Find the area of hexagon

Let IJKLMN be a hexagon with side lengths IJ = LM = 2, JK = MN = 3, and KL = NI = 8.

Also, all the interior angles of the hexagon are equal.

Find the area of hexagon IJKLMN.

\(\text{Let the area of of hexagon $IJKLMN = A $} \\ \text{Let the area of $\triangle KNP = $ area of $ \triangle KNQ = A_1 $} \\ \text{Let the area of $\triangle LMP = $ area of $ \triangle IJQ = A_2 $}\)

The interior angles of the hexagon is \(\dfrac{4\cdot 180^\circ}{6}=\mathbf{120^\circ}\)

\(\begin{array}{|rcll|} \hline A &=& 2A_1-2A_2 \\ A &=& 2(A_1-A_2) \\\\ A &=& 2\left( \dfrac{(8+2)(2+3)\sin(60^\circ)} {2} - \dfrac{2\cdot 2\sin(60^\circ)} {2} \right) \\\\ A &=& 2\left( \dfrac{10\cdot 5\sin(60^\circ)} {2} - \dfrac{2\cdot 2\sin(60^\circ)} {2} \right) \\\\ A &=& \dfrac{2\sin(60^\circ)}{2} \left( 10\cdot 5 -2\cdot 2 \right) \\\\ A &=& \dfrac{2\sin(60^\circ)}{2} \cdot 46 \\\\ A &=& 2\sin(60^\circ) \cdot 23 \quad |\quad \sin(60^\circ)=\dfrac{\sqrt{3} }{2} \\\\ A &=& 2\cdot \dfrac{\sqrt{3} }{2} \cdot 23 \\\\ \mathbf{A} &=& \mathbf{23\sqrt{3}} \\ \mathbf{A} &=& \mathbf{39.8371685741~ u^2} \\ \hline \end{array}\)

The area of hexagon IJKLMN is \(\mathbf{\approx 40~u^2}\)