# Rhombus - Rectangle

• Oct 31st 2010, 12:20 PM
MATNTRNG
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.
• Oct 31st 2010, 04:40 PM
Soroban
Hello, MATNTRNG!

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

Quote:

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

$BEDF\text{ is a rhombus.}$

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

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

$\text{In right triangle }EGF\!:$

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

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

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

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

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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!