Cuboid shortest distance

Geometry Level pending

A cuboid has side lengths of 2 , 3 , 4 2, 3, 4 . Find the shortest distance between two opposite vertices (i.e. end points of a space diagonal) if your path between the two vertices has to be on the surface of the cuboid.

53 \sqrt{53} 45 \sqrt{45} 41 \sqrt{41}

Hosam Hajjir by Hosam Hajjir , 56, Canada

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Cut the top face of the cube and unfold the remaining open top cuboid. Then we get two minimum distances between the body diagonally opposite vertices :

( 2 + 3 ) 2 + 4 2 = 41 , ( 2 + 4 ) 2 + 3 2 = 45 \sqrt {(2+3)^2+4^2}=\sqrt {41}, \sqrt {(2+4)^2+3^2}=\sqrt {45} .

Hence the shortest such distance is 41 \boxed {\sqrt {41}} units.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...