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In a rectangle, $GHIJ$, where $E$ is on $GH$ and $F$ is on $JI$ in such a way that $GEIF$ form a rhombus.
Determine the following: $1)$ $x=FI$ in terms of $a=GH$ and $b=HI$ and
$2)$calculate $y=EF$ in terms of $a$ and $b$.

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What have you tried as of yet? –  Tomarinator May 14 '12 at 3:46
I know that GE is equal to x and that BE is equal to a-x, since we are given that GEIF is a rhombus and that x=FC. –  user31284 May 14 '12 at 4:01
I fell that making minor edits in half-a-dozen questions to bring them back to the front page is anti-social behavior. Your desperation is not sufficient justification for trying to take over this website! –  Gerry Myerson May 16 '12 at 4:18
@Gerry: Do you know how to delete questions then? –  user31284 May 16 '12 at 4:36
I can't imagine why you would want to delete any of your questions. If you do want to delete one, I think the delete option is there, under the text of your question; if not, then you could contact a moderator (use the "contact us" link at the bottom of this page) and ask for a question to be deleted. But it is considered bad form to delete a question when other people have written answers or made other contributions, since those contributions get deleted, too. –  Gerry Myerson May 16 '12 at 5:53
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3 Answers

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We have $GHIJ$ is rectangle and $GEIF$ is a rhombus and also $GH = a = IJ$ and $HI = b=GJ$

We have to find $x=FI$ (side of rhombus) and $y=EF$ (one of the diagonals of rhombus)

(Here I have to draw picture of your problem. I know diagram. But i am not able to draw a picture in mac OS X. You can draw diagram easily)

$$x= FI = EI = GE = FG$$ (since sides of the rhombus are equal)

$$EH = GH-GE = a-x$$

$\triangle EHI$ is right angle triangle. That means

$$\begin{align*}(EI)^2 &= (EH)^2 + (IH)^2\\ X^2 &= (a-x)^2 + b^2\end{align*}$$

From above, we will get $x = \dfrac{a^2 + b^2}{2a}$

Draw rectangle $CEDF$ such that $GE\perp FC$ and $ED\perp FI$.

$EF$ is the diagonal of rectangle $CEDF$ and also $ED = HI = b$

$$FD = FI-DI = FI-(EH)$$ (since $DI = EH$)

$$ x - a + x = 2x - a = \frac{a^2 + b^2}{a} - a = \frac{b^2}{a}$$

$\triangle EDF$ is right angle triangle.

That means

$$\begin{align*}y^2 &= (EF)^2 = (ED)^2 + (FD)^2\\ &= b^2 + \left({\frac{b^2}{a^2}}\right)^2\\ &= \frac{b^2(a^2 + b^2)}{a^2} \end{align*}$$

then you can easily get value of y.

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Look at the triangle $FGJ$. It's a right-triangle, you know one side, you can express another in terms of $a$ and $x$, and the hypotenuse in terms of $x$. You should be able to use that to get an expression for $x$ in terms of $a$ and $b$.

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I don't know - there isn't any $D$ anywhere that I can see. –  Gerry Myerson May 14 '12 at 4:56
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enter image description here

Since its a rhombus, GE=EI+FI=GF


this equation gives you x

Finding EF

take the point J as origin, then the coordinates of E are $(x,b)$ i.e. $(\frac{a^2+b^2}{2a},b)$ and that of F are $(a-x,0)$ i.e $(\frac{a^2-b^2}{2a},0)$

$EF=\sqrt{(a-2x)^2 +b^2} $

this equation gives you EF after you plug in the value of x.

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You did notice that this was tagged "homework", right? Did you really want to give a complete solution? –  Gerry Myerson May 14 '12 at 4:56
Sometimes its hard to draw the line , as the how further you should go while solving homework tagged questions. I just edited the post to delete the final answers though. –  Tomarinator May 14 '12 at 5:38
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