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This was from a past exam we were given. I am stuck, but here is what I have so far.

To find the area of EFGH, we can find the individual coordinates of E,F,G,H so we can find the distance between each vertex and finally compute the area.

According to hint, I will centre $B(0,0)$. Since $|AB| =2$, then I label $A(0,2)$. We were also given that $E$ has a height of $1$. So I label $E(x,1)$, where x is some unknown x coordinate. I was able to obtain x by using that the triangle ABE is equilateral, so all three sides are equal. Hence $|AB| = |BE| = 2$. Also $|EG| = 1$. Then the triangle BEG is a right angled triangle, so by pythagorean $x = \sqrt{3}$. Hence $E(\sqrt{3}, 1)$. Following this, we also have $G(\sqrt{3},0)$.

But now, I am stuck on finding $H$ and $F$, as it seems they are not really given anything in the question to work with.

Edit: I can try to find the equation of the line connecting B and E. That would give $y = \frac{1}{\sqrt{3}}x$. However, it doesn't seem like we know either x or y coordinates.


5 Answers 5


$H$ can be obtained as the intersection of lines $AG$ and $BE$. The line $AG$ can be found using the points $A(0,2)$ and $G(\sqrt3,0)$.

Set the two equations equal to solve for the coordinates of $H$. I hope that helps!

  • $\begingroup$ Yeah, thanks I was able to figure it out using this. $\endgroup$
    – IcedTea
    Apr 21 at 22:58

Let $L$, $M$, $N$ be the midpoints of $\overline{AB}$, $\overline{BG}$, $\overline{GC}$, and note that $\overline{AG}\parallel\overline{LM}\parallel\overline{EN}$.

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Since the parallels trisect transversal $\overline{BN}$, they also trisect $\overline{BE}$. We conclude that $|\triangle EHG|=\frac13|\triangle EBG|$. (Why?) Consequently, $$|\square EFGH|=2\,|\triangle EHG|=\frac23\,|\triangle EBG|=\frac23\cdot\frac12|EG||BG|=\frac23\cdot\frac12\cdot 1\cdot \sqrt{3}\;=\;\frac{1}{\sqrt{3}}$$

Addendum. @Saeed's answer suggests a far simpler setup.

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Since $\triangle EGH \sim \triangle BAH$ with scale factor $1/2$, we have $|EH|=\frac12|BH|$, so that $|EH|=\frac13|EB|$ and thus also $|\triangle EHG|=\frac13|\triangle EBG|$. The final calculation proceeds as before.

  • 2
    $\begingroup$ A possibly simpler setup: Draw the horizontal through F and H, and to AB. It too gets divided in thirds, and is the altitude of the triangle EGH whose base is 1. $\endgroup$
    – David G.
    Apr 22 at 7:10

You've made a good start. However, note this answer shows a method using similar triangles and areas of triangles rather than coordinates. The following is the top part of your diagram, with the lines $EG$ and $HF$, intersecting at $J$, as well as the the horizontal line $BGC$, and the vertical line down from $H$, intersecting $BC$ at $K$, all having been added. Also, $\lvert KG\rvert = x$.

OP diagram with several lines and a length added

The quadrilateral $EFGH$ area is the sum of the areas of $\triangle EGH$ and $\triangle EFG$, i.e.,

$$\frac{\lvert EG\rvert\lvert HJ\rvert}{2} + \frac{\lvert EG\rvert\lvert JF\rvert}{2} = \frac{\lvert EG\rvert}{2}(\lvert HJ\rvert + \lvert JF\rvert) = \frac{1}{2}(x + x) = x$$

where it uses $\lvert GE\rvert = 1$ that you've determined, and symmetry to get $\lvert HJ\rvert = \lvert JF\rvert = x$. To determine $x$, first use that $\triangle HKG \sim \triangle ABG$ and $\lvert BG\rvert = \sqrt{3}$, as you've already stated, to get

$$\frac{\lvert HK\rvert}{x} = \frac{\lvert AB\vert}{\lvert BG\rvert} = \frac{2}{\sqrt{3}} \;\;\to\;\; \sqrt{3}\,\lvert HK\rvert = 2x$$

Next, using that $\triangle HKB \sim \triangle DCB$, and the above result, we have

$$\begin{equation}\begin{aligned} \frac{\lvert HK\rvert}{\sqrt{3}-x} = \frac{\lvert DC\vert}{\lvert BC\rvert} & = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \\ \sqrt{3}\, \lvert HK\rvert & = \sqrt{3} - x \\ 2x & = \sqrt{3} - x \\ x & = \frac{1}{\sqrt{3}} \end{aligned}\end{equation}$$

Thus, the area of quadrilateral $EFGH$ is $\frac{1}{\sqrt{3}}$.


enter image description here You can do without calculating coordinates. Connect $B$ to $C$.

Each of $\triangle{ABC}$ and $\triangle{DBC}$ is a 30-60-90 right triangle (why?). It Follows that $\triangle{EGB}$ and $\triangle{EGC}$ are also 30-60-90 right triangles, $EG \parallel AB \parallel DC$, $GB = GC$, and $EG$ is the vertical axis of symmetry of the figure (why?) As a result, $S_{\triangle{HEG}} = S_{\triangle{FEG}} = \frac12 S_{EFGH}$. We will set out to find $S_{\triangle{HEG}}$.

Note that $\triangle{HEG}$ and $\triangle{HBA}$ are similar, and therefore $S_{\triangle{HBA}} = 4S_{\triangle{HEG}}$, $BH = 2EH$, $S_{\triangle{HBG}} = 2S_{\triangle{HEG}}$ (why?).

Now calculate the area of trapezoid $ABGE$, note that this area is the sum of areas of 4 triangles formed by intersection of its diagonals, and we know each of those smaller areas as a multiple of $S_{\triangle{HEG}}$.

  • 1
    $\begingroup$ Observing the similarity of $\triangle HEG$ and $\triangle HBA$ is cleaner than my approach. As you note, $|BH|=2|EH|$. From there, we have $|EH|=\frac13|EB|$, so that $|\triangle EHG|=\frac13|\triangle EBG|$, with the latter area being easily calculated (as in my answer). $\endgroup$
    – Blue
    Apr 22 at 3:22

This is a two-liner if you know the following result (it's quite useful):

Lemma: In a trapezoid $ABCD$ ($AB \ ||\ DC$), if the diagonals intersect at $O$, then $$\mathrm{Ar}(\triangle AOB) : \mathrm{Ar}(\triangle AOD) : \mathrm{Ar}(\triangle DOC) = 1:r:r^2 = \mathrm{Ar}(\triangle AOB) : \mathrm{Ar}(\triangle BOC) : \mathrm{Ar}(\triangle DOC) $$for some positive real $r$.
Proof: Basically, we want to show the red and pink areas are equal, and areas of the green, red (or pink) and blue triangles are in a geometric progression.

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Since $\triangle AOB \sim \triangle COD$, let the former have area $a$, and the latter have area $ar^2$, where $r = CO/OA = DO/OB = CD/AB$.
Then since $\triangle AOD$ and $\triangle COD$ lie on the same base and share the same height, their areas are proportional with the same ratio. This means $\mathrm{Ar}(\triangle AOD) : \mathrm{Ar}(\triangle DOC) = 1:r$, and hence $\mathrm{Ar}(\triangle AOD) = ar$. A similar argument follows for $\triangle BOC$. $\blacksquare$

Now, the question is trivial (I solved it within a minute) :

  • $AEGB$ is a trapezoid.
  • Let $\mathrm{Ar}(\triangle EGH) = a$, $\mathrm{Ar}(\triangle AEH) = \mathrm{Ar}(\triangle GBH) = ar$ and $\mathrm{Ar}(\triangle ABH) = ar^2$.
  • You know $ar^2+ar=\mathrm{Ar}(\triangle ABE) = \sqrt 3$ and $a+ar = \mathrm{Ar}(\triangle BEG) = \sqrt 3/2$.
  • Dividing yields $r = 2$, so that $3a = \sqrt 3/2 \iff 2a = \boxed{1/\sqrt 3}$. Note that $2a$ was what we were looking for, by symmetry.

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