<p><span style="color:#8e44ad;"><span style="font-size:20px;">Deductions from the Given Information</span></span></p>
<p>Using the Pythagorean Theorem, we can work on the lengths of the sides $x$ and $y$:</p>
<p><img alt="" src="https://gregmatapi.s3.amazonaws.com/media/misc/files/_class_pp1-shorter-quant-section-2-medium-q1.png" style="width: 400px; height: 149px;" /></p>
<p>Let's start with side x, which is the base of a right triangle with a height of $4$ and a hypotenuse of $8$:</p>
<ul>
<li>$4^2+x^2=8^2$</li>
<li>$x^2=8^2-4^2$</li>
<li>$x=\sqrt{8^2-4^2}$</li>
<li>$x=\sqrt{48}$</li>
</ul>
<p>Now, we can work out side $y$ which is the hypotenuse of a right triangle with a height of $4$ and a base of $4$:</p>
<ul>
<li>$4^2+4^2=y^2$</li>
<li>$16+16=y^2$</li>
<li>$32=y^2$</li>
<li>$\sqrt{32}=y$</li>
</ul>
<p><span style="color:#27ae60;"><span style="font-size:20px;">Solving the Problem </span></span></p>
<p>Since we know have the values for x and y, solving the problem is as simple as just comparing them. We know that since both $32$ and $48$ are bigger than $1$, $\sqrt{48}$ will definitely be bigger than $\sqrt{32}$, since $48$ is bigger than $32$.</p>
<p>So, the answer is <span style="color:#27ae60;">A</span>.</p>