Answer
$8\sqrt{5}$ in $\times$ $4\sqrt{5}$ in
Work Step by Step
Let $x$ in and $y$ in be the length and width of the rectangle, respectively.
Since the diameter of the circular is $20$ in, the diagonal of the rectangle is $20$ in and so $x^2+y^2=20^2$.
Since the area of the rectangle is $160$ in$^2$, we have $xy=160$.
Now, the system corresponding the problem is given by
$x^2+y^2=400$
$xy=160$
Multiplying the first equation by $x^2$ and taking the square in the second equation, we have
$x^4+x^2y^2=400x^2$
$x^2y^2=160^2$
Substituting the second equation to the first one,
$x^4+160^2=400x^2$
$x^4-400x^2+160^2=0$
$x^4-(80+320)x^2+80\cdot 320=0$
$(x^2-80)(x^2-320)=0$
$x^2=80$ or $x^2=320$
$x=4\sqrt{5}$ or $x=8\sqrt{5}$
If $x=4\sqrt{5}$, then $y=\frac{160}{4\sqrt{5}}=\frac{40}{\sqrt{5}}=8\sqrt{5}$.
If $x=8\sqrt{5}$, then $y=\frac{160}{8\sqrt{5}}=\frac{20}{\sqrt{5}}=4\sqrt{5}$.
So, the rectangle has the dimensions of $8\sqrt{5}$ in $\times$ $4\sqrt{5}$ in.
