Represent lengths rectangle using given terms 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 $b$.
 A: 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 $b$. 
A: 
Since its a rhombus, GE=EI+FI=GF
$GF=\sqrt{b^2+(a-x)^2}=GE=x$
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.
A: 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.
