Question

Find the area of the figure.
Hint: Separate into regular shapes and combine the areas!
Shapes: Rectangle and a triangle
Triangle 1: 8 in
Triangle 2: 8 in
Square: 12 in, 12 in and 12 in
A = [ ? ] in²
Area Formulas: B x H and B x H/2

Answer 1

To find the area of the figure as described, we can separate it into the regular shapes mentioned: two triangles and a square.

1. **Square**:
- Length of each side = 12 in
- Area of the square \(A_{square} = \text{side} \times \text{side} = 12 \, \text{in} \times 12 \, \text{in} = 144 \, \text{in}^2\)

2. **Triangle 1**:
- Base = 8 in (assuming this is the base)
- Height = 8 in
- Area of Triangle 1 \(A_{triangle1} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \, \text{in} \times 8 \, \text{in} = \frac{1}{2} \times 64 \, \text{in}^2 = 32 \, \text{in}^2\)

3. **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 \(\boxed{208 \, \text{in}^2}\).