Chapter 10 - Exponents and Radicals - 10.7 The Distance Formula, the Midpoint Formula, and Other Applications - 10.7 Exercise Set - Page 676: 16

$\sqrt{11}\approx3.317 \text{ }ft$

Let the right triangle have $a$ and $b$ as the legs and $c$ as the hypotenuse. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, with $ c=\sqrt{15} $ and $ a=2 ,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 2^2+b^2=(\sqrt{15})^2 \\\\ 4+b^2=15 \\\\ b^2=15-4 \\\\ b^2=11 \\\\ b=\sqrt{11} .\end{array} Hence, the other leg is $ \sqrt{11}\approx3.317 \text{ }ft $ long.