ then $ W=\pm 6$ and if ${{W}^{2}}=64$ then, $ W=\pm 8$. The width cannot be negative so, $ W=6,\ 8$.
If $ W=6$ then,
$\begin{align}
& L=\frac{48}{6} \\
& =8
\end{align}$
If $ W=8$
Then,
$\begin{align}
& L=\frac{48}{8} \\
& =6
\end{align}$
Thus, the dimensions are 8 inches by 6 inches.
