Let IJKLMN be a hexagon with side lengths IJ = LM = 3, JK = LM = 4, and KL = IN = 6. Also, all the interior angles of the hexagon are equal. Find the area of hexagon IJKLMN.

themia Oct 19, 2024

#1**0 **

We are tasked with finding the area of a regular hexagon \(IJKLMN\) where the interior angles are all equal. However, the given side lengths of the hexagon vary, suggesting that this is not a regular hexagon, but rather an **equiangular hexagon** (a hexagon with equal angles but varying side lengths).

### **Solution By Steps**

**Step 1: Properties of an equiangular hexagon**

For any equiangular hexagon, the sum of the interior angles is \(720^\circ\), and since all the angles are equal, each interior angle is:

\[

\frac{720^\circ}{6} = 120^\circ

\]

This is characteristic of any equiangular hexagon. Additionally, an important property of equiangular hexagons is that opposite sides of the hexagon are parallel. We can utilize this property to find relationships between the sides and the area.

**Step 2: Application of the area formula for an equiangular hexagon**

There is a known formula for the area of an equiangular hexagon with side lengths \(a\), \(b\), and \(c\) such that opposite sides are equal:

\[

\text{Area} = \frac{3\sqrt{3}}{2} \times (a^2 + b^2 + c^2)

\]

where \(a\), \(b\), and \(c\) are the three distinct side lengths of the hexagon.

From the problem, we know:

- \(IJ = LM = 3\) (opposite sides are equal),

- \(JK = MN = 4\) (opposite sides are equal),

- \(KL = IN = 6\) (opposite sides are equal).

Thus, we have:

- \(a = 3\),

- \(b = 4\),

- \(c = 6\).

**Step 3: Calculate the area using the formula**

Now, we substitute the values of \(a\), \(b\), and \(c\) into the area formula:

\[

\text{Area} = \frac{3\sqrt{3}}{2} \times (3^2 + 4^2 + 6^2)

\]

First, compute the squares of the side lengths:

\[

3^2 = 9, \quad 4^2 = 16, \quad 6^2 = 36

\]

Now, sum these values:

\[

9 + 16 + 36 = 61

\]

Substitute this into the formula:

\[

\text{Area} = \frac{3\sqrt{3}}{2} \times 61

\]

Simplify:

\[

\text{Area} = \frac{3\sqrt{3} \times 61}{2} = \frac{183\sqrt{3}}{2}

\]

### **Final Answer**

The area of the hexagon \(IJKLMN\) is:

\[

\frac{183\sqrt{3}}{2} \text{ square units}.

\]

Pythagorearn Oct 19, 2024