Three squares of area $4, 9$ and $36$ are inscribed in the triangle as shown in the figure. Find the area of the big triangle
[img]https://cdn.artofproblemsolving.com/attachments/9/7/3e904a9c78307e1be169ec0b95b1d3d24c1aa2.png[/img]
1. **Identify the side lengths of the squares:**
   - The side length of the square with area $4$ is $\sqrt{4} = 2$.
   - The side length of the square with area $9$ is $\sqrt{9} = 3$.
   - The side length of the square with area $36$ is $\sqrt{36} = 6$.

2. **Set up the problem using similar triangles:**
   - Let the vertices of the triangle be $X, Y, Z$.
   - Place the squares inside the triangle such that the square with side length $2$ is at the top, the square with side length $3$ is in the middle, and the square with side length $6$ is at the bottom.

3. **Use the properties of similar triangles:**
   - Consider the triangle formed by the top vertex of the triangle $X$ and the vertices of the square with side length $2$.
   - Let the height of the triangle from $X$ to the base be $h$ and the base be $b$.

4. **Calculate the height and base of the triangle:**
   - The height of the triangle can be divided into three segments: the height of the square with side length $2$, the height of the square with side length $3$, and the height of the square with side length $6$.
   - Therefore, the total height $h$ is $2 + 3 + 6 = 11$.

5. **Calculate the base of the triangle:**
   - The base of the triangle can be divided into three segments: the base of the square with side length $2$, the base of the square with side length $3$, and the base of the square with side length $6$.
   - Therefore, the total base $b$ is $2 + 3 + 6 = 11$.

6. **Calculate the area of the triangle:**
   - The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.
   - Substituting the values, we get:
     \[
     \text{Area} = \frac{1}{2} \times 11 \times 11 = \frac{121}{2} = 60.5
     \]

The final answer is $\boxed{60.5}$