## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$4\sqrt{2}\approx5.657 \text{ } ft$
The ramp forms a right triangle. Let $c$ be the hypotenuse and $a,b$ be the legs of the right triangle. Using $a^2+b^2=c^2$ or the Pythagorean Theorem, where $c=6$ and $a=2,$ then \begin{array}{l}\require{cancel} a^2+b^2=c^2 \\\\ 2^2+b^2=6^2 \\\\ 4+b^2=36 \\\\ b^2=36-4 \\\\ b^2=32 \\\\ b=\sqrt{32} \\\\ b=\sqrt{16\cdot2} \\\\ b=\sqrt{(4)^2\cdot2} \\\\ b=4\sqrt{2} .\end{array} Hence, the base of the ramp (which has the same value as $b$) is equal to $4\sqrt{2}\approx5.657 \text{ } ft .$