The answer choices are not very legible to be, but I would solve it as follows
Now min lenght from a vertice to a pt on the sphere will be when moving diagonally from a vertice & the diameter of sphere is 10
diagonal of base of square is rt(2) *10
so diagonal from side is rt(10^2 + 10*rt(2) ^2)
= 10* rt(3)
so min lenght is (10*rt(3) - 10)/2 = 5(rt(3) -1)
Sphere Problem
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
-
- Legendary Member
- Posts: 645
- Joined: Wed Sep 05, 2007 4:37 am
- Location: India
- Thanked: 34 times
- Followed by:5 members
-
- Legendary Member
- Posts: 645
- Joined: Wed Sep 05, 2007 4:37 am
- Location: India
- Thanked: 34 times
- Followed by:5 members
-
- Master | Next Rank: 500 Posts
- Posts: 460
- Joined: Sun Mar 25, 2007 7:42 am
- Thanked: 27 times
Hey Amitava the figures mentioned in the Q are 3D figures, ie cube, sphere, & diagonal of a cube is rt(3)*s (lenght of side)
Regards
Samir
Samir
-
- Master | Next Rank: 500 Posts
- Posts: 195
- Joined: Sun Oct 21, 2007 4:33 am
- Thanked: 10 times
The distance between 2 opposite vertices of the cube =
SqRt(10^2 + 10^2 + 10^2) = 10*SqRt(3)
Since the Sphere is inscribed in the cude, radius = 5 and the centre will be the midpoint of the line joining the two opposite vertices i.e. [10SqRt(3)/2] = 5SqRt(3)
So, Shortest distance between the surface of Sphere and vertex of square = 5SqRt(3) - 5 = 5[SqRt(3)-1]
Hope that helps!
SqRt(10^2 + 10^2 + 10^2) = 10*SqRt(3)
Since the Sphere is inscribed in the cude, radius = 5 and the centre will be the midpoint of the line joining the two opposite vertices i.e. [10SqRt(3)/2] = 5SqRt(3)
So, Shortest distance between the surface of Sphere and vertex of square = 5SqRt(3) - 5 = 5[SqRt(3)-1]
Hope that helps!