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The volume of a cube is 1 cubic ft . The distance from any vertex to t
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18 Jan 2025, 00:36
18 Jan 2025, 00:36
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Question Stats:
50% (00:57) correct
50% (01:26) wrong based on 2 sessions
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The volume of a cube is 1 cubic ft . The distance from any vertex to the centre of the cube (in feet) is
(A) $\(\frac{1}{2}\)$
(B) $\(\frac{1}{\sqrt{2\)$
(C) $\(\sqrt{2}\)$
(D) $\(\frac{\sqrt{3}}{2}\)$
(E) $\(\sqrt{3}\)$
(A) $\(\frac{1}{2}\)$
(B) $\(\frac{1}{\sqrt{2\)$
(C) $\(\sqrt{2}\)$
(D) $\(\frac{\sqrt{3}}{2}\)$
(E) $\(\sqrt{3}\)$
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The volume of a cube is 1 cubic ft . The distance from any vertex to t
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23 Feb 2025, 02:17
23 Feb 2025, 02:17
Expert Reply
The volume of a cube is Side $\(^3=1\)$ cubic feet $\\(Rightarrow\)$ Side $\(=\sqrt[3]{1}=1\)$ feet
Now, the distance from any vertex to the centre of the cube is same as $\(\frac{1}{2}\)$ the length of the diagonal i.e. $\(\frac{1}{2}\)$ of $\(\sqrt{3}=\frac{\sqrt{3}}{2}\)$
$(\(\because\)$ Diagonal in a cube is $\(\sqrt{3} \times\)$ Side $)$
Hence the answer is (D).
Now, the distance from any vertex to the centre of the cube is same as $\(\frac{1}{2}\)$ the length of the diagonal i.e. $\(\frac{1}{2}\)$ of $\(\sqrt{3}=\frac{\sqrt{3}}{2}\)$
$(\(\because\)$ Diagonal in a cube is $\(\sqrt{3} \times\)$ Side $)$
Hence the answer is (D).
gmatclubot
The volume of a cube is 1 cubic ft . The distance from any vertex to t [#permalink]
23 Feb 2025, 02:17
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