Welcome! Check out our free B-School Guides to learn how you compare with other applicants.
Login or Register

Diagonal of a cube

This topic has 1 expert reply and 3 member replies
AleksandrM GMAT Destroyer!
Joined
04 Jan 2008
Posted:
566 messages
Thanked:
29 times
Test Date:
September 8, 2008
Target GMAT Score:
650
GMAT Score:
640
Diagonal of a cube Post Wed Jun 11, 2008 9:27 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    The diagonal of a cube of side x is xsqroot3.

    This can be found by applying the Pythagorean Theorem twice (first to find the diagonal of a face of the cube, xsqroot2, and then to find the diagonal through the center, xsqroot3).

    Can someone please demonstrate for me the latter part (xsqroot3).

    Assume we are dealing with a cube with side 4.

    Thanks.

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    AleksandrM GMAT Destroyer!
    Joined
    04 Jan 2008
    Posted:
    566 messages
    Thanked:
    29 times
    Test Date:
    September 8, 2008
    Target GMAT Score:
    650
    GMAT Score:
    640
    Post Wed Jun 11, 2008 9:31 am
    Never mind I got it. Just imagine a cube that has been cut through the side

    You have base 4 and hight 4sqroot2:

    (4sqroot2)^2 + 4^2 = x^2 diagonal from one end of cube to ther other

    x = 4sqroot3

    egybs Really wants to Beat The GMAT! Default Avatar
    Joined
    14 May 2008
    Posted:
    177 messages
    Thanked:
    16 times
    Test Date:
    08/2008
    GMAT Score:
    99%+
    Post Wed Jun 11, 2008 9:46 am
    Diag1 = Diagonal of one side of the cube
    Diag2 = Diagonal of the entire cube


    Diag1^2 = x^2+x^2
    Diag1 = (2x^2)^.5 = x(2^.5)
    Diag2^2 = (x(2^.5))^2 + x^2
    Diag2^2 = 2x^2 + x^2 = 3x^2
    Diag2 = (3x^2)^.5 = x(3^.5)

    GMAT/MBA Expert

    Ian Stewart GMAT Instructor
    Joined
    02 Jun 2008
    Posted:
    2178 messages
    Followed by:
    295 members
    Thanked:
    998 times
    GMAT Score:
    780
    Post Wed Jun 11, 2008 12:09 pm
    You could also use the three-dimensional version of the Pythagorean Theorem:

    d^2 = x^2 + y^2 + z^2

    where d is the diagonal inside a box ('rectangular prism', in math terms) of dimensions x by y by z.

    Thus in the special case of a cube measuring x by x by x,

    d^2 = x^2 + x^2 + x^2
    d^2 = 3x^2
    d = root(3)*x

    This won't be useful very often on the GMAT, but occasionally it does help.

    Thanked by: joiedevivre
    egybs Really wants to Beat The GMAT! Default Avatar
    Joined
    14 May 2008
    Posted:
    177 messages
    Thanked:
    16 times
    Test Date:
    08/2008
    GMAT Score:
    99%+
    Post Wed Jun 11, 2008 12:24 pm
    Heh good point Ian... that's a lot nicer!


    Ian Stewart wrote:
    You could also use the three-dimensional version of the Pythagorean Theorem:

    d^2 = x^2 + y^2 + z^2

    where d is the diagonal inside a box ('rectangular prism', in math terms) of dimensions x by y by z.

    Thus in the special case of a cube measuring x by x by x,

    d^2 = x^2 + x^2 + x^2
    d^2 = 3x^2
    d = root(3)*x

    This won't be useful very often on the GMAT, but occasionally it does help.

    Best Conversation Starters

    1 j_shreyans 82 topics
    2 aditya8062 43 topics
    3 anksm22 35 topics
    4 kamalakarthi 29 topics
    5 RiyaR 25 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description GMATGuruNY

    The Princeton Review Teacher

    175 posts
    2 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    141 posts
    3 image description MBAPrepAdvantage

    MBAPrepAdvantage

    92 posts
    4 image description CriticalSquareMBA

    Critical Square

    72 posts
    5 image description Jon@Admissionado

    Admissionado

    46 posts
    See More Top Beat The GMAT Experts