Answer to Question #148943 in Geometry for solid mensuration

Question

__17. A rectangular solid whose dimensions are 4cm, 8cm, and 12cm is inscribed in a sphere. What is the diagonal of the rectangular solid?

_18. In question 17, what is the surface area of the sphere in ¹?

Answer 1

17. Let $a=$ the length of a rectangular solid, $b=$ its width, a $c=$ its height.

The the diagonal $d$ of a rectangular solid will be

$$d=\sqrt{a^2 +b^2+c^2}$$ $$d=\sqrt{4^2+8^2+12^2}=4\sqrt{14}(cm)\approx14.97(cm)$$

18. If the rectangular solid is inscribed in a sphere, the center of the sphere is the point of intersection of the diagonals of the rectangular solid. Then

$$R_{sphere}=\dfrac{1}{2}d=\dfrac{1}{2}(4\sqrt{14}cm)=2\sqrt{14}\ cm$$

The surface area of the sphere is

$$A=4\pi R_{sphere}^2=4\pi (2\sqrt{14}cm)^2$$ $$=224\pi\ cm^2\approx703.72\ cm^2$$