17.
d = a 2 + b 2 + c 2 d=\sqrt{a^2 +b^2+c^2} d = a 2 + b 2 + c 2
d = 4 2 + 8 2 + 1 2 2 = 4 14 ( c m ) ≈ 14.97 ( c m ) d=\sqrt{4^2+8^2+12^2}=4\sqrt{14}(cm)\approx14.97(cm) d = 4 2 + 8 2 + 1 2 2 = 4 14 ( c m ) ≈ 14.97 ( c m ) d) 14.97 cm
18. The center of the sphere is the point of intersection of the diagonals of the rectangular solid. Then
R s p h e r e = 1 2 d = 1 2 ( 4 14 c m ) = 2 14 c m R_{sphere}=\dfrac{1}{2}d=\dfrac{1}{2}(4\sqrt{14}cm)=2\sqrt{14}\ cm R s p h ere = 2 1 d = 2 1 ( 4 14 c m ) = 2 14 c m The surface area of the sphere is
A = 4 π R s p h e r e 2 = 4 π ( 2 14 c m ) 2 A=4\pi R_{sphere}^2=4\pi (2\sqrt{14}cm)^2 A = 4 π R s p h ere 2 = 4 π ( 2 14 c m ) 2
= 224 π c m 2 ≈ 703.72 c m 2 =224\pi\ cm^2\approx703.72\ cm^2 = 224 π c m 2 ≈ 703.72 c m 2 Rough approximation
A = 4 π ( d 2 ) 2 = π d 2 ≈ π ( 14.97 c m ) 2 ≈ 704.23 c m 2 A=4\pi (\dfrac{d}{2})^2=\pi d^2\approx\pi(14.97\ cm)^2\approx704.23\ cm^2 A = 4 π ( 2 d ) 2 = π d 2 ≈ π ( 14.97 c m ) 2 ≈ 704.23 c m 2 c) 704.03
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