Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $b$.

share|cite|improve this question
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
up vote 0 down vote accepted

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.

share|cite|improve this answer

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 $b$.

share|cite|improve this answer
I don't know - there isn't any $D$ anywhere that I can see. – Gerry Myerson May 14 '12 at 4:56

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.