To find the area of the figure as described, we can separate it into the regular shapes mentioned: two triangles and a square.
-
Square:
- Length of each side = 12 in
- Area of the square \(A_{square} = \text{side} \times \text{side} = 12 , \text{in} \times 12 , \text{in} = 144 , \text{in}^2\)
-
Triangle 1:
- Base = 8 in (assuming this is the base)
- Height = 8 in
- Area of Triangle 1 \(A_{triangle1} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 , \text{in} \times 8 , \text{in} = \frac{1}{2} \times 64 , \text{in}^2 = 32 , \text{in}^2\)
-
Triangle 2:
- Base = 8 in (assumed same as Triangle 1)
- Height = 8 in
- Area of Triangle 2 \(A_{triangle2} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 , \text{in} \times 8 , \text{in} = \frac{1}{2} \times 64 , \text{in}^2 = 32 , \text{in}^2\)
Now combine the areas of all the shapes: \[ A = A_{square} + A_{triangle1} + A_{triangle2} \] \[ A = 144 , \text{in}^2 + 32 , \text{in}^2 + 32 , \text{in}^2 \] \[ A = 144 , \text{in}^2 + 64 , \text{in}^2 \] \[ A = 208 , \text{in}^2 \]
Thus, the total area of the figure is \(\boxed{208 , \text{in}^2}\).