1. **State the problem:** We have a right square pyramid with an altitude (height) of 10 units and a base side length of 6 units. We need to find the distance from the apex (top) to a vertex of the base, denoted as $x$. 2. **Understand the geometry:** The apex is directly above the center of the square base. The distance from the center of the base to a vertex is half the diagonal of the square base. 3. **Calculate the diagonal of the base:** The diagonal $d$ of a square with side length $s$ is given by $$d = s\sqrt{2}$$ Here, $s=6$, so $$d = 6\sqrt{2}$$ 4. **Calculate the distance from the center to a vertex:** This is half the diagonal, so $$\frac{d}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$ 5. **Use the Pythagorean theorem:** The distance $x$ from the apex to a vertex forms a right triangle with the altitude and the half-diagonal as legs: $$x = \sqrt{(\text{altitude})^2 + (\text{half diagonal})^2} = \sqrt{10^2 + (3\sqrt{2})^2}$$ 6. **Simplify inside the square root:** $$(3\sqrt{2})^2 = 3^2 \times (\sqrt{2})^2 = 9 \times 2 = 18$$ So, $$x = \sqrt{100 + 18} = \sqrt{118}$$ 7. **Calculate the numerical value:** $$x \approx 10.8627$$ 8. **Round to the nearest tenth:** $$x \approx 10.9$$ **Final answer:** The distance from the apex to each vertex of the base is approximately $10.9$ units.