Answer to Question #148752 in Math 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?

a) 7.49 cm
b) 8.94 cm
c) 14.42 cm
d) 14.97 cm

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

a) 176.24
b) 251.09
c) 704.03
d) 1,405.25

Answer 1

17.

$$d=\sqrt{a^2 +b^2+c^2}$$

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

d) 14.97 cm

18. 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

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

$$=224\pi\ cm^2\approx703.72\ cm^2$$

Rough approximation

$$A=4\pi (\dfrac{d}{2})^2=\pi d^2\approx\pi(14.97\ cm)^2\approx704.23\ cm^2$$

c) 704.03