In parallelogram $ABCD$, the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$. If the area of $XYZW$ is $10$, find the area of $ABCD$
1. **Define the sides and angles of the parallelogram:**
   Let $\overline{AB}$ be the longer side of the parallelogram $ABCD$, and let $\overline{AD}$ be the shorter side. Given that the longer side is twice the shorter side, we have:
   \[
   \overline{AB} = 2\overline{AD}
   \]
   Let $\angle DAB = 2\theta$.

2. **Determine the shape of $XYZW$:**
   The quadrilateral $XYZW$ is formed by the internal bisectors of the angles of $ABCD$. By angle chasing, we can show that $XYZW$ is a rectangle. This is because the internal angle bisectors of a parallelogram form a rectangle.

3. **Calculate the side lengths of $XYZW$:**
   Let $B$ bisector meet $C$ bisector at $X$ and $Y$ be the intersection of $A$ bisector and $B$ bisector. From $\triangle XBC$, we have:
   \[
   \overline{XB} = x \sin(\theta)
   \]
   From $\triangle AYB$, we have:
   \[
   \overline{XY} = 2x \sin(\theta)
   \]
   Since $X$ is the midpoint of $\overline{YB}$, the side lengths of the rectangle $XYZW$ are $x \sin(\theta)$ and $x \cos(\theta)$.

4. **Relate the area of $XYZW$ to the given area:**
   The area of rectangle $XYZW$ is given by:
   \[
   \text{Area of } XYZW = x \sin(\theta) \cdot x \cos(\theta) = 10
   \]
   Using the double-angle identity $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$, we get:
   \[
   x^2 \sin(\theta) \cos(\theta) = 10 \implies \frac{1}{2} x^2 \sin(2\theta) = 10 \implies x^2 \sin(2\theta) = 20
   \]

5. **Calculate the area of parallelogram $ABCD$:**
   The area of parallelogram $ABCD$ can be calculated using the formula for the area of a triangle:
   \[
   \text{Area of } ABCD = 2 \times \text{Area of } \triangle BCD
   \]
   Using the formula for the area of a triangle, $\frac{1}{2}ab\sin(C)$, we get:
   \[
   \text{Area of } \triangle BCD = \frac{1}{2} \cdot x \cdot 2x \cdot \sin(2\theta) = x^2 \sin(2\theta)
   \]
   Therefore, the area of parallelogram $ABCD$ is:
   \[
   \text{Area of } ABCD = 2 \times x^2 \sin(2\theta) = 2 \times 20 = 40
   \]

The final answer is $\boxed{40}$.