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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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    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.2012-05-14

3 Answers 3

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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.2012-05-14
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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.