Question #148752
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
1
Expert's answer
2020-12-07T20:50:11-0500

17.


d=a2+b2+c2d=\sqrt{a^2 +b^2+c^2}

d=42+82+122=414(cm)14.97(cm)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


Rsphere=12d=12(414cm)=214 cmR_{sphere}=\dfrac{1}{2}d=\dfrac{1}{2}(4\sqrt{14}cm)=2\sqrt{14}\ cm

The surface area of the sphere is


A=4πRsphere2=4π(214cm)2A=4\pi R_{sphere}^2=4\pi (2\sqrt{14}cm)^2

=224π cm2703.72 cm2=224\pi\ cm^2\approx703.72\ cm^2

Rough approximation


A=4π(d2)2=πd2π(14.97 cm)2704.23 cm2A=4\pi (\dfrac{d}{2})^2=\pi d^2\approx\pi(14.97\ cm)^2\approx704.23\ cm^2

c) 704.03



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