Question

Find the area of the figure. Hint: Separate into regular shapes and combine the areas!
Shapes: 1 rectangle and 2 triangles
Triangle 1: 5 in and 3 in
Triangle 2: 5 in and 3 in
Rectangle: 8 in and 2 in
A = [ ? ] in²
Area Formulas: B x H and B x H/2

Answer 1

To find the area of the figure, we will calculate the areas of the rectangle and the two triangles separately and then combine them.

1. **Area of the Rectangle:**
The formula for the area of a rectangle is:
\[
A = B \times H
\]
where \( B \) is the base and \( H \) is the height.

For the rectangle:
- Base \( B = 8 \, \text{in} \)
- Height \( H = 2 \, \text{in} \)

So, the area of the rectangle is:
\[
A_{\text{rectangle}} = 8 \, \text{in} \times 2 \, \text{in} = 16 \, \text{in}^2
\]

2. **Area of Triangle 1:**
The formula for the area of a triangle is:
\[
A = \frac{B \times H}{2}
\]
where \( B \) is the base and \( H \) is the height.

For Triangle 1:
- Base \( B = 5 \, \text{in} \)
- Height \( H = 3 \, \text{in} \)

So, the area of Triangle 1 is:
\[
A_{\text{triangle1}} = \frac{5 \, \text{in} \times 3 \, \text{in}}{2} = \frac{15 \, \text{in}^2}{2} = 7.5 \, \text{in}^2
\]

3. **Area of Triangle 2:**
Triangle 2 has the same dimensions as Triangle 1 (5 in and 3 in).

So, the area of Triangle 2 is also:
\[
A_{\text{triangle2}} = \frac{5 \, \text{in} \times 3 \, \text{in}}{2} = \frac{15 \, \text{in}^2}{2} = 7.5 \, \text{in}^2
\]

4. **Total Area:**
Now, we combine all the areas:
\[
A_{\text{total}} = A_{\text{rectangle}} + A_{\text{triangle1}} + A_{\text{triangle2}}
\]
\[
A_{\text{total}} = 16 \, \text{in}^2 + 7.5 \, \text{in}^2 + 7.5 \, \text{in}^2
\]
\[
A_{\text{total}} = 16 \, \text{in}^2 + 15 \, \text{in}^2
\]
\[
A_{\text{total}} = 31 \, \text{in}^2
\]

Thus, the total area of the figure is:
\[
\boxed{31 \, \text{in}^2}
\]