Will a truck that is 12 leet wide carrying a load that reaches 11 feet above the ground clear the semielliptical arch on the one-way road that passes under the bnige shown in the figure on the night? It (Round to two decimal places as needed.)

We model the semi‐elliptical arch by writing its equation in the form \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] so that the arch (the upper half of the ellipse) is \[ y=b\sqrt{1-\frac{x^2}{a^2}}, \] where \(2a\) is the width of the arch and \(b\) is its maximum height at the center. A common version of this problem assumes that the arch has a width of 24 feet (so that \(a=12\)) and a maximum height of 13 feet (so that \(b=13\)). With these values the equation becomes \[ y=13\sqrt{1-\frac{x^2}{144}}. \] A truck 12 feet wide means that its edges are 6 feet from the midline (i.e. \(x=\pm6\)). To see whether a truck carrying a load that reaches 11 feet above the ground will clear the arch, we must check the arch’s height when \(x=6\). Step 1. Substitute \(x=6\) into the equation: \[ y=13\sqrt{1-\frac{6^2}{144}}. \] Step 2. Compute the fraction: \[ \frac{6^2}{144}=\frac{36}{144}=\frac{1}{4}. \] Step 3. Then \[ y=13\sqrt{1-\frac{1}{4}}=13\sqrt{\frac{3}{4}}=13\cdot \frac{\sqrt{3}}{2}=\frac{13\sqrt{3}}{2}. \] Step 4. Evaluating numerically (recall that \(\sqrt{3}\approx1.732\)): \[ y\approx\frac{13\times1.732}{2}\approx\frac{22.516}{2}\approx11.26~\text{ft}. \] Since the load reaches 11 feet, the clearance at the edge is \[ 11.26-11=0.26~\text{ft}. \] Thus, the truck will just clear the arch (by about 0.26 feet).