Results 1 to 2 of 2

Math Help - Rhombus - Rectangle

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    144

    Rhombus - Rectangle

    ABCD is a rectangle with sides AB = a and AD = b. BEDF is a rhombus. Find EF in terms of a and b.
    Rhombus - Rectangle-pyth.jpg
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,740
    Thanks
    645
    Hello, MATNTRNG!

    I solved it, but I'm certain there is a more elegant way.


    ABCD\text{ is a rectangle with sides }AB = a\text{ and }AD = b.

    BEDF\text{ is a rhombus.}

    \text{Find }EF\text{ in terms of }a\text{ and }b.

    Code:
          :   x   H  b-2x E   x   :
        B *-------*-------*-------* C
          |*      :      *:*      |
          | *     :     * : *     |
          |  *    :    *  :  *    |
        a |   *   :   *   :a  *   | a
          |    *  :  *    :    *  |
          |     * : *     :     * |
          |      *:*      :      *|
        A *-------*-------*-------* D
          :   x   F  b-2x G   x   :
          : - - - - - b - - - - - :

    Draw EG \perp AD. .Draw FH \perp BC.

    Let x \,=\, AF \quad\Rightarrow\quad FD \:=\: b-x \:=\:BF .[1]

    In right triangle BAF\!:\;BF \:=\:\sqrt{x^2+a^2} .[2]

    Equate [2] and [1]: . \sqrt{x^2+a^2} \:=\:b-x

    Square both sides: . x^2+a^2 \:=\:b^2 - 2bx + x^2 \quad\Rightarrow\quad 2bx \:=\:b^2-a^2

    . . Hence: . x \:=\:\dfrac{b^2-a^2}{2b}


    \text{In right triangle }EGF\!:

    . . EF^2 \;=\;a^2 + (b-2x)^2 \;=\;a^2 + \left(b - \dfrac{b^2-a^2}{b}\right)^2

    . . . . . . =\; a^2 + \left(\dfrac{b^2-b^2+a^2}{b}\right)^2 \;=\;a^2 + \left(\dfrac{a^2}{b}\right)^2

    . . . . . . =\;a^2 + \dfrac{a^4}{b^2} \;=\;\dfrac{a^2b^2 + a^4}{b^2} \;=\;\dfrac{a^2}{b^2}(a^2+b^2)


    \text{Therefore: }\;EF \;=\;\dfrac{a}{b}\sqrt{a^2+b^2}


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Fascinating!

    The answer is the ratio of the width and length
    . . times the length of the diagonal.

    Does this suggest a more direct solution?
    . . I don't know!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Rectangle and Rhombus
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 8th 2011, 09:02 AM
  2. Replies: 6
    Last Post: June 2nd 2011, 04:56 PM
  3. Replies: 3
    Last Post: November 8th 2010, 06:16 PM
  4. Replies: 1
    Last Post: November 23rd 2009, 06:41 PM
  5. Rhombus inside Rectangle : testing coordinates
    Posted in the Math Software Forum
    Replies: 1
    Last Post: October 11th 2008, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum