GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A sphere is inscribed in a cube with an edge of 10. What is the
[#permalink]
06 Mar 2021, 04:57
06 Mar 2021, 04:57
1
7
Bookmarks
00:00
Question Stats:
51% (02:01) correct
48% (01:47) wrong based on 68 sessions
Hide Show timer Statistics
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
A. \(5(\sqrt{2} – 1)\)
B. \(5(\sqrt{3} – 1)\)
C. \(5\)
D. \(10(\sqrt{2} – 1)\)
E. \(10(\sqrt{3} – 1)\)
A. \(5(\sqrt{2} – 1)\)
B. \(5(\sqrt{3} – 1)\)
C. \(5\)
D. \(10(\sqrt{2} – 1)\)
E. \(10(\sqrt{3} – 1)\)
A sphere is inscribed in a cube with an edge of 10. What is the
[#permalink]
06 Mar 2021, 17:37
06 Mar 2021, 17:37
2
1
The shortest distance would be the (diagonal of the cube - diameter of the sphere)/2
Diagonal = \(10 \sqrt{3}\)
Diameter = 10
The shortest distance = \(\frac{10 \sqrt{3} - 10}{2} = \frac{10 (\sqrt{3} - 1)}{2} = 5(\sqrt{3} - 1)\)
Option B
Arun Kumar
Diagonal = \(10 \sqrt{3}\)
Diameter = 10
The shortest distance = \(\frac{10 \sqrt{3} - 10}{2} = \frac{10 (\sqrt{3} - 1)}{2} = 5(\sqrt{3} - 1)\)
Option B
Arun Kumar
General Discussion
A sphere is inscribed in a cube with an edge of 10. What is the
[#permalink]
07 Aug 2021, 23:38
07 Aug 2021, 23:38
2
Subtract the diameter of the sphere from the length of the diagonal and divide by two.
\(diameter-of-the-sphere = 10\) (since the sphere is inscribed inside the cube, it touches the faces and thus the diameter of the sphere = edge of the cube
\(length-of-the-diagonal\) = \(\sqrt{3}\)\( * 10\) (since diagonal of the cube is \(\sqrt{3}\)\( * (edge-of-the-cube)\))
\((\sqrt{3}*10 - 10)/2\)
\(10*(\sqrt{3} - 1)/2\)
\(5*(\sqrt{3} - 1)\)
Choice B.
If you don't know the formula for the diagonal of the cube, you can derive it from Pythagoras theorem on the fly.
Diagonal of the cube (one of them) is from bottom right vertice to the top left vertice on the opposite face and forms the hypotenuse of the triangle whose adjacent starts from the same bottom vertice and ends in the bottom vertice below the top vertice where the hypotenuse ends. The opposite side is the edge constituted between the end-points of the hypotenuse and the adjacent.
If \(x\) is the length of the edge, then \(adjacent^2 = x^2 + x^2\)
Now the \((diagonal-of-the-cube)^2 = adjacent^2 + opposite^2\)
=> \((diagonal-of-the-cube)^2 = x^2 + x^2 + opposite^2\)
=> \((diagonal-of-the-cube)^2 = x^2 + x^2 + x^2\)
(since opposite = length of the edge of the cube =\( x\))
therefore,
=> \((diagonal-of-the-cube)^2 = 3x^2\)
=> \((diagonal-of-the-cube)\) = \(\sqrt{3x^2}\)
=> \((diagonal-of-the-cube)\) = \(\sqrt{3}\)\(x\)
\(diameter-of-the-sphere = 10\) (since the sphere is inscribed inside the cube, it touches the faces and thus the diameter of the sphere = edge of the cube
\(length-of-the-diagonal\) = \(\sqrt{3}\)\( * 10\) (since diagonal of the cube is \(\sqrt{3}\)\( * (edge-of-the-cube)\))
\((\sqrt{3}*10 - 10)/2\)
\(10*(\sqrt{3} - 1)/2\)
\(5*(\sqrt{3} - 1)\)
Choice B.
If you don't know the formula for the diagonal of the cube, you can derive it from Pythagoras theorem on the fly.
Diagonal of the cube (one of them) is from bottom right vertice to the top left vertice on the opposite face and forms the hypotenuse of the triangle whose adjacent starts from the same bottom vertice and ends in the bottom vertice below the top vertice where the hypotenuse ends. The opposite side is the edge constituted between the end-points of the hypotenuse and the adjacent.
If \(x\) is the length of the edge, then \(adjacent^2 = x^2 + x^2\)
Now the \((diagonal-of-the-cube)^2 = adjacent^2 + opposite^2\)
=> \((diagonal-of-the-cube)^2 = x^2 + x^2 + opposite^2\)
=> \((diagonal-of-the-cube)^2 = x^2 + x^2 + x^2\)
(since opposite = length of the edge of the cube =\( x\))
therefore,
=> \((diagonal-of-the-cube)^2 = 3x^2\)
=> \((diagonal-of-the-cube)\) = \(\sqrt{3x^2}\)
=> \((diagonal-of-the-cube)\) = \(\sqrt{3}\)\(x\)
