1. **Problem statement:** We have a solid hemisphere with center $A$ at coordinates $(5,-2,-3)$ and a point $B$ on the curved surface at $(3,-6,-9)$. We need to find: a) The exact radius of the hemisphere. b) The total surface area of the solid hemisphere to the nearest tenth. 2. **Formula and rules:** - The radius $r$ of the hemisphere is the distance between the center $A$ and point $B$ on the surface. - Distance between two points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$ is given by: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$ - The total surface area $S$ of a solid hemisphere is the curved surface area plus the base area: $$S = 2\pi r^2 + \pi r^2 = 3\pi r^2$$ 3. **Calculate the radius $r$:** $$r = \sqrt{(3 - 5)^2 + (-6 + 2)^2 + (-9 + 3)^2}$$ $$= \sqrt{(-2)^2 + (-4)^2 + (-6)^2}$$ $$= \sqrt{4 + 16 + 36}$$ $$= \sqrt{56}$$ $$= \sqrt{4 \times 14} = 2\sqrt{14}$$ So, the exact radius is $r = 2\sqrt{14}$. 4. **Calculate the total surface area $S$:** $$S = 3\pi r^2 = 3\pi (2\sqrt{14})^2$$ $$= 3\pi \times 4 \times 14$$ $$= 3 \times 56 \pi = 168\pi$$ 5. **Numerical approximation:** Using $\pi \approx 3.1416$, $$S \approx 168 \times 3.1416 = 527.787$$ Rounded to the nearest tenth: $$S \approx 527.8$$ **Final answers:** a) Radius $r = 2\sqrt{14}$ b) Total surface area $S \approx 527.8$ (to nearest tenth)