In the image, $ABCD$ is a square of area $1$ and $\triangle DCE$ is an equilateral triangle. If $F$ and $G$ are the midpoints of $AD$ and $BC$ respectively, find the area of the quadrilateral $HJEI$.

I did this problem with a pretty tedious method. I put the figure in the plane and found the coordinates of the vertices of the quadrilateral, and then applied the Shoelace formula for the area of any polygon in the plane. With this method, I found that the answer is $\frac{54\sqrt3 - 93}{88}$.

Is there a better way to do this problem? I think there are plenty of ways but I couldn't find a better way than this.

enter image description here

  • 2
    $\begingroup$ I would find the vertical distance between $E$ and $H$. Then, find the horizontal distance to $J$ from the midline. The area is then the product of these two numbers due to symmetry. $\endgroup$ – Michael Burr Dec 16 '19 at 5:17

enter image description here


$$EH = PH - (PQ-EQ) = \frac14 - (1-\frac{\sqrt3}2)=\frac{2\sqrt3-3}4=a$$

and the triangles $\triangle HIE$ and $\triangle ADI$ are similar, which leads to,

$$\frac{EI}{ED-EI}=\frac{EH}{AD} \implies EI = \frac a{1+a}$$

The area of the quadrilateral $HJEI$ is twice the area of $\triangle HIE$ with $\angle IEH = 30^\circ$. Thus,

$$A = EH\cdot EI \cdot \sin30^\circ = \frac12 \frac {a^2}{1+a} =\frac12 \frac{\left(\frac{2\sqrt3-3}4\right)^2}{1+\frac{2\sqrt3-3}4}=\frac{54\sqrt3-93}{88}$$

  • $\begingroup$ Nice. Alternatively, without involving $\sin 30^\circ,$ from the similar triangles conclude that the altitude of $\triangle HIE$ on the base $EI$ is $\frac12 a/(1+a)$ and apply Michael Burr's comment under the question. $\endgroup$ – David K Dec 16 '19 at 6:48
  • $\begingroup$ @DavidK - True. Actual, $EI = IJ$ due to equilateral $△EIJ$ $\endgroup$ – Quanto Dec 16 '19 at 14:17

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